{"id":339,"date":"2017-09-08T02:38:20","date_gmt":"2017-09-08T02:38:20","guid":{"rendered":"http:\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/chapter\/additional-information-and-full-hypothesis-test-examples\/"},"modified":"2017-09-08T02:38:20","modified_gmt":"2017-09-08T02:38:20","slug":"additional-information-and-full-hypothesis-test-examples","status":"publish","type":"chapter","link":"https:\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/chapter\/additional-information-and-full-hypothesis-test-examples\/","title":{"raw":"Additional Information and Full Hypothesis Test Examples","rendered":"Additional Information and Full Hypothesis Test Examples"},"content":{"raw":"[latexpage]<ul><li>In a <span data-type=\"term\">hypothesis test<\/span> problem, you may see words such as \"the level of significance is 1%.\" The \"1%\" is the preconceived or preset <em data-effect=\"italics\">\u03b1<\/em>.<\/li><li>The statistician setting up the hypothesis test selects the value of <em data-effect=\"italics\">\u03b1<\/em> to use <strong>before<\/strong> collecting the sample data.<\/li><li><strong>If no level of significance is given, a common standard to use is <em data-effect=\"italics\">\u03b1<\/em> = 0.05.<\/strong><\/li><li>When you calculate the <em data-effect=\"italics\">p<\/em>-value and draw the picture, the <em data-effect=\"italics\">p<\/em>-value is the area in the left tail, the right tail, or split evenly between the two tails. For this reason, we call the hypothesis test left, right, or two tailed.<\/li><li>The <strong>alternative hypothesis<\/strong>, \\({H}_{a}\\), tells you if the test is left, right, or two-tailed. It is the <strong>key<\/strong> to conducting the appropriate test.<\/li><li><em data-effect=\"italics\">H<sub>a<\/sub><\/em><strong>never<\/strong> has a symbol that contains an equal sign.<\/li><li><strong>Thinking about the meaning of the<\/strong><span data-type=\"term\"><em data-effect=\"italics\">p<\/em>-value<\/span>: A data analyst (and anyone else) should have more confidence that he made the correct decision to reject the null hypothesis with a smaller <em data-effect=\"italics\">p<\/em>-value (for example, 0.001 as opposed to 0.04) even if using the 0.05 level for alpha. Similarly, for a large <em data-effect=\"italics\">p<\/em>-value such as 0.4, as opposed to a <em data-effect=\"italics\">p<\/em>-value of 0.056 (alpha = 0.05 is less than either number), a data analyst should have more confidence that she made the correct decision in not rejecting the null hypothesis.  This makes the data analyst use judgment rather than mindlessly applying rules.<\/li><\/ul><p id=\"fs-idp42898080\">The following examples illustrate a left-, right-, and two-tailed test.<\/p><div data-type=\"example\" class=\"textbox examples\" id=\"fs-idp41487520\"><p id=\"fs-idp41487648\"><em data-effect=\"italics\">H<sub>o<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> = 5, <em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &lt; 5<\/p><p>Test of a single population mean. <em data-effect=\"italics\">H<sub>a<\/sub><\/em> tells you the test is left-tailed. The picture of the <em data-effect=\"italics\">p<\/em>-value is as follows:<\/p><div class=\"bc-figure figure\" id=\"fs-idm151356816\"><span data-type=\"media\" id=\"id14703034\" data-alt=\"Normal distribution curve of a single population mean with a value of 5 on the x-axis and the p-value points to the area on the left tail of the curve.\"><img src=\"http:\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/fig-ch09_09_01-02.jpg\" data-media-type=\"image\/jpg\" alt=\"Normal distribution curve of a single population mean with a value of 5 on the x-axis and the p-value points to the area on the left tail of the curve.\" width=\"380\" \/><\/span><\/div><\/div><div data-type=\"note\" data-has-label=\"true\" id=\"fs-idp42408528\" class=\"note statistics try\" data-label=\"\"><div data-type=\"title\" class=\"title\">Try It<\/div><div data-type=\"exercise\" class=\"exercise\" id=\"fs-idp67352928\"><div data-type=\"problem\" class=\"problem\"><p id=\"eip-idm132446832\"><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> = 10, <em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &lt; 10<\/p><p id=\"eip-idp3861728\">Assume the <em data-effect=\"italics\">p<\/em>-value is 0.0935. What type of test is this? Draw the picture of the <em data-effect=\"italics\">p<\/em>-value.<\/p><\/div><div data-type=\"solution\" class=\"solution\" id=\"fs-idp1251488\"><p>left-tailed test<\/p><div class=\"bc-figure figure\" id=\"fs-idm70716944\"><span data-type=\"media\" id=\"eip-idp9289536\" data-alt=\"\"><img src=\"http:\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/CNX_Stats_C09_M09_tryit002annoN.jpg\" data-media-type=\"image\/jpeg\" alt=\"\" width=\"380\" \/><\/span><\/div><\/div><\/div><\/div><div data-type=\"example\" class=\"textbox examples\"><p><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2264 0.2\u2003\u2003<em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> &gt; 0.2<\/p><p id=\"fs-idp117549824\">This is a test of a single population proportion. <em data-effect=\"italics\">H<sub>a<\/sub><\/em> tells you the test is <strong>right-tailed<\/strong>. The picture of the <em data-effect=\"italics\">p<\/em>-value is as follows:<\/p><div class=\"bc-figure figure\" id=\"fs-idm2304848\"><span data-type=\"media\" id=\"id14703170\" data-alt=\"Normal distribution curve of a single population proportion with the value of 0.2 on the x-axis. The p-value points to the area on the right tail of the curve.\"><img src=\"http:\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/fig-ch09_09_02a.jpg\" data-media-type=\"image\/jpg\" alt=\"Normal distribution curve of a single population proportion with the value of 0.2 on the x-axis. The p-value points to the area on the right tail of the curve.\" width=\"380\" \/><\/span><\/div><\/div><div data-type=\"note\" data-has-label=\"true\" id=\"fs-idm126098448\" class=\"note statistics try\" data-label=\"\"><div data-type=\"title\" class=\"title\">Try It<\/div><div data-type=\"exercise\" class=\"exercise\" id=\"eip-417\"><div data-type=\"problem\" class=\"problem\" id=\"fs-idp131553552\"><p id=\"eip-idm89526848\"><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> \u2264 1, <em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &gt; 1<\/p><p id=\"eip-idm24425280\">Assume the <em data-effect=\"italics\">p<\/em>-value is 0.1243. What type of test is this? Draw the picture of the <em data-effect=\"italics\">p<\/em>-value.<\/p><\/div><div data-type=\"solution\" class=\"solution\"><p>right-tailed test<\/p><div class=\"bc-figure figure\" id=\"fs-idm53526160\"><span data-type=\"media\" id=\"eip-idm27872752\" data-alt=\"\"><img src=\"http:\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/CNX_Stats_C09_M09_tryit001annoN.jpg\" data-media-type=\"image\/jpeg\" alt=\"\" width=\"380\" \/><\/span><\/div><\/div><\/div><\/div><div data-type=\"example\" class=\"textbox examples\"><p id=\"element-38a\"><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> = 50\u2003\u2003<em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2260 50<\/p><p>This is a test of a single population mean. <em data-effect=\"italics\">H<sub>a<\/sub><\/em> tells you the test is <strong>two-tailed<\/strong>. The picture of the <em data-effect=\"italics\">p<\/em>-value is as follows.<\/p><div class=\"bc-figure figure\" id=\"fs-idm117972816\"><span data-type=\"media\" id=\"id4829633\" data-alt=\"Normal distribution curve of a single population mean with a value of 50 on the x-axis. The p-value formulas, 1\/2(p-value), for a two-tailed test is shown for the areas on the left and right tails of the curve.\"><img src=\"http:\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/fig-ch09_09_03N.jpg\" data-media-type=\"image\/jpg\" alt=\"Normal distribution curve of a single population mean with a value of 50 on the x-axis. The p-value formulas, 1\/2(p-value), for a two-tailed test is shown for the areas on the left and right tails of the curve.\" width=\"380\" \/><\/span><\/div><\/div><div data-type=\"note\" data-has-label=\"true\" id=\"fs-idm63250432\" class=\"note statistics try\" data-label=\"\"><div data-type=\"title\" class=\"title\">Try It<\/div><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\"><p id=\"eip-idp41687312\"><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> = 0.5, <em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2260 0.5<\/p><p id=\"eip-idm5668432\">Assume the <em data-effect=\"italics\">p<\/em>-value is 0.2564. What type of test is this? Draw the picture of the <em data-effect=\"italics\">p<\/em>-value.<\/p><\/div><div data-type=\"solution\" class=\"solution\"><p>two-tailed test<\/p><div class=\"bc-figure figure\" id=\"fs-idm126590640\"><span data-type=\"media\" id=\"eip-idm113874560\" data-alt=\"\"><img src=\"http:\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/CNX_Stats_C09_M09_tryit003annoN.jpg\" data-media-type=\"image\/jpeg\" alt=\"\" width=\"380\" \/><\/span><\/div><\/div><\/div><\/div><div class=\"bc-section section\" data-depth=\"1\" id=\"fs-idm12892576\"><h1 data-type=\"title\">Full Hypothesis Test Examples<\/h1><div data-type=\"example\" class=\"textbox examples\"><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\" id=\"id5300933\"><p>Jeffrey, as an eight-year old, <strong>established a mean time of 16.43 seconds<\/strong> for swimming the 25-yard freestyle, with a <strong>standard deviation of 0.8 seconds<\/strong>. His dad, Frank, thought that Jeffrey could swim the 25-yard freestyle faster using goggles. Frank bought Jeffrey a new pair of expensive goggles and timed Jeffrey for <strong>15 25-yard freestyle swims<\/strong>. For the 15 swims, <strong>Jeffrey's mean time was 16 seconds. Frank thought that the goggles helped Jeffrey to swim faster than the 16.43 seconds.<\/strong> Conduct a hypothesis test using a preset <em data-effect=\"italics\">\u03b1<\/em> = 0.05. Assume that the swim times for the 25-yard freestyle are normal.<\/p><\/div><div data-type=\"solution\" class=\"solution\" id=\"id5036614\"><p>Set up the Hypothesis Test:<\/p><p>Since the problem is about a mean, this is a <strong>test of a single population mean<\/strong>.<\/p><p><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> = 16.43\u2003\u2003<em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &lt; 16.43<\/p><p id=\"element-241\">For Jeffrey to swim faster, his time will be less than 16.43 seconds. The \"&lt;\" tells you this is left-tailed.<\/p><p>Determine the distribution needed:<\/p><p><strong>Random variable: <\/strong>\\(\\overline{X}\\) = the mean time to swim the 25-yard freestyle.<\/p><p><strong> Distribution for the test: <\/strong>\\(\\overline{X}\\) is normal (population <span data-type=\"term\">standard deviation<\/span> is known: <em data-effect=\"italics\">\u03c3<\/em> = 0.8)<\/p><p>\\(\\overline{X}~N\\left(\\mu ,\\frac{{\\sigma }_{X}}{\\sqrt{n}}\\right)\\)\n\nTherefore, \\(\\overline{X}~N\\left(16.43,\\frac{0.8}{\\sqrt{15}}\\right)\\)<\/p><p><em data-effect=\"italics\">\u03bc<\/em> = 16.43 comes from <em data-effect=\"italics\">H<sub>0<\/sub><\/em> and not the data. <em data-effect=\"italics\">\u03c3<\/em> = 0.8, and <em data-effect=\"italics\">n<\/em> = 15.<\/p><p>Calculate the <em data-effect=\"italics\">p<\/em>-value using the normal distribution for a mean:<\/p><p id=\"element-706\"><em data-effect=\"italics\">p<\/em>-value = <em data-effect=\"italics\">P<\/em>(\\(\\overline{x}\\) &lt; 16) = 0.0187 where the sample mean in the problem is given as 16.<\/p><p id=\"element-656\"><em data-effect=\"italics\">p<\/em>-value = 0.0187 (This is called the <strong>actual level of significance<\/strong>.) The <em data-effect=\"italics\">p<\/em>-value is the area to the left of the sample mean is given as 16.<\/p><p><strong>Graph:<\/strong><\/p><div class=\"bc-figure figure\" id=\"hyptest11_ex1\"><span data-type=\"media\" id=\"id4282005\" data-alt=\"Normal distribution curve for the average time to swim the 25-yard freestyle with values 16, as the sample mean, and 16.43 on the x-axis. A vertical upward line extends from 16 on the x-axis to the curve. An arrow points to the left tail of the curve.\"><img src=\"http:\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/fig-ch09_11_01N.jpg\" data-media-type=\"image\/jpg\" alt=\"Normal distribution curve for the average time to swim the 25-yard freestyle with values 16, as the sample mean, and 16.43 on the x-axis. A vertical upward line extends from 16 on the x-axis to the curve. An arrow points to the left tail of the curve.\" width=\"380\" \/><\/span><\/div><p><em data-effect=\"italics\">\u03bc<\/em> = 16.43 comes from <em data-effect=\"italics\">H<sub>0<\/sub><\/em>. Our assumption is <em data-effect=\"italics\">\u03bc<\/em> = 16.43.<\/p><p><strong>Interpretation of the <em data-effect=\"italics\">p<\/em>-value: If <em data-effect=\"italics\">H<sub>0<\/sub><\/em> is true<\/strong>, there is a 0.0187 probability (1.87%)that Jeffrey's mean time to swim the 25-yard freestyle is 16 seconds or less. Because a 1.87% chance is small, the mean time of 16 seconds or less is unlikely to have happened randomly. It is a rare event.<\/p><p>Compare <em data-effect=\"italics\">\u03b1<\/em> and the <em data-effect=\"italics\">p<\/em>-value:<\/p><p id=\"element-890\"><em data-effect=\"italics\">\u03b1<\/em> = 0.05 <em data-effect=\"italics\">p<\/em>-value = 0.0187 <em data-effect=\"italics\">\u03b1<\/em> &gt; <em data-effect=\"italics\">p<\/em>-value<\/p><p id=\"element-543\"><strong>Make a decision:<\/strong> Since <em data-effect=\"italics\">\u03b1<\/em> &gt; <em data-effect=\"italics\">p<\/em>-value, reject <em data-effect=\"italics\">H<sub>0<\/sub><\/em>.<\/p>&lt;!-- CONTINUE HERE --&gt;<p>This means that you reject <em data-effect=\"italics\">\u03bc<\/em> = 16.43. In other words, you do not think Jeffrey swims the\n25-yard freestyle in 16.43 seconds but faster with the new goggles.<\/p><p><strong>Conclusion:<\/strong> At the 5% significance level, we conclude that Jeffrey swims faster using the new goggles. The sample data show there is sufficient evidence that Jeffrey's mean time to swim the 25-yard freestyle is less than 16.43 seconds.<\/p><p id=\"element-252\">The <em data-effect=\"italics\">p<\/em>-value can easily be calculated.<\/p><div data-type=\"note\" data-has-label=\"true\" id=\"fs-idm31641280\" class=\"note statistics calculator\" data-label=\"\"><p id=\"element-362\">Press <code>STAT<\/code> and arrow over to <code>TESTS<\/code>. Press <code>1:Z-Test<\/code>. Arrow over to <code>Stats<\/code> and press <code>ENTER<\/code>. Arrow down and enter 16.43 for <em data-effect=\"italics\">\u03bc<sub>0<\/sub><\/em> (null hypothesis), .8 for <em data-effect=\"italics\">\u03c3<\/em>, 16 for the sample mean, and 15 for <em data-effect=\"italics\">n<\/em>. Arrow down to <em data-effect=\"italics\">\u03bc<\/em> : (alternate hypothesis) and arrow over to &lt; <em data-effect=\"italics\">\u03bc<sub>0<\/sub><\/em>. Press <code>ENTER<\/code>. Arrow down to <code>Calculate<\/code> and press <code>ENTER<\/code>. The calculator not only calculates the <em data-effect=\"italics\">p<\/em>-value (<em data-effect=\"italics\">p<\/em> = 0.0187) but it also calculates the test statistic (<em data-effect=\"italics\">z<\/em>-score) for the sample mean. <em data-effect=\"italics\">\u03bc<\/em> &lt; 16.43 is the alternative hypothesis. Do this set of instructions again except arrow to <code>Draw<\/code>(instead of <code>Calculate<\/code>). Press <code>ENTER<\/code>. A shaded graph appears with <em data-effect=\"italics\">z<\/em> = -2.08 (test statistic) and <em data-effect=\"italics\">p<\/em> = 0.0187 (<em data-effect=\"italics\">p<\/em>-value). Make sure when you use <code>Draw<\/code> that no other equations are highlighted in <em data-effect=\"italics\">Y<\/em> = and the plots are turned off.<\/p><\/div><p>When the calculator does a <em data-effect=\"italics\">Z<\/em>-Test, the <code>Z-Test<\/code> function finds the <em data-effect=\"italics\">p<\/em>-value by doing a normal probability calculation using the <span data-type=\"term\">central limit theorem<\/span>:<\/p><p id=\"fs-idp159988784\">\\(P\\left(\\overline{x}&lt;16\\right)=\\)<code>2nd DISTR normcdf<\/code>\\(\\left(-10^99,16,16.43,0.8\/\\sqrt{15}\\right)\\)\n.<\/p><p>The Type I and Type II errors for this problem are as follows:<\/p><p id=\"element-385\">The Type I error is to conclude that Jeffrey swims the 25-yard freestyle, on average, in\nless than 16.43 seconds when, in fact, he actually swims the 25-yard freestyle, on\naverage, in 16.43 seconds. (Reject the null hypothesis when the null hypothesis is true.)<\/p><p>The Type II error is that there is not evidence to conclude that Jeffrey swims the 25-yard free-style, on average, in less than 16.43 seconds when, in fact, he actually does swim the 25-yard free-style, on average, in less than 16.43 seconds. (Do not reject the null hypothesis when the null hypothesis is false.)<\/p><\/div><\/div><\/div><div data-type=\"note\" data-has-label=\"true\" id=\"fs-idm112426112\" class=\"note statistics try\" data-label=\"\"><div data-type=\"title\" class=\"title\">Try It<\/div><div data-type=\"exercise\" class=\"exercise\" id=\"fs-idp78508304\"><div data-type=\"problem\" class=\"problem\" id=\"fs-idp117367632\"><p>The mean throwing distance of a football for a Marco, a high school freshman quarterback, is 40 yards, with a standard deviation of two yards. The team coach tells Marco to adjust his grip to get more distance. The coach records the distances for 20 throws. For the 20 throws, Marco\u2019s mean distance was 45 yards. The coach thought the different grip helped Marco throw farther than 40 yards. Conduct a hypothesis test using a preset <em data-effect=\"italics\">\u03b1<\/em> = 0.05. Assume the throw distances for footballs are normal.<\/p><p id=\"eip-idp30205504\">First, determine what type of test this is, set up the hypothesis test, find the <em data-effect=\"italics\">p<\/em>-value, sketch the graph, and state your conclusion.<\/p><div data-type=\"note\" data-has-label=\"true\" id=\"fs-idm24767312\" class=\"note statistics calculator\" data-label=\"\"><p id=\"eip-idm96892800\">Press STAT and arrow over to TESTS. Press 1:Z-Test. Arrow over to Stats and press ENTER. Arrow down and enter 40 for <em data-effect=\"italics\">\u03bc<\/em>0 (null hypothesis), 2 for <em data-effect=\"italics\">\u03c3<\/em>, 45 for the sample mean, and 20 for <em data-effect=\"italics\">n<\/em>. Arrow down to <em data-effect=\"italics\">\u03bc<\/em>: (alternative hypothesis) and set it either as &lt;, \u2260, or &gt;. Press ENTER. Arrow down to Calculate and press ENTER. The calculator not only calculates the <em data-effect=\"italics\">p<\/em>-value but it also calculates the test statistic (<em data-effect=\"italics\">z<\/em>-score) for the sample mean. Select &lt;, \u2260, or &gt; for the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate). Press ENTER. A shaded graph appears with test statistic and <em data-effect=\"italics\">p<\/em>-value. Make sure when you use Draw that no other equations are highlighted in <em data-effect=\"italics\">Y<\/em> = and the plots are turned off.<\/p><\/div><\/div><div data-type=\"solution\" class=\"solution\"><p>\nSince the problem is about a mean, this is a test of a single population mean.\n  <\/p><p id=\"eip-idm108002480\"><em data-effect=\"italics\">H<sub>0<\/sub><\/em> : <em data-effect=\"italics\">\u03bc<\/em> = 40<\/p><p id=\"eip-idm216373056\"><em data-effect=\"italics\">H<sub>a<\/sub><\/em> : <em data-effect=\"italics\">\u03bc<\/em> &gt; 40<\/p><p id=\"eip-idm58633744\"><em data-effect=\"italics\">p<\/em> = 0.0062<\/p><div class=\"bc-figure figure\" id=\"fs-idm152825344\"><span data-type=\"media\" id=\"eip-idm58633360\" data-alt=\"\"><img src=\"http:\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/CNX_Stats_C09_M11_tryit001annoN.jpg\" data-media-type=\"image\/jpeg\" alt=\"\" width=\"380\" \/><\/span><\/div><p id=\"eip-idm24608304\">Because <em data-effect=\"italics\">p<\/em> &lt; <em data-effect=\"italics\">\u03b1<\/em>, we reject the null hypothesis. There is sufficient evidence to suggest that the change in grip improved Marco\u2019s throwing distance.<\/p><\/div><\/div><\/div><\/div><div data-type=\"note\" class=\"note\" data-has-label=\"true\" data-label=\"\"><div data-type=\"title\" class=\"title\">Historical Note (<a href=\"#fs-idp41487520\" class=\"autogenerated-content\">[link]<\/a>)<\/div><p id=\"eip-idp233123200\">The traditional way to compare the two probabilities, <em data-effect=\"italics\">\u03b1<\/em> and the <em data-effect=\"italics\">p<\/em>-value, is to compare the critical value (<em data-effect=\"italics\">z<\/em>-score from <em data-effect=\"italics\">\u03b1<\/em>) to the test statistic (<em data-effect=\"italics\">z<\/em>-score from data). The calculated test statistic for the <em data-effect=\"italics\">p<\/em>-value is \u20132.08. (From the Central Limit Theorem, the test statistic formula is \\(z=\\frac{\\overline{x}-{\\mu }_{X}}{\\left(\\frac{{\\sigma }_{X}}{\\sqrt{n}}\\right)}\\). \n\n&lt;!--LINK THIS PLEASE --&gt;\nFor this problem, \\(\\overline{x}\\) = 16, <em data-effect=\"italics\">\u03bc<sub>X<\/sub><\/em> = 16.43 from the null hypothes is, <em data-effect=\"italics\">\u03c3<sub>X<\/sub><\/em> = 0.8, and <em data-effect=\"italics\">n<\/em> = 15.) You can find the critical value for <em data-effect=\"italics\">\u03b1<\/em> = 0.05 in the normal table (see <strong>15.Tables<\/strong> in the Table of Contents). The <em data-effect=\"italics\">z<\/em>-score for an area to the left equal to 0.05 is midway between \u20131.65 and \u20131.64 (0.05 is midway between 0.0505 and 0.0495). The <em data-effect=\"italics\">z<\/em>-score is \u20131.645. Since \u20131.645 &gt; \u20132.08 (which demonstrates that \u03b1 &gt; <em data-effect=\"italics\">p<\/em>-value), reject <em data-effect=\"italics\">H<sub>0<\/sub><\/em>. Traditionally, the decision\nto reject or not reject was done in this way. Today, comparing the two probabilities <em data-effect=\"italics\">\u03b1<\/em> and the <em data-effect=\"italics\">p<\/em>-value is very common. For this problem, the <em data-effect=\"italics\">p<\/em>-value, 0.0187 is considerably smaller than <em data-effect=\"italics\">\u03b1<\/em>, 0.05. You can be confident about your decision to reject. The graph shows <em data-effect=\"italics\">\u03b1<\/em>, the <em data-effect=\"italics\">p<\/em>-value, and the test statistics and the critical value.\n<\/p><div class=\"bc-figure figure\" id=\"eip-idm47467424\"><span data-type=\"media\" id=\"eip-idp126239824\" data-alt=\"Distribution curve comparing the &#x3b1; to the p-value. Values of -2.15 and -1.645 are on the x-axis. Vertical upward lines extend from both of these values to the curve. The p-value is equal to 0.0158 and points to the area to the left of -2.15. &#x3b1; is equal to 0.05 and points to the area between the values of -2.15 and -1.645.\"><img src=\"http:\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/fig-ch09_11_02.png\" data-media-type=\"image\/png\" alt=\"Distribution curve comparing the &#x3b1; to the p-value. Values of -2.15 and -1.645 are on the x-axis. Vertical upward lines extend from both of these values to the curve. The p-value is equal to 0.0158 and points to the area to the left of -2.15. &#x3b1; is equal to 0.05 and points to the area between the values of -2.15 and -1.645.\" width=\"380\" \/><\/span><\/div><\/div><div data-type=\"example\" class=\"textbox examples\"><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\" id=\"id5291254\"><p>A college football coach thought that his players could bench press a <strong>mean weight of 275 pounds<\/strong>. It is known that the <strong>standard deviation is 55\npounds<\/strong>. Three of his players thought that the mean weight was <strong>more than<\/strong> that amount.\nThey asked <strong>30<\/strong> of their teammates for their estimated maximum lift on the bench\npress exercise. The data ranged from 205 pounds to 385 pounds. The actual\ndifferent weights were (frequencies are in parentheses) \n<span data-type=\"list\" data-list-type=\"labeled-item\" id=\"set-1\" data-display=\"inline\"><span data-type=\"item\">205(3)<\/span><span data-type=\"item\">215(3)<\/span><span data-type=\"item\">225(1)<\/span><span data-type=\"item\">241(2)<\/span><span data-type=\"item\">252(2)<\/span><span data-type=\"item\">265(2)<\/span><span data-type=\"item\">275(2)<\/span><span data-type=\"item\">313(2)<\/span><span data-type=\"item\">316(5)<\/span><span data-type=\"item\">338(2)<\/span><span data-type=\"item\">341(1)<\/span><span data-type=\"item\">345(2)<\/span><span data-type=\"item\">368(2)<\/span><span data-type=\"item\">385(1)<\/span><\/span>.<\/p><p>Conduct a hypothesis test using a 2.5% level of significance to determine if the bench press mean is <strong>more than 275 pounds<\/strong>.<\/p><\/div><div data-type=\"solution\" class=\"solution\" id=\"id5342777\"><p>Set up the Hypothesis Test:<\/p><p>Since the problem is about a mean weight, this is a <strong>test of a single population mean<\/strong>.<\/p><p><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> = 275<\/p><div data-type=\"newline\"><br \/><\/div><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &gt; 275<div data-type=\"newline\"><br \/><\/div>This is a right-tailed test.<p>Calculating the distribution needed:<\/p><p id=\"element-975\">Random variable: \\(\\overline{X}\\) = the mean weight, in pounds, lifted by the football players.<\/p><p><strong>Distribution for the test:<\/strong> It is normal because <em data-effect=\"italics\">\u03c3<\/em> is known.<\/p><p>\\(\\overline{X}~N\\left(275,\\frac{55}{\\sqrt{30}}\\right)\\)<\/p><p>\\(\\overline{x}=286.2\\) pounds (from the data).<\/p><p><em data-effect=\"italics\">\u03c3<\/em> = 55 pounds <strong>(Always use <em data-effect=\"italics\">\u03c3<\/em> if you know it.)<\/strong> We assume <em data-effect=\"italics\">\u03bc<\/em> = 275 pounds unless our data shows us otherwise. <\/p><p>Calculate the <em data-effect=\"italics\">p<\/em>-value using the normal distribution for a mean and using the sample mean as input (see <a href=\"\/contents\/d0ba1833-f0d2-4195-8765-3c436745f0fb@4\" class=\"autogenerated-content\">[link]<\/a> for using the data as input):<\/p><p>\\(p\\text{-value}=P\\left(\\overline{x}&gt;286.2\\right)=0.1323\\).<\/p><p><strong>Interpretation of the <em data-effect=\"italics\">p<\/em>-value:<\/strong> If <em data-effect=\"italics\">H<sub>0<\/sub><\/em> is true, then there is a 0.1331 probability (13.23%) that the football players can lift a mean weight of 286.2 pounds or more. Because a 13.23% chance is large enough, a mean weight lift of 286.2 pounds or more is not a rare event.<\/p><div class=\"bc-figure figure\" id=\"hyptest11_ex3\"><span data-type=\"media\" id=\"id4317824\" data-alt=\"Normal distribution curve of the average weight lifted by football players with values of 275 and 286.2 on the x-axis. A vertical upward line extends from 286.2 to the curve. The p-value points to the area to the right of 286.2.\"><img src=\"http:\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/fig-ch09_11_03N.jpg\" data-media-type=\"image\/jpg\" alt=\"Normal distribution curve of the average weight lifted by football players with values of 275 and 286.2 on the x-axis. A vertical upward line extends from 286.2 to the curve. The p-value points to the area to the right of 286.2.\" width=\"380\" \/><\/span><\/div><p>Compare <em data-effect=\"italics\">\u03b1<\/em> and the <em data-effect=\"italics\">p<\/em>-value:<\/p><p><em data-effect=\"italics\">\u03b1<\/em> = 0.025 <em data-effect=\"italics\">p<\/em>-value = 0.1323<\/p><p><strong>Make a decision:<\/strong> Since <em data-effect=\"italics\">\u03b1<\/em> &lt;<em data-effect=\"italics\">p<\/em>-value, do not reject <em data-effect=\"italics\">H<sub>0<\/sub><\/em>.<\/p><p><strong>Conclusion:<\/strong> At the 2.5% level of significance, from the sample data, there is not sufficient evidence to conclude that the true mean weight lifted is more than 275 pounds.<\/p><p>The <em data-effect=\"italics\">p<\/em>-value can easily be calculated.<\/p><div data-type=\"note\" data-has-label=\"true\" id=\"fs-idm79549936\" class=\"note statistics calculator\" data-label=\"\"><p id=\"element-233\">Put the data and frequencies into lists. Press <code>STAT<\/code> and arrow over to <code>TESTS<\/code>. Press <code>1:Z-Test<\/code>. Arrow over to <code>Data<\/code> and press <code>ENTER<\/code>. Arrow down and enter 275 for <em data-effect=\"italics\">\u03bc<sub>0<\/sub><\/em>, 55 for <em data-effect=\"italics\">\u03c3<\/em>, the name of the list where you put the data, and the name of the list where you put the frequencies. Arrow down to <em data-effect=\"italics\">\u03bc:<\/em> and arrow over to &gt; <em data-effect=\"italics\">\u03bc<sub>0<\/sub><\/em>. Press <code>ENTER<\/code>. Arrow down to <code>Calculate<\/code> and press <code>ENTER<\/code>. The calculator not only calculates the <em data-effect=\"italics\">p<\/em>-value (<em data-effect=\"italics\">p<\/em> = 0.1331, a little different from the previous calculation - in it we used the sample mean rounded to one decimal place instead of the data) but it also calculates the test statistic (<em data-effect=\"italics\">z<\/em>-score) for the sample mean, the sample mean, and the sample standard deviation. <em data-effect=\"italics\">\u03bc<\/em> &gt; 275 is the alternative hypothesis. Do this set of instructions again except arrow to <code>Draw<\/code> (instead of <code>Calculate<\/code>). Press <code>ENTER<\/code>. A shaded graph appears with <em data-effect=\"italics\">z<\/em> = 1.112 (test statistic) and <em data-effect=\"italics\">p<\/em> = 0.1331 (<em data-effect=\"italics\">p<\/em>-value). Make sure when you use <code>Draw<\/code> that no other equations are highlighted in <em data-effect=\"italics\">Y<\/em> = and the plots are turned off.<\/p><\/div><\/div><\/div><\/div><div data-type=\"example\" class=\"textbox examples\" id=\"element-207\"><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\" id=\"id4319892\"><p id=\"fs-idp90231232\">Statistics students believe that the mean score on the first statistics test is 65. A statistics instructor thinks the mean score is higher than 65. He samples ten statistics students and obtains the scores <span data-type=\"list\" data-list-type=\"labeled-item\" id=\"set-2\" data-display=\"inline\"><span data-type=\"item\">65<\/span><span data-type=\"item\">65<\/span><span data-type=\"item\">70<\/span><span data-type=\"item\">67<\/span><span data-type=\"item\">66<\/span><span data-type=\"item\">63<\/span><span data-type=\"item\">63<\/span><span data-type=\"item\">68<\/span><span data-type=\"item\">72<\/span><span data-type=\"item\">71<\/span><\/span>. He performs a hypothesis test using a 5% level of significance. The data are assumed to be from a normal distribution.<\/p><\/div><div data-type=\"solution\" class=\"solution\" id=\"id5147242\"><p>Set up the hypothesis test:<\/p><p>A 5% level of significance means that <em data-effect=\"italics\">\u03b1<\/em> = 0.05. This is a test of a <strong>single population mean<\/strong>.<\/p><p><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> = 65\u2003\u2003<em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &gt; 65<\/p><p>Since the instructor thinks the average score is higher, use a \"&gt;\". The \"&gt;\" means the test is right-tailed.<\/p><p id=\"fs-idp165624976\">Determine the distribution needed:<\/p><p><strong>Random variable:<\/strong>\\(\\overline{X}\\) = average score on the first statistics test.<\/p><p><strong> Distribution for the test:<\/strong> If you read the problem carefully, you will notice that there is\n<strong>no population standard deviation given<\/strong>. You are only given <em data-effect=\"italics\">n<\/em> = 10 sample data values. Notice also that the data come from a normal distribution. This means that the distribution for the test is a student's <em data-effect=\"italics\">t<\/em>.<\/p><p id=\"element-240\">Use <em data-effect=\"italics\">t<\/em><sub>df<\/sub>. Therefore, the distribution for the test is <em data-effect=\"italics\">t<\/em><sub>9<\/sub> where <em data-effect=\"italics\">n<\/em> = 10 and <em data-effect=\"italics\">df<\/em> = 10 - 1 = 9.<\/p><p>Calculate the <em data-effect=\"italics\">p<\/em>-value using the Student's <em data-effect=\"italics\">t<\/em>-distribution:<\/p><p><em data-effect=\"italics\">p<\/em>-value = <em data-effect=\"italics\">P<\/em>(\\(\\overline{x}\\) &gt; 67) = 0.0396 where the sample mean and sample standard deviation are calculated as 67 and 3.1972 from the data.<\/p><p><strong>Interpretation of the <em data-effect=\"italics\">p<\/em>-value:<\/strong> If the null hypothesis is true, then there is a 0.0396\nprobability (3.96%) that the sample mean is 65 or more.<\/p><div class=\"bc-figure figure\" id=\"hyptest11_ex4\"><span data-type=\"media\" id=\"id5241154\" data-alt=\"Normal distribution curve of average scores on the first statistic tests with 65 and 67 values on the x-axis. A vertical upward line extends from 67 to the curve. The p-value points to the area to the right of 67.\"><img src=\"http:\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/fig-ch09_11_04N.jpg\" data-media-type=\"image\/jpg\" alt=\"Normal distribution curve of average scores on the first statistic tests with 65 and 67 values on the x-axis. A vertical upward line extends from 67 to the curve. The p-value points to the area to the right of 67.\" width=\"380\" \/><\/span><\/div><p>Compare <em data-effect=\"italics\">\u03b1<\/em> and the <em data-effect=\"italics\">p<\/em>-value:<\/p><p>Since <em data-effect=\"italics\">\u03b1<\/em> = 0.05 and <em data-effect=\"italics\">p<\/em>-value = 0.0396. <em data-effect=\"italics\">\u03b1<\/em> &gt; <em data-effect=\"italics\">p<\/em>-value.<\/p><p><strong>Make a decision:<\/strong> Since <em data-effect=\"italics\">\u03b1<\/em> &gt; <em data-effect=\"italics\">p<\/em>-value, reject <em data-effect=\"italics\">H<sub>0<\/sub><\/em>.<\/p><p id=\"fs-idm21057040\">This means you reject <em data-effect=\"italics\">\u03bc<\/em> = 65. In other words, you believe the average test score is more than 65.<\/p><p><strong>Conclusion:<\/strong> At a 5% level of significance, the sample data show sufficient evidence that the mean (average) test score is more than 65, just as the math instructor thinks.<\/p><p>The <em data-effect=\"italics\">p<\/em>-value can easily be calculated.<\/p><div data-type=\"note\" data-has-label=\"true\" id=\"fs-idm14472336\" class=\"note statistics calculator\" data-label=\"\"><p>Put the data into a list. Press <code>STAT<\/code> and arrow over to <code>TESTS<\/code>. Press <code>2:T-Test<\/code>. Arrow over to <code>Data<\/code> and press <code>ENTER<\/code>. Arrow down and enter 65 for <em data-effect=\"italics\">\u03bc<\/em><sub>0<\/sub>, the name of the list where you put the data, and 1 for <code>Freq:<\/code>. Arrow down to <em data-effect=\"italics\">\u03bc<\/em>: and arrow over to &gt; <em data-effect=\"italics\">\u03bc<\/em><sub>0<\/sub>. Press <code>ENTER<\/code>. Arrow down to <code>Calculate<\/code> and press <code>ENTER<\/code>. The calculator not only calculates the <em data-effect=\"italics\">p<\/em>-value (<em data-effect=\"italics\">p<\/em> = 0.0396) but it also calculates the test statistic (<em data-effect=\"italics\">t<\/em>-score) for the sample mean, the sample mean, and the sample standard deviation. <em data-effect=\"italics\">\u03bc<\/em> &gt; 65 is the alternative hypothesis. Do this set of instructions again except arrow to <code>Draw<\/code> (instead of <code>Calculate<\/code>). Press <code>ENTER<\/code>. A shaded graph appears with <em data-effect=\"italics\">t<\/em> = 1.9781 (test statistic) and <em data-effect=\"italics\">p<\/em> = 0.0396 (<em data-effect=\"italics\">p<\/em>-value). Make sure when you use <code>Draw<\/code> that no other equations are highlighted in <em data-effect=\"italics\">Y<\/em> =  and the plots are turned off.<\/p><\/div><\/div><\/div><\/div><div data-type=\"note\" data-has-label=\"true\" id=\"fs-idm186895712\" class=\"note statistics try\" data-label=\"\"><div data-type=\"title\" class=\"title\">Try It<\/div><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\"><p id=\"eip-idm59977568\">It is believed that a stock price for a particular company will grow at a rate of &#x1f4b2;5 per week with a standard deviation of &#x1f4b2;1. An investor believes the stock won\u2019t grow as quickly. The changes in stock price is recorded for ten weeks and are as follows: &#x1f4b2;4, &#x1f4b2;3, &#x1f4b2;2, &#x1f4b2;3, &#x1f4b2;1, &#x1f4b2;7, &#x1f4b2;2, &#x1f4b2;1, &#x1f4b2;1, &#x1f4b2;2. Perform a hypothesis test using a 5% level of significance. State the null and alternative hypotheses, find the <em data-effect=\"italics\">p<\/em>-value, state your conclusion, and identify the Type I and Type II errors.<\/p><\/div><div data-type=\"solution\" class=\"solution\" id=\"eip-348\"><p id=\"eip-idp45715760\"><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> = 5<\/p><p id=\"eip-idp96462720\"><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &lt; 5<\/p><p id=\"eip-idp8662576\"><em data-effect=\"italics\">p<\/em> = 0.0082<\/p><p id=\"eip-idp17544256\">Because <em data-effect=\"italics\">p<\/em> &lt; <em data-effect=\"italics\">\u03b1<\/em>, we reject the null hypothesis. There is sufficient evidence to suggest that the stock price of the company grows at a rate less than &#x1f4b2;5 a week.<\/p><p id=\"eip-idm775584\">Type I Error: To conclude that the stock price is growing slower than &#x1f4b2;5 a week when, in fact, the stock price is growing at &#x1f4b2;5 a week (reject the null hypothesis when the null hypothesis is true).<\/p><p id=\"eip-idp7466800\">Type II Error: To conclude that the stock price is growing at a rate of &#x1f4b2;5 a week when, in fact, the stock price is growing slower than &#x1f4b2;5 a week (do not reject the null hypothesis when the null hypothesis is false).<\/p><\/div><\/div><\/div><div data-type=\"example\" class=\"textbox examples\"><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\" id=\"id5304801\"><p>Joon believes that 50% of first-time brides in the United States are younger than their grooms. She performs a hypothesis test to determine if the percentage is <strong>the same or different from 50%<\/strong>. Joon samples <strong>100 first-time brides<\/strong> and <strong>53<\/strong> reply that they are younger than their grooms. For the hypothesis test, she uses a 1% level of\nsignificance.<\/p><\/div><div data-type=\"solution\" class=\"solution\" id=\"id4150001\"><p id=\"element-399\">Set up the hypothesis test:<\/p><p>The 1% level of significance means that <em data-effect=\"italics\">\u03b1<\/em> = 0.01. This is a <strong>test of a single population proportion<\/strong>.<\/p><p><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> = 0.50\u2003\u2003<em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2260 0.50<\/p><p>The words <strong>\"is the same or different from\"<\/strong> tell you this is a two-tailed test.<\/p><p>Calculate the distribution needed:<\/p><p id=\"element-783\"><strong>Random variable:<\/strong><em data-effect=\"italics\">P\u2032<\/em> = the percent of of first-time brides who are younger than their grooms.<\/p><p id=\"element-581\"><strong>Distribution for the test:<\/strong> The problem contains no mention of a mean. The\ninformation is given in terms of percentages. Use the distribution for <em data-effect=\"italics\">P\u2032<\/em>, the estimated proportion.<\/p><p id=\"fs-idm81154768\">\\({P}^{\\prime }~N\\left(p,\\sqrt{\\frac{p\\cdot q}{n}}\\right)\\) Therefore, \\({P}^{\\prime }~N\\left(0.5,\\sqrt{\\frac{0.5\\cdot 0.5}{100}}\\right)\\)<\/p><p id=\"fs-idm107085552\">where <em data-effect=\"italics\">p<\/em> = 0.50, <em data-effect=\"italics\">q<\/em> = 1\u2212<em data-effect=\"italics\">p<\/em> = 0.50, and <em data-effect=\"italics\">n<\/em> = 100<\/p><p id=\"fs-idp14246672\">Calculate the <em data-effect=\"italics\">p<\/em>-value using the normal distribution for proportions:<\/p><p><em data-effect=\"italics\">p<\/em>-value = <em data-effect=\"italics\">P<\/em> (<em data-effect=\"italics\">p\u2032<\/em> &lt; 0.47 or <em data-effect=\"italics\">p\u2032<\/em> &gt; 0.53) = 0.5485<\/p><p id=\"fs-idp133040416\">where <em data-effect=\"italics\">x<\/em> = 53, <em data-effect=\"italics\">p\u2032<\/em> = \\(\\frac{x}{n}\\text{\u00a0=\u00a0}\\frac{\\text{53}}{\\text{100}}\\) = 0.53.<\/p><p><strong>Interpretation of the <em data-effect=\"italics\">p<\/em>-value:<\/strong> If the null hypothesis is true, there is 0.5485 probability\n(54.85%) that the sample (estimated) proportion \\(p\\text{'}\\) is 0.53 or more OR 0.47 or less (see\nthe graph in <a href=\"#hyptest11_ex5\" class=\"autogenerated-content\">[link]<\/a>).<\/p><div class=\"bc-figure figure\" id=\"hyptest11_ex5\"><span data-type=\"media\" id=\"id5140475\" data-alt=\"Normal distribution curve of the percent of first time brides who are younger than the groom with values of 0.47, 0.50, and 0.53 on the x-axis. Vertical upward lines extend from 0.47 and 0.53 to the curve. 1\/2(p-values) are calculated for the areas on outsides of 0.47 and 0.53.\"><img src=\"http:\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/fig-ch09_11_05N.jpg\" data-media-type=\"image\/jpg\" alt=\"Normal distribution curve of the percent of first time brides who are younger than the groom with values of 0.47, 0.50, and 0.53 on the x-axis. Vertical upward lines extend from 0.47 and 0.53 to the curve. 1\/2(p-values) are calculated for the areas on outsides of 0.47 and 0.53.\" width=\"380\" \/><\/span><\/div><p id=\"element-327\"><em data-effect=\"italics\">\u03bc<\/em> = <em data-effect=\"italics\">p<\/em> = 0.50 comes from <em data-effect=\"italics\">H<sub>0<\/sub><\/em>, the null hypothesis.<\/p><p><em data-effect=\"italics\">p\u2032<\/em> = 0.53. Since the curve is symmetrical and the test is two-tailed, the <em data-effect=\"italics\">p\u2032<\/em> for the left tail is equal to 0.50 \u2013 0.03 = 0.47 where <em data-effect=\"italics\">\u03bc<\/em> = <em data-effect=\"italics\">p<\/em> = 0.50. (0.03 is the difference between 0.53 and 0.50.)<\/p><p>Compare <em data-effect=\"italics\">\u03b1<\/em> and the <em data-effect=\"italics\">p<\/em>-value:<\/p><p>Since <em data-effect=\"italics\">\u03b1<\/em> = 0.01 and <em data-effect=\"italics\">p<\/em>-value = 0.5485. <em data-effect=\"italics\">\u03b1<\/em> &lt; <em data-effect=\"italics\">p<\/em>-value.<\/p><p><strong>Make a decision:<\/strong> Since <em data-effect=\"italics\">\u03b1<\/em> &lt; <em data-effect=\"italics\">p<\/em>-value, you cannot reject <em data-effect=\"italics\">H<sub>0<\/sub><\/em>.<\/p><p><strong>Conclusion:<\/strong> At the 1% level of significance, the sample data do not show sufficient evidence that the percentage of first-time brides who are younger than their grooms is different from 50%.<\/p><p>The <em data-effect=\"italics\">p<\/em>-value can easily be calculated.<\/p><div data-type=\"note\" data-has-label=\"true\" id=\"fs-idm35225600\" class=\"note statistics calculator\" data-label=\"\"><p>Press <code>STAT<\/code> and arrow over to <code>TESTS<\/code>. Press <code>5:1-PropZTest<\/code>. Enter .5 for <em data-effect=\"italics\">p<\/em><sub>0<\/sub>, 53 for <em data-effect=\"italics\">x<\/em> and 100 for <em data-effect=\"italics\">n<\/em>. Arrow down to <code>Prop<\/code> and arrow to <code>not equals<\/code> <em data-effect=\"italics\">p<\/em><sub>0<\/sub>. Press <code>ENTER<\/code>. Arrow down to <code>Calculate<\/code> and press <code>ENTER<\/code>. The calculator calculates the <em data-effect=\"italics\">p<\/em>-value (<em data-effect=\"italics\">p<\/em> = 0.5485) and the test statistic (<em data-effect=\"italics\">z<\/em>-score). <code>Prop not equals<\/code> .5 is the alternate hypothesis. Do this set of instructions again except arrow to <code>Draw<\/code> (instead of <code>Calculate<\/code>). Press <code>ENTER<\/code>. A shaded graph appears with <em data-effect=\"italics\">z<\/em> = 0.6 (test statistic) and <em data-effect=\"italics\">p<\/em> = 0.5485 (<em data-effect=\"italics\">p<\/em>-value). Make sure when you use <code>Draw<\/code> that no other equations are highlighted in <em data-effect=\"italics\">Y<\/em> = and the plots are turned off.<\/p><\/div><p>The Type I and Type II errors are as follows:<\/p><p>The Type I error is to conclude that the proportion of first-time brides who are younger than their grooms is different from 50% when, in fact, the proportion is actually 50%. (Reject the null hypothesis when the null hypothesis is true). <\/p><p id=\"fs-idp134008544\">The Type II error is there is not enough evidence to conclude that the proportion of first time brides who are younger than their grooms differs from 50% when, in fact, the proportion does differ from 50%. (Do not reject the null hypothesis when the null hypothesis is false.)<\/p><\/div><\/div><\/div><div data-type=\"note\" data-has-label=\"true\" id=\"fs-idp8711616\" class=\"note statistics try\" data-label=\"\"><div data-type=\"title\" class=\"title\">Try It<\/div><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\"><p id=\"eip-idp103490704\">A teacher believes that 85% of students in the class will want to go on a field trip to the local zoo. She performs a hypothesis test to determine if the percentage is the same or different from 85%. The teacher samples 50 students and 39 reply that they would want to go to the zoo. For the hypothesis test, use a 1% level of significance.<\/p><p id=\"eip-idp10599024\">First, determine what type of test this is, set up the hypothesis test, find the <em data-effect=\"italics\">p<\/em>-value, sketch the graph, and state your conclusion.<\/p><\/div><div data-type=\"solution\" class=\"solution\"><p id=\"eip-idp139727615275152\">Since the problem is about percentages, this is a test of single population proportions.<\/p><p id=\"eip-idp139727615275648\"><em data-effect=\"italics\">H<sub>0<\/sub><\/em> : <em data-effect=\"italics\">p<\/em> = 0.85<\/p><p id=\"eip-idp139727615276416\"><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2260 0.85<\/p><p id=\"eip-idp139727615277216\"><em data-effect=\"italics\">p<\/em> = 0.7554<\/p><div class=\"bc-figure figure\" id=\"fs-idp11874016\"><span data-type=\"media\" id=\"eip-idp139727615277600\" data-alt=\"\"><img src=\"http:\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/CNX_Stats_C09_M11_tryit002annoN.jpg\" data-media-type=\"image\/jpeg\" alt=\"\" width=\"380\" \/><\/span><\/div><p id=\"eip-idp95887808\">Because <em data-effect=\"italics\">p<\/em> &gt; <em data-effect=\"italics\">\u03b1<\/em>, we fail to reject the null hypothesis. There is not sufficient evidence to suggest that the proportion of students that want to go to the zoo is not 85%.<\/p><\/div><\/div><\/div><div data-type=\"example\" class=\"textbox examples\"><div data-type=\"exercise\" class=\"exercise\" id=\"element-312\"><div data-type=\"problem\" class=\"problem\" id=\"id5318755\"><p>Suppose a consumer group suspects that the proportion of households that have three cell phones is 30%. A cell phone  company has reason to believe that the proportion  is not 30%. Before they start a big advertising campaign, they conduct a hypothesis test. Their marketing people survey 150 households with the result that 43 of the households have three cell phones.<\/p><\/div><div data-type=\"solution\" class=\"solution\" id=\"id4213559\"><p>Set up the Hypothesis Test:<\/p><p><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> = 0.30 <em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2260 0.30<\/p><p>Determine the distribution needed:<\/p><p>The <strong>random variable<\/strong> is <em data-effect=\"italics\">P\u2032<\/em> = proportion of households that have three cell phones.<\/p><p>The <strong>distribution<\/strong> for the hypothesis test is \\(P\\text{'}~N\\left(0.30,\\sqrt{\\frac{\\left(0.30\\right)\\cdot \\left(0.70\\right)}{150}}\\right)\\)<\/p><div data-type=\"newline\" data-count=\"2\"><br \/><br \/><\/div><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\" id=\"id4146441\"><p>a. The value that helps determine the <em data-effect=\"italics\">p<\/em>-value is <em data-effect=\"italics\">p\u2032<\/em>. Calculate <em data-effect=\"italics\">p\u2032<\/em>.<\/p><\/div><div data-type=\"solution\" class=\"solution\" id=\"id4274849\" data-print-placement=\"end\"><p>a. <em data-effect=\"italics\">p\u2032<\/em> = \\(\\frac{x}{n}\\) where <em data-effect=\"italics\">x<\/em> is the number of successes and <em data-effect=\"italics\">n<\/em> is the total number in the sample.<\/p><p><em data-effect=\"italics\">x<\/em> = 43, <em data-effect=\"italics\">n<\/em> = 150<\/p><p><em data-effect=\"italics\">p\u2032<\/em> = \\(\\frac{\\text{43}}{\\text{150}}\\)\n<\/p><div data-type=\"newline\" data-count=\"2\"><br \/><br \/><\/div><\/div><\/div><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\" id=\"id5150803\"><p>b. What is a <strong>success<\/strong> for this problem?<\/p><\/div><div data-type=\"solution\" class=\"solution\" id=\"id5036464\" data-print-placement=\"end\"><p>b. A success is having three cell phones in a household.<\/p><div data-type=\"newline\" data-count=\"2\"><br \/><br \/><\/div><\/div><\/div><div data-type=\"exercise\" class=\"exercise\" id=\"element-519\"><div data-type=\"problem\" class=\"problem\" id=\"id4292096\"><p id=\"element-781\">c. What is the level of significance?<\/p><\/div><div data-type=\"solution\" class=\"solution\" id=\"id5242701\" data-print-placement=\"end\"><p>c. The level of significance is the preset <em data-effect=\"italics\">\u03b1<\/em>. Since <em data-effect=\"italics\">\u03b1<\/em> is not given, assume that <em data-effect=\"italics\">\u03b1<\/em> = 0.05.<\/p><div data-type=\"newline\" data-count=\"2\"><br \/><br \/><\/div><\/div><\/div><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\" id=\"id4673389\"><p>d. Draw the graph for this problem. Draw the horizontal axis. Label and shade appropriately.<\/p><div data-type=\"newline\"><br \/><\/div>Calculate the <em data-effect=\"italics\">p<\/em>-value.<\/div><div data-type=\"solution\" class=\"solution\" id=\"id5212164\" data-print-placement=\"end\"><p>d. <em data-effect=\"italics\">p<\/em>-value = 0.7216<\/p><div data-type=\"newline\" data-count=\"2\"><br \/><br \/><\/div>\n  <\/div><\/div><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\" id=\"id5104601\"><p>e. Make a decision. _____________(Reject\/Do not reject) <em data-effect=\"italics\">H<sub>0<\/sub><\/em> because____________.<\/p><\/div><div data-type=\"solution\" class=\"solution\" id=\"id5137008\" data-print-placement=\"end\"><p>e. Assuming that <em data-effect=\"italics\">\u03b1<\/em> = 0.05, <em data-effect=\"italics\">\u03b1<\/em> &lt; <em data-effect=\"italics\">p<\/em>-value. The decision is do not reject <em data-effect=\"italics\">H<sub>0<\/sub><\/em> because there is not sufficient evidence to conclude that the proportion of households that have three cell phones is not 30%.<\/p><\/div><\/div><\/div><\/div><\/div><div data-type=\"note\" data-has-label=\"true\" id=\"fs-idm14663600\" class=\"note statistics try\" data-label=\"\"><div data-type=\"title\" class=\"title\">Try It<\/div><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\"><p id=\"eip-idm17000432\">Marketers believe that 92% of adults in the United States own a cell phone. A cell phone manufacturer believes that number is actually lower. 200 American adults are surveyed, of which, 174 report having cell phones. Use a 5% level of significance. State the null and alternative hypothesis, find the <em data-effect=\"italics\">p<\/em>-value, state your conclusion, and identify the Type I and Type II errors.<\/p><\/div><div data-type=\"solution\" class=\"solution\"><p id=\"eip-idm64882736\"><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> = 0.92<\/p><p id=\"eip-idm90844256\"><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> &lt; 0.92<\/p><p id=\"eip-idm90843872\"><em data-effect=\"italics\">p<\/em>-value = 0.0046<\/p><p id=\"eip-idm90843488\">Because <em data-effect=\"italics\">p<\/em> &lt; 0.05, we reject the null hypothesis. There is sufficient evidence to conclude that fewer than 92% of American adults own cell phones.<\/p><p id=\"eip-idm6499552\">Type I Error: To conclude that fewer than 92% of American adults own cell phones when, in fact, 92% of American adults do own cell phones (reject the null hypothesis when the null hypothesis is true).<\/p><p id=\"eip-idm80852000\">Type II Error: To conclude that 92% of American adults own cell phones when, in fact, fewer than 92% of American adults own cell phones (do not reject the null hypothesis when the null hypothesis is false).<\/p><\/div><\/div><\/div><p id=\"fs-idp32935184\">The next example is a poem written by a statistics student named Nicole Hart. The solution to the problem follows the poem. Notice that the hypothesis test is for a single population proportion. This means that the null and alternate hypotheses use the parameter <em data-effect=\"italics\">p<\/em>. The distribution for the test is normal. The estimated proportion <em data-effect=\"italics\">p<\/em>\u2032 is the proportion of fleas killed to the total fleas found on Fido. This is sample information. The problem gives a preconceived <em data-effect=\"italics\">\u03b1<\/em> = 0.01, for comparison, and a 95% confidence interval computation. The poem is clever and humorous, so please enjoy it!<\/p><div data-type=\"example\" class=\"textbox examples\"><div data-type=\"exercise\" class=\"exercise\" id=\"fs-idp65427488\"><div data-type=\"problem\" class=\"problem\" id=\"id5257658\"><p id=\"eip-idm151173424\">My dog has so many fleas,<\/p><div data-type=\"newline\"><br \/><\/div>\nThey do not come off with ease.<div data-type=\"newline\"><br \/><\/div>\nAs for shampoo, I have tried many types<div data-type=\"newline\"><br \/><\/div>\nEven one called Bubble Hype,<div data-type=\"newline\"><br \/><\/div>\nWhich only killed 25% of the fleas,<div data-type=\"newline\"><br \/><\/div>\nUnfortunately I was not pleased.<div data-type=\"newline\" data-count=\"2\"><br \/><br \/><\/div>\nI've used all kinds of soap,<div data-type=\"newline\"><br \/><\/div>\nUntil I had given up hope<div data-type=\"newline\"><br \/><\/div>\nUntil one day I saw<div data-type=\"newline\"><br \/><\/div>\nAn ad that put me in awe.<div data-type=\"newline\" data-count=\"2\"><br \/><br \/><\/div>\nA shampoo used for dogs<div data-type=\"newline\"><br \/><\/div>\nCalled GOOD ENOUGH to Clean a Hog<div data-type=\"newline\"><br \/><\/div>\nGuaranteed to kill more fleas.<div data-type=\"newline\" data-count=\"2\"><br \/><br \/><\/div>\nI gave Fido a bath<div data-type=\"newline\"><br \/><\/div>\nAnd after doing the math<div data-type=\"newline\"><br \/><\/div>\nHis number of fleas<div data-type=\"newline\"><br \/><\/div>\nStarted dropping by 3's!<div data-type=\"newline\" data-count=\"2\"><br \/><br \/><\/div>\nBefore his shampoo<div data-type=\"newline\"><br \/><\/div>\nI counted 42.<div data-type=\"newline\"><br \/><\/div>\nAt the end of his bath,<div data-type=\"newline\"><br \/><\/div>\nI redid the math<div data-type=\"newline\"><br \/><\/div>\nAnd the new shampoo had killed 17 fleas.<div data-type=\"newline\"><br \/><\/div>\nSo now I was pleased.<div data-type=\"newline\" data-count=\"2\"><br \/><br \/><\/div>\nNow it is time for you to have some fun<div data-type=\"newline\"><br \/><\/div>\nWith the level of significance being .01,<div data-type=\"newline\"><br \/><\/div>\nYou must help me figure out<div data-type=\"newline\"><br \/><\/div>\nUse the new shampoo or go without?<\/div><div data-type=\"solution\" class=\"solution\" id=\"id5100583\"><p>Set up the hypothesis test:<\/p><p id=\"element-278\"><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2264 0.25 \u2003\u2003<em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> &gt; 0.25<\/p><p>Determine the distribution needed:<\/p><p>In words, CLEARLY state what your random variable \\(\\overline{X}\\) or <em data-effect=\"italics\">P\u2032<\/em> represents.<\/p><p id=\"fs-idp88897968\"><em data-effect=\"italics\">P\u2032<\/em> = The proportion of fleas that are killed by the new shampoo<\/p><p>State the distribution to use for the test.<\/p><p><strong>Normal:<\/strong>\\(N\\left(0.25,\\sqrt{\\frac{\\left(0.25\\right)\\left(1-0.25\\right)}{42}}\\right)\\)<\/p><p><strong>Test Statistic:<\/strong><em data-effect=\"italics\">z<\/em> = 2.3163<\/p><p>Calculate the <em data-effect=\"italics\">p<\/em>-value using the normal distribution for proportions:<\/p><p><em data-effect=\"italics\">p<\/em>-value = 0.0103<\/p><p>In one to two complete sentences, explain what the <em data-effect=\"italics\">p<\/em>-value means for this problem.<\/p><p>If the null hypothesis is true (the proportion is 0.25), then there is a 0.0103 probability that the sample (estimated) proportion is 0.4048 \\(\\left(\\frac{17}{42}\\right)\\) or more.<\/p><p>Use the previous information to sketch a picture of this situation. CLEARLY, label and scale the horizontal axis and\nshade the region(s) corresponding to the <em data-effect=\"italics\">p<\/em>-value.<\/p><div class=\"bc-figure figure\" id=\"hyptest11_ex6\"><span data-type=\"media\" id=\"id4295819\" data-alt=\"Normal distribution graph of the proportion of fleas killed by the new shampoo with values of 0.25 and 0.4048 on the x-axis. A vertical upward line extends from 0.4048 to the curve and the area to the left of this is shaded in. The test statistic of the sample proportion is listed.\"><img src=\"http:\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/fig-ch09_11_06.jpg\" data-media-type=\"image\/jpg\" alt=\"Normal distribution graph of the proportion of fleas killed by the new shampoo with values of 0.25 and 0.4048 on the x-axis. A vertical upward line extends from 0.4048 to the curve and the area to the left of this is shaded in. The test statistic of the sample proportion is listed.\" width=\"380\" \/><\/span><\/div><p id=\"element-372\">Compare <em data-effect=\"italics\">\u03b1<\/em> and the <em data-effect=\"italics\">p<\/em>-value:<\/p><p>Indicate the correct decision (\u201creject\u201d or \u201cdo not reject\u201d the null hypothesis), the reason for it, and write an appropriate conclusion, using complete sentences.<\/p><table summary=\"\"><thead valign=\"middle\"><tr><th data-align=\"center\">alpha<\/th><th data-align=\"center\">decision<\/th><th data-align=\"center\">reason for decision<\/th><\/tr><\/thead><tbody><tr><td data-align=\"center\">0.01<\/td><td data-align=\"center\">Do not reject \\({H}_{0}\\)<\/td><td data-align=\"center\"><em data-effect=\"italics\">\u03b1<\/em> &lt; <em data-effect=\"italics\">p<\/em>-value<\/td><\/tr><\/tbody><\/table><p><strong>Conclusion:<\/strong> At the 1% level of significance, the sample data do not show sufficient evidence that the percentage of fleas that are killed by the new shampoo is more than 25%.<\/p><p>Construct a 95% confidence interval for the true mean or proportion. Include a sketch of the graph of the situation.\nLabel the point estimate and the lower and upper bounds of the confidence interval.<\/p><p><\/p><div class=\"bc-figure figure\" id=\"hyptest11_ex7\"><span data-type=\"media\" id=\"id5244822\" data-alt=\"Normal distribution graph of the proportion of fleas killed by the new shampoo with values of 0.26, 17\/42, and 0.55 on the x-axis. A vertical upward line extends from 0.26 and 0.55. The area between these two points is equal to 0.95.\"><img src=\"http:\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/fig-ch09_11_07.jpg\" data-media-type=\"image\/jpg\" alt=\"Normal distribution graph of the proportion of fleas killed by the new shampoo with values of 0.26, 17\/42, and 0.55 on the x-axis. A vertical upward line extends from 0.26 and 0.55. The area between these two points is equal to 0.95.\" width=\"380\" \/><\/span><\/div><p><strong>Confidence Interval:<\/strong> (0.26,0.55) We are 95% confident that the true population proportion <em data-effect=\"italics\">p<\/em> of fleas that are killed by the new shampoo is between 26% and 55%.<\/p><div data-type=\"note\" class=\"note\" data-has-label=\"true\" id=\"id5328435\" data-label=\"\"><div data-type=\"title\" class=\"title\">Note<\/div><p id=\"eip-idm81920704\">This test result is not very definitive since the <em data-effect=\"italics\">p<\/em>-value is very close to alpha. In reality, one would probably do more tests by giving the dog another bath after the fleas have had a chance to return.<\/p><\/div><\/div><\/div><\/div><div data-type=\"example\" class=\"textbox examples\"><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\"><p id=\"eip-idm29000496\">The National Institute of Standards and Technology provides exact data on conductivity properties of materials. Following are conductivity measurements for 11 randomly selected pieces of a particular type of glass.<\/p><p id=\"eip-idm46067920\">1.11; 1.07; 1.11; 1.07; 1.12; 1.08; .98; .98 1.02; .95; .95\n<\/p><div data-type=\"newline\"><br \/><\/div>Is there convincing evidence that the average conductivity of this type of glass is greater than one? Use a significance level of 0.05. Assume the population is normal.<\/div><div data-type=\"solution\" class=\"solution\"><p id=\"eip-idp69766080\">Let\u2019s follow a four-step process to answer this statistical question.\n<\/p><div data-type=\"list\" data-list-type=\"enumerated\" id=\"eip-idp69766512\" data-number-style=\"arabic\"><div data-type=\"item\"><strong>State the Question<\/strong>: We need to determine if, at a 0.05 significance level, the average conductivity of the selected glass is greater than one. Our hypotheses will be\n<div data-type=\"list\" data-list-type=\"enumerated\" id=\"eip-idp129554720\" data-number-style=\"lower-alpha\"><div data-type=\"item\"><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> \u2264 1<\/div><div data-type=\"item\"><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &gt; 1<\/div><\/div><\/div><div data-type=\"item\"><strong>Plan<\/strong>: We are testing a sample mean without a known population standard deviation. Therefore, we need to use a Student's-t distribution. Assume the underlying population is normal.<\/div><div data-type=\"item\"><strong>Do the calculations<\/strong>: We will input the sample data into the TI-83 as follows.\n\n<div class=\"bc-figure figure\" id=\"eip-idp122053552\"><span data-type=\"media\" id=\"eip-idp167220912\" data-alt=\"\"><img src=\"http:\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/CNX_Stats_C09_M09_001a.jpg\" data-media-type=\"image\/jpeg\" alt=\"\" width=\"250\" \/><\/span><\/div><div class=\"bc-figure figure\" id=\"eip-idp29927184\"><span data-type=\"media\" id=\"eip-idp29927312\" data-alt=\"\"><img src=\"http:\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/CNX_Stats_C09_M09_001b.jpg\" data-media-type=\"image\/jpeg\" alt=\"\" width=\"250\" \/><\/span><\/div><div class=\"bc-figure figure\" id=\"eip-idp126675984\"><span data-type=\"media\" id=\"eip-idp156983600\" data-alt=\"\"><img src=\"http:\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/CNX_Stats_C09_M09_001c.jpg\" data-media-type=\"image\/jpeg\" alt=\"\" width=\"250\" \/><\/span><\/div><div class=\"bc-figure figure\" id=\"eip-idp154420944\"><span data-type=\"media\" id=\"eip-idp154421072\" data-alt=\"\"><img src=\"http:\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/CNX_Stats_C09_M09_001d.jpg\" data-media-type=\"image\/jpeg\" alt=\"\" width=\"250\" \/><\/span><\/div><\/div><div data-type=\"item\"><strong>State the Conclusions<\/strong>: Since the <em data-effect=\"italics\">p<\/em>-value* (<em data-effect=\"italics\">p<\/em> = 0.036) is less than our alpha value, we will reject the null hypothesis. It is reasonable to state that the data supports the claim that the average conductivity level is greater than one.<\/div><\/div>\n<\/div><\/div><\/div><div data-type=\"example\" class=\"textbox examples\" id=\"eip-767\"><div data-type=\"exercise\" class=\"exercise\" id=\"eip-178\"><div data-type=\"problem\" class=\"problem\"><p id=\"eip-idp173526416\">In a study of 420,019 cell phone users, 172 of the subjects developed brain cancer. Test the claim that cell phone users developed brain cancer at a greater rate than that for non-cell phone users (the rate of brain cancer for non-cell phone users is 0.0340%). Since this is a critical issue, use a 0.005 significance level. Explain why the significance level should be so low in terms of a Type I error.<\/p><\/div><div data-type=\"solution\" class=\"solution\"><p id=\"eip-idp3953440\">We will follow the four-step process.<\/p><ol id=\"eip-idp191218624\" data-number-style=\"arabic\"><li>We need to conduct a hypothesis test on the claimed cancer rate. Our hypotheses will be\n<ol id=\"eip-idp171226608\" data-number-style=\"lower-alpha\"><li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2264 0.00034<\/li><li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> &gt; 0.00034<\/li><\/ol><p id=\"eip-idm39752176\" class=\"finger\">If we commit a Type I error, we are essentially accepting a false claim. Since the claim describes cancer-causing environments, we want to minimize the chances of incorrectly identifying causes of cancer.<\/p>\n<\/li><li>We will be testing a sample proportion with <em data-effect=\"italics\">x<\/em> = 172 and <em data-effect=\"italics\">n<\/em> = 420,019. The sample is sufficiently large because we have <em data-effect=\"italics\">np<\/em> = 420,019(0.00034) = 142.8, <em data-effect=\"italics\">nq<\/em> = 420,019(0.99966) = 419,876.2, two independent outcomes, and a fixed probability of success <em data-effect=\"italics\">p<\/em> = 0.00034. Thus we will be able to generalize our results to the population.<\/li><li>The associated TI results are \n<div data-type=\"newline\"><br \/><\/div>\n<div class=\"bc-figure figure\" id=\"fs-idm80776672\"><span data-type=\"media\" id=\"eip-idp39356816\" data-alt=\"\"><img src=\"http:\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/CNX_Stats_C09_M09_002a.jpg\" data-media-type=\"image\/jpeg\" alt=\"\" width=\"250\" \/><\/span><\/div><div class=\"bc-figure figure\" id=\"eip-idp38098208\"><span data-type=\"media\" id=\"eip-idp38098336\" data-alt=\"\"><img src=\"http:\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/CNX_Stats_C09_M09_002b.jpg\" data-media-type=\"image\/jpeg\" alt=\"\" width=\"250\" \/><\/span><\/div><\/li><li>Since the <em data-effect=\"italics\">p<\/em>-value = 0.0073 is greater than our alpha value = 0.005, we cannot reject the null. Therefore, we conclude that there is not enough evidence to support the claim of higher brain cancer rates for the cell phone users.<\/li><\/ol><\/div><\/div><\/div><div data-type=\"example\" class=\"textbox examples\"><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\"><p id=\"eip-idp47042256\">According to the US Census there are approximately 268,608,618 residents aged 12 and older. Statistics from the Rape, Abuse, and Incest National Network indicate that, on average, 207,754 rapes occur each year (male and female) for persons aged 12 and older. This translates into a percentage of sexual assaults of 0.078%. In Daviess County, KY, there were reported 11 rapes for a population of 37,937. Conduct an appropriate hypothesis test to determine if there is a statistically significant difference between the local sexual assault percentage and the national sexual assault percentage. Use a significance level of 0.01.<\/p><\/div><div data-type=\"solution\" class=\"solution\"><p>We will follow the four-step plan.<\/p><ol id=\"eip-idp113522624\" data-number-style=\"arabic\"><li>We need to test whether the proportion of sexual assaults in Daviess County, KY is significantly different from the national average.<\/li><li>Since we are presented with proportions, we will use a one-proportion <em data-effect=\"italics\">z<\/em>-test. The hypotheses for the test will be\n<ol id=\"eip-idp17854992\" data-number-style=\"lower-alpha\"><li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> = 0.00078<\/li><li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2260 0.00078<\/li><\/ol><\/li><li>The following screen shots display the summary statistics from the hypothesis test.\n\n<div class=\"bc-figure figure\" id=\"fs-idm3544720\"><span data-type=\"media\" id=\"eip-idp26791072\" data-alt=\"\"><img src=\"http:\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/CNX_Stats_C09_M09_003a.jpg\" data-media-type=\"image\/jpeg\" alt=\"\" width=\"250\" \/><\/span><\/div><div class=\"bc-figure figure\" id=\"eip-idp38471600\"><span data-type=\"media\" id=\"eip-idp38471728\" data-alt=\"\"><img src=\"http:\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/CNX_Stats_C09_M09_003b.jpg\" data-media-type=\"image\/jpeg\" alt=\"\" width=\"250\" \/><\/span><\/div><\/li><li>Since the <em data-effect=\"italics\">p<\/em>-value, <em data-effect=\"italics\">p<\/em> = 0.00063, is less than the alpha level of 0.01, the sample data indicates that we should reject the null hypothesis. In conclusion, the sample data support the claim that the proportion of sexual assaults in Daviess County, Kentucky is different from the national average proportion.<\/li><\/ol><\/div><\/div><\/div><div class=\"summary\" data-depth=\"1\" id=\"fs-idp69608272\"><h1 data-type=\"title\">Chapter Review<\/h1><p id=\"fs-idp168110512\">The <span data-type=\"term\">hypothesis test<\/span> itself has an established process. This can be summarized as follows:\n<\/p><div data-type=\"list\" data-list-type=\"enumerated\"><div data-type=\"item\">Determine <em data-effect=\"italics\">H<sub>0<\/sub><\/em> and <em data-effect=\"italics\">H<sub>a<\/sub><\/em>. Remember, they are contradictory.<\/div><div data-type=\"item\">Determine the random variable.<\/div><div data-type=\"item\">Determine the distribution for the test.<\/div><div data-type=\"item\">Draw a graph, calculate the test statistic, and use the test statistic to calculate the\n<em data-effect=\"italics\">p<\/em>-value. (A <em data-effect=\"italics\">z<\/em>-score and a <em data-effect=\"italics\">t<\/em>-score are examples of test statistics.)<\/div><div data-type=\"item\">Compare the preconceived <em data-effect=\"italics\">\u03b1<\/em> with the <em data-effect=\"italics\">p<\/em>-value, make a decision (reject or do not reject <em data-effect=\"italics\">H<sub>0<\/sub><\/em>), and write a clear conclusion using English sentences.<\/div><\/div><p>Notice that in performing the hypothesis test, you use <em data-effect=\"italics\">\u03b1<\/em> and not <em data-effect=\"italics\">\u03b2<\/em>. <em data-effect=\"italics\">\u03b2<\/em> is needed to help determine the sample size of the data that is used in calculating the <em data-effect=\"italics\">p<\/em>-value. Remember that the quantity 1 \u2013 <em data-effect=\"italics\">\u03b2<\/em> is called the <strong>Power of the Test<\/strong>. A high power is desirable. If the power is too low, statisticians typically increase the sample size while keeping <em data-effect=\"italics\">\u03b1<\/em> the same.If the power is low, the null hypothesis might not be rejected when it should be.<\/p><\/div><div class=\"practice\" data-depth=\"1\"><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\" id=\"eip-135\"><p>Assume <em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> = 9 and <em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &lt; 9. Is this a left-tailed, right-tailed, or two-tailed test?<\/p><\/div><div data-type=\"solution\" class=\"solution\"><p>This is a left-tailed test.<\/p><\/div><\/div><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\" id=\"eip-286\"><p>Assume <em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> \u2264 6 and <em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &gt; 6. Is this a left-tailed, right-tailed, or two-tailed test?<\/p><\/div><\/div><div data-type=\"exercise\" class=\"exercise\" id=\"fs-idp45725152\"><div data-type=\"problem\" class=\"problem\" id=\"eip-116\"><p>Assume <em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> = 0.25 and <em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2260 0.25. Is this a left-tailed, right-tailed, or two-tailed test?<\/p><\/div><div data-type=\"solution\" class=\"solution\"><p>This is a two-tailed test.<\/p><\/div><\/div><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\"><p>Draw the general graph of a left-tailed test.<\/p><\/div><\/div><div data-type=\"exercise\" class=\"exercise\" id=\"eip-18\"><div data-type=\"problem\" class=\"problem\"><p>\nDraw the graph of a two-tailed test.\n  <\/p><\/div><div data-type=\"solution\" class=\"solution\"><div class=\"bc-figure figure\" id=\"fs-idm149132800\"><span data-type=\"media\" id=\"eip-idp62459136\" data-alt=\"\"><img src=\"http:\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/CNX_Stats_C09_M09_item002annoN.jpg\" data-media-type=\"image\/jpeg\" alt=\"\" width=\"380\" \/><\/span><\/div><\/div><\/div><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\" id=\"eip-648\"><p>A bottle of water is labeled as containing 16 fluid ounces of water. You believe it is less than that. What type of test would you use?<\/p><\/div><\/div><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\"><p>Your friend claims that his mean golf score is 63. You want to show that it is higher than that. What type of test would you use?<\/p><\/div><div data-type=\"solution\" class=\"solution\"><p id=\"eip-68\">a right-tailed test<\/p><\/div><\/div><div data-type=\"exercise\" class=\"exercise\" id=\"eip-729\"><div data-type=\"problem\" class=\"problem\" id=\"eip-491\"><p>A bathroom scale claims to be able to identify correctly any weight within a pound. You think that it cannot be that accurate. What type of test would you use?<\/p><\/div><\/div><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\"><p id=\"eip-458\">You flip a coin and record whether it shows heads or tails. You know the probability of getting heads is 50%, but you think it is less for this particular coin. What type of test would you use?<\/p><\/div><div data-type=\"solution\" class=\"solution\" id=\"fs-idp66689312\"><p id=\"eip-139\">a left-tailed test<\/p><\/div><\/div><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\" id=\"eip-519\"><p>If the alternative hypothesis has a not equals ( \u2260 ) symbol, you know to use which type of test?<\/p><\/div><\/div><div data-type=\"exercise\" class=\"exercise\" id=\"eip-504\"><div data-type=\"problem\" class=\"problem\" id=\"eip-634\"><p>Assume the null hypothesis states that the mean is at least 18. Is this a left-tailed, right-tailed, or two-tailed test?<\/p><\/div><div data-type=\"solution\" class=\"solution\" id=\"eip-625\"><p>This is a left-tailed test.<\/p><\/div><\/div><div data-type=\"exercise\" class=\"exercise\" id=\"eip-320\"><div data-type=\"problem\" class=\"problem\"><p id=\"eip-621\">Assume the null hypothesis states that the mean is at most 12. Is this a left-tailed, right-tailed, or two-tailed test?<\/p><\/div><\/div><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\"><p>Assume the null hypothesis states that the mean is equal to 88. The alternative hypothesis states that the mean is not equal to 88. Is this a left-tailed, right-tailed, or two-tailed test?<\/p><\/div><div data-type=\"solution\" class=\"solution\"><p>This is a two-tailed test.<\/p><\/div><\/div>&lt;!--moved from 9.3--&gt;<\/div><div class=\"free-response\" data-depth=\"1\" id=\"fs-idp139727497473120\"><h1 data-type=\"title\">Homework<\/h1><p id=\"eip-idm30094416\"><em data-effect=\"italics\">For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in <a href=\"\/contents\/c0449c55-aa47-4f1c-bd5f-0521652f0e82@5\" class=\"autogenerated-content\">[link]<\/a>. Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files.<\/em><\/p><div data-type=\"note\" class=\"note\" data-has-label=\"true\" id=\"id8395739\" data-label=\"\"><div data-type=\"title\" class=\"title\">Note<\/div><p id=\"eip-idm120632832\">If you are using a Student's-<em data-effect=\"italics\">t<\/em> distribution for one of the following homework problems, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, however.)<\/p><\/div><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\" id=\"id8395759\"><p id=\"fs-idp44498592\">A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8,000. A survey of owners of that tire design is conducted. From the 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9,800 miles. Using alpha = 0.05, is the data highly inconsistent with the claim?<\/p><\/div><div data-type=\"solution\" class=\"solution\" id=\"fs-idm164410224\"><ol id=\"fs-idp21859504\" data-number-style=\"lower-alpha\"><li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> \u2265 50,000<\/li><li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &lt; 50,000<\/li><li>Let \\(\\overline{X}\\) = the average lifespan of a brand of tires.<\/li><li>normal distribution<\/li><li><em data-effect=\"italics\">z<\/em> = -2.315<\/li><li><em data-effect=\"italics\">p<\/em>-value = 0.0103<\/li><li>Check student\u2019s solution.<\/li><li><ol id=\"fs-idm113459088\" data-number-style=\"lower-roman\"><li>alpha: 0.05<\/li><li>Decision: Reject the null hypothesis.<\/li><li>Reason for decision: The <em data-effect=\"italics\">p<\/em>-value is less than 0.05.<\/li><li>Conclusion: There is sufficient evidence to conclude that the mean lifespan of the tires is less than 50,000 miles.<\/li><\/ol><\/li><li>(43,537, 49,463)<\/li><\/ol><\/div><\/div><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\" id=\"id8395793\"><p>From generation to generation, the mean age when smokers first start to smoke varies. However, the standard deviation of that age remains constant of around 2.1 years. A survey of 40 smokers of this generation was done to see if the mean starting age is at least 19. The sample mean was 18.1 with a sample standard deviation of 1.3. Do the data support the claim at the 5% level?\n  <\/p><\/div><\/div><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\" id=\"id9424298\"><p>The cost of a daily newspaper varies from city to city. However, the variation among prices remains steady with a standard deviation of 20\u00a2. A study was done to test the claim that the mean cost of a daily newspaper is &#x1f4b2;1.00. Twelve costs yield a mean cost of 95\u00a2 with a standard deviation of 18\u00a2. Do the data support the claim at the 1% level?\n  <\/p><\/div><div data-type=\"solution\" class=\"solution\" id=\"fs-idm17421856\"><ol id=\"fs-idm17421600\" data-number-style=\"lower-alpha\"><li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> = &#x1f4b2;1.00<\/li><li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> \u2260 &#x1f4b2;1.00<\/li><li>Let \\(\\overline{X}\\) = the average cost of a daily newspaper.<\/li><li>normal distribution<\/li><li><em data-effect=\"italics\">z<\/em> = \u20130.866<\/li><li><em data-effect=\"italics\">p<\/em>-value = 0.3865<\/li><li>Check student\u2019s solution.<\/li><li><ol id=\"fs-idm69728368\" data-number-style=\"lower-roman\"><li>Alpha: 0.01<\/li><li>Decision: Do not reject the null hypothesis.<\/li><li>Reason for decision: The <em data-effect=\"italics\">p<\/em>-value is greater than 0.01.<\/li><li>Conclusion: There is sufficient evidence to support the claim that the mean cost of daily papers is &#x1f4b2;1. The mean cost could be &#x1f4b2;1.<\/li><\/ol><\/li><li>(&#x1f4b2;0.84, &#x1f4b2;1.06)<\/li><\/ol><\/div><\/div><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\" id=\"id9424338\"><p id=\"element-977\">An article in the <em data-effect=\"italics\">San Jose Mercury News<\/em> stated that students in the California state university system take 4.5 years, on average, to finish their undergraduate degrees. Suppose you believe that the mean time is longer. You conduct a survey of 49 students and obtain a sample mean of 5.1 with a sample standard deviation of 1.2. Do the data support your claim at the 1% level?<\/p><\/div><\/div><div data-type=\"exercise\" class=\"exercise\" id=\"element-932a\"><div data-type=\"problem\" class=\"problem\" id=\"id9424572\"><p>The mean number of sick days an employee takes per year is believed to be about ten. Members of a personnel department do not believe this figure. They randomly survey eight employees. The number of sick days they took for the past year are as follows: 12; 4; 15; 3; 11; 8; 6; 8. Let <em data-effect=\"italics\">x<\/em> = the number of sick days they took for the past year. Should the personnel team believe that the mean number is ten?  <\/p><\/div><div data-type=\"solution\" class=\"solution\" id=\"fs-idm143535568\"><ol id=\"fs-idm8405344\" data-number-style=\"lower-alpha\"><li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> = 10<\/li><li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> \u2260 10<\/li><li>Let \\(\\overline{X}\\) the mean number of sick days an employee takes per year.<\/li><li>Student\u2019s <em data-effect=\"italics\">t<\/em>-distribution<\/li><li><em data-effect=\"italics\">t<\/em> = \u20131.12<\/li><li><em data-effect=\"italics\">p<\/em>-value = 0.300<\/li><li>Check student\u2019s solution.<\/li><li><ol id=\"fs-idm68554224\" data-number-style=\"lower-roman\"><li>Alpha: 0.05<\/li><li>Decision: Do not reject the null hypothesis.<\/li><li>Reason for decision: The <em data-effect=\"italics\">p<\/em>-value is greater than 0.05.<\/li><li>Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the mean number of sick days is not ten.<\/li><\/ol><\/li><li>(4.9443, 11.806)<\/li><\/ol><\/div><\/div><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\" id=\"id9424624\"><p>In 1955, <em data-effect=\"italics\">Life Magazine<\/em> reported that the 25 year-old mother of three worked, on average, an 80 hour week. Recently, many groups have been studying whether or not the women's movement has, in fact, resulted in an increase in the average work week for women (combining employment and at-home work). Suppose a study was done to determine if the mean work week has increased. 81 women were surveyed with the following results. The sample mean was 83; the sample standard deviation was ten. Does it appear that the mean work week has increased for women at the 5% level?\n  <\/p><\/div><\/div><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\" id=\"id9400577\"><p>\n    Your statistics instructor claims that 60 percent of the students who take her Elementary Statistics class go through life feeling more enriched. For some reason that she can't quite figure out, most people don't believe her. You decide to check this out on your own. You randomly survey 64 of her past Elementary Statistics students and find that 34 feel more enriched as a result of her class. Now, what do you think?\n  <\/p><\/div><div data-type=\"solution\" class=\"solution\" id=\"fs-idm68403536\"><ol id=\"fs-idm150879872\" data-number-style=\"lower-alpha\"><li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2265 0.6<\/li><li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> &lt; 0.6<\/li><li>Let <em data-effect=\"italics\">P\u2032<\/em> = the proportion of students who feel more enriched as a result of taking Elementary Statistics.<\/li><li>normal for a single proportion<\/li><li>1.12<\/li><li><em data-effect=\"italics\">p<\/em>-value =  0.1308<\/li><li>Check student\u2019s solution.<\/li><li><ol id=\"fs-idp14435968\" data-number-style=\"lower-roman\"><li>Alpha: 0.05<\/li><li>Decision: Do not reject the null hypothesis.<\/li><li>Reason for decision: The <em data-effect=\"italics\">p<\/em>-value is greater than 0.05.<\/li><li>Conclusion: There is insufficient evidence to conclude that less than 60 percent of her students feel more enriched.<\/li><\/ol><\/li><li>Confidence Interval: (0.409,  0.654)<div data-type=\"newline\"><br \/><\/div>The \u201cplus-4s\u201d confidence interval is (0.411, 0.648)<\/li><\/ol><\/div><\/div><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\" id=\"id9400611\"><p>A Nissan Motor Corporation advertisement read, \u201cThe average man\u2019s I.Q. is 107. The average brown trout\u2019s I.Q. is 4. So why can\u2019t man catch brown trout?\u201d Suppose you believe that the brown trout\u2019s mean I.Q. is greater than four. You catch 12 brown trout. A fish psychologist determines the I.Q.s as follows: 5; 4; 7; 3; 6; 4; 5; 3; 6; 3; 8; 5. Conduct a hypothesis test of your belief.<\/p><\/div><\/div><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\" id=\"id9400954\"><p>Refer to <a href=\"#element-410\">Exercise 9.119<\/a>. Conduct a hypothesis test to see if your decision and conclusion would change if your belief were that the brown trout\u2019s mean I.Q. is <strong>not<\/strong> four.<\/p><\/div><div data-type=\"solution\" class=\"solution\" id=\"fs-idm153124944\"><ol id=\"fs-idm153124688\" data-number-style=\"lower-alpha\"><li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> = 4<\/li><li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> \u2260 4<\/li><li>Let \\(\\overline{X}\\) the average I.Q. of a set of brown trout.<\/li><li>two-tailed Student's t-test<\/li><li><em data-effect=\"italics\">t<\/em> = 1.95<\/li><li><em data-effect=\"italics\">p<\/em>-value = 0.076<\/li><li>Check student\u2019s solution.<\/li><li><ol id=\"fs-idp21135088\" data-number-style=\"lower-roman\"><li>Alpha: 0.05<\/li><li>Decision: Reject the null hypothesis.<\/li><li>Reason for decision: The <em data-effect=\"italics\">p<\/em>-value is greater than 0.05<\/li><li>Conclusion: There is insufficient evidence to conclude that the average IQ of brown trout is not four. <\/li><\/ol><\/li><li>(3.8865,5.9468)<\/li><\/ol><\/div><\/div><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\" id=\"id9400994\"><p>According to an article in <em data-effect=\"italics\">Newsweek<\/em>, the natural ratio of girls to boys is 100:105. In China, the birth ratio is 100: 114 (46.7% girls). Suppose you don\u2019t believe the reported figures of the percent of girls born in China. You conduct a study. In this study, you count the number of girls and boys born in 150 randomly chosen recent births. There are 60 girls and 90 boys born of the 150. Based on your study, do you believe that the percent of girls born in China is 46.7?<\/p><\/div><\/div><div data-type=\"exercise\" class=\"exercise\" id=\"element-658\"><div data-type=\"problem\" class=\"problem\" id=\"id9401185\"><p>A poll done for <em data-effect=\"italics\">Newsweek<\/em> found that 13% of Americans have seen or sensed the presence of an angel. A contingent doubts that the percent is really that high. It conducts its own survey. Out of 76 Americans surveyed, only two had seen or sensed the presence of an angel. As a result of the contingent\u2019s survey, would you agree with the <em data-effect=\"italics\">Newsweek<\/em> poll? In complete sentences, also give three reasons why the two polls might give different results.\n  <\/p><\/div><div data-type=\"solution\" class=\"solution\" id=\"fs-idm3310512\"><ol id=\"fs-idm3310256\" data-number-style=\"lower-alpha\"><li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2265 0.13<\/li><li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> &lt; 0.13<\/li><li>Let <em data-effect=\"italics\">P\u2032<\/em> = the proportion of Americans who have seen or sensed angels<\/li><li>normal for a single proportion<\/li><li>\u20132.688<\/li><li><em data-effect=\"italics\">p<\/em>-value = 0.0036<\/li><li>Check student\u2019s solution.<\/li><li><ol id=\"fs-idm15778400\" data-number-style=\"lower-roman\"><li>alpha: 0.05<\/li><li>Decision: Reject the null hypothesis.<\/li><li>Reason for decision: The <em data-effect=\"italics\">p<\/em>-value is less than 0.05.<\/li><li>Conclusion: There is sufficient evidence to conclude that the percentage of Americans who have seen or sensed an angel is less than 13%.<\/li><\/ol><\/li><li>(0, 0.0623).<div data-type=\"newline\"><br \/><\/div>\n      The\u201cplus-4s\u201d confidence interval is (0.0022, 0.0978)<\/li><\/ol><\/div><\/div><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\" id=\"id9401240\"><p id=\"element-140a\">The mean work week for engineers in a start-up company is believed to be about 60 hours. A newly hired engineer hopes that it\u2019s shorter. She asks ten engineering friends in start-ups for the lengths of their mean work weeks. Based on the results that follow, should she count on the mean work week to be shorter than 60 hours?<\/p><p id=\"fs-idp187110944\">Data (length of mean work week): 70; 45; 55; 60; 65; 55; 55; 60; 50; 55.<\/p><\/div><\/div>&lt;!-- LINK --&gt;<div data-type=\"exercise\" class=\"exercise\" id=\"element-761\"><div data-type=\"problem\" class=\"problem\" id=\"id9401488\"><p id=\"element-926\">Use the \u201cLap time\u201d data for Lap 4 (see <a href=\"\/contents\/3ef830bc-5247-460a-9007-e3fd762e5e93@3\" class=\"autogenerated-content\">[link]<\/a>) to test the claim that Terri finishes Lap 4, on average, in less than 129 seconds. Use all twenty races given.<\/p><\/div><div data-type=\"solution\" class=\"solution\" id=\"fs-idp13518384\"><ol id=\"fs-idp13518640\" data-number-style=\"lower-alpha\"><li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> \u2265 129<\/li><li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &lt; 129<\/li><li>Let \\(\\overline{X}\\) = the average time in seconds that Terri finishes Lap 4.<\/li><li>Student's <em data-effect=\"italics\">t<\/em>-distribution<\/li><li><em data-effect=\"italics\">t<\/em> = 1.209<\/li><li>0.8792<\/li><li>Check student\u2019s solution.<\/li><li><ol id=\"fs-idm130980080\" data-number-style=\"lower-roman\"><li>Alpha: 0.05<\/li><li>Decision: Do not reject the null hypothesis.<\/li><li>Reason for decision: The <em data-effect=\"italics\">p<\/em>-value is greater than 0.05.<\/li><li>Conclusion: There is insufficient evidence to conclude that Terri\u2019s mean lap time is less than 129 seconds.<\/li><\/ol><\/li><li>(128.63, 130.37)<\/li><\/ol><\/div><\/div><div data-type=\"exercise\" class=\"exercise\" id=\"fs-idp191961952\"><div data-type=\"problem\" class=\"problem\" id=\"id9401521\">&lt;!-- LINK --&gt;<p id=\"element-564a\">Use the \u201cInitial Public Offering\u201d data (see <a href=\"\/contents\/3ef830bc-5247-460a-9007-e3fd762e5e93@3\" class=\"autogenerated-content\">[link]<\/a>) to test the claim that the mean offer price was &#x1f4b2;18 per share. Do not use all the data. Use your random number generator to randomly survey 15 prices.<\/p><\/div><\/div><div data-type=\"note\" class=\"note\" data-has-label=\"true\" id=\"id9401550\" data-label=\"\"><div data-type=\"title\" class=\"title\">Note<\/div><p id=\"eip-idp174015456\">The following questions were written by past students. They are excellent problems!<\/p><\/div><div data-type=\"exercise\" class=\"exercise\" id=\"ex1\"><div data-type=\"problem\" class=\"problem\" id=\"id9401566\"><p id=\"id10961970\">\"Asian Family Reunion,\" by Chau Nguyen<\/p><p id=\"eip-idp174081328\">Every two years it comes around.<\/p><p id=\"eip-idp174081712\">We all get together from different towns.<\/p><p id=\"eip-idp174082096\">In my honest opinion,<\/p><p id=\"eip-idp174082512\">It's not a typical family reunion.<\/p><p id=\"eip-idp174082896\">Not forty, or fifty, or sixty,<\/p><p id=\"eip-idp73534816\">But how about seventy companions!<\/p><p id=\"eip-idp73535200\">The kids would play, scream, and shout<\/p><p id=\"eip-idp73535584\">One minute they're happy, another they'll pout.<\/p><p id=\"eip-idp73535968\">The teenagers would look, stare, and compare<\/p><p id=\"eip-idp73536352\">From how they look to what they wear.<\/p><p id=\"eip-idp73536736\">The men would chat about their business<\/p><p id=\"eip-idp73537120\">That they make more, but never less.<\/p><p id=\"eip-idp73537504\">Money is always their subject<\/p><p id=\"eip-idp73537888\">And there's always talk of more new projects.<\/p><p id=\"eip-idp73538272\">The women get tired from all of the chats<\/p><p id=\"eip-idp123573408\">They head to the kitchen to set out the mats.<\/p><p id=\"eip-idp123573792\">Some would sit and some would stand<\/p><p id=\"eip-idp123574176\">Eating and talking with plates in their hands.<\/p><p id=\"eip-idp123574560\">Then come the games and the songs<\/p><p id=\"eip-idp123574944\">And suddenly, everyone gets along!<\/p><p id=\"eip-idp123575328\">With all that laughter, it's sad to say<\/p><p id=\"eip-idp123575712\">That it always ends in the same old way.<\/p><p id=\"eip-idp123576096\">They hug and kiss and say \"good-bye\"<\/p><p id=\"eip-idp123576480\">And then they all begin to cry!<\/p><p id=\"eip-idp123576864\">I say that 60 percent shed their tears<\/p><p id=\"eip-idp38133824\">But my mom counted 35 people this year.<\/p><p id=\"eip-idp38134208\">She said that boys and men will always have their pride,<\/p><p id=\"eip-idp38134592\">So we won't ever see them cry.<\/p><p id=\"eip-idp38134976\">I myself don't think she's correct,<\/p><p id=\"eip-idp38135360\">So could you please try this problem to see if you object?<\/p><\/div><div data-type=\"solution\" class=\"solution\" id=\"fs-idm114330544\"><ol id=\"fs-idm114330288\" data-number-style=\"lower-alpha\"><li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> = 0.60<\/li><li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> &lt; 0.60<\/li><li>Let <em data-effect=\"italics\">P\u2032<\/em> = the proportion of family members who shed tears at a reunion.<\/li><li>normal for a single proportion<\/li><li>\u20131.71<\/li><li>0.0438<\/li><li>Check student\u2019s solution.<\/li><li><ol id=\"fs-idm44642224\" data-number-style=\"lower-roman\"><li>alpha: 0.05<\/li><li>Decision: Reject the null hypothesis.<\/li><li>Reason for decision: <em data-effect=\"italics\">p<\/em>-value &lt; alpha<\/li><li>Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the proportion of family members who shed tears at a reunion is less than 0.60. However, the test is weak because the <em data-effect=\"italics\">p<\/em>-value and alpha are quite close, so other tests should be done.<\/li><\/ol><\/li><li>We are 95% confident that between 38.29% and 61.71% of family members will shed tears at a family reunion. (0.3829, 0.6171). The\u201cplus-4s\u201d confidence interval (see chapter 8) is (0.3861, 0.6139)<\/li><\/ol><p id=\"fs-idm62390992\">Note that here the  \u201clarge-sample\u201d 1 \u2013 PropZTest provides the approximate <em data-effect=\"italics\">p<\/em>-value of 0.0438. Whenever a <em data-effect=\"italics\">p<\/em>-value based on a normal approximation is close to the level of significance, the exact <em data-effect=\"italics\">p<\/em>-value based on binomial probabilities should be calculated whenever possible. This is beyond the scope of this course.<\/p><\/div><\/div><div data-type=\"exercise\" class=\"exercise\" id=\"ex2\"><div data-type=\"problem\" class=\"problem\" id=\"id9401620\"><p id=\"id5063144\">\"The Problem with Angels,\" by Cyndy Dowling<\/p><p id=\"eip-idm27993696\">Although this problem is wholly mine,<\/p><p id=\"eip-idm27993312\">The catalyst came from the magazine, Time.<\/p><p id=\"eip-idm27992928\">On the magazine cover I did find<\/p><p id=\"eip-idm27992544\">The realm of angels tickling my mind.<\/p><div data-type=\"newline\" data-count=\"1\"><br \/><\/div><p id=\"eip-idm27991792\">Inside, 69% I found to be<\/p><p id=\"eip-idm27991408\">In angels, Americans do believe.<\/p><div data-type=\"newline\" data-count=\"1\"><br \/><\/div><p id=\"eip-idp10519280\">Then, it was time to rise to the task,<\/p><p id=\"eip-idp10519664\">Ninety-five high school and college students I did ask.<\/p><p id=\"eip-idp10520048\">Viewing all as one group,<\/p><p id=\"eip-idp10520432\">Random sampling to get the scoop.<\/p><div data-type=\"newline\" data-count=\"1\"><br \/><\/div><p id=\"eip-idp10521184\">So, I asked each to be true,<\/p><p id=\"eip-idp10521568\">\"Do you believe in angels?\"  Tell me, do!<\/p><div data-type=\"newline\" data-count=\"1\"><br \/><\/div><p id=\"eip-idp10522320\">Hypothesizing at the start,<\/p><p id=\"eip-idm24780496\">Totally believing in my heart<\/p><p id=\"eip-idm24780112\">That the proportion who said yes<\/p><p id=\"eip-idm24779728\">Would be equal on this test.<\/p><div data-type=\"newline\" data-count=\"1\"><br \/><\/div><p id=\"eip-idm24778976\">Lo and behold, seventy-three did arrive,<\/p><p id=\"eip-idm24778592\">Out of the sample of ninety-five.<\/p><p id=\"eip-idm24778208\">Now your job has just begun,<\/p><p id=\"eip-idm24777824\">Solve this problem and have some fun.<\/p><\/div><\/div><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\" id=\"id9401850\"><p id=\"id9415993\">\"Blowing Bubbles,\" by Sondra Prull<\/p><p id=\"eip-idp3753920\">Studying stats just made me tense,<\/p><p id=\"eip-idp3754352\">I had to find some sane defense.<\/p><p id=\"eip-idp3754784\">Some light and lifting simple play<\/p><p id=\"eip-idp3755216\">To float my math anxiety away.<\/p><div data-type=\"newline\" data-count=\"1\"><br \/><\/div><p id=\"eip-idp3756016\">Blowing bubbles lifts me high<\/p><p id=\"eip-idp3756448\">Takes my troubles to the sky.<\/p><p id=\"eip-idp3756880\">POIK! They're gone, with all my stress<\/p><p id=\"eip-idp3757312\">Bubble therapy is the best.<\/p><div data-type=\"newline\" data-count=\"1\"><br \/><\/div><p id=\"eip-idm21778064\">The label said each time I blew<\/p><p id=\"eip-idm21777632\">The average number of bubbles would be at least 22.<\/p><p id=\"eip-idm21777248\">I blew and blew and this I found<\/p><p id=\"eip-idm21776816\">From 64 blows, they all are round!<\/p><div data-type=\"newline\" data-count=\"1\"><br \/><\/div><p id=\"eip-idm21776016\">But the number of bubbles in 64 blows<\/p><p id=\"eip-idm21775584\">Varied widely, this I know.<\/p><p id=\"eip-idm21775152\">20 per blow became the mean<\/p><p id=\"eip-idm35613136\">They deviated by 6, and not 16.<\/p><div data-type=\"newline\" data-count=\"1\"><br \/><\/div><p id=\"eip-idm35612336\">From counting bubbles, I sure did relax<\/p><p id=\"eip-idm35611904\">But now I give to you your task.<\/p><p id=\"eip-idm35611472\">Was 22 a reasonable guess?<\/p><p id=\"eip-idm35611040\">Find the answer and pass this test!<\/p><\/div><div data-type=\"solution\" class=\"solution\" id=\"fs-idm33073568\"><ol id=\"fs-idm33073312\" data-number-style=\"lower-alpha\"><li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> \u2265 22<\/li><li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &lt; 22<\/li><li>Let \\(\\overline{X}\\) = the mean number of bubbles per blow.<\/li><li>Student's <em data-effect=\"italics\">t<\/em>-distribution<\/li><li>\u20132.667<\/li><li><em data-effect=\"italics\">p<\/em>-value = 0.00486<\/li><li>Check student\u2019s solution.<\/li><li><ol id=\"fs-idm27368736\" data-number-style=\"lower-roman\"><li>Alpha: 0.05<\/li><li>Decision: Reject the null hypothesis.<\/li><li>Reason for decision: The <em data-effect=\"italics\">p<\/em>-value is less than 0.05.<\/li><li>Conclusion: There is sufficient evidence to conclude that the mean number of bubbles per blow is less than 22.<\/li><\/ol><\/li><li>(18.501, 21.499)<\/li><\/ol><\/div><\/div><div data-type=\"exercise\" class=\"exercise\" id=\"ex3\"><div data-type=\"problem\" class=\"problem\" id=\"id9401931\"><p id=\"id11754724\">\"Dalmatian Darnation,\" by Kathy Sparling<\/p><p id=\"eip-idm41372368\">A greedy dog breeder named Spreckles<\/p><p id=\"eip-idm41371936\">Bred puppies with numerous freckles<\/p><p id=\"eip-idm41371504\">The Dalmatians he sought<\/p><p id=\"eip-idm41371072\">Possessed spot upon spot<\/p><p id=\"eip-idm41370640\">The more spots, he thought, the more shekels.<\/p><div data-type=\"newline\" data-count=\"1\"><br \/><\/div><p id=\"eip-idm41369888\">His competitors did not agree<\/p><p id=\"eip-idm41369456\">That freckles would increase the fee.<\/p><p id=\"eip-idm41369024\">They said, \u201cSpots are quite nice<\/p><p id=\"eip-idm98226224\">But they don't affect price;<\/p><p id=\"eip-idm98225792\">One should breed for improved pedigree.\u201d<\/p><div data-type=\"newline\" data-count=\"1\"><br \/><\/div><p id=\"eip-idm98224944\">The breeders decided to prove<\/p><p id=\"eip-idm98224512\">This strategy was a wrong move.<\/p><p id=\"eip-idm98224080\">Breeding only for spots<\/p><p id=\"eip-idm98223696\">Would wreak havoc, they thought.<\/p><p id=\"eip-idm98223264\">His theory they want to disprove.<\/p><div data-type=\"newline\" data-count=\"1\"><br \/><\/div><p id=\"eip-idm209854448\">They proposed a contest to Spreckles<\/p><p id=\"eip-idm209854016\">Comparing dog prices to freckles.<\/p><p id=\"eip-idm209853584\">In records they looked up<\/p><p id=\"eip-idm209853152\">One hundred one pups:<\/p><p id=\"eip-idm209852768\">Dalmatians that fetched the most shekels.<\/p><div data-type=\"newline\" data-count=\"1\"><br \/><\/div><p id=\"eip-idm209852016\">They asked Mr. Spreckles to name<\/p><p id=\"eip-idm209851584\">An average spot count he'd claim<\/p><p id=\"eip-idm209851152\">To bring in big bucks.<\/p><p id=\"eip-idm209850768\">Said Spreckles, \u201cWell, shucks,<\/p><p id=\"eip-idm105107920\">It's for one hundred one that I aim.\u201d<\/p><div data-type=\"newline\" data-count=\"1\"><br \/><\/div><p id=\"eip-idm105107184\">Said an amateur statistician<\/p><p id=\"eip-idm105106752\">Who wanted to help with this mission.<\/p><p id=\"eip-idm105106320\">\u201cTwenty-one for the sample<\/p><p id=\"eip-idm105105936\">Standard deviation's ample:<\/p><div data-type=\"newline\" data-count=\"1\"><br \/><\/div><p id=\"eip-idm105105136\">They examined one hundred and one<\/p><p id=\"eip-idm105104704\">Dalmatians that fetched a good sum.<\/p><p id=\"eip-idm202236592\">They counted each spot,<\/p><p id=\"eip-idm202236208\">Mark, freckle and dot<\/p><p id=\"eip-idm202235824\">And tallied up every one.<\/p><div data-type=\"newline\" data-count=\"1\"><br \/><\/div><p id=\"eip-idm202235024\">Instead of one hundred one spots<\/p><p id=\"eip-idm202234592\">They averaged ninety six dots<\/p><p id=\"eip-idm202234160\">Can they muzzle Spreckles\u2019<\/p><p id=\"eip-idm202233696\">Obsession with freckles<\/p><p id=\"eip-idm202233312\">Based on all the dog data they've got?<\/p><\/div><\/div><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\" id=\"id9402199\"><p id=\"id9870720\">\"Macaroni and Cheese, please!!\" by Nedda Misherghi and Rachelle Hall<\/p><p id=\"id11671377\">As a poor starving student I don't have much money to spend for even the bare necessities. So my favorite and main staple food is macaroni and cheese. It's high in taste and low in cost and nutritional value.<\/p><p id=\"id8780562\">One day, as I sat down to determine the meaning of life, I got a serious craving for this, oh, so important, food of my life. So I went down the street to Greatway to get a box of macaroni and cheese, but it was SO expensive! &#x1f4b2;2.02 !!! Can you believe it? It made me stop and think. The world is changing fast. I had thought that the mean cost of a box (the normal size, not some super-gigantic-family-value-pack) was at most &#x1f4b2;1, but now I wasn't so sure. However, I was determined to find out. I went to 53 of the closest grocery stores and surveyed the prices of macaroni and cheese. Here are the data I wrote in my notebook:<\/p><div data-type=\"list\" id=\"element-357\"><div data-type=\"title\">Price per box of Mac and Cheese:<\/div><ul><li>5 stores @ &#x1f4b2;2.02<\/li><li>15 stores @ &#x1f4b2;0.25<\/li><li>3 stores @ &#x1f4b2;1.29<\/li><li>6 stores @ &#x1f4b2;0.35<\/li><li>4 stores @ &#x1f4b2;2.27<\/li><li>7 stores @ &#x1f4b2;1.50<\/li><li>5 stores @ &#x1f4b2;1.89<\/li><li>8 stores @ 0.75.<\/li><\/ul><\/div><p id=\"id5072942\">I could see that the cost varied but I had to sit down to figure out whether or not I was right. If it does turn out that this mouth-watering dish is at most &#x1f4b2;1, then I'll throw a big cheesy party in our next statistics lab, with enough macaroni and cheese for just me. (After all, as a poor starving student I can't be expected to feed our class of animals!)<\/p><\/div><div data-type=\"solution\" class=\"solution\" id=\"fs-idp45964640\"><ol id=\"fs-idp45964896\" data-number-style=\"lower-alpha\"><li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> \u2264 1<\/li><li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &gt; 1<\/li><li>Let \\(\\overline{X}\\) = the mean cost in dollars of macaroni and cheese in a certain town.<\/li><li>Student's <em data-effect=\"italics\">t<\/em>-distribution<\/li><li><em data-effect=\"italics\">t<\/em> = 0.340<\/li><li><em data-effect=\"italics\">p<\/em>-value = 0.36756<\/li><li>Check student\u2019s solution.<\/li><li><ol id=\"fs-idm78282592\" data-number-style=\"lower-roman\"><li>Alpha: 0.05<\/li><li>Decision: Do not reject the null hypothesis.<\/li><li>Reason for decision: The <em data-effect=\"italics\">p<\/em>-value is greater than 0.05<\/li><li>Conclusion: The mean cost could be &#x1f4b2;1, or less.  At the 5% significance level, there is insufficient evidence to conclude that the mean price of a box of macaroni and cheese is more than &#x1f4b2;1.<\/li><\/ol><\/li><li>(0.8291, 1.241)<\/li><\/ol><\/div><\/div><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\" id=\"id9402361\"><p id=\"id8933019\">\"William Shakespeare: The Tragedy of Hamlet, Prince of Denmark,\" by Jacqueline Ghodsi<\/p><div data-type=\"list\" id=\"element-392\"><div data-type=\"title\">THE CHARACTERS (in order of appearance):<\/div><ul><li>HAMLET, Prince of Denmark and student of Statistics<\/li><li>POLONIUS, Hamlet\u2019s tutor<\/li><li>HOROTIO, friend to Hamlet and fellow student<\/li><\/ul><\/div><p>Scene: The great library of the castle, in which Hamlet does his lessons<\/p><p id=\"id10572210\">Act I<\/p><p id=\"id10572215\">(The day is fair, but the face of Hamlet is clouded. He paces the large room. His tutor, Polonius, is reprimanding Hamlet regarding the latter\u2019s recent experience. Horatio is seated at the large table at right stage.)<\/p><p id=\"id10603284\">POLONIUS: My Lord, how cans\u2019t thou admit that thou hast seen a ghost! It is but a figment of your imagination!<\/p><p id=\"id4997547\">HAMLET: I beg to differ; I know of a certainty that five-and-seventy in one hundred of us, condemned to the whips and scorns of time as we are, have gazed upon a spirit of health, or goblin damn\u2019d, be their intents wicked or charitable.<\/p><p id=\"id9031598\">POLONIUS If thou doest insist upon thy wretched vision then let me invest your time; be true to thy work and speak to me through the reason of the null and alternate hypotheses. (He turns to Horatio.) Did not Hamlet himself say, \u201cWhat piece of work is man, how noble in reason, how infinite in faculties? Then let not this foolishness persist. Go, Horatio, make a survey of three-and-sixty and discover what the true proportion be. For my part, I will never succumb to this fantasy, but deem man to be devoid of all reason should thy proposal of at least five-and-seventy in one hundred hold true.<\/p><p id=\"id10821885\">HORATIO (to Hamlet): What should we do, my Lord?<\/p><p id=\"id10821892\">HAMLET: Go to thy purpose, Horatio.<\/p><p id=\"id9142272\">HORATIO: To what end, my Lord?<\/p><p id=\"id9142279\">HAMLET: That you must teach me. But let me conjure you by the rights of our fellowship, by the consonance of our youth, but the obligation of our ever-preserved love, be even and direct with me, whether I am right or no.<\/p><p id=\"id5011881\">(Horatio exits, followed by Polonius, leaving Hamlet to ponder alone.)<\/p><p id=\"id10626816\">Act II<\/p><p id=\"id10626821\">(The next day, Hamlet awaits anxiously the presence of his friend, Horatio. Polonius enters and places some books upon the table just a moment before Horatio enters.)<\/p><p id=\"id11671359\">POLONIUS: So, Horatio, what is it thou didst reveal through thy deliberations?<\/p><p id=\"id11671367\">HORATIO: In a random survey, for which purpose thou thyself sent me forth, I did discover that one-and-forty believe fervently that the spirits of the dead walk with us. Before my God, I might not this believe, without the sensible and true avouch of mine own eyes.<\/p><p id=\"id4945509\">POLONIUS: Give thine own thoughts no tongue, Horatio. (Polonius turns to Hamlet.) But look to\u2019t I charge you, my Lord. Come Horatio, let us go together, for this is not our test. (Horatio and Polonius leave together.)<\/p><p id=\"id4986106\">HAMLET: To reject, or not reject, that is the question: whether \u2018tis nobler in the mind to suffer the slings and arrows of outrageous statistics, or to take arms against a sea of data, and, by opposing, end them. (Hamlet resignedly attends to his task.)<\/p><p id=\"id10766013\">(Curtain falls)<\/p><\/div><\/div><div data-type=\"exercise\" class=\"exercise\" id=\"exer7\"><div data-type=\"problem\" class=\"problem\" id=\"id9402839\"><p id=\"id8293070\">\"Untitled,\" by Stephen Chen<\/p><p id=\"id5050494\">I've often wondered how software is released and sold to the public. Ironically, I work for a company that sells products with known problems. Unfortunately, most of the problems are difficult to create, which makes them difficult to fix. I usually use the test program X, which tests the product, to try to create a specific problem. When the test program is run to make an error occur, the likelihood of generating an error is 1%.<\/p><p id=\"id11215303\">So, armed with this knowledge, I wrote a new test program Y that will generate the same error that test program X creates, but more often. To find out if my test program is better than the original, so that I can convince the management that I'm right, I ran my test program to find out how often I can generate the same error. When I ran my test program 50 times, I generated the error twice. While this may not seem much better, I think that I can convince the management to use my test program instead of the original test program. Am I right?<\/p><\/div><div data-type=\"solution\" class=\"solution\" id=\"fs-idp10825072\"><ol id=\"fs-idp10825200\" data-number-style=\"lower-alpha\"><li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> = 0.01<\/li><li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> &gt; 0.01<\/li><li>Let <em data-effect=\"italics\">P\u2032<\/em> = the proportion of errors generated<\/li><li>Normal for a single proportion<\/li><li>2.13<\/li><li>0.0165<\/li><li>Check student\u2019s solution.<\/li><li><ol id=\"fs-idm61739936\" data-number-style=\"lower-roman\"><li>Alpha: 0.05<\/li><li>Decision: Reject the null hypothesis<\/li><li>Reason for decision: The <em data-effect=\"italics\">p<\/em>-value is less than 0.05.<\/li><li>Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the proportion of errors generated is more than 0.01.<\/li><\/ol><\/li><li>Confidence interval: (0, 0.094).<div data-type=\"newline\"><br \/><\/div>The\u201cplus-4s\u201d confidence interval is (0.004, 0.144).<\/li><\/ol><\/div><\/div><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\" id=\"id9402904\"><p id=\"id10126776\">\"Japanese Girls\u2019 Names\"<\/p><p id=\"id9977266\">by Kumi Furuichi<\/p><p id=\"id5031080\">It used to be very typical for Japanese girls\u2019 names to end with \u201cko.\u201d (The trend might have started around my grandmothers\u2019 generation and its peak might have been around my mother\u2019s generation.) \u201cKo\u201d means \u201cchild\u201d in Chinese characters. Parents would name their daughters with \u201cko\u201d attaching to other Chinese characters which have meanings that they want their daughters to become, such as Sachiko\u2014happy child, Yoshiko\u2014a good child, Yasuko\u2014a healthy child, and so on.<\/p><p id=\"id4930619\">However, I noticed recently that only two out of nine of my Japanese girlfriends at this school have names which end with \u201cko.\u201d More and more, parents seem to have become creative, modernized, and, sometimes, westernized in naming their children.<\/p><p id=\"id5048749\">I have a feeling that, while 70 percent or more of my mother\u2019s generation would have names with \u201cko\u201d at the end, the proportion has dropped among my peers. I wrote down all my Japanese friends\u2019, ex-classmates\u2019, co-workers, and acquaintances\u2019 names that I could remember. Following are the names. (Some are repeats.) Test to see if the proportion has dropped for this generation.<\/p><p id=\"id12545176\">Ai, Akemi, Akiko, Ayumi, Chiaki, Chie, Eiko, Eri, Eriko, Fumiko, Harumi, Hitomi, Hiroko, Hiroko, Hidemi, Hisako, Hinako, Izumi, Izumi, Junko, Junko, Kana, Kanako, Kanayo, Kayo, Kayoko, Kazumi, Keiko, Keiko, Kei, Kumi, Kumiko, Kyoko, Kyoko, Madoka, Maho, Mai, Maiko, Maki, Miki, Miki, Mikiko, Mina, Minako, Miyako, Momoko, Nana, Naoko, Naoko, Naoko, Noriko, Rieko, Rika, Rika, Rumiko, Rei, Reiko, Reiko, Sachiko, Sachiko, Sachiyo, Saki, Sayaka, Sayoko, Sayuri, Seiko, Shiho, Shizuka, Sumiko, Takako, Takako, Tomoe, Tomoe, Tomoko, Touko, Yasuko, Yasuko, Yasuyo, Yoko, Yoko, Yoko, Yoshiko, Yoshiko, Yoshiko, Yuka, Yuki, Yuki, Yukiko, Yuko, Yuko.<\/p><\/div><\/div><div data-type=\"exercise\" class=\"exercise\" id=\"exerrr\"><div data-type=\"problem\" class=\"problem\" id=\"id9403278\"><p id=\"id4925098\">\"Phillip\u2019s Wish,\" by Suzanne Osorio<\/p><p id=\"eip-idm86797472\">My nephew likes to play<\/p><p id=\"eip-idm258943856\">Chasing the girls makes his day.<\/p><p id=\"eip-idm258943424\">He asked his mother<\/p><p id=\"eip-idm258943008\">If it is okay<\/p><p id=\"eip-idm258942624\">To get his ear pierced.<\/p><p id=\"eip-idm258942208\">She said, \u201cNo way!\u201d<\/p><p id=\"eip-idm258941776\">To poke a hole through your ear,<\/p><p id=\"eip-idm258941344\">Is not what I want for you, dear.<\/p><p id=\"eip-idm258940912\">He argued his point quite well,<\/p><p id=\"eip-idm258940480\">Says even my macho pal,  Mel,<\/p><p id=\"eip-idm2866384\">Has gotten this done.<\/p><p id=\"eip-idm2866000\">It\u2019s all just for fun.<\/p><p id=\"eip-idm2865536\">C\u2019mon please, mom, please, what the hell.<\/p><p id=\"eip-idm2865040\">Again Phillip complained to his mother,<\/p><p id=\"eip-idm2864608\">Saying half his friends (including their brothers)<\/p><p id=\"eip-idm2864160\">Are piercing their ears<\/p><p id=\"eip-idm2863744\">And they have no fears<\/p><p id=\"eip-idm2863328\">He wants to be like the others.<\/p><p id=\"eip-idm2862896\">She said, \u201cI think it\u2019s much less.<\/p><p id=\"eip-idm2862416\">We must do a hypothesis test.<\/p><p id=\"eip-idm190635664\">And if you are right,<\/p><p id=\"eip-idm190635280\">I won\u2019t put up a fight.<\/p><p id=\"eip-idm190634816\">But, if not, then my case will rest.\u201d<\/p><p id=\"eip-idm190634336\">We proceeded to call fifty guys<\/p><p id=\"eip-idm190633904\">To see whose prediction would fly.<\/p><p id=\"eip-idm190633472\">Nineteen of the fifty<\/p><p id=\"eip-idm190633056\">Said piercing was nifty<\/p><p id=\"eip-idm190632640\">And earrings they\u2019d occasionally buy.<\/p><p id=\"eip-idm190632192\">Then there\u2019s the other thirty-one,<\/p><p id=\"eip-idm96693360\">Who said they\u2019d never have this done.<\/p><p id=\"eip-idm96692848\">So now this poem\u2019s finished.<\/p><p id=\"eip-idm96692416\">Will his hopes be diminished,<\/p><p id=\"eip-idm96692032\">Or will my nephew have his fun?<\/p><\/div><div data-type=\"solution\" class=\"solution\" id=\"fs-idm61462416\"><ol id=\"fs-idp31538240\" data-number-style=\"lower-alpha\"><li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> = 0.50<\/li><li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> &lt; 0.50<\/li><li>Let <em data-effect=\"italics\">P\u2032<\/em> = the proportion of friends that has a pierced ear.<\/li><li>normal for a single proportion<\/li><li>\u20131.70<\/li><li><em data-effect=\"italics\">p<\/em>-value = 0.0448<\/li><li>Check student\u2019s solution.<\/li><li><ol id=\"fs-idm172480592\" data-number-style=\"lower-roman\"><li>Alpha: 0.05<\/li><li>Decision: Reject the null hypothesis<\/li><li>Reason for decision: The <em data-effect=\"italics\">p<\/em>-value is less than 0.05. (However, they are very close.)<\/li><li>Conclusion: There is sufficient evidence to support the claim that less than 50% of his friends have pierced ears.<\/li><\/ol><\/li><li>Confidence Interval: (0.245, 0.515): The \u201cplus-4s\u201d confidence interval is (0.259, 0.519).<\/li><\/ol><\/div><\/div><div data-type=\"exercise\" class=\"exercise\" id=\"exercise\"><div data-type=\"problem\" class=\"problem\" id=\"id9403352\"><p id=\"id8450089\">\"The Craven,\" by Mark Salangsang<\/p><p id=\"eip-idm8295648\">Once upon a morning dreary<\/p><p id=\"eip-idm8295216\">In stats class I was weak and weary.<\/p><p id=\"eip-idm8294784\">Pondering over last night\u2019s homework<\/p><p id=\"eip-idp24682128\">Whose answers were now on the board<\/p><p id=\"eip-idp24682560\">This I did and nothing more.<\/p><div data-type=\"newline\" data-count=\"1\"><br \/><\/div><p id=\"eip-idp24683360\">While I nodded nearly napping<\/p><p id=\"eip-idp24683792\">Suddenly, there came a tapping.<\/p><p id=\"eip-idp24684224\">As someone gently rapping,<\/p><p id=\"eip-idp24684656\">Rapping my head as I snore.<\/p><p id=\"eip-idp24685088\">Quoth the teacher, \u201cSleep no more.\u201d<\/p><div data-type=\"newline\" data-count=\"1\"><br \/><\/div><p id=\"eip-idp24685904\">\u201cIn every class you fall asleep,\u201d<\/p><p id=\"eip-idm21424304\">The teacher said, his voice was deep.<\/p><p id=\"eip-idm21423872\">\u201cSo a tally I\u2019ve begun to keep<\/p><p id=\"eip-idm21423376\">Of every class you nap and snore.<\/p><p id=\"eip-idm21422944\">The percentage being forty-four.\u201d<\/p><div data-type=\"newline\" data-count=\"1\"><br \/><\/div><p id=\"eip-idm21422112\">\u201cMy dear teacher I must confess,<\/p><p id=\"eip-idm21421648\">While sleeping is what I do best.  <\/p><p id=\"eip-idm21421216\">The percentage, I think, must be less,<\/p><p id=\"eip-idm21420784\">A percentage less than forty-four.\u201d<\/p><p id=\"eip-idm57340704\">This I said and nothing more.<\/p><div data-type=\"newline\" data-count=\"1\"><br \/><\/div><p id=\"eip-idm57339904\">\u201cWe\u2019ll see,\u201d he said and walked away,<\/p><p id=\"eip-idm57339376\">And fifty classes from that day<\/p><p id=\"eip-idm57338944\">He counted till the month of May<\/p><p id=\"eip-idm57338512\">The classes in which I napped and snored.<\/p><p id=\"eip-idm57338128\">The number he found was twenty-four.<\/p><div data-type=\"newline\" data-count=\"1\"><br \/><\/div><p id=\"eip-idm57337328\">At a significance level of 0.05,<\/p><p id=\"eip-idm57336896\">Please tell me am I still alive?<\/p><p id=\"eip-idm11006496\">Or did my grade just take a dive<\/p><p id=\"eip-idm11006112\">Plunging down beneath the floor?<\/p><p id=\"eip-idm11005680\">Upon thee I hereby implore.<\/p><\/div><\/div><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\" id=\"id9403596\"><p>Toastmasters International cites a report by Gallop Poll that 40% of Americans fear public speaking. A student believes that less than 40% of students at her school fear public speaking. She randomly surveys 361 schoolmates and finds that 135 report they fear public speaking. Conduct a hypothesis test to determine if the percent at her school is less than 40%.\n  <\/p><\/div><div data-type=\"solution\" class=\"solution\" id=\"fs-idm62899616\"><ol id=\"fs-idm62899488\" data-number-style=\"lower-alpha\"><li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> = 0.40<\/li><li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> &lt; 0.40<\/li><li>Let <em data-effect=\"italics\">P\u2032<\/em> = the proportion of schoolmates who fear public speaking.<\/li><li>normal for a single proportion<\/li><li>\u20131.01<\/li><li><em data-effect=\"italics\">p<\/em>-value = 0.1563<\/li><li>Check student\u2019s solution.<\/li><li><ol id=\"fs-idm92461088\" data-number-style=\"lower-roman\"><li>Alpha: 0.05<\/li><li>Decision: Do not reject the null hypothesis.<\/li><li>Reason for decision: The <em data-effect=\"italics\">p<\/em>-value is greater than 0.05.<\/li><li>Conclusion: There is insufficient evidence to support the claim that less than 40% of students at the school fear public speaking.<\/li><\/ol><\/li><li>Confidence Interval: (0.3241, 0.4240): The \u201cplus-4s\u201d confidence interval is (0.3257, 0.4250).<\/li><\/ol><\/div><\/div><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\" id=\"id9403657\"><p>Sixty-eight percent of online courses taught at community colleges nationwide were taught by full-time faculty. To test if 68% also represents California\u2019s percent for full-time faculty teaching the online classes, Long Beach City College (LBCC) in California, was randomly selected for comparison. In the same year, 34 of the 44 online courses LBCC offered were taught by full-time faculty. Conduct a hypothesis test to determine if 68% represents California. NOTE: For more accurate results, use more California community colleges and this past year's data.<\/p><\/div><\/div><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\" id=\"id9403858\"><p>According to an article in <em data-effect=\"italics\">Bloomberg Businessweek<\/em>, New York City's most recent adult smoking rate is 14%. Suppose that a survey is conducted to determine this year\u2019s rate. Nine out of 70 randomly chosen N.Y. City residents reply that they smoke. Conduct a hypothesis test to determine if the rate is still 14% or if it has decreased.<\/p><\/div><div data-type=\"solution\" class=\"solution\" id=\"fs-idp21365712\"><ol id=\"fs-idp21365968\" data-number-style=\"lower-alpha\"><li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> = 0.14<\/li><li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> &lt; 0.14<\/li><li>Let <em data-effect=\"italics\">P\u2032<\/em> = the proportion of NYC residents that smoke.<\/li><li>normal for a single proportion<\/li><li>\u20130.2756<\/li><li><em data-effect=\"italics\">p<\/em>-value = 0.3914<\/li><li>Check student\u2019s solution.<\/li><li><ol id=\"fs-idm186262784\" data-number-style=\"lower-roman\"><li>alpha: 0.05<\/li><li>Decision: Do not reject the null hypothesis.<\/li><li>Reason for decision: The <em data-effect=\"italics\">p<\/em>-value is greater than 0.05.<\/li><li>At the 5% significance level, there is insufficient evidence to conclude that the proportion of NYC residents who smoke is less than 0.14.<\/li><\/ol><\/li><li>Confidence Interval: (0.0502, 0.2070): The \u201cplus-4s\u201d confidence interval (see chapter 8) is (0.0676, 0.2297).<\/li><\/ol><\/div><\/div><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\" id=\"id9403905\"><p>The mean age of De Anza College students in a previous term was 26.6 years old. An instructor thinks the mean age for online students is older than 26.6. She randomly surveys 56 online students and finds that the sample mean is 29.4 with a standard deviation of 2.1. Conduct a hypothesis test.<\/p><\/div><\/div><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\" id=\"id9404099\"><p>Registered nurses earned an average annual salary of &#x1f4b2;69,110. For that same year, a survey was conducted of 41 California registered nurses to determine if the annual salary is higher than &#x1f4b2;69,110 for California nurses. The sample average was &#x1f4b2;71,121 with a sample standard deviation of &#x1f4b2;7,489. Conduct a hypothesis test.\n  <\/p><\/div><div data-type=\"solution\" class=\"solution\" id=\"fs-idm11266464\"><ol id=\"fs-idm126211024\" data-number-style=\"lower-alpha\"><li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> = 69,110<\/li><li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &gt; 69,110<\/li><li>Let \\(\\overline{X}\\) = the mean salary in dollars for California registered nurses.<\/li><li>Student's <em data-effect=\"italics\">t<\/em>-distribution<\/li><li><em data-effect=\"italics\">t<\/em> = 1.719<\/li><li><em data-effect=\"italics\">p<\/em>-value: 0.0466<\/li><li>Check student\u2019s solution.<\/li><li><ol id=\"fs-idm11928480\" data-number-style=\"lower-roman\"><li>Alpha: 0.05<\/li><li>Decision: Reject the null hypothesis.<\/li><li>Reason for decision: The <em data-effect=\"italics\">p<\/em>-value is less than 0.05.<\/li><li>Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean salary of California registered nurses exceeds &#x1f4b2;69,110.<\/li><\/ol><\/li><li>(&#x1f4b2;68,757, &#x1f4b2;73,485) <\/li><\/ol><\/div><\/div><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\" id=\"id9404155\"><p>La Leche League International reports that the mean age of weaning a child from breastfeeding is age four to five worldwide. In America, most nursing mothers wean their children much earlier. Suppose a random survey is conducted of 21 U.S. mothers who recently weaned their children. The mean weaning age was nine months (3\/4 year) with a standard deviation of 4 months. Conduct a hypothesis test to determine if the mean weaning age in the U.S. is less than four years old.<\/p><\/div><\/div><div data-type=\"exercise\" class=\"exercise\" id=\"exer12\"><div data-type=\"problem\" class=\"problem\" id=\"id8184194\"><p id=\"id10291861\">Over the past few decades, public health officials have examined the link between weight concerns and teen girls' smoking. Researchers surveyed a group of 273 randomly selected teen girls living in Massachusetts (between 12 and 15 years old). After four years the girls were surveyed again. Sixty-three said they smoked to stay thin. Is there good evidence that more than thirty percent of the teen girls smoke to stay thin?<\/p><div data-type=\"newline\"><br \/><\/div>After conducting the test, your decision and conclusion are<ol id=\"id10291869\" data-number-style=\"lower-alpha\"><li>Reject <em data-effect=\"italics\">H<sub>0<\/sub><\/em>: There is sufficient evidence to conclude that more than 30% of teen girls smoke to stay thin.<\/li><li>Do not reject <em data-effect=\"italics\">H<sub>0<\/sub><\/em>: There is not sufficient evidence to conclude that less than 30% of teen girls smoke to stay thin.<\/li><li>Do not reject <em data-effect=\"italics\">H<sub>0<\/sub><\/em>: There is not sufficient evidence to conclude that more than 30% of teen girls smoke to stay thin.<\/li><li>Reject <em data-effect=\"italics\">H<sub>0<\/sub><\/em>: There is sufficient evidence to conclude that less than 30% of teen girls smoke to stay thin.<\/li><\/ol><\/div><div data-type=\"solution\" class=\"solution\" id=\"id8184383\"><p id=\"sulution\">c<\/p><\/div><\/div><div data-type=\"exercise\" class=\"exercise\" id=\"exer14\"><div data-type=\"problem\" class=\"problem\" id=\"id8184753\"><p id=\"eip-idp66540992\">A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening night midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 of them attended the midnight showing.\n<\/p><div data-type=\"newline\"><br \/><\/div>At a 1% level of significance, an appropriate conclusion is:<ol id=\"id6241780\" data-number-style=\"lower-alpha\"><li>There is insufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is less than 20%.<\/li><li>There is sufficient evidence to conclude that the  percent of EVC students who attended the midnight showing of Harry Potter is more than 20%.<\/li><li>There is sufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is less than 20%.<\/li><li>There is insufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is at least 20%.<\/li><\/ol><\/div><\/div><div data-type=\"exercise\" class=\"exercise\" id=\"exer19\"><div data-type=\"problem\" class=\"problem\" id=\"id8186196\"><p id=\"id11679199\">Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test.\n<\/p><div data-type=\"newline\"><br \/><\/div>\n<div data-type=\"newline\"><br \/><\/div>At a significance level of \n<em data-effect=\"italics\">a<\/em> = 0.05, what is the correct conclusion?<ol id=\"id4986877\" data-number-style=\"lower-alpha\"><li>There is enough evidence to conclude that the mean number of hours is more than 4.75<\/li><li>There is enough evidence to conclude that the mean number of hours is more than 4.5<\/li><li>There is not enough evidence to conclude that the mean number of hours is more than 4.5<\/li><li>There is not enough evidence to conclude that the mean number of hours is more than 4.75<\/li><\/ol><\/div><div data-type=\"solution\" class=\"solution\" id=\"id8186317\"><p id=\"paragraph123\">c<\/p><\/div><\/div><p id=\"eip-964\">Instructions: For the following ten exercises,\n<\/p><div data-type=\"newline\"><br \/><\/div>Hypothesis testing: For the following ten exercises, answer each question.<ol id=\"eip-76\" data-number-style=\"lower-alpha\"><li>State the null and alternate hypothesis.<\/li><li>State the <em data-effect=\"italics\">p<\/em>-value.<\/li><li>State alpha.<\/li><li>What is your decision?<\/li><li>Write a conclusion.<\/li><li>Answer any other questions asked in the problem.<\/li><\/ol><div data-type=\"exercise\" class=\"exercise\" id=\"eip-69\"><div data-type=\"problem\" class=\"problem\"><p id=\"eip-971\">According to the Center for Disease Control website, in 2011 at least 18% of high school students have smoked a cigarette. An Introduction to Statistics class in Davies County, KY conducted a hypothesis test at the local high school (a medium sized\u2013approximately 1,200 students\u2013small city demographic) to determine if the local high school\u2019s percentage was lower. One hundred fifty students were chosen at random and surveyed. Of the 150 students surveyed, 82 have smoked. Use a significance level of 0.05 and using appropriate statistical evidence, conduct a hypothesis test and state the conclusions.<\/p><\/div><\/div><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\"><p>A recent survey in the <em data-effect=\"italics\">N.Y. Times Almanac<\/em> indicated that 48.8% of families own stock. A broker wanted to determine if this survey could be valid. He surveyed a random sample of 250 families and found that 142 owned some type of stock. At the 0.05 significance level, can the survey be considered to be accurate?<\/p><\/div><div data-type=\"solution\" class=\"solution\"><ol id=\"eip-idm39194080\" data-number-style=\"lower-alpha\"><li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> = 0.488 <em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2260 0.488<\/li><li><em data-effect=\"italics\">p<\/em>-value = 0.0114<\/li><li>alpha = 0.05<\/li><li>Reject the null hypothesis.<\/li><li>At the 5% level of significance, there is enough evidence to conclude that 48.8% of families own stocks.<\/li><li>The survey does not appear to be accurate.<\/li><\/ol><\/div><\/div><div data-type=\"exercise\" class=\"exercise\" id=\"eip-59\"><div data-type=\"problem\" class=\"problem\" id=\"eip-900\"><p>Driver error can be listed as the cause of approximately 54% of all fatal auto accidents, according to the American Automobile Association. Thirty randomly selected fatal accidents are examined, and it is determined that 14 were caused by driver error. Using \u03b1 = 0.05, is the AAA proportion accurate?<\/p><\/div><\/div><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\"><p>The US Department of Energy reported that 51.7% of homes were heated by natural gas. A random sample of 221 homes in Kentucky found that 115 were heated by natural gas. Does the evidence support the claim for Kentucky at the \u03b1 = 0.05 level in Kentucky? Are the results applicable across the country? Why?<\/p><\/div><div data-type=\"solution\" class=\"solution\"><ol id=\"eip-idp117547184\" data-number-style=\"lower-alpha\"><li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> = 0.517 <em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2260 0.517<\/li><li><em data-effect=\"italics\">p<\/em>-value = 0.9203.<\/li><li>alpha = 0.05.<\/li><li>Do not reject the null hypothesis.<\/li><li>At the 5% significance level, there is not enough evidence to conclude that the proportion of homes in Kentucky that are heated by natural gas is 0.517.<\/li><li>However, we cannot generalize this result to the entire nation. First, the sample\u2019s population is only the state of Kentucky. Second, it is reasonable to assume that homes in the extreme north and south will have extreme high usage and low usage, respectively. We would need to expand our sample base to include these possibilities if we wanted to generalize this claim to the entire nation.<\/li><\/ol><\/div><\/div><div data-type=\"exercise\" class=\"exercise\" id=\"eip-835\"><div data-type=\"problem\" class=\"problem\"><p>For Americans using library services, the American Library Association claims that at most 67% of patrons borrow books. The library director in Owensboro, Kentucky feels this is not true, so she asked a local college statistic class to conduct a survey. The class randomly selected 100 patrons and found that 82 borrowed books. Did the class demonstrate that the percentage was higher in Owensboro, KY?  Use \u03b1 = 0.01 level of significance. What is the possible proportion of patrons that do borrow books from the Owensboro Library?<\/p><\/div><\/div><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\"><p>The Weather Underground reported that the mean amount of summer rainfall for the northeastern US is at least 11.52 inches. Ten cities in the northeast are randomly selected and the mean rainfall amount is calculated to be 7.42 inches with a standard deviation of 1.3 inches. At the \u03b1 = 0.05 level, can it be concluded that the mean rainfall was below the reported average? What if \u03b1 = 0.01? Assume the amount of summer rainfall follows a normal distribution.<\/p><\/div><div data-type=\"solution\" class=\"solution\"><ol id=\"eip-idp46967856\" data-number-style=\"lower-alpha\"><li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u00b5<\/em> \u2265 11.52  <em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u00b5<\/em> &lt; 11.52<\/li><li><em data-effect=\"italics\">p<\/em>-value = 0.000002 which is almost 0.<\/li><li>alpha = 0.05.<\/li><li>Reject the null hypothesis.<\/li><li>At the 5% significance level, there is enough evidence to conclude that the mean amount of summer rain in the northeaster US is less than 11.52 inches, on average.<\/li><li>We would make the same conclusion if alpha was 1% because the <em data-effect=\"italics\">p<\/em>-value is almost 0.<\/li><\/ol><\/div><\/div><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\" id=\"eip-444\"><p>A survey in the <em data-effect=\"italics\">N.Y. Times Almanac<\/em> finds the mean commute time (one way) is 25.4 minutes for the 15 largest US cities. The Austin, TX chamber of commerce feels that Austin\u2019s commute time is less and wants to publicize this fact.  The mean for 25 randomly selected commuters is 22.1 minutes with a standard deviation of 5.3 minutes. At the \u03b1 = 0.10 level, is the Austin, TX commute significantly less than the mean commute time for the 15 largest US cities?<\/p><\/div><\/div><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\" id=\"eip-426\"><p>A report by the Gallup Poll found that a woman visits her doctor, on average, at most 5.8 times each year. A random sample of 20 women results in these yearly visit totals<\/p><p id=\"eip-idm37849296\"><span data-type=\"list\" data-list-type=\"labeled-item\" data-display=\"inline\" data-mark-suffix=\"\"><span data-type=\"item\">3<\/span><span data-type=\"item\">2<\/span><span data-type=\"item\">1<\/span><span data-type=\"item\">3<\/span><span data-type=\"item\">7<\/span><span data-type=\"item\">2<\/span><span data-type=\"item\">9<\/span><span data-type=\"item\">4<\/span><span data-type=\"item\">6<\/span><span data-type=\"item\">6<\/span><span data-type=\"item\">8<\/span><span data-type=\"item\">0<\/span><span data-type=\"item\">5<\/span><span data-type=\"item\">6<\/span><span data-type=\"item\">4<\/span><span data-type=\"item\">2<\/span><span data-type=\"item\">1<\/span><span data-type=\"item\">3<\/span><span data-type=\"item\">4<\/span><span data-type=\"item\">1<\/span><\/span><\/p><div data-type=\"newline\"><br \/><\/div>At the <em data-effect=\"italics\">\u03b1<\/em> = 0.05 level can it be concluded that the sample mean is higher than 5.8 visits per year?\n<\/div><div data-type=\"solution\" class=\"solution\"><ol id=\"eip-idp114932976\" data-number-style=\"lower-alpha\"><li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u00b5<\/em> \u2264 5.8\t<em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u00b5<\/em> &gt; 5.8<\/li><li><em data-effect=\"italics\">p<\/em>-value =  0.9987<\/li><li>alpha = 0.05<\/li><li>Do not reject the null hypothesis.<\/li><li>At the 5% level of significance, there is not enough evidence to conclude that a woman visits her doctor, on average, more than 5.8 times a year.<\/li><\/ol><\/div><\/div><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\" id=\"eip-484\"><p>According to the <em data-effect=\"italics\">N.Y. Times Almanac<\/em> the mean family size in the U.S. is 3.18. A sample of a college math class resulted in the following family sizes:\n<\/p><div data-type=\"newline\"><br \/><\/div>\n<span data-type=\"list\" data-list-type=\"labeled-item\" data-display=\"inline\"><span data-type=\"item\">5<\/span><span data-type=\"item\">4<\/span><span data-type=\"item\">5<\/span><span data-type=\"item\">4<\/span><span data-type=\"item\">4<\/span><span data-type=\"item\">3<\/span><span data-type=\"item\">6<\/span><span data-type=\"item\">4<\/span><span data-type=\"item\">3<\/span><span data-type=\"item\">3<\/span><span data-type=\"item\">5<\/span><span data-type=\"item\">5<\/span><span data-type=\"item\">6<\/span><span data-type=\"item\">3<\/span><span data-type=\"item\">3<\/span><span data-type=\"item\">2<\/span><span data-type=\"item\">7<\/span><span data-type=\"item\">4<\/span><span data-type=\"item\">5<\/span><span data-type=\"item\">2<\/span><span data-type=\"item\">2<\/span><span data-type=\"item\">2<\/span><span data-type=\"item\">3<\/span><span data-type=\"item\">2<\/span><\/span>\n<div data-type=\"newline\"><br \/><\/div>At <em data-effect=\"italics\">\u03b1<\/em> = 0.05 level, is the class\u2019 mean family size greater than the national average? Does the Almanac result remain valid? Why?\n<\/div><\/div><div data-type=\"exercise\" class=\"exercise\"><div data-type=\"problem\" class=\"problem\"><p>The student academic group on a college campus claims that freshman students study at least 2.5 hours per day, on average. One Introduction to Statistics class was skeptical. The class took a random sample of 30 freshman students and found a mean study time of 137 minutes with a standard deviation of 45 minutes. At <em data-effect=\"italics\">\u03b1<\/em> = 0.01 level, is the student academic group\u2019s claim correct?<\/p><\/div><div data-type=\"solution\" class=\"solution\"><ol id=\"eip-idp64977040\" data-number-style=\"lower-alpha\"><li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u00b5<\/em> \u2265 150\t<em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u00b5<\/em> &lt; 150<\/li><li><em data-effect=\"italics\">p<\/em>-value = 0.0622<\/li><li>alpha = 0.01<\/li><li>Do not reject the null hypothesis.<\/li><li>At the 1% significance level, there is not enough evidence to conclude that freshmen students study less than 2.5 hours per day, on average.<\/li><li>The student academic group\u2019s claim appears to be correct.<\/li><\/ol><\/div><\/div><\/div><div class=\"references\" data-depth=\"1\" id=\"fs-idp118682672\"><h1 data-type=\"title\">References<\/h1><p>Data from Amit Schitai. Director of Instructional Technology and Distance Learning. LBCC.<\/p><p>Data from <em data-effect=\"italics\">Bloomberg Businessweek<\/em>. Available online at http:\/\/www.businessweek.com\/news\/2011-\n09-15\/nyc-smoking-rate-falls-to-record-low-of-14-bloomberg-says.html.<\/p><p>Data from energy.gov. Available online at http:\/\/energy.gov (accessed June 27. 2013).<\/p><p>Data from Gallup\u00ae. Available online at www.gallup.com (accessed June 27, 2013).<\/p><p>Data from <em data-effect=\"italics\">Growing by Degrees<\/em> by Allen and Seaman.<\/p><p>Data from La Leche League International. Available online at http:\/\/www.lalecheleague.org\/Law\/BAFeb01.html.<\/p><p>Data from the American Automobile Association. Available online at www.aaa.com (accessed June 27, 2013).<\/p><p>Data from the American Library Association. Available online at www.ala.org (accessed June 27, 2013).<\/p><p>Data from the Bureau of Labor Statistics. Available online at http:\/\/www.bls.gov\/oes\/current\/oes291111.htm.<\/p><p id=\"fs-idp1\">Data from the Centers for Disease Control and Prevention. Available online at www.cdc.gov (accessed June 27, 2013)<\/p><p>Data from the U.S. Census Bureau, available online at http:\/\/quickfacts.census.gov\/qfd\/states\/00000.html (accessed June 27, 2013).<\/p><p>Data from the United States Census Bureau. Available online at http:\/\/www.census.gov\/hhes\/socdemo\/language\/.<\/p><p id=\"eip-79\">Data from Toastmasters International. Available online at http:\/\/toastmasters.org\/artisan\/detail.asp?CategoryID=1&amp;SubCategoryID=10&amp;ArticleID=429&amp;Page=1.<\/p><p>Data from Weather Underground. Available online at www.wunderground.com (accessed June 27, 2013).<\/p><p>Federal Bureau of Investigations. \u201cUniform Crime Reports and Index of Crime in Daviess in the State of Kentucky enforced by Daviess County from 1985 to 2005.\u201d Available online at http:\/\/www.disastercenter.com\/kentucky\/crime\/3868.htm (accessed June 27, 2013).<\/p><p>\u201cFoothill-De Anza Community College District.\u201d De Anza College, Winter 2006. Available online at http:\/\/research.fhda.edu\/factbook\/DAdemofs\/Fact_sheet_da_2006w.pdf.<\/p><p>Johansen, C., J. Boice, Jr., J. McLaughlin, J. Olsen. \u201cCellular Telephones and Cancer\u2014a Nationwide Cohort Study in Denmark.\u201d Institute of Cancer Epidemiology and the Danish Cancer Society, 93(3):203-7. Available online at http:\/\/www.ncbi.nlm.nih.gov\/pubmed\/11158188 (accessed June 27, 2013).<\/p><p>Rape, Abuse &amp; Incest National Network. \u201cHow often does sexual assault occur?\u201d RAINN, 2009. Available online at http:\/\/www.rainn.org\/get-information\/statistics\/frequency-of-sexual-assault (accessed June 27, 2013).<\/p><\/div><div data-type=\"glossary\" class=\"textbox shaded\"><h2 data-type=\"glossary-title\">Glossary<\/h2><dl class=\"definition\"><dt>Central Limit Theorem<\/dt><dd>\n     Given a random variable (RV) with known mean \\(\\mu \\) and known standard deviation \u03c3. We are sampling with size <em data-effect=\"italics\">n<\/em> and we are interested in two new RVs - the sample mean, \\(\\overline{X}\\), and the sample sum, \\(\\Sigma X\\). \nIf the size <em data-effect=\"italics\">n<\/em> of the sample is sufficiently large, then \\(\\overline{X}~N\\left(\\mu \\text{,}\\frac{\\sigma }{\\sqrt{n}}\\right)\\) and \\(\\Sigma X~N\\left(n\\mu ,\\sqrt{n}\\sigma \\right)\\). If the size <em data-effect=\"italics\">n<\/em> of the sample is sufficiently large, then the distribution of the sample means and the distribution of the sample sums will approximate a normal distribution regardless of the shape of the population. The mean of the sample  means will equal the population mean and the mean of the sample sums will equal <em data-effect=\"italics\">n<\/em> times the population mean. The standard deviation of the distribution of the sample means, \\(\\frac{\\sigma }{\\sqrt{n}}\\), is called the standard error of the mean.<\/dd><\/dl><\/div>","rendered":"<p>[latexpage]<\/p>\n<ul>\n<li>In a <span data-type=\"term\">hypothesis test<\/span> problem, you may see words such as &#8220;the level of significance is 1%.&#8221; The &#8220;1%&#8221; is the preconceived or preset <em data-effect=\"italics\">\u03b1<\/em>.<\/li>\n<li>The statistician setting up the hypothesis test selects the value of <em data-effect=\"italics\">\u03b1<\/em> to use <strong>before<\/strong> collecting the sample data.<\/li>\n<li><strong>If no level of significance is given, a common standard to use is <em data-effect=\"italics\">\u03b1<\/em> = 0.05.<\/strong><\/li>\n<li>When you calculate the <em data-effect=\"italics\">p<\/em>-value and draw the picture, the <em data-effect=\"italics\">p<\/em>-value is the area in the left tail, the right tail, or split evenly between the two tails. For this reason, we call the hypothesis test left, right, or two tailed.<\/li>\n<li>The <strong>alternative hypothesis<\/strong>, \\({H}_{a}\\), tells you if the test is left, right, or two-tailed. It is the <strong>key<\/strong> to conducting the appropriate test.<\/li>\n<li><em data-effect=\"italics\">H<sub>a<\/sub><\/em><strong>never<\/strong> has a symbol that contains an equal sign.<\/li>\n<li><strong>Thinking about the meaning of the<\/strong><span data-type=\"term\"><em data-effect=\"italics\">p<\/em>-value<\/span>: A data analyst (and anyone else) should have more confidence that he made the correct decision to reject the null hypothesis with a smaller <em data-effect=\"italics\">p<\/em>-value (for example, 0.001 as opposed to 0.04) even if using the 0.05 level for alpha. Similarly, for a large <em data-effect=\"italics\">p<\/em>-value such as 0.4, as opposed to a <em data-effect=\"italics\">p<\/em>-value of 0.056 (alpha = 0.05 is less than either number), a data analyst should have more confidence that she made the correct decision in not rejecting the null hypothesis.  This makes the data analyst use judgment rather than mindlessly applying rules.<\/li>\n<\/ul>\n<p id=\"fs-idp42898080\">The following examples illustrate a left-, right-, and two-tailed test.<\/p>\n<div data-type=\"example\" class=\"textbox examples\" id=\"fs-idp41487520\">\n<p id=\"fs-idp41487648\"><em data-effect=\"italics\">H<sub>o<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> = 5, <em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &lt; 5<\/p>\n<p>Test of a single population mean. <em data-effect=\"italics\">H<sub>a<\/sub><\/em> tells you the test is left-tailed. The picture of the <em data-effect=\"italics\">p<\/em>-value is as follows:<\/p>\n<div class=\"bc-figure figure\" id=\"fs-idm151356816\"><span data-type=\"media\" id=\"id14703034\" data-alt=\"Normal distribution curve of a single population mean with a value of 5 on the x-axis and the p-value points to the area on the left tail of the curve.\"><img decoding=\"async\" src=\"\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/fig-ch09_09_01-02.jpg\" data-media-type=\"image\/jpg\" alt=\"Normal distribution curve of a single population mean with a value of 5 on the x-axis and the p-value points to the area on the left tail of the curve.\" width=\"380\" \/><\/span><\/div>\n<\/div>\n<div data-type=\"note\" data-has-label=\"true\" id=\"fs-idp42408528\" class=\"note statistics try\" data-label=\"\">\n<div data-type=\"title\" class=\"title\">Try It<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-idp67352928\">\n<div data-type=\"problem\" class=\"problem\">\n<p id=\"eip-idm132446832\"><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> = 10, <em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &lt; 10<\/p>\n<p id=\"eip-idp3861728\">Assume the <em data-effect=\"italics\">p<\/em>-value is 0.0935. What type of test is this? Draw the picture of the <em data-effect=\"italics\">p<\/em>-value.<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-idp1251488\">\n<p>left-tailed test<\/p>\n<div class=\"bc-figure figure\" id=\"fs-idm70716944\"><span data-type=\"media\" id=\"eip-idp9289536\" data-alt=\"\"><img decoding=\"async\" src=\"\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/CNX_Stats_C09_M09_tryit002annoN.jpg\" data-media-type=\"image\/jpeg\" alt=\"\" width=\"380\" \/><\/span><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"example\" class=\"textbox examples\">\n<p><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2264 0.2\u2003\u2003<em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> &gt; 0.2<\/p>\n<p id=\"fs-idp117549824\">This is a test of a single population proportion. <em data-effect=\"italics\">H<sub>a<\/sub><\/em> tells you the test is <strong>right-tailed<\/strong>. The picture of the <em data-effect=\"italics\">p<\/em>-value is as follows:<\/p>\n<div class=\"bc-figure figure\" id=\"fs-idm2304848\"><span data-type=\"media\" id=\"id14703170\" data-alt=\"Normal distribution curve of a single population proportion with the value of 0.2 on the x-axis. The p-value points to the area on the right tail of the curve.\"><img decoding=\"async\" src=\"\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/fig-ch09_09_02a.jpg\" data-media-type=\"image\/jpg\" alt=\"Normal distribution curve of a single population proportion with the value of 0.2 on the x-axis. The p-value points to the area on the right tail of the curve.\" width=\"380\" \/><\/span><\/div>\n<\/div>\n<div data-type=\"note\" data-has-label=\"true\" id=\"fs-idm126098448\" class=\"note statistics try\" data-label=\"\">\n<div data-type=\"title\" class=\"title\">Try It<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"eip-417\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-idp131553552\">\n<p id=\"eip-idm89526848\"><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> \u2264 1, <em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &gt; 1<\/p>\n<p id=\"eip-idm24425280\">Assume the <em data-effect=\"italics\">p<\/em>-value is 0.1243. What type of test is this? Draw the picture of the <em data-effect=\"italics\">p<\/em>-value.<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\">\n<p>right-tailed test<\/p>\n<div class=\"bc-figure figure\" id=\"fs-idm53526160\"><span data-type=\"media\" id=\"eip-idm27872752\" data-alt=\"\"><img decoding=\"async\" src=\"\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/CNX_Stats_C09_M09_tryit001annoN.jpg\" data-media-type=\"image\/jpeg\" alt=\"\" width=\"380\" \/><\/span><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"example\" class=\"textbox examples\">\n<p id=\"element-38a\"><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> = 50\u2003\u2003<em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2260 50<\/p>\n<p>This is a test of a single population mean. <em data-effect=\"italics\">H<sub>a<\/sub><\/em> tells you the test is <strong>two-tailed<\/strong>. The picture of the <em data-effect=\"italics\">p<\/em>-value is as follows.<\/p>\n<div class=\"bc-figure figure\" id=\"fs-idm117972816\"><span data-type=\"media\" id=\"id4829633\" data-alt=\"Normal distribution curve of a single population mean with a value of 50 on the x-axis. The p-value formulas, 1\/2(p-value), for a two-tailed test is shown for the areas on the left and right tails of the curve.\"><img decoding=\"async\" src=\"\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/fig-ch09_09_03N.jpg\" data-media-type=\"image\/jpg\" alt=\"Normal distribution curve of a single population mean with a value of 50 on the x-axis. The p-value formulas, 1\/2(p-value), for a two-tailed test is shown for the areas on the left and right tails of the curve.\" width=\"380\" \/><\/span><\/div>\n<\/div>\n<div data-type=\"note\" data-has-label=\"true\" id=\"fs-idm63250432\" class=\"note statistics try\" data-label=\"\">\n<div data-type=\"title\" class=\"title\">Try It<\/div>\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\">\n<p id=\"eip-idp41687312\"><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> = 0.5, <em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2260 0.5<\/p>\n<p id=\"eip-idm5668432\">Assume the <em data-effect=\"italics\">p<\/em>-value is 0.2564. What type of test is this? Draw the picture of the <em data-effect=\"italics\">p<\/em>-value.<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\">\n<p>two-tailed test<\/p>\n<div class=\"bc-figure figure\" id=\"fs-idm126590640\"><span data-type=\"media\" id=\"eip-idm113874560\" data-alt=\"\"><img decoding=\"async\" src=\"\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/CNX_Stats_C09_M09_tryit003annoN.jpg\" data-media-type=\"image\/jpeg\" alt=\"\" width=\"380\" \/><\/span><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bc-section section\" data-depth=\"1\" id=\"fs-idm12892576\">\n<h1 data-type=\"title\">Full Hypothesis Test Examples<\/h1>\n<div data-type=\"example\" class=\"textbox examples\">\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\" id=\"id5300933\">\n<p>Jeffrey, as an eight-year old, <strong>established a mean time of 16.43 seconds<\/strong> for swimming the 25-yard freestyle, with a <strong>standard deviation of 0.8 seconds<\/strong>. His dad, Frank, thought that Jeffrey could swim the 25-yard freestyle faster using goggles. Frank bought Jeffrey a new pair of expensive goggles and timed Jeffrey for <strong>15 25-yard freestyle swims<\/strong>. For the 15 swims, <strong>Jeffrey&#8217;s mean time was 16 seconds. Frank thought that the goggles helped Jeffrey to swim faster than the 16.43 seconds.<\/strong> Conduct a hypothesis test using a preset <em data-effect=\"italics\">\u03b1<\/em> = 0.05. Assume that the swim times for the 25-yard freestyle are normal.<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"id5036614\">\n<p>Set up the Hypothesis Test:<\/p>\n<p>Since the problem is about a mean, this is a <strong>test of a single population mean<\/strong>.<\/p>\n<p><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> = 16.43\u2003\u2003<em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &lt; 16.43<\/p>\n<p id=\"element-241\">For Jeffrey to swim faster, his time will be less than 16.43 seconds. The &#8220;&lt;&#8221; tells you this is left-tailed.<\/p>\n<p>Determine the distribution needed:<\/p>\n<p><strong>Random variable: <\/strong>\\(\\overline{X}\\) = the mean time to swim the 25-yard freestyle.<\/p>\n<p><strong> Distribution for the test: <\/strong>\\(\\overline{X}\\) is normal (population <span data-type=\"term\">standard deviation<\/span> is known: <em data-effect=\"italics\">\u03c3<\/em> = 0.8)<\/p>\n<p>\\(\\overline{X}~N\\left(\\mu ,\\frac{{\\sigma }_{X}}{\\sqrt{n}}\\right)\\)<\/p>\n<p>Therefore, \\(\\overline{X}~N\\left(16.43,\\frac{0.8}{\\sqrt{15}}\\right)\\)<\/p>\n<p><em data-effect=\"italics\">\u03bc<\/em> = 16.43 comes from <em data-effect=\"italics\">H<sub>0<\/sub><\/em> and not the data. <em data-effect=\"italics\">\u03c3<\/em> = 0.8, and <em data-effect=\"italics\">n<\/em> = 15.<\/p>\n<p>Calculate the <em data-effect=\"italics\">p<\/em>-value using the normal distribution for a mean:<\/p>\n<p id=\"element-706\"><em data-effect=\"italics\">p<\/em>-value = <em data-effect=\"italics\">P<\/em>(\\(\\overline{x}\\) &lt; 16) = 0.0187 where the sample mean in the problem is given as 16.<\/p>\n<p id=\"element-656\"><em data-effect=\"italics\">p<\/em>-value = 0.0187 (This is called the <strong>actual level of significance<\/strong>.) The <em data-effect=\"italics\">p<\/em>-value is the area to the left of the sample mean is given as 16.<\/p>\n<p><strong>Graph:<\/strong><\/p>\n<div class=\"bc-figure figure\" id=\"hyptest11_ex1\"><span data-type=\"media\" id=\"id4282005\" data-alt=\"Normal distribution curve for the average time to swim the 25-yard freestyle with values 16, as the sample mean, and 16.43 on the x-axis. A vertical upward line extends from 16 on the x-axis to the curve. An arrow points to the left tail of the curve.\"><img decoding=\"async\" src=\"\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/fig-ch09_11_01N.jpg\" data-media-type=\"image\/jpg\" alt=\"Normal distribution curve for the average time to swim the 25-yard freestyle with values 16, as the sample mean, and 16.43 on the x-axis. A vertical upward line extends from 16 on the x-axis to the curve. An arrow points to the left tail of the curve.\" width=\"380\" \/><\/span><\/div>\n<p><em data-effect=\"italics\">\u03bc<\/em> = 16.43 comes from <em data-effect=\"italics\">H<sub>0<\/sub><\/em>. Our assumption is <em data-effect=\"italics\">\u03bc<\/em> = 16.43.<\/p>\n<p><strong>Interpretation of the <em data-effect=\"italics\">p<\/em>-value: If <em data-effect=\"italics\">H<sub>0<\/sub><\/em> is true<\/strong>, there is a 0.0187 probability (1.87%)that Jeffrey&#8217;s mean time to swim the 25-yard freestyle is 16 seconds or less. Because a 1.87% chance is small, the mean time of 16 seconds or less is unlikely to have happened randomly. It is a rare event.<\/p>\n<p>Compare <em data-effect=\"italics\">\u03b1<\/em> and the <em data-effect=\"italics\">p<\/em>-value:<\/p>\n<p id=\"element-890\"><em data-effect=\"italics\">\u03b1<\/em> = 0.05 <em data-effect=\"italics\">p<\/em>-value = 0.0187 <em data-effect=\"italics\">\u03b1<\/em> &gt; <em data-effect=\"italics\">p<\/em>-value<\/p>\n<p id=\"element-543\"><strong>Make a decision:<\/strong> Since <em data-effect=\"italics\">\u03b1<\/em> &gt; <em data-effect=\"italics\">p<\/em>-value, reject <em data-effect=\"italics\">H<sub>0<\/sub><\/em>.<\/p>\n<p>&lt;!&#8211; CONTINUE HERE &#8211;&gt;<\/p>\n<p>This means that you reject <em data-effect=\"italics\">\u03bc<\/em> = 16.43. In other words, you do not think Jeffrey swims the<br \/>\n25-yard freestyle in 16.43 seconds but faster with the new goggles.<\/p>\n<p><strong>Conclusion:<\/strong> At the 5% significance level, we conclude that Jeffrey swims faster using the new goggles. The sample data show there is sufficient evidence that Jeffrey&#8217;s mean time to swim the 25-yard freestyle is less than 16.43 seconds.<\/p>\n<p id=\"element-252\">The <em data-effect=\"italics\">p<\/em>-value can easily be calculated.<\/p>\n<div data-type=\"note\" data-has-label=\"true\" id=\"fs-idm31641280\" class=\"note statistics calculator\" data-label=\"\">\n<p id=\"element-362\">Press <code>STAT<\/code> and arrow over to <code>TESTS<\/code>. Press <code>1:Z-Test<\/code>. Arrow over to <code>Stats<\/code> and press <code>ENTER<\/code>. Arrow down and enter 16.43 for <em data-effect=\"italics\">\u03bc<sub>0<\/sub><\/em> (null hypothesis), .8 for <em data-effect=\"italics\">\u03c3<\/em>, 16 for the sample mean, and 15 for <em data-effect=\"italics\">n<\/em>. Arrow down to <em data-effect=\"italics\">\u03bc<\/em> : (alternate hypothesis) and arrow over to &lt; <em data-effect=\"italics\">\u03bc<sub>0<\/sub><\/em>. Press <code>ENTER<\/code>. Arrow down to <code>Calculate<\/code> and press <code>ENTER<\/code>. The calculator not only calculates the <em data-effect=\"italics\">p<\/em>-value (<em data-effect=\"italics\">p<\/em> = 0.0187) but it also calculates the test statistic (<em data-effect=\"italics\">z<\/em>-score) for the sample mean. <em data-effect=\"italics\">\u03bc<\/em> &lt; 16.43 is the alternative hypothesis. Do this set of instructions again except arrow to <code>Draw<\/code>(instead of <code>Calculate<\/code>). Press <code>ENTER<\/code>. A shaded graph appears with <em data-effect=\"italics\">z<\/em> = -2.08 (test statistic) and <em data-effect=\"italics\">p<\/em> = 0.0187 (<em data-effect=\"italics\">p<\/em>-value). Make sure when you use <code>Draw<\/code> that no other equations are highlighted in <em data-effect=\"italics\">Y<\/em> = and the plots are turned off.<\/p>\n<\/div>\n<p>When the calculator does a <em data-effect=\"italics\">Z<\/em>-Test, the <code>Z-Test<\/code> function finds the <em data-effect=\"italics\">p<\/em>-value by doing a normal probability calculation using the <span data-type=\"term\">central limit theorem<\/span>:<\/p>\n<p id=\"fs-idp159988784\">\\(P\\left(\\overline{x}&lt;16\\right)=\\)<code>2nd DISTR normcdf<\/code>\\(\\left(-10^99,16,16.43,0.8\/\\sqrt{15}\\right)\\)<br \/>\n.<\/p>\n<p>The Type I and Type II errors for this problem are as follows:<\/p>\n<p id=\"element-385\">The Type I error is to conclude that Jeffrey swims the 25-yard freestyle, on average, in<br \/>\nless than 16.43 seconds when, in fact, he actually swims the 25-yard freestyle, on<br \/>\naverage, in 16.43 seconds. (Reject the null hypothesis when the null hypothesis is true.)<\/p>\n<p>The Type II error is that there is not evidence to conclude that Jeffrey swims the 25-yard free-style, on average, in less than 16.43 seconds when, in fact, he actually does swim the 25-yard free-style, on average, in less than 16.43 seconds. (Do not reject the null hypothesis when the null hypothesis is false.)<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"note\" data-has-label=\"true\" id=\"fs-idm112426112\" class=\"note statistics try\" data-label=\"\">\n<div data-type=\"title\" class=\"title\">Try It<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-idp78508304\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-idp117367632\">\n<p>The mean throwing distance of a football for a Marco, a high school freshman quarterback, is 40 yards, with a standard deviation of two yards. The team coach tells Marco to adjust his grip to get more distance. The coach records the distances for 20 throws. For the 20 throws, Marco\u2019s mean distance was 45 yards. The coach thought the different grip helped Marco throw farther than 40 yards. Conduct a hypothesis test using a preset <em data-effect=\"italics\">\u03b1<\/em> = 0.05. Assume the throw distances for footballs are normal.<\/p>\n<p id=\"eip-idp30205504\">First, determine what type of test this is, set up the hypothesis test, find the <em data-effect=\"italics\">p<\/em>-value, sketch the graph, and state your conclusion.<\/p>\n<div data-type=\"note\" data-has-label=\"true\" id=\"fs-idm24767312\" class=\"note statistics calculator\" data-label=\"\">\n<p id=\"eip-idm96892800\">Press STAT and arrow over to TESTS. Press 1:Z-Test. Arrow over to Stats and press ENTER. Arrow down and enter 40 for <em data-effect=\"italics\">\u03bc<\/em>0 (null hypothesis), 2 for <em data-effect=\"italics\">\u03c3<\/em>, 45 for the sample mean, and 20 for <em data-effect=\"italics\">n<\/em>. Arrow down to <em data-effect=\"italics\">\u03bc<\/em>: (alternative hypothesis) and set it either as &lt;, \u2260, or &gt;. Press ENTER. Arrow down to Calculate and press ENTER. The calculator not only calculates the <em data-effect=\"italics\">p<\/em>-value but it also calculates the test statistic (<em data-effect=\"italics\">z<\/em>-score) for the sample mean. Select &lt;, \u2260, or &gt; for the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate). Press ENTER. A shaded graph appears with test statistic and <em data-effect=\"italics\">p<\/em>-value. Make sure when you use Draw that no other equations are highlighted in <em data-effect=\"italics\">Y<\/em> = and the plots are turned off.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"solution\" class=\"solution\">\n<p>\nSince the problem is about a mean, this is a test of a single population mean.\n  <\/p>\n<p id=\"eip-idm108002480\"><em data-effect=\"italics\">H<sub>0<\/sub><\/em> : <em data-effect=\"italics\">\u03bc<\/em> = 40<\/p>\n<p id=\"eip-idm216373056\"><em data-effect=\"italics\">H<sub>a<\/sub><\/em> : <em data-effect=\"italics\">\u03bc<\/em> &gt; 40<\/p>\n<p id=\"eip-idm58633744\"><em data-effect=\"italics\">p<\/em> = 0.0062<\/p>\n<div class=\"bc-figure figure\" id=\"fs-idm152825344\"><span data-type=\"media\" id=\"eip-idm58633360\" data-alt=\"\"><img decoding=\"async\" src=\"\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/CNX_Stats_C09_M11_tryit001annoN.jpg\" data-media-type=\"image\/jpeg\" alt=\"\" width=\"380\" \/><\/span><\/div>\n<p id=\"eip-idm24608304\">Because <em data-effect=\"italics\">p<\/em> &lt; <em data-effect=\"italics\">\u03b1<\/em>, we reject the null hypothesis. There is sufficient evidence to suggest that the change in grip improved Marco\u2019s throwing distance.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"note\" class=\"note\" data-has-label=\"true\" data-label=\"\">\n<div data-type=\"title\" class=\"title\">Historical Note (<a href=\"#fs-idp41487520\" class=\"autogenerated-content\">[link]<\/a>)<\/div>\n<p id=\"eip-idp233123200\">The traditional way to compare the two probabilities, <em data-effect=\"italics\">\u03b1<\/em> and the <em data-effect=\"italics\">p<\/em>-value, is to compare the critical value (<em data-effect=\"italics\">z<\/em>-score from <em data-effect=\"italics\">\u03b1<\/em>) to the test statistic (<em data-effect=\"italics\">z<\/em>-score from data). The calculated test statistic for the <em data-effect=\"italics\">p<\/em>-value is \u20132.08. (From the Central Limit Theorem, the test statistic formula is \\(z=\\frac{\\overline{x}-{\\mu }_{X}}{\\left(\\frac{{\\sigma }_{X}}{\\sqrt{n}}\\right)}\\). <\/p>\n<p>&lt;!&#8211;LINK THIS PLEASE &#8211;&gt;<br \/>\nFor this problem, \\(\\overline{x}\\) = 16, <em data-effect=\"italics\">\u03bc<sub>X<\/sub><\/em> = 16.43 from the null hypothes is, <em data-effect=\"italics\">\u03c3<sub>X<\/sub><\/em> = 0.8, and <em data-effect=\"italics\">n<\/em> = 15.) You can find the critical value for <em data-effect=\"italics\">\u03b1<\/em> = 0.05 in the normal table (see <strong>15.Tables<\/strong> in the Table of Contents). The <em data-effect=\"italics\">z<\/em>-score for an area to the left equal to 0.05 is midway between \u20131.65 and \u20131.64 (0.05 is midway between 0.0505 and 0.0495). The <em data-effect=\"italics\">z<\/em>-score is \u20131.645. Since \u20131.645 &gt; \u20132.08 (which demonstrates that \u03b1 &gt; <em data-effect=\"italics\">p<\/em>-value), reject <em data-effect=\"italics\">H<sub>0<\/sub><\/em>. Traditionally, the decision<br \/>\nto reject or not reject was done in this way. Today, comparing the two probabilities <em data-effect=\"italics\">\u03b1<\/em> and the <em data-effect=\"italics\">p<\/em>-value is very common. For this problem, the <em data-effect=\"italics\">p<\/em>-value, 0.0187 is considerably smaller than <em data-effect=\"italics\">\u03b1<\/em>, 0.05. You can be confident about your decision to reject. The graph shows <em data-effect=\"italics\">\u03b1<\/em>, the <em data-effect=\"italics\">p<\/em>-value, and the test statistics and the critical value.\n<\/p>\n<div class=\"bc-figure figure\" id=\"eip-idm47467424\"><span data-type=\"media\" id=\"eip-idp126239824\" data-alt=\"Distribution curve comparing the &#x3b1; to the p-value. Values of -2.15 and -1.645 are on the x-axis. Vertical upward lines extend from both of these values to the curve. The p-value is equal to 0.0158 and points to the area to the left of -2.15. &#x3b1; is equal to 0.05 and points to the area between the values of -2.15 and -1.645.\"><img decoding=\"async\" src=\"\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/fig-ch09_11_02.png\" data-media-type=\"image\/png\" alt=\"Distribution curve comparing the &#x3b1; to the p-value. Values of -2.15 and -1.645 are on the x-axis. Vertical upward lines extend from both of these values to the curve. The p-value is equal to 0.0158 and points to the area to the left of -2.15. &#x3b1; is equal to 0.05 and points to the area between the values of -2.15 and -1.645.\" width=\"380\" \/><\/span><\/div>\n<\/div>\n<div data-type=\"example\" class=\"textbox examples\">\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\" id=\"id5291254\">\n<p>A college football coach thought that his players could bench press a <strong>mean weight of 275 pounds<\/strong>. It is known that the <strong>standard deviation is 55<br \/>\npounds<\/strong>. Three of his players thought that the mean weight was <strong>more than<\/strong> that amount.<br \/>\nThey asked <strong>30<\/strong> of their teammates for their estimated maximum lift on the bench<br \/>\npress exercise. The data ranged from 205 pounds to 385 pounds. The actual<br \/>\ndifferent weights were (frequencies are in parentheses)<br \/>\n<span data-type=\"list\" data-list-type=\"labeled-item\" id=\"set-1\" data-display=\"inline\"><span data-type=\"item\">205(3)<\/span><span data-type=\"item\">215(3)<\/span><span data-type=\"item\">225(1)<\/span><span data-type=\"item\">241(2)<\/span><span data-type=\"item\">252(2)<\/span><span data-type=\"item\">265(2)<\/span><span data-type=\"item\">275(2)<\/span><span data-type=\"item\">313(2)<\/span><span data-type=\"item\">316(5)<\/span><span data-type=\"item\">338(2)<\/span><span data-type=\"item\">341(1)<\/span><span data-type=\"item\">345(2)<\/span><span data-type=\"item\">368(2)<\/span><span data-type=\"item\">385(1)<\/span><\/span>.<\/p>\n<p>Conduct a hypothesis test using a 2.5% level of significance to determine if the bench press mean is <strong>more than 275 pounds<\/strong>.<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"id5342777\">\n<p>Set up the Hypothesis Test:<\/p>\n<p>Since the problem is about a mean weight, this is a <strong>test of a single population mean<\/strong>.<\/p>\n<p><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> = 275<\/p>\n<div data-type=\"newline\"><\/div>\n<p><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &gt; 275<\/p>\n<div data-type=\"newline\"><\/div>\n<p>This is a right-tailed test.<\/p>\n<p>Calculating the distribution needed:<\/p>\n<p id=\"element-975\">Random variable: \\(\\overline{X}\\) = the mean weight, in pounds, lifted by the football players.<\/p>\n<p><strong>Distribution for the test:<\/strong> It is normal because <em data-effect=\"italics\">\u03c3<\/em> is known.<\/p>\n<p>\\(\\overline{X}~N\\left(275,\\frac{55}{\\sqrt{30}}\\right)\\)<\/p>\n<p>\\(\\overline{x}=286.2\\) pounds (from the data).<\/p>\n<p><em data-effect=\"italics\">\u03c3<\/em> = 55 pounds <strong>(Always use <em data-effect=\"italics\">\u03c3<\/em> if you know it.)<\/strong> We assume <em data-effect=\"italics\">\u03bc<\/em> = 275 pounds unless our data shows us otherwise. <\/p>\n<p>Calculate the <em data-effect=\"italics\">p<\/em>-value using the normal distribution for a mean and using the sample mean as input (see <a href=\"\/contents\/d0ba1833-f0d2-4195-8765-3c436745f0fb@4\" class=\"autogenerated-content\">[link]<\/a> for using the data as input):<\/p>\n<p>\\(p\\text{-value}=P\\left(\\overline{x}&gt;286.2\\right)=0.1323\\).<\/p>\n<p><strong>Interpretation of the <em data-effect=\"italics\">p<\/em>-value:<\/strong> If <em data-effect=\"italics\">H<sub>0<\/sub><\/em> is true, then there is a 0.1331 probability (13.23%) that the football players can lift a mean weight of 286.2 pounds or more. Because a 13.23% chance is large enough, a mean weight lift of 286.2 pounds or more is not a rare event.<\/p>\n<div class=\"bc-figure figure\" id=\"hyptest11_ex3\"><span data-type=\"media\" id=\"id4317824\" data-alt=\"Normal distribution curve of the average weight lifted by football players with values of 275 and 286.2 on the x-axis. A vertical upward line extends from 286.2 to the curve. The p-value points to the area to the right of 286.2.\"><img decoding=\"async\" src=\"\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/fig-ch09_11_03N.jpg\" data-media-type=\"image\/jpg\" alt=\"Normal distribution curve of the average weight lifted by football players with values of 275 and 286.2 on the x-axis. A vertical upward line extends from 286.2 to the curve. The p-value points to the area to the right of 286.2.\" width=\"380\" \/><\/span><\/div>\n<p>Compare <em data-effect=\"italics\">\u03b1<\/em> and the <em data-effect=\"italics\">p<\/em>-value:<\/p>\n<p><em data-effect=\"italics\">\u03b1<\/em> = 0.025 <em data-effect=\"italics\">p<\/em>-value = 0.1323<\/p>\n<p><strong>Make a decision:<\/strong> Since <em data-effect=\"italics\">\u03b1<\/em> &lt;<em data-effect=\"italics\">p<\/em>-value, do not reject <em data-effect=\"italics\">H<sub>0<\/sub><\/em>.<\/p>\n<p><strong>Conclusion:<\/strong> At the 2.5% level of significance, from the sample data, there is not sufficient evidence to conclude that the true mean weight lifted is more than 275 pounds.<\/p>\n<p>The <em data-effect=\"italics\">p<\/em>-value can easily be calculated.<\/p>\n<div data-type=\"note\" data-has-label=\"true\" id=\"fs-idm79549936\" class=\"note statistics calculator\" data-label=\"\">\n<p id=\"element-233\">Put the data and frequencies into lists. Press <code>STAT<\/code> and arrow over to <code>TESTS<\/code>. Press <code>1:Z-Test<\/code>. Arrow over to <code>Data<\/code> and press <code>ENTER<\/code>. Arrow down and enter 275 for <em data-effect=\"italics\">\u03bc<sub>0<\/sub><\/em>, 55 for <em data-effect=\"italics\">\u03c3<\/em>, the name of the list where you put the data, and the name of the list where you put the frequencies. Arrow down to <em data-effect=\"italics\">\u03bc:<\/em> and arrow over to &gt; <em data-effect=\"italics\">\u03bc<sub>0<\/sub><\/em>. Press <code>ENTER<\/code>. Arrow down to <code>Calculate<\/code> and press <code>ENTER<\/code>. The calculator not only calculates the <em data-effect=\"italics\">p<\/em>-value (<em data-effect=\"italics\">p<\/em> = 0.1331, a little different from the previous calculation &#8211; in it we used the sample mean rounded to one decimal place instead of the data) but it also calculates the test statistic (<em data-effect=\"italics\">z<\/em>-score) for the sample mean, the sample mean, and the sample standard deviation. <em data-effect=\"italics\">\u03bc<\/em> &gt; 275 is the alternative hypothesis. Do this set of instructions again except arrow to <code>Draw<\/code> (instead of <code>Calculate<\/code>). Press <code>ENTER<\/code>. A shaded graph appears with <em data-effect=\"italics\">z<\/em> = 1.112 (test statistic) and <em data-effect=\"italics\">p<\/em> = 0.1331 (<em data-effect=\"italics\">p<\/em>-value). Make sure when you use <code>Draw<\/code> that no other equations are highlighted in <em data-effect=\"italics\">Y<\/em> = and the plots are turned off.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"example\" class=\"textbox examples\" id=\"element-207\">\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\" id=\"id4319892\">\n<p id=\"fs-idp90231232\">Statistics students believe that the mean score on the first statistics test is 65. A statistics instructor thinks the mean score is higher than 65. He samples ten statistics students and obtains the scores <span data-type=\"list\" data-list-type=\"labeled-item\" id=\"set-2\" data-display=\"inline\"><span data-type=\"item\">65<\/span><span data-type=\"item\">65<\/span><span data-type=\"item\">70<\/span><span data-type=\"item\">67<\/span><span data-type=\"item\">66<\/span><span data-type=\"item\">63<\/span><span data-type=\"item\">63<\/span><span data-type=\"item\">68<\/span><span data-type=\"item\">72<\/span><span data-type=\"item\">71<\/span><\/span>. He performs a hypothesis test using a 5% level of significance. The data are assumed to be from a normal distribution.<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"id5147242\">\n<p>Set up the hypothesis test:<\/p>\n<p>A 5% level of significance means that <em data-effect=\"italics\">\u03b1<\/em> = 0.05. This is a test of a <strong>single population mean<\/strong>.<\/p>\n<p><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> = 65\u2003\u2003<em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &gt; 65<\/p>\n<p>Since the instructor thinks the average score is higher, use a &#8220;&gt;&#8221;. The &#8220;&gt;&#8221; means the test is right-tailed.<\/p>\n<p id=\"fs-idp165624976\">Determine the distribution needed:<\/p>\n<p><strong>Random variable:<\/strong>\\(\\overline{X}\\) = average score on the first statistics test.<\/p>\n<p><strong> Distribution for the test:<\/strong> If you read the problem carefully, you will notice that there is<br \/>\n<strong>no population standard deviation given<\/strong>. You are only given <em data-effect=\"italics\">n<\/em> = 10 sample data values. Notice also that the data come from a normal distribution. This means that the distribution for the test is a student&#8217;s <em data-effect=\"italics\">t<\/em>.<\/p>\n<p id=\"element-240\">Use <em data-effect=\"italics\">t<\/em><sub>df<\/sub>. Therefore, the distribution for the test is <em data-effect=\"italics\">t<\/em><sub>9<\/sub> where <em data-effect=\"italics\">n<\/em> = 10 and <em data-effect=\"italics\">df<\/em> = 10 &#8211; 1 = 9.<\/p>\n<p>Calculate the <em data-effect=\"italics\">p<\/em>-value using the Student&#8217;s <em data-effect=\"italics\">t<\/em>-distribution:<\/p>\n<p><em data-effect=\"italics\">p<\/em>-value = <em data-effect=\"italics\">P<\/em>(\\(\\overline{x}\\) &gt; 67) = 0.0396 where the sample mean and sample standard deviation are calculated as 67 and 3.1972 from the data.<\/p>\n<p><strong>Interpretation of the <em data-effect=\"italics\">p<\/em>-value:<\/strong> If the null hypothesis is true, then there is a 0.0396<br \/>\nprobability (3.96%) that the sample mean is 65 or more.<\/p>\n<div class=\"bc-figure figure\" id=\"hyptest11_ex4\"><span data-type=\"media\" id=\"id5241154\" data-alt=\"Normal distribution curve of average scores on the first statistic tests with 65 and 67 values on the x-axis. A vertical upward line extends from 67 to the curve. The p-value points to the area to the right of 67.\"><img decoding=\"async\" src=\"\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/fig-ch09_11_04N.jpg\" data-media-type=\"image\/jpg\" alt=\"Normal distribution curve of average scores on the first statistic tests with 65 and 67 values on the x-axis. A vertical upward line extends from 67 to the curve. The p-value points to the area to the right of 67.\" width=\"380\" \/><\/span><\/div>\n<p>Compare <em data-effect=\"italics\">\u03b1<\/em> and the <em data-effect=\"italics\">p<\/em>-value:<\/p>\n<p>Since <em data-effect=\"italics\">\u03b1<\/em> = 0.05 and <em data-effect=\"italics\">p<\/em>-value = 0.0396. <em data-effect=\"italics\">\u03b1<\/em> &gt; <em data-effect=\"italics\">p<\/em>-value.<\/p>\n<p><strong>Make a decision:<\/strong> Since <em data-effect=\"italics\">\u03b1<\/em> &gt; <em data-effect=\"italics\">p<\/em>-value, reject <em data-effect=\"italics\">H<sub>0<\/sub><\/em>.<\/p>\n<p id=\"fs-idm21057040\">This means you reject <em data-effect=\"italics\">\u03bc<\/em> = 65. In other words, you believe the average test score is more than 65.<\/p>\n<p><strong>Conclusion:<\/strong> At a 5% level of significance, the sample data show sufficient evidence that the mean (average) test score is more than 65, just as the math instructor thinks.<\/p>\n<p>The <em data-effect=\"italics\">p<\/em>-value can easily be calculated.<\/p>\n<div data-type=\"note\" data-has-label=\"true\" id=\"fs-idm14472336\" class=\"note statistics calculator\" data-label=\"\">\n<p>Put the data into a list. Press <code>STAT<\/code> and arrow over to <code>TESTS<\/code>. Press <code>2:T-Test<\/code>. Arrow over to <code>Data<\/code> and press <code>ENTER<\/code>. Arrow down and enter 65 for <em data-effect=\"italics\">\u03bc<\/em><sub>0<\/sub>, the name of the list where you put the data, and 1 for <code>Freq:<\/code>. Arrow down to <em data-effect=\"italics\">\u03bc<\/em>: and arrow over to &gt; <em data-effect=\"italics\">\u03bc<\/em><sub>0<\/sub>. Press <code>ENTER<\/code>. Arrow down to <code>Calculate<\/code> and press <code>ENTER<\/code>. The calculator not only calculates the <em data-effect=\"italics\">p<\/em>-value (<em data-effect=\"italics\">p<\/em> = 0.0396) but it also calculates the test statistic (<em data-effect=\"italics\">t<\/em>-score) for the sample mean, the sample mean, and the sample standard deviation. <em data-effect=\"italics\">\u03bc<\/em> &gt; 65 is the alternative hypothesis. Do this set of instructions again except arrow to <code>Draw<\/code> (instead of <code>Calculate<\/code>). Press <code>ENTER<\/code>. A shaded graph appears with <em data-effect=\"italics\">t<\/em> = 1.9781 (test statistic) and <em data-effect=\"italics\">p<\/em> = 0.0396 (<em data-effect=\"italics\">p<\/em>-value). Make sure when you use <code>Draw<\/code> that no other equations are highlighted in <em data-effect=\"italics\">Y<\/em> =  and the plots are turned off.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"note\" data-has-label=\"true\" id=\"fs-idm186895712\" class=\"note statistics try\" data-label=\"\">\n<div data-type=\"title\" class=\"title\">Try It<\/div>\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\">\n<p id=\"eip-idm59977568\">It is believed that a stock price for a particular company will grow at a rate of &#x1f4b2;5 per week with a standard deviation of &#x1f4b2;1. An investor believes the stock won\u2019t grow as quickly. The changes in stock price is recorded for ten weeks and are as follows: &#x1f4b2;4, &#x1f4b2;3, &#x1f4b2;2, &#x1f4b2;3, &#x1f4b2;1, &#x1f4b2;7, &#x1f4b2;2, &#x1f4b2;1, &#x1f4b2;1, &#x1f4b2;2. Perform a hypothesis test using a 5% level of significance. State the null and alternative hypotheses, find the <em data-effect=\"italics\">p<\/em>-value, state your conclusion, and identify the Type I and Type II errors.<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"eip-348\">\n<p id=\"eip-idp45715760\"><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> = 5<\/p>\n<p id=\"eip-idp96462720\"><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &lt; 5<\/p>\n<p id=\"eip-idp8662576\"><em data-effect=\"italics\">p<\/em> = 0.0082<\/p>\n<p id=\"eip-idp17544256\">Because <em data-effect=\"italics\">p<\/em> &lt; <em data-effect=\"italics\">\u03b1<\/em>, we reject the null hypothesis. There is sufficient evidence to suggest that the stock price of the company grows at a rate less than &#x1f4b2;5 a week.<\/p>\n<p id=\"eip-idm775584\">Type I Error: To conclude that the stock price is growing slower than &#x1f4b2;5 a week when, in fact, the stock price is growing at &#x1f4b2;5 a week (reject the null hypothesis when the null hypothesis is true).<\/p>\n<p id=\"eip-idp7466800\">Type II Error: To conclude that the stock price is growing at a rate of &#x1f4b2;5 a week when, in fact, the stock price is growing slower than &#x1f4b2;5 a week (do not reject the null hypothesis when the null hypothesis is false).<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"example\" class=\"textbox examples\">\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\" id=\"id5304801\">\n<p>Joon believes that 50% of first-time brides in the United States are younger than their grooms. She performs a hypothesis test to determine if the percentage is <strong>the same or different from 50%<\/strong>. Joon samples <strong>100 first-time brides<\/strong> and <strong>53<\/strong> reply that they are younger than their grooms. For the hypothesis test, she uses a 1% level of<br \/>\nsignificance.<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"id4150001\">\n<p id=\"element-399\">Set up the hypothesis test:<\/p>\n<p>The 1% level of significance means that <em data-effect=\"italics\">\u03b1<\/em> = 0.01. This is a <strong>test of a single population proportion<\/strong>.<\/p>\n<p><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> = 0.50\u2003\u2003<em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2260 0.50<\/p>\n<p>The words <strong>&#8220;is the same or different from&#8221;<\/strong> tell you this is a two-tailed test.<\/p>\n<p>Calculate the distribution needed:<\/p>\n<p id=\"element-783\"><strong>Random variable:<\/strong><em data-effect=\"italics\">P\u2032<\/em> = the percent of of first-time brides who are younger than their grooms.<\/p>\n<p id=\"element-581\"><strong>Distribution for the test:<\/strong> The problem contains no mention of a mean. The<br \/>\ninformation is given in terms of percentages. Use the distribution for <em data-effect=\"italics\">P\u2032<\/em>, the estimated proportion.<\/p>\n<p id=\"fs-idm81154768\">\\({P}^{\\prime }~N\\left(p,\\sqrt{\\frac{p\\cdot q}{n}}\\right)\\) Therefore, \\({P}^{\\prime }~N\\left(0.5,\\sqrt{\\frac{0.5\\cdot 0.5}{100}}\\right)\\)<\/p>\n<p id=\"fs-idm107085552\">where <em data-effect=\"italics\">p<\/em> = 0.50, <em data-effect=\"italics\">q<\/em> = 1\u2212<em data-effect=\"italics\">p<\/em> = 0.50, and <em data-effect=\"italics\">n<\/em> = 100<\/p>\n<p id=\"fs-idp14246672\">Calculate the <em data-effect=\"italics\">p<\/em>-value using the normal distribution for proportions:<\/p>\n<p><em data-effect=\"italics\">p<\/em>-value = <em data-effect=\"italics\">P<\/em> (<em data-effect=\"italics\">p\u2032<\/em> &lt; 0.47 or <em data-effect=\"italics\">p\u2032<\/em> &gt; 0.53) = 0.5485<\/p>\n<p id=\"fs-idp133040416\">where <em data-effect=\"italics\">x<\/em> = 53, <em data-effect=\"italics\">p\u2032<\/em> = \\(\\frac{x}{n}\\text{\u00a0=\u00a0}\\frac{\\text{53}}{\\text{100}}\\) = 0.53.<\/p>\n<p><strong>Interpretation of the <em data-effect=\"italics\">p<\/em>-value:<\/strong> If the null hypothesis is true, there is 0.5485 probability<br \/>\n(54.85%) that the sample (estimated) proportion \\(p\\text{&#8216;}\\) is 0.53 or more OR 0.47 or less (see<br \/>\nthe graph in <a href=\"#hyptest11_ex5\" class=\"autogenerated-content\">[link]<\/a>).<\/p>\n<div class=\"bc-figure figure\" id=\"hyptest11_ex5\"><span data-type=\"media\" id=\"id5140475\" data-alt=\"Normal distribution curve of the percent of first time brides who are younger than the groom with values of 0.47, 0.50, and 0.53 on the x-axis. Vertical upward lines extend from 0.47 and 0.53 to the curve. 1\/2(p-values) are calculated for the areas on outsides of 0.47 and 0.53.\"><img decoding=\"async\" src=\"\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/fig-ch09_11_05N.jpg\" data-media-type=\"image\/jpg\" alt=\"Normal distribution curve of the percent of first time brides who are younger than the groom with values of 0.47, 0.50, and 0.53 on the x-axis. Vertical upward lines extend from 0.47 and 0.53 to the curve. 1\/2(p-values) are calculated for the areas on outsides of 0.47 and 0.53.\" width=\"380\" \/><\/span><\/div>\n<p id=\"element-327\"><em data-effect=\"italics\">\u03bc<\/em> = <em data-effect=\"italics\">p<\/em> = 0.50 comes from <em data-effect=\"italics\">H<sub>0<\/sub><\/em>, the null hypothesis.<\/p>\n<p><em data-effect=\"italics\">p\u2032<\/em> = 0.53. Since the curve is symmetrical and the test is two-tailed, the <em data-effect=\"italics\">p\u2032<\/em> for the left tail is equal to 0.50 \u2013 0.03 = 0.47 where <em data-effect=\"italics\">\u03bc<\/em> = <em data-effect=\"italics\">p<\/em> = 0.50. (0.03 is the difference between 0.53 and 0.50.)<\/p>\n<p>Compare <em data-effect=\"italics\">\u03b1<\/em> and the <em data-effect=\"italics\">p<\/em>-value:<\/p>\n<p>Since <em data-effect=\"italics\">\u03b1<\/em> = 0.01 and <em data-effect=\"italics\">p<\/em>-value = 0.5485. <em data-effect=\"italics\">\u03b1<\/em> &lt; <em data-effect=\"italics\">p<\/em>-value.<\/p>\n<p><strong>Make a decision:<\/strong> Since <em data-effect=\"italics\">\u03b1<\/em> &lt; <em data-effect=\"italics\">p<\/em>-value, you cannot reject <em data-effect=\"italics\">H<sub>0<\/sub><\/em>.<\/p>\n<p><strong>Conclusion:<\/strong> At the 1% level of significance, the sample data do not show sufficient evidence that the percentage of first-time brides who are younger than their grooms is different from 50%.<\/p>\n<p>The <em data-effect=\"italics\">p<\/em>-value can easily be calculated.<\/p>\n<div data-type=\"note\" data-has-label=\"true\" id=\"fs-idm35225600\" class=\"note statistics calculator\" data-label=\"\">\n<p>Press <code>STAT<\/code> and arrow over to <code>TESTS<\/code>. Press <code>5:1-PropZTest<\/code>. Enter .5 for <em data-effect=\"italics\">p<\/em><sub>0<\/sub>, 53 for <em data-effect=\"italics\">x<\/em> and 100 for <em data-effect=\"italics\">n<\/em>. Arrow down to <code>Prop<\/code> and arrow to <code>not equals<\/code> <em data-effect=\"italics\">p<\/em><sub>0<\/sub>. Press <code>ENTER<\/code>. Arrow down to <code>Calculate<\/code> and press <code>ENTER<\/code>. The calculator calculates the <em data-effect=\"italics\">p<\/em>-value (<em data-effect=\"italics\">p<\/em> = 0.5485) and the test statistic (<em data-effect=\"italics\">z<\/em>-score). <code>Prop not equals<\/code> .5 is the alternate hypothesis. Do this set of instructions again except arrow to <code>Draw<\/code> (instead of <code>Calculate<\/code>). Press <code>ENTER<\/code>. A shaded graph appears with <em data-effect=\"italics\">z<\/em> = 0.6 (test statistic) and <em data-effect=\"italics\">p<\/em> = 0.5485 (<em data-effect=\"italics\">p<\/em>-value). Make sure when you use <code>Draw<\/code> that no other equations are highlighted in <em data-effect=\"italics\">Y<\/em> = and the plots are turned off.<\/p>\n<\/div>\n<p>The Type I and Type II errors are as follows:<\/p>\n<p>The Type I error is to conclude that the proportion of first-time brides who are younger than their grooms is different from 50% when, in fact, the proportion is actually 50%. (Reject the null hypothesis when the null hypothesis is true). <\/p>\n<p id=\"fs-idp134008544\">The Type II error is there is not enough evidence to conclude that the proportion of first time brides who are younger than their grooms differs from 50% when, in fact, the proportion does differ from 50%. (Do not reject the null hypothesis when the null hypothesis is false.)<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"note\" data-has-label=\"true\" id=\"fs-idp8711616\" class=\"note statistics try\" data-label=\"\">\n<div data-type=\"title\" class=\"title\">Try It<\/div>\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\">\n<p id=\"eip-idp103490704\">A teacher believes that 85% of students in the class will want to go on a field trip to the local zoo. She performs a hypothesis test to determine if the percentage is the same or different from 85%. The teacher samples 50 students and 39 reply that they would want to go to the zoo. For the hypothesis test, use a 1% level of significance.<\/p>\n<p id=\"eip-idp10599024\">First, determine what type of test this is, set up the hypothesis test, find the <em data-effect=\"italics\">p<\/em>-value, sketch the graph, and state your conclusion.<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\">\n<p id=\"eip-idp139727615275152\">Since the problem is about percentages, this is a test of single population proportions.<\/p>\n<p id=\"eip-idp139727615275648\"><em data-effect=\"italics\">H<sub>0<\/sub><\/em> : <em data-effect=\"italics\">p<\/em> = 0.85<\/p>\n<p id=\"eip-idp139727615276416\"><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2260 0.85<\/p>\n<p id=\"eip-idp139727615277216\"><em data-effect=\"italics\">p<\/em> = 0.7554<\/p>\n<div class=\"bc-figure figure\" id=\"fs-idp11874016\"><span data-type=\"media\" id=\"eip-idp139727615277600\" data-alt=\"\"><img decoding=\"async\" src=\"\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/CNX_Stats_C09_M11_tryit002annoN.jpg\" data-media-type=\"image\/jpeg\" alt=\"\" width=\"380\" \/><\/span><\/div>\n<p id=\"eip-idp95887808\">Because <em data-effect=\"italics\">p<\/em> &gt; <em data-effect=\"italics\">\u03b1<\/em>, we fail to reject the null hypothesis. There is not sufficient evidence to suggest that the proportion of students that want to go to the zoo is not 85%.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"example\" class=\"textbox examples\">\n<div data-type=\"exercise\" class=\"exercise\" id=\"element-312\">\n<div data-type=\"problem\" class=\"problem\" id=\"id5318755\">\n<p>Suppose a consumer group suspects that the proportion of households that have three cell phones is 30%. A cell phone  company has reason to believe that the proportion  is not 30%. Before they start a big advertising campaign, they conduct a hypothesis test. Their marketing people survey 150 households with the result that 43 of the households have three cell phones.<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"id4213559\">\n<p>Set up the Hypothesis Test:<\/p>\n<p><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> = 0.30 <em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2260 0.30<\/p>\n<p>Determine the distribution needed:<\/p>\n<p>The <strong>random variable<\/strong> is <em data-effect=\"italics\">P\u2032<\/em> = proportion of households that have three cell phones.<\/p>\n<p>The <strong>distribution<\/strong> for the hypothesis test is \\(P\\text{&#8216;}~N\\left(0.30,\\sqrt{\\frac{\\left(0.30\\right)\\cdot \\left(0.70\\right)}{150}}\\right)\\)<\/p>\n<div data-type=\"newline\" data-count=\"2\">\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\" id=\"id4146441\">\n<p>a. The value that helps determine the <em data-effect=\"italics\">p<\/em>-value is <em data-effect=\"italics\">p\u2032<\/em>. Calculate <em data-effect=\"italics\">p\u2032<\/em>.<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"id4274849\" data-print-placement=\"end\">\n<p>a. <em data-effect=\"italics\">p\u2032<\/em> = \\(\\frac{x}{n}\\) where <em data-effect=\"italics\">x<\/em> is the number of successes and <em data-effect=\"italics\">n<\/em> is the total number in the sample.<\/p>\n<p><em data-effect=\"italics\">x<\/em> = 43, <em data-effect=\"italics\">n<\/em> = 150<\/p>\n<p><em data-effect=\"italics\">p\u2032<\/em> = \\(\\frac{\\text{43}}{\\text{150}}\\)\n<\/p>\n<div data-type=\"newline\" data-count=\"2\">\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\" id=\"id5150803\">\n<p>b. What is a <strong>success<\/strong> for this problem?<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"id5036464\" data-print-placement=\"end\">\n<p>b. A success is having three cell phones in a household.<\/p>\n<div data-type=\"newline\" data-count=\"2\">\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"element-519\">\n<div data-type=\"problem\" class=\"problem\" id=\"id4292096\">\n<p id=\"element-781\">c. What is the level of significance?<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"id5242701\" data-print-placement=\"end\">\n<p>c. The level of significance is the preset <em data-effect=\"italics\">\u03b1<\/em>. Since <em data-effect=\"italics\">\u03b1<\/em> is not given, assume that <em data-effect=\"italics\">\u03b1<\/em> = 0.05.<\/p>\n<div data-type=\"newline\" data-count=\"2\">\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\" id=\"id4673389\">\n<p>d. Draw the graph for this problem. Draw the horizontal axis. Label and shade appropriately.<\/p>\n<div data-type=\"newline\"><\/div>\n<p>Calculate the <em data-effect=\"italics\">p<\/em>-value.<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"id5212164\" data-print-placement=\"end\">\n<p>d. <em data-effect=\"italics\">p<\/em>-value = 0.7216<\/p>\n<div data-type=\"newline\" data-count=\"2\">\n<\/div><\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\" id=\"id5104601\">\n<p>e. Make a decision. _____________(Reject\/Do not reject) <em data-effect=\"italics\">H<sub>0<\/sub><\/em> because____________.<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"id5137008\" data-print-placement=\"end\">\n<p>e. Assuming that <em data-effect=\"italics\">\u03b1<\/em> = 0.05, <em data-effect=\"italics\">\u03b1<\/em> &lt; <em data-effect=\"italics\">p<\/em>-value. The decision is do not reject <em data-effect=\"italics\">H<sub>0<\/sub><\/em> because there is not sufficient evidence to conclude that the proportion of households that have three cell phones is not 30%.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"note\" data-has-label=\"true\" id=\"fs-idm14663600\" class=\"note statistics try\" data-label=\"\">\n<div data-type=\"title\" class=\"title\">Try It<\/div>\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\">\n<p id=\"eip-idm17000432\">Marketers believe that 92% of adults in the United States own a cell phone. A cell phone manufacturer believes that number is actually lower. 200 American adults are surveyed, of which, 174 report having cell phones. Use a 5% level of significance. State the null and alternative hypothesis, find the <em data-effect=\"italics\">p<\/em>-value, state your conclusion, and identify the Type I and Type II errors.<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\">\n<p id=\"eip-idm64882736\"><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> = 0.92<\/p>\n<p id=\"eip-idm90844256\"><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> &lt; 0.92<\/p>\n<p id=\"eip-idm90843872\"><em data-effect=\"italics\">p<\/em>-value = 0.0046<\/p>\n<p id=\"eip-idm90843488\">Because <em data-effect=\"italics\">p<\/em> &lt; 0.05, we reject the null hypothesis. There is sufficient evidence to conclude that fewer than 92% of American adults own cell phones.<\/p>\n<p id=\"eip-idm6499552\">Type I Error: To conclude that fewer than 92% of American adults own cell phones when, in fact, 92% of American adults do own cell phones (reject the null hypothesis when the null hypothesis is true).<\/p>\n<p id=\"eip-idm80852000\">Type II Error: To conclude that 92% of American adults own cell phones when, in fact, fewer than 92% of American adults own cell phones (do not reject the null hypothesis when the null hypothesis is false).<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-idp32935184\">The next example is a poem written by a statistics student named Nicole Hart. The solution to the problem follows the poem. Notice that the hypothesis test is for a single population proportion. This means that the null and alternate hypotheses use the parameter <em data-effect=\"italics\">p<\/em>. The distribution for the test is normal. The estimated proportion <em data-effect=\"italics\">p<\/em>\u2032 is the proportion of fleas killed to the total fleas found on Fido. This is sample information. The problem gives a preconceived <em data-effect=\"italics\">\u03b1<\/em> = 0.01, for comparison, and a 95% confidence interval computation. The poem is clever and humorous, so please enjoy it!<\/p>\n<div data-type=\"example\" class=\"textbox examples\">\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-idp65427488\">\n<div data-type=\"problem\" class=\"problem\" id=\"id5257658\">\n<p id=\"eip-idm151173424\">My dog has so many fleas,<\/p>\n<div data-type=\"newline\"><\/div>\n<p>They do not come off with ease.<\/p>\n<div data-type=\"newline\"><\/div>\n<p>As for shampoo, I have tried many types<\/p>\n<div data-type=\"newline\"><\/div>\n<p>Even one called Bubble Hype,<\/p>\n<div data-type=\"newline\"><\/div>\n<p>Which only killed 25% of the fleas,<\/p>\n<div data-type=\"newline\"><\/div>\n<p>Unfortunately I was not pleased.<\/p>\n<div data-type=\"newline\" data-count=\"2\">\n<\/div>\n<p>I&#8217;ve used all kinds of soap,<\/p>\n<div data-type=\"newline\"><\/div>\n<p>Until I had given up hope<\/p>\n<div data-type=\"newline\"><\/div>\n<p>Until one day I saw<\/p>\n<div data-type=\"newline\"><\/div>\n<p>An ad that put me in awe.<\/p>\n<div data-type=\"newline\" data-count=\"2\">\n<\/div>\n<p>A shampoo used for dogs<\/p>\n<div data-type=\"newline\"><\/div>\n<p>Called GOOD ENOUGH to Clean a Hog<\/p>\n<div data-type=\"newline\"><\/div>\n<p>Guaranteed to kill more fleas.<\/p>\n<div data-type=\"newline\" data-count=\"2\">\n<\/div>\n<p>I gave Fido a bath<\/p>\n<div data-type=\"newline\"><\/div>\n<p>And after doing the math<\/p>\n<div data-type=\"newline\"><\/div>\n<p>His number of fleas<\/p>\n<div data-type=\"newline\"><\/div>\n<p>Started dropping by 3&#8217;s!<\/p>\n<div data-type=\"newline\" data-count=\"2\">\n<\/div>\n<p>Before his shampoo<\/p>\n<div data-type=\"newline\"><\/div>\n<p>I counted 42.<\/p>\n<div data-type=\"newline\"><\/div>\n<p>At the end of his bath,<\/p>\n<div data-type=\"newline\"><\/div>\n<p>I redid the math<\/p>\n<div data-type=\"newline\"><\/div>\n<p>And the new shampoo had killed 17 fleas.<\/p>\n<div data-type=\"newline\"><\/div>\n<p>So now I was pleased.<\/p>\n<div data-type=\"newline\" data-count=\"2\">\n<\/div>\n<p>Now it is time for you to have some fun<\/p>\n<div data-type=\"newline\"><\/div>\n<p>With the level of significance being .01,<\/p>\n<div data-type=\"newline\"><\/div>\n<p>You must help me figure out<\/p>\n<div data-type=\"newline\"><\/div>\n<p>Use the new shampoo or go without?<\/p><\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"id5100583\">\n<p>Set up the hypothesis test:<\/p>\n<p id=\"element-278\"><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2264 0.25 \u2003\u2003<em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> &gt; 0.25<\/p>\n<p>Determine the distribution needed:<\/p>\n<p>In words, CLEARLY state what your random variable \\(\\overline{X}\\) or <em data-effect=\"italics\">P\u2032<\/em> represents.<\/p>\n<p id=\"fs-idp88897968\"><em data-effect=\"italics\">P\u2032<\/em> = The proportion of fleas that are killed by the new shampoo<\/p>\n<p>State the distribution to use for the test.<\/p>\n<p><strong>Normal:<\/strong>\\(N\\left(0.25,\\sqrt{\\frac{\\left(0.25\\right)\\left(1-0.25\\right)}{42}}\\right)\\)<\/p>\n<p><strong>Test Statistic:<\/strong><em data-effect=\"italics\">z<\/em> = 2.3163<\/p>\n<p>Calculate the <em data-effect=\"italics\">p<\/em>-value using the normal distribution for proportions:<\/p>\n<p><em data-effect=\"italics\">p<\/em>-value = 0.0103<\/p>\n<p>In one to two complete sentences, explain what the <em data-effect=\"italics\">p<\/em>-value means for this problem.<\/p>\n<p>If the null hypothesis is true (the proportion is 0.25), then there is a 0.0103 probability that the sample (estimated) proportion is 0.4048 \\(\\left(\\frac{17}{42}\\right)\\) or more.<\/p>\n<p>Use the previous information to sketch a picture of this situation. CLEARLY, label and scale the horizontal axis and<br \/>\nshade the region(s) corresponding to the <em data-effect=\"italics\">p<\/em>-value.<\/p>\n<div class=\"bc-figure figure\" id=\"hyptest11_ex6\"><span data-type=\"media\" id=\"id4295819\" data-alt=\"Normal distribution graph of the proportion of fleas killed by the new shampoo with values of 0.25 and 0.4048 on the x-axis. A vertical upward line extends from 0.4048 to the curve and the area to the left of this is shaded in. The test statistic of the sample proportion is listed.\"><img decoding=\"async\" src=\"\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/fig-ch09_11_06.jpg\" data-media-type=\"image\/jpg\" alt=\"Normal distribution graph of the proportion of fleas killed by the new shampoo with values of 0.25 and 0.4048 on the x-axis. A vertical upward line extends from 0.4048 to the curve and the area to the left of this is shaded in. The test statistic of the sample proportion is listed.\" width=\"380\" \/><\/span><\/div>\n<p id=\"element-372\">Compare <em data-effect=\"italics\">\u03b1<\/em> and the <em data-effect=\"italics\">p<\/em>-value:<\/p>\n<p>Indicate the correct decision (\u201creject\u201d or \u201cdo not reject\u201d the null hypothesis), the reason for it, and write an appropriate conclusion, using complete sentences.<\/p>\n<table summary=\"\">\n<thead valign=\"middle\">\n<tr>\n<th data-align=\"center\">alpha<\/th>\n<th data-align=\"center\">decision<\/th>\n<th data-align=\"center\">reason for decision<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td data-align=\"center\">0.01<\/td>\n<td data-align=\"center\">Do not reject \\({H}_{0}\\)<\/td>\n<td data-align=\"center\"><em data-effect=\"italics\">\u03b1<\/em> &lt; <em data-effect=\"italics\">p<\/em>-value<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong>Conclusion:<\/strong> At the 1% level of significance, the sample data do not show sufficient evidence that the percentage of fleas that are killed by the new shampoo is more than 25%.<\/p>\n<p>Construct a 95% confidence interval for the true mean or proportion. Include a sketch of the graph of the situation.<br \/>\nLabel the point estimate and the lower and upper bounds of the confidence interval.<\/p>\n<\/p>\n<div class=\"bc-figure figure\" id=\"hyptest11_ex7\"><span data-type=\"media\" id=\"id5244822\" data-alt=\"Normal distribution graph of the proportion of fleas killed by the new shampoo with values of 0.26, 17\/42, and 0.55 on the x-axis. A vertical upward line extends from 0.26 and 0.55. The area between these two points is equal to 0.95.\"><img decoding=\"async\" src=\"\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/fig-ch09_11_07.jpg\" data-media-type=\"image\/jpg\" alt=\"Normal distribution graph of the proportion of fleas killed by the new shampoo with values of 0.26, 17\/42, and 0.55 on the x-axis. A vertical upward line extends from 0.26 and 0.55. The area between these two points is equal to 0.95.\" width=\"380\" \/><\/span><\/div>\n<p><strong>Confidence Interval:<\/strong> (0.26,0.55) We are 95% confident that the true population proportion <em data-effect=\"italics\">p<\/em> of fleas that are killed by the new shampoo is between 26% and 55%.<\/p>\n<div data-type=\"note\" class=\"note\" data-has-label=\"true\" id=\"id5328435\" data-label=\"\">\n<div data-type=\"title\" class=\"title\">Note<\/div>\n<p id=\"eip-idm81920704\">This test result is not very definitive since the <em data-effect=\"italics\">p<\/em>-value is very close to alpha. In reality, one would probably do more tests by giving the dog another bath after the fleas have had a chance to return.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"example\" class=\"textbox examples\">\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\">\n<p id=\"eip-idm29000496\">The National Institute of Standards and Technology provides exact data on conductivity properties of materials. Following are conductivity measurements for 11 randomly selected pieces of a particular type of glass.<\/p>\n<p id=\"eip-idm46067920\">1.11; 1.07; 1.11; 1.07; 1.12; 1.08; .98; .98 1.02; .95; .95\n<\/p>\n<div data-type=\"newline\"><\/div>\n<p>Is there convincing evidence that the average conductivity of this type of glass is greater than one? Use a significance level of 0.05. Assume the population is normal.<\/p><\/div>\n<div data-type=\"solution\" class=\"solution\">\n<p id=\"eip-idp69766080\">Let\u2019s follow a four-step process to answer this statistical question.\n<\/p>\n<div data-type=\"list\" data-list-type=\"enumerated\" id=\"eip-idp69766512\" data-number-style=\"arabic\">\n<div data-type=\"item\"><strong>State the Question<\/strong>: We need to determine if, at a 0.05 significance level, the average conductivity of the selected glass is greater than one. Our hypotheses will be<\/p>\n<div data-type=\"list\" data-list-type=\"enumerated\" id=\"eip-idp129554720\" data-number-style=\"lower-alpha\">\n<div data-type=\"item\"><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> \u2264 1<\/div>\n<div data-type=\"item\"><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &gt; 1<\/div>\n<\/div>\n<\/div>\n<div data-type=\"item\"><strong>Plan<\/strong>: We are testing a sample mean without a known population standard deviation. Therefore, we need to use a Student&#8217;s-t distribution. Assume the underlying population is normal.<\/div>\n<div data-type=\"item\"><strong>Do the calculations<\/strong>: We will input the sample data into the TI-83 as follows.<\/p>\n<div class=\"bc-figure figure\" id=\"eip-idp122053552\"><span data-type=\"media\" id=\"eip-idp167220912\" data-alt=\"\"><img decoding=\"async\" src=\"\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/CNX_Stats_C09_M09_001a.jpg\" data-media-type=\"image\/jpeg\" alt=\"\" width=\"250\" \/><\/span><\/div>\n<div class=\"bc-figure figure\" id=\"eip-idp29927184\"><span data-type=\"media\" id=\"eip-idp29927312\" data-alt=\"\"><img decoding=\"async\" src=\"\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/CNX_Stats_C09_M09_001b.jpg\" data-media-type=\"image\/jpeg\" alt=\"\" width=\"250\" \/><\/span><\/div>\n<div class=\"bc-figure figure\" id=\"eip-idp126675984\"><span data-type=\"media\" id=\"eip-idp156983600\" data-alt=\"\"><img decoding=\"async\" src=\"\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/CNX_Stats_C09_M09_001c.jpg\" data-media-type=\"image\/jpeg\" alt=\"\" width=\"250\" \/><\/span><\/div>\n<div class=\"bc-figure figure\" id=\"eip-idp154420944\"><span data-type=\"media\" id=\"eip-idp154421072\" data-alt=\"\"><img decoding=\"async\" src=\"\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/CNX_Stats_C09_M09_001d.jpg\" data-media-type=\"image\/jpeg\" alt=\"\" width=\"250\" \/><\/span><\/div>\n<\/div>\n<div data-type=\"item\"><strong>State the Conclusions<\/strong>: Since the <em data-effect=\"italics\">p<\/em>-value* (<em data-effect=\"italics\">p<\/em> = 0.036) is less than our alpha value, we will reject the null hypothesis. It is reasonable to state that the data supports the claim that the average conductivity level is greater than one.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"example\" class=\"textbox examples\" id=\"eip-767\">\n<div data-type=\"exercise\" class=\"exercise\" id=\"eip-178\">\n<div data-type=\"problem\" class=\"problem\">\n<p id=\"eip-idp173526416\">In a study of 420,019 cell phone users, 172 of the subjects developed brain cancer. Test the claim that cell phone users developed brain cancer at a greater rate than that for non-cell phone users (the rate of brain cancer for non-cell phone users is 0.0340%). Since this is a critical issue, use a 0.005 significance level. Explain why the significance level should be so low in terms of a Type I error.<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\">\n<p id=\"eip-idp3953440\">We will follow the four-step process.<\/p>\n<ol id=\"eip-idp191218624\" data-number-style=\"arabic\">\n<li>We need to conduct a hypothesis test on the claimed cancer rate. Our hypotheses will be\n<ol id=\"eip-idp171226608\" data-number-style=\"lower-alpha\">\n<li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2264 0.00034<\/li>\n<li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> &gt; 0.00034<\/li>\n<\/ol>\n<p id=\"eip-idm39752176\" class=\"finger\">If we commit a Type I error, we are essentially accepting a false claim. Since the claim describes cancer-causing environments, we want to minimize the chances of incorrectly identifying causes of cancer.<\/p>\n<\/li>\n<li>We will be testing a sample proportion with <em data-effect=\"italics\">x<\/em> = 172 and <em data-effect=\"italics\">n<\/em> = 420,019. The sample is sufficiently large because we have <em data-effect=\"italics\">np<\/em> = 420,019(0.00034) = 142.8, <em data-effect=\"italics\">nq<\/em> = 420,019(0.99966) = 419,876.2, two independent outcomes, and a fixed probability of success <em data-effect=\"italics\">p<\/em> = 0.00034. Thus we will be able to generalize our results to the population.<\/li>\n<li>The associated TI results are\n<div data-type=\"newline\"><\/div>\n<div class=\"bc-figure figure\" id=\"fs-idm80776672\"><span data-type=\"media\" id=\"eip-idp39356816\" data-alt=\"\"><img decoding=\"async\" src=\"\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/CNX_Stats_C09_M09_002a.jpg\" data-media-type=\"image\/jpeg\" alt=\"\" width=\"250\" \/><\/span><\/div>\n<div class=\"bc-figure figure\" id=\"eip-idp38098208\"><span data-type=\"media\" id=\"eip-idp38098336\" data-alt=\"\"><img decoding=\"async\" src=\"\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/CNX_Stats_C09_M09_002b.jpg\" data-media-type=\"image\/jpeg\" alt=\"\" width=\"250\" \/><\/span><\/div>\n<\/li>\n<li>Since the <em data-effect=\"italics\">p<\/em>-value = 0.0073 is greater than our alpha value = 0.005, we cannot reject the null. Therefore, we conclude that there is not enough evidence to support the claim of higher brain cancer rates for the cell phone users.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"example\" class=\"textbox examples\">\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\">\n<p id=\"eip-idp47042256\">According to the US Census there are approximately 268,608,618 residents aged 12 and older. Statistics from the Rape, Abuse, and Incest National Network indicate that, on average, 207,754 rapes occur each year (male and female) for persons aged 12 and older. This translates into a percentage of sexual assaults of 0.078%. In Daviess County, KY, there were reported 11 rapes for a population of 37,937. Conduct an appropriate hypothesis test to determine if there is a statistically significant difference between the local sexual assault percentage and the national sexual assault percentage. Use a significance level of 0.01.<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\">\n<p>We will follow the four-step plan.<\/p>\n<ol id=\"eip-idp113522624\" data-number-style=\"arabic\">\n<li>We need to test whether the proportion of sexual assaults in Daviess County, KY is significantly different from the national average.<\/li>\n<li>Since we are presented with proportions, we will use a one-proportion <em data-effect=\"italics\">z<\/em>-test. The hypotheses for the test will be\n<ol id=\"eip-idp17854992\" data-number-style=\"lower-alpha\">\n<li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> = 0.00078<\/li>\n<li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2260 0.00078<\/li>\n<\/ol>\n<\/li>\n<li>The following screen shots display the summary statistics from the hypothesis test.\n<div class=\"bc-figure figure\" id=\"fs-idm3544720\"><span data-type=\"media\" id=\"eip-idp26791072\" data-alt=\"\"><img decoding=\"async\" src=\"\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/CNX_Stats_C09_M09_003a.jpg\" data-media-type=\"image\/jpeg\" alt=\"\" width=\"250\" \/><\/span><\/div>\n<div class=\"bc-figure figure\" id=\"eip-idp38471600\"><span data-type=\"media\" id=\"eip-idp38471728\" data-alt=\"\"><img decoding=\"async\" src=\"\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/CNX_Stats_C09_M09_003b.jpg\" data-media-type=\"image\/jpeg\" alt=\"\" width=\"250\" \/><\/span><\/div>\n<\/li>\n<li>Since the <em data-effect=\"italics\">p<\/em>-value, <em data-effect=\"italics\">p<\/em> = 0.00063, is less than the alpha level of 0.01, the sample data indicates that we should reject the null hypothesis. In conclusion, the sample data support the claim that the proportion of sexual assaults in Daviess County, Kentucky is different from the national average proportion.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"summary\" data-depth=\"1\" id=\"fs-idp69608272\">\n<h1 data-type=\"title\">Chapter Review<\/h1>\n<p id=\"fs-idp168110512\">The <span data-type=\"term\">hypothesis test<\/span> itself has an established process. This can be summarized as follows:\n<\/p>\n<div data-type=\"list\" data-list-type=\"enumerated\">\n<div data-type=\"item\">Determine <em data-effect=\"italics\">H<sub>0<\/sub><\/em> and <em data-effect=\"italics\">H<sub>a<\/sub><\/em>. Remember, they are contradictory.<\/div>\n<div data-type=\"item\">Determine the random variable.<\/div>\n<div data-type=\"item\">Determine the distribution for the test.<\/div>\n<div data-type=\"item\">Draw a graph, calculate the test statistic, and use the test statistic to calculate the<br \/>\n<em data-effect=\"italics\">p<\/em>-value. (A <em data-effect=\"italics\">z<\/em>-score and a <em data-effect=\"italics\">t<\/em>-score are examples of test statistics.)<\/div>\n<div data-type=\"item\">Compare the preconceived <em data-effect=\"italics\">\u03b1<\/em> with the <em data-effect=\"italics\">p<\/em>-value, make a decision (reject or do not reject <em data-effect=\"italics\">H<sub>0<\/sub><\/em>), and write a clear conclusion using English sentences.<\/div>\n<\/div>\n<p>Notice that in performing the hypothesis test, you use <em data-effect=\"italics\">\u03b1<\/em> and not <em data-effect=\"italics\">\u03b2<\/em>. <em data-effect=\"italics\">\u03b2<\/em> is needed to help determine the sample size of the data that is used in calculating the <em data-effect=\"italics\">p<\/em>-value. Remember that the quantity 1 \u2013 <em data-effect=\"italics\">\u03b2<\/em> is called the <strong>Power of the Test<\/strong>. A high power is desirable. If the power is too low, statisticians typically increase the sample size while keeping <em data-effect=\"italics\">\u03b1<\/em> the same.If the power is low, the null hypothesis might not be rejected when it should be.<\/p>\n<\/div>\n<div class=\"practice\" data-depth=\"1\">\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\" id=\"eip-135\">\n<p>Assume <em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> = 9 and <em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &lt; 9. Is this a left-tailed, right-tailed, or two-tailed test?<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\">\n<p>This is a left-tailed test.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\" id=\"eip-286\">\n<p>Assume <em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> \u2264 6 and <em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &gt; 6. Is this a left-tailed, right-tailed, or two-tailed test?<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-idp45725152\">\n<div data-type=\"problem\" class=\"problem\" id=\"eip-116\">\n<p>Assume <em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> = 0.25 and <em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2260 0.25. Is this a left-tailed, right-tailed, or two-tailed test?<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\">\n<p>This is a two-tailed test.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\">\n<p>Draw the general graph of a left-tailed test.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"eip-18\">\n<div data-type=\"problem\" class=\"problem\">\n<p>\nDraw the graph of a two-tailed test.\n  <\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\">\n<div class=\"bc-figure figure\" id=\"fs-idm149132800\"><span data-type=\"media\" id=\"eip-idp62459136\" data-alt=\"\"><img decoding=\"async\" src=\"\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-content\/uploads\/sites\/28\/2017\/09\/CNX_Stats_C09_M09_item002annoN.jpg\" data-media-type=\"image\/jpeg\" alt=\"\" width=\"380\" \/><\/span><\/div>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\" id=\"eip-648\">\n<p>A bottle of water is labeled as containing 16 fluid ounces of water. You believe it is less than that. What type of test would you use?<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\">\n<p>Your friend claims that his mean golf score is 63. You want to show that it is higher than that. What type of test would you use?<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\">\n<p id=\"eip-68\">a right-tailed test<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"eip-729\">\n<div data-type=\"problem\" class=\"problem\" id=\"eip-491\">\n<p>A bathroom scale claims to be able to identify correctly any weight within a pound. You think that it cannot be that accurate. What type of test would you use?<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\">\n<p id=\"eip-458\">You flip a coin and record whether it shows heads or tails. You know the probability of getting heads is 50%, but you think it is less for this particular coin. What type of test would you use?<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-idp66689312\">\n<p id=\"eip-139\">a left-tailed test<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\" id=\"eip-519\">\n<p>If the alternative hypothesis has a not equals ( \u2260 ) symbol, you know to use which type of test?<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"eip-504\">\n<div data-type=\"problem\" class=\"problem\" id=\"eip-634\">\n<p>Assume the null hypothesis states that the mean is at least 18. Is this a left-tailed, right-tailed, or two-tailed test?<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"eip-625\">\n<p>This is a left-tailed test.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"eip-320\">\n<div data-type=\"problem\" class=\"problem\">\n<p id=\"eip-621\">Assume the null hypothesis states that the mean is at most 12. Is this a left-tailed, right-tailed, or two-tailed test?<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\">\n<p>Assume the null hypothesis states that the mean is equal to 88. The alternative hypothesis states that the mean is not equal to 88. Is this a left-tailed, right-tailed, or two-tailed test?<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\">\n<p>This is a two-tailed test.<\/p>\n<\/div>\n<\/div>\n<p>&lt;!&#8211;moved from 9.3&#8211;&gt;<\/p><\/div>\n<div class=\"free-response\" data-depth=\"1\" id=\"fs-idp139727497473120\">\n<h1 data-type=\"title\">Homework<\/h1>\n<p id=\"eip-idm30094416\"><em data-effect=\"italics\">For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in <a href=\"\/contents\/c0449c55-aa47-4f1c-bd5f-0521652f0e82@5\" class=\"autogenerated-content\">[link]<\/a>. Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files.<\/em><\/p>\n<div data-type=\"note\" class=\"note\" data-has-label=\"true\" id=\"id8395739\" data-label=\"\">\n<div data-type=\"title\" class=\"title\">Note<\/div>\n<p id=\"eip-idm120632832\">If you are using a Student&#8217;s-<em data-effect=\"italics\">t<\/em> distribution for one of the following homework problems, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, however.)<\/p>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\" id=\"id8395759\">\n<p id=\"fs-idp44498592\">A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8,000. A survey of owners of that tire design is conducted. From the 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9,800 miles. Using alpha = 0.05, is the data highly inconsistent with the claim?<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-idm164410224\">\n<ol id=\"fs-idp21859504\" data-number-style=\"lower-alpha\">\n<li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> \u2265 50,000<\/li>\n<li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &lt; 50,000<\/li>\n<li>Let \\(\\overline{X}\\) = the average lifespan of a brand of tires.<\/li>\n<li>normal distribution<\/li>\n<li><em data-effect=\"italics\">z<\/em> = -2.315<\/li>\n<li><em data-effect=\"italics\">p<\/em>-value = 0.0103<\/li>\n<li>Check student\u2019s solution.<\/li>\n<li>\n<ol id=\"fs-idm113459088\" data-number-style=\"lower-roman\">\n<li>alpha: 0.05<\/li>\n<li>Decision: Reject the null hypothesis.<\/li>\n<li>Reason for decision: The <em data-effect=\"italics\">p<\/em>-value is less than 0.05.<\/li>\n<li>Conclusion: There is sufficient evidence to conclude that the mean lifespan of the tires is less than 50,000 miles.<\/li>\n<\/ol>\n<\/li>\n<li>(43,537, 49,463)<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\" id=\"id8395793\">\n<p>From generation to generation, the mean age when smokers first start to smoke varies. However, the standard deviation of that age remains constant of around 2.1 years. A survey of 40 smokers of this generation was done to see if the mean starting age is at least 19. The sample mean was 18.1 with a sample standard deviation of 1.3. Do the data support the claim at the 5% level?\n  <\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\" id=\"id9424298\">\n<p>The cost of a daily newspaper varies from city to city. However, the variation among prices remains steady with a standard deviation of 20\u00a2. A study was done to test the claim that the mean cost of a daily newspaper is &#x1f4b2;1.00. Twelve costs yield a mean cost of 95\u00a2 with a standard deviation of 18\u00a2. Do the data support the claim at the 1% level?\n  <\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-idm17421856\">\n<ol id=\"fs-idm17421600\" data-number-style=\"lower-alpha\">\n<li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> = &#x1f4b2;1.00<\/li>\n<li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> \u2260 &#x1f4b2;1.00<\/li>\n<li>Let \\(\\overline{X}\\) = the average cost of a daily newspaper.<\/li>\n<li>normal distribution<\/li>\n<li><em data-effect=\"italics\">z<\/em> = \u20130.866<\/li>\n<li><em data-effect=\"italics\">p<\/em>-value = 0.3865<\/li>\n<li>Check student\u2019s solution.<\/li>\n<li>\n<ol id=\"fs-idm69728368\" data-number-style=\"lower-roman\">\n<li>Alpha: 0.01<\/li>\n<li>Decision: Do not reject the null hypothesis.<\/li>\n<li>Reason for decision: The <em data-effect=\"italics\">p<\/em>-value is greater than 0.01.<\/li>\n<li>Conclusion: There is sufficient evidence to support the claim that the mean cost of daily papers is &#x1f4b2;1. The mean cost could be &#x1f4b2;1.<\/li>\n<\/ol>\n<\/li>\n<li>(&#x1f4b2;0.84, &#x1f4b2;1.06)<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\" id=\"id9424338\">\n<p id=\"element-977\">An article in the <em data-effect=\"italics\">San Jose Mercury News<\/em> stated that students in the California state university system take 4.5 years, on average, to finish their undergraduate degrees. Suppose you believe that the mean time is longer. You conduct a survey of 49 students and obtain a sample mean of 5.1 with a sample standard deviation of 1.2. Do the data support your claim at the 1% level?<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"element-932a\">\n<div data-type=\"problem\" class=\"problem\" id=\"id9424572\">\n<p>The mean number of sick days an employee takes per year is believed to be about ten. Members of a personnel department do not believe this figure. They randomly survey eight employees. The number of sick days they took for the past year are as follows: 12; 4; 15; 3; 11; 8; 6; 8. Let <em data-effect=\"italics\">x<\/em> = the number of sick days they took for the past year. Should the personnel team believe that the mean number is ten?  <\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-idm143535568\">\n<ol id=\"fs-idm8405344\" data-number-style=\"lower-alpha\">\n<li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> = 10<\/li>\n<li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> \u2260 10<\/li>\n<li>Let \\(\\overline{X}\\) the mean number of sick days an employee takes per year.<\/li>\n<li>Student\u2019s <em data-effect=\"italics\">t<\/em>-distribution<\/li>\n<li><em data-effect=\"italics\">t<\/em> = \u20131.12<\/li>\n<li><em data-effect=\"italics\">p<\/em>-value = 0.300<\/li>\n<li>Check student\u2019s solution.<\/li>\n<li>\n<ol id=\"fs-idm68554224\" data-number-style=\"lower-roman\">\n<li>Alpha: 0.05<\/li>\n<li>Decision: Do not reject the null hypothesis.<\/li>\n<li>Reason for decision: The <em data-effect=\"italics\">p<\/em>-value is greater than 0.05.<\/li>\n<li>Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the mean number of sick days is not ten.<\/li>\n<\/ol>\n<\/li>\n<li>(4.9443, 11.806)<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\" id=\"id9424624\">\n<p>In 1955, <em data-effect=\"italics\">Life Magazine<\/em> reported that the 25 year-old mother of three worked, on average, an 80 hour week. Recently, many groups have been studying whether or not the women&#8217;s movement has, in fact, resulted in an increase in the average work week for women (combining employment and at-home work). Suppose a study was done to determine if the mean work week has increased. 81 women were surveyed with the following results. The sample mean was 83; the sample standard deviation was ten. Does it appear that the mean work week has increased for women at the 5% level?\n  <\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\" id=\"id9400577\">\n<p>\n    Your statistics instructor claims that 60 percent of the students who take her Elementary Statistics class go through life feeling more enriched. For some reason that she can&#8217;t quite figure out, most people don&#8217;t believe her. You decide to check this out on your own. You randomly survey 64 of her past Elementary Statistics students and find that 34 feel more enriched as a result of her class. Now, what do you think?\n  <\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-idm68403536\">\n<ol id=\"fs-idm150879872\" data-number-style=\"lower-alpha\">\n<li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2265 0.6<\/li>\n<li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> &lt; 0.6<\/li>\n<li>Let <em data-effect=\"italics\">P\u2032<\/em> = the proportion of students who feel more enriched as a result of taking Elementary Statistics.<\/li>\n<li>normal for a single proportion<\/li>\n<li>1.12<\/li>\n<li><em data-effect=\"italics\">p<\/em>-value =  0.1308<\/li>\n<li>Check student\u2019s solution.<\/li>\n<li>\n<ol id=\"fs-idp14435968\" data-number-style=\"lower-roman\">\n<li>Alpha: 0.05<\/li>\n<li>Decision: Do not reject the null hypothesis.<\/li>\n<li>Reason for decision: The <em data-effect=\"italics\">p<\/em>-value is greater than 0.05.<\/li>\n<li>Conclusion: There is insufficient evidence to conclude that less than 60 percent of her students feel more enriched.<\/li>\n<\/ol>\n<\/li>\n<li>Confidence Interval: (0.409,  0.654)\n<div data-type=\"newline\"><\/div>\n<p>The \u201cplus-4s\u201d confidence interval is (0.411, 0.648)<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\" id=\"id9400611\">\n<p>A Nissan Motor Corporation advertisement read, \u201cThe average man\u2019s I.Q. is 107. The average brown trout\u2019s I.Q. is 4. So why can\u2019t man catch brown trout?\u201d Suppose you believe that the brown trout\u2019s mean I.Q. is greater than four. You catch 12 brown trout. A fish psychologist determines the I.Q.s as follows: 5; 4; 7; 3; 6; 4; 5; 3; 6; 3; 8; 5. Conduct a hypothesis test of your belief.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\" id=\"id9400954\">\n<p>Refer to <a href=\"#element-410\">Exercise 9.119<\/a>. Conduct a hypothesis test to see if your decision and conclusion would change if your belief were that the brown trout\u2019s mean I.Q. is <strong>not<\/strong> four.<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-idm153124944\">\n<ol id=\"fs-idm153124688\" data-number-style=\"lower-alpha\">\n<li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> = 4<\/li>\n<li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> \u2260 4<\/li>\n<li>Let \\(\\overline{X}\\) the average I.Q. of a set of brown trout.<\/li>\n<li>two-tailed Student&#8217;s t-test<\/li>\n<li><em data-effect=\"italics\">t<\/em> = 1.95<\/li>\n<li><em data-effect=\"italics\">p<\/em>-value = 0.076<\/li>\n<li>Check student\u2019s solution.<\/li>\n<li>\n<ol id=\"fs-idp21135088\" data-number-style=\"lower-roman\">\n<li>Alpha: 0.05<\/li>\n<li>Decision: Reject the null hypothesis.<\/li>\n<li>Reason for decision: The <em data-effect=\"italics\">p<\/em>-value is greater than 0.05<\/li>\n<li>Conclusion: There is insufficient evidence to conclude that the average IQ of brown trout is not four. <\/li>\n<\/ol>\n<\/li>\n<li>(3.8865,5.9468)<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\" id=\"id9400994\">\n<p>According to an article in <em data-effect=\"italics\">Newsweek<\/em>, the natural ratio of girls to boys is 100:105. In China, the birth ratio is 100: 114 (46.7% girls). Suppose you don\u2019t believe the reported figures of the percent of girls born in China. You conduct a study. In this study, you count the number of girls and boys born in 150 randomly chosen recent births. There are 60 girls and 90 boys born of the 150. Based on your study, do you believe that the percent of girls born in China is 46.7?<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"element-658\">\n<div data-type=\"problem\" class=\"problem\" id=\"id9401185\">\n<p>A poll done for <em data-effect=\"italics\">Newsweek<\/em> found that 13% of Americans have seen or sensed the presence of an angel. A contingent doubts that the percent is really that high. It conducts its own survey. Out of 76 Americans surveyed, only two had seen or sensed the presence of an angel. As a result of the contingent\u2019s survey, would you agree with the <em data-effect=\"italics\">Newsweek<\/em> poll? In complete sentences, also give three reasons why the two polls might give different results.\n  <\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-idm3310512\">\n<ol id=\"fs-idm3310256\" data-number-style=\"lower-alpha\">\n<li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2265 0.13<\/li>\n<li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> &lt; 0.13<\/li>\n<li>Let <em data-effect=\"italics\">P\u2032<\/em> = the proportion of Americans who have seen or sensed angels<\/li>\n<li>normal for a single proportion<\/li>\n<li>\u20132.688<\/li>\n<li><em data-effect=\"italics\">p<\/em>-value = 0.0036<\/li>\n<li>Check student\u2019s solution.<\/li>\n<li>\n<ol id=\"fs-idm15778400\" data-number-style=\"lower-roman\">\n<li>alpha: 0.05<\/li>\n<li>Decision: Reject the null hypothesis.<\/li>\n<li>Reason for decision: The <em data-effect=\"italics\">p<\/em>-value is less than 0.05.<\/li>\n<li>Conclusion: There is sufficient evidence to conclude that the percentage of Americans who have seen or sensed an angel is less than 13%.<\/li>\n<\/ol>\n<\/li>\n<li>(0, 0.0623).\n<div data-type=\"newline\"><\/div>\n<p>      The\u201cplus-4s\u201d confidence interval is (0.0022, 0.0978)<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\" id=\"id9401240\">\n<p id=\"element-140a\">The mean work week for engineers in a start-up company is believed to be about 60 hours. A newly hired engineer hopes that it\u2019s shorter. She asks ten engineering friends in start-ups for the lengths of their mean work weeks. Based on the results that follow, should she count on the mean work week to be shorter than 60 hours?<\/p>\n<p id=\"fs-idp187110944\">Data (length of mean work week): 70; 45; 55; 60; 65; 55; 55; 60; 50; 55.<\/p>\n<\/div>\n<\/div>\n<p>&lt;!&#8211; LINK &#8211;&gt;<\/p>\n<div data-type=\"exercise\" class=\"exercise\" id=\"element-761\">\n<div data-type=\"problem\" class=\"problem\" id=\"id9401488\">\n<p id=\"element-926\">Use the \u201cLap time\u201d data for Lap 4 (see <a href=\"\/contents\/3ef830bc-5247-460a-9007-e3fd762e5e93@3\" class=\"autogenerated-content\">[link]<\/a>) to test the claim that Terri finishes Lap 4, on average, in less than 129 seconds. Use all twenty races given.<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-idp13518384\">\n<ol id=\"fs-idp13518640\" data-number-style=\"lower-alpha\">\n<li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> \u2265 129<\/li>\n<li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &lt; 129<\/li>\n<li>Let \\(\\overline{X}\\) = the average time in seconds that Terri finishes Lap 4.<\/li>\n<li>Student&#8217;s <em data-effect=\"italics\">t<\/em>-distribution<\/li>\n<li><em data-effect=\"italics\">t<\/em> = 1.209<\/li>\n<li>0.8792<\/li>\n<li>Check student\u2019s solution.<\/li>\n<li>\n<ol id=\"fs-idm130980080\" data-number-style=\"lower-roman\">\n<li>Alpha: 0.05<\/li>\n<li>Decision: Do not reject the null hypothesis.<\/li>\n<li>Reason for decision: The <em data-effect=\"italics\">p<\/em>-value is greater than 0.05.<\/li>\n<li>Conclusion: There is insufficient evidence to conclude that Terri\u2019s mean lap time is less than 129 seconds.<\/li>\n<\/ol>\n<\/li>\n<li>(128.63, 130.37)<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-idp191961952\">\n<div data-type=\"problem\" class=\"problem\" id=\"id9401521\">&lt;!&#8211; LINK &#8211;&gt;<\/p>\n<p id=\"element-564a\">Use the \u201cInitial Public Offering\u201d data (see <a href=\"\/contents\/3ef830bc-5247-460a-9007-e3fd762e5e93@3\" class=\"autogenerated-content\">[link]<\/a>) to test the claim that the mean offer price was &#x1f4b2;18 per share. Do not use all the data. Use your random number generator to randomly survey 15 prices.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"note\" class=\"note\" data-has-label=\"true\" id=\"id9401550\" data-label=\"\">\n<div data-type=\"title\" class=\"title\">Note<\/div>\n<p id=\"eip-idp174015456\">The following questions were written by past students. They are excellent problems!<\/p>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"ex1\">\n<div data-type=\"problem\" class=\"problem\" id=\"id9401566\">\n<p id=\"id10961970\">&#8220;Asian Family Reunion,&#8221; by Chau Nguyen<\/p>\n<p id=\"eip-idp174081328\">Every two years it comes around.<\/p>\n<p id=\"eip-idp174081712\">We all get together from different towns.<\/p>\n<p id=\"eip-idp174082096\">In my honest opinion,<\/p>\n<p id=\"eip-idp174082512\">It&#8217;s not a typical family reunion.<\/p>\n<p id=\"eip-idp174082896\">Not forty, or fifty, or sixty,<\/p>\n<p id=\"eip-idp73534816\">But how about seventy companions!<\/p>\n<p id=\"eip-idp73535200\">The kids would play, scream, and shout<\/p>\n<p id=\"eip-idp73535584\">One minute they&#8217;re happy, another they&#8217;ll pout.<\/p>\n<p id=\"eip-idp73535968\">The teenagers would look, stare, and compare<\/p>\n<p id=\"eip-idp73536352\">From how they look to what they wear.<\/p>\n<p id=\"eip-idp73536736\">The men would chat about their business<\/p>\n<p id=\"eip-idp73537120\">That they make more, but never less.<\/p>\n<p id=\"eip-idp73537504\">Money is always their subject<\/p>\n<p id=\"eip-idp73537888\">And there&#8217;s always talk of more new projects.<\/p>\n<p id=\"eip-idp73538272\">The women get tired from all of the chats<\/p>\n<p id=\"eip-idp123573408\">They head to the kitchen to set out the mats.<\/p>\n<p id=\"eip-idp123573792\">Some would sit and some would stand<\/p>\n<p id=\"eip-idp123574176\">Eating and talking with plates in their hands.<\/p>\n<p id=\"eip-idp123574560\">Then come the games and the songs<\/p>\n<p id=\"eip-idp123574944\">And suddenly, everyone gets along!<\/p>\n<p id=\"eip-idp123575328\">With all that laughter, it&#8217;s sad to say<\/p>\n<p id=\"eip-idp123575712\">That it always ends in the same old way.<\/p>\n<p id=\"eip-idp123576096\">They hug and kiss and say &#8220;good-bye&#8221;<\/p>\n<p id=\"eip-idp123576480\">And then they all begin to cry!<\/p>\n<p id=\"eip-idp123576864\">I say that 60 percent shed their tears<\/p>\n<p id=\"eip-idp38133824\">But my mom counted 35 people this year.<\/p>\n<p id=\"eip-idp38134208\">She said that boys and men will always have their pride,<\/p>\n<p id=\"eip-idp38134592\">So we won&#8217;t ever see them cry.<\/p>\n<p id=\"eip-idp38134976\">I myself don&#8217;t think she&#8217;s correct,<\/p>\n<p id=\"eip-idp38135360\">So could you please try this problem to see if you object?<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-idm114330544\">\n<ol id=\"fs-idm114330288\" data-number-style=\"lower-alpha\">\n<li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> = 0.60<\/li>\n<li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> &lt; 0.60<\/li>\n<li>Let <em data-effect=\"italics\">P\u2032<\/em> = the proportion of family members who shed tears at a reunion.<\/li>\n<li>normal for a single proportion<\/li>\n<li>\u20131.71<\/li>\n<li>0.0438<\/li>\n<li>Check student\u2019s solution.<\/li>\n<li>\n<ol id=\"fs-idm44642224\" data-number-style=\"lower-roman\">\n<li>alpha: 0.05<\/li>\n<li>Decision: Reject the null hypothesis.<\/li>\n<li>Reason for decision: <em data-effect=\"italics\">p<\/em>-value &lt; alpha<\/li>\n<li>Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the proportion of family members who shed tears at a reunion is less than 0.60. However, the test is weak because the <em data-effect=\"italics\">p<\/em>-value and alpha are quite close, so other tests should be done.<\/li>\n<\/ol>\n<\/li>\n<li>We are 95% confident that between 38.29% and 61.71% of family members will shed tears at a family reunion. (0.3829, 0.6171). The\u201cplus-4s\u201d confidence interval (see chapter 8) is (0.3861, 0.6139)<\/li>\n<\/ol>\n<p id=\"fs-idm62390992\">Note that here the  \u201clarge-sample\u201d 1 \u2013 PropZTest provides the approximate <em data-effect=\"italics\">p<\/em>-value of 0.0438. Whenever a <em data-effect=\"italics\">p<\/em>-value based on a normal approximation is close to the level of significance, the exact <em data-effect=\"italics\">p<\/em>-value based on binomial probabilities should be calculated whenever possible. This is beyond the scope of this course.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"ex2\">\n<div data-type=\"problem\" class=\"problem\" id=\"id9401620\">\n<p id=\"id5063144\">&#8220;The Problem with Angels,&#8221; by Cyndy Dowling<\/p>\n<p id=\"eip-idm27993696\">Although this problem is wholly mine,<\/p>\n<p id=\"eip-idm27993312\">The catalyst came from the magazine, Time.<\/p>\n<p id=\"eip-idm27992928\">On the magazine cover I did find<\/p>\n<p id=\"eip-idm27992544\">The realm of angels tickling my mind.<\/p>\n<div data-type=\"newline\" data-count=\"1\"><\/div>\n<p id=\"eip-idm27991792\">Inside, 69% I found to be<\/p>\n<p id=\"eip-idm27991408\">In angels, Americans do believe.<\/p>\n<div data-type=\"newline\" data-count=\"1\"><\/div>\n<p id=\"eip-idp10519280\">Then, it was time to rise to the task,<\/p>\n<p id=\"eip-idp10519664\">Ninety-five high school and college students I did ask.<\/p>\n<p id=\"eip-idp10520048\">Viewing all as one group,<\/p>\n<p id=\"eip-idp10520432\">Random sampling to get the scoop.<\/p>\n<div data-type=\"newline\" data-count=\"1\"><\/div>\n<p id=\"eip-idp10521184\">So, I asked each to be true,<\/p>\n<p id=\"eip-idp10521568\">&#8220;Do you believe in angels?&#8221;  Tell me, do!<\/p>\n<div data-type=\"newline\" data-count=\"1\"><\/div>\n<p id=\"eip-idp10522320\">Hypothesizing at the start,<\/p>\n<p id=\"eip-idm24780496\">Totally believing in my heart<\/p>\n<p id=\"eip-idm24780112\">That the proportion who said yes<\/p>\n<p id=\"eip-idm24779728\">Would be equal on this test.<\/p>\n<div data-type=\"newline\" data-count=\"1\"><\/div>\n<p id=\"eip-idm24778976\">Lo and behold, seventy-three did arrive,<\/p>\n<p id=\"eip-idm24778592\">Out of the sample of ninety-five.<\/p>\n<p id=\"eip-idm24778208\">Now your job has just begun,<\/p>\n<p id=\"eip-idm24777824\">Solve this problem and have some fun.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\" id=\"id9401850\">\n<p id=\"id9415993\">&#8220;Blowing Bubbles,&#8221; by Sondra Prull<\/p>\n<p id=\"eip-idp3753920\">Studying stats just made me tense,<\/p>\n<p id=\"eip-idp3754352\">I had to find some sane defense.<\/p>\n<p id=\"eip-idp3754784\">Some light and lifting simple play<\/p>\n<p id=\"eip-idp3755216\">To float my math anxiety away.<\/p>\n<div data-type=\"newline\" data-count=\"1\"><\/div>\n<p id=\"eip-idp3756016\">Blowing bubbles lifts me high<\/p>\n<p id=\"eip-idp3756448\">Takes my troubles to the sky.<\/p>\n<p id=\"eip-idp3756880\">POIK! They&#8217;re gone, with all my stress<\/p>\n<p id=\"eip-idp3757312\">Bubble therapy is the best.<\/p>\n<div data-type=\"newline\" data-count=\"1\"><\/div>\n<p id=\"eip-idm21778064\">The label said each time I blew<\/p>\n<p id=\"eip-idm21777632\">The average number of bubbles would be at least 22.<\/p>\n<p id=\"eip-idm21777248\">I blew and blew and this I found<\/p>\n<p id=\"eip-idm21776816\">From 64 blows, they all are round!<\/p>\n<div data-type=\"newline\" data-count=\"1\"><\/div>\n<p id=\"eip-idm21776016\">But the number of bubbles in 64 blows<\/p>\n<p id=\"eip-idm21775584\">Varied widely, this I know.<\/p>\n<p id=\"eip-idm21775152\">20 per blow became the mean<\/p>\n<p id=\"eip-idm35613136\">They deviated by 6, and not 16.<\/p>\n<div data-type=\"newline\" data-count=\"1\"><\/div>\n<p id=\"eip-idm35612336\">From counting bubbles, I sure did relax<\/p>\n<p id=\"eip-idm35611904\">But now I give to you your task.<\/p>\n<p id=\"eip-idm35611472\">Was 22 a reasonable guess?<\/p>\n<p id=\"eip-idm35611040\">Find the answer and pass this test!<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-idm33073568\">\n<ol id=\"fs-idm33073312\" data-number-style=\"lower-alpha\">\n<li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> \u2265 22<\/li>\n<li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &lt; 22<\/li>\n<li>Let \\(\\overline{X}\\) = the mean number of bubbles per blow.<\/li>\n<li>Student&#8217;s <em data-effect=\"italics\">t<\/em>-distribution<\/li>\n<li>\u20132.667<\/li>\n<li><em data-effect=\"italics\">p<\/em>-value = 0.00486<\/li>\n<li>Check student\u2019s solution.<\/li>\n<li>\n<ol id=\"fs-idm27368736\" data-number-style=\"lower-roman\">\n<li>Alpha: 0.05<\/li>\n<li>Decision: Reject the null hypothesis.<\/li>\n<li>Reason for decision: The <em data-effect=\"italics\">p<\/em>-value is less than 0.05.<\/li>\n<li>Conclusion: There is sufficient evidence to conclude that the mean number of bubbles per blow is less than 22.<\/li>\n<\/ol>\n<\/li>\n<li>(18.501, 21.499)<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"ex3\">\n<div data-type=\"problem\" class=\"problem\" id=\"id9401931\">\n<p id=\"id11754724\">&#8220;Dalmatian Darnation,&#8221; by Kathy Sparling<\/p>\n<p id=\"eip-idm41372368\">A greedy dog breeder named Spreckles<\/p>\n<p id=\"eip-idm41371936\">Bred puppies with numerous freckles<\/p>\n<p id=\"eip-idm41371504\">The Dalmatians he sought<\/p>\n<p id=\"eip-idm41371072\">Possessed spot upon spot<\/p>\n<p id=\"eip-idm41370640\">The more spots, he thought, the more shekels.<\/p>\n<div data-type=\"newline\" data-count=\"1\"><\/div>\n<p id=\"eip-idm41369888\">His competitors did not agree<\/p>\n<p id=\"eip-idm41369456\">That freckles would increase the fee.<\/p>\n<p id=\"eip-idm41369024\">They said, \u201cSpots are quite nice<\/p>\n<p id=\"eip-idm98226224\">But they don&#8217;t affect price;<\/p>\n<p id=\"eip-idm98225792\">One should breed for improved pedigree.\u201d<\/p>\n<div data-type=\"newline\" data-count=\"1\"><\/div>\n<p id=\"eip-idm98224944\">The breeders decided to prove<\/p>\n<p id=\"eip-idm98224512\">This strategy was a wrong move.<\/p>\n<p id=\"eip-idm98224080\">Breeding only for spots<\/p>\n<p id=\"eip-idm98223696\">Would wreak havoc, they thought.<\/p>\n<p id=\"eip-idm98223264\">His theory they want to disprove.<\/p>\n<div data-type=\"newline\" data-count=\"1\"><\/div>\n<p id=\"eip-idm209854448\">They proposed a contest to Spreckles<\/p>\n<p id=\"eip-idm209854016\">Comparing dog prices to freckles.<\/p>\n<p id=\"eip-idm209853584\">In records they looked up<\/p>\n<p id=\"eip-idm209853152\">One hundred one pups:<\/p>\n<p id=\"eip-idm209852768\">Dalmatians that fetched the most shekels.<\/p>\n<div data-type=\"newline\" data-count=\"1\"><\/div>\n<p id=\"eip-idm209852016\">They asked Mr. Spreckles to name<\/p>\n<p id=\"eip-idm209851584\">An average spot count he&#8217;d claim<\/p>\n<p id=\"eip-idm209851152\">To bring in big bucks.<\/p>\n<p id=\"eip-idm209850768\">Said Spreckles, \u201cWell, shucks,<\/p>\n<p id=\"eip-idm105107920\">It&#8217;s for one hundred one that I aim.\u201d<\/p>\n<div data-type=\"newline\" data-count=\"1\"><\/div>\n<p id=\"eip-idm105107184\">Said an amateur statistician<\/p>\n<p id=\"eip-idm105106752\">Who wanted to help with this mission.<\/p>\n<p id=\"eip-idm105106320\">\u201cTwenty-one for the sample<\/p>\n<p id=\"eip-idm105105936\">Standard deviation&#8217;s ample:<\/p>\n<div data-type=\"newline\" data-count=\"1\"><\/div>\n<p id=\"eip-idm105105136\">They examined one hundred and one<\/p>\n<p id=\"eip-idm105104704\">Dalmatians that fetched a good sum.<\/p>\n<p id=\"eip-idm202236592\">They counted each spot,<\/p>\n<p id=\"eip-idm202236208\">Mark, freckle and dot<\/p>\n<p id=\"eip-idm202235824\">And tallied up every one.<\/p>\n<div data-type=\"newline\" data-count=\"1\"><\/div>\n<p id=\"eip-idm202235024\">Instead of one hundred one spots<\/p>\n<p id=\"eip-idm202234592\">They averaged ninety six dots<\/p>\n<p id=\"eip-idm202234160\">Can they muzzle Spreckles\u2019<\/p>\n<p id=\"eip-idm202233696\">Obsession with freckles<\/p>\n<p id=\"eip-idm202233312\">Based on all the dog data they&#8217;ve got?<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\" id=\"id9402199\">\n<p id=\"id9870720\">&#8220;Macaroni and Cheese, please!!&#8221; by Nedda Misherghi and Rachelle Hall<\/p>\n<p id=\"id11671377\">As a poor starving student I don&#8217;t have much money to spend for even the bare necessities. So my favorite and main staple food is macaroni and cheese. It&#8217;s high in taste and low in cost and nutritional value.<\/p>\n<p id=\"id8780562\">One day, as I sat down to determine the meaning of life, I got a serious craving for this, oh, so important, food of my life. So I went down the street to Greatway to get a box of macaroni and cheese, but it was SO expensive! &#x1f4b2;2.02 !!! Can you believe it? It made me stop and think. The world is changing fast. I had thought that the mean cost of a box (the normal size, not some super-gigantic-family-value-pack) was at most &#x1f4b2;1, but now I wasn&#8217;t so sure. However, I was determined to find out. I went to 53 of the closest grocery stores and surveyed the prices of macaroni and cheese. Here are the data I wrote in my notebook:<\/p>\n<div data-type=\"list\" id=\"element-357\">\n<div data-type=\"title\">Price per box of Mac and Cheese:<\/div>\n<ul>\n<li>5 stores @ &#x1f4b2;2.02<\/li>\n<li>15 stores @ &#x1f4b2;0.25<\/li>\n<li>3 stores @ &#x1f4b2;1.29<\/li>\n<li>6 stores @ &#x1f4b2;0.35<\/li>\n<li>4 stores @ &#x1f4b2;2.27<\/li>\n<li>7 stores @ &#x1f4b2;1.50<\/li>\n<li>5 stores @ &#x1f4b2;1.89<\/li>\n<li>8 stores @ 0.75.<\/li>\n<\/ul>\n<\/div>\n<p id=\"id5072942\">I could see that the cost varied but I had to sit down to figure out whether or not I was right. If it does turn out that this mouth-watering dish is at most &#x1f4b2;1, then I&#8217;ll throw a big cheesy party in our next statistics lab, with enough macaroni and cheese for just me. (After all, as a poor starving student I can&#8217;t be expected to feed our class of animals!)<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-idp45964640\">\n<ol id=\"fs-idp45964896\" data-number-style=\"lower-alpha\">\n<li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> \u2264 1<\/li>\n<li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &gt; 1<\/li>\n<li>Let \\(\\overline{X}\\) = the mean cost in dollars of macaroni and cheese in a certain town.<\/li>\n<li>Student&#8217;s <em data-effect=\"italics\">t<\/em>-distribution<\/li>\n<li><em data-effect=\"italics\">t<\/em> = 0.340<\/li>\n<li><em data-effect=\"italics\">p<\/em>-value = 0.36756<\/li>\n<li>Check student\u2019s solution.<\/li>\n<li>\n<ol id=\"fs-idm78282592\" data-number-style=\"lower-roman\">\n<li>Alpha: 0.05<\/li>\n<li>Decision: Do not reject the null hypothesis.<\/li>\n<li>Reason for decision: The <em data-effect=\"italics\">p<\/em>-value is greater than 0.05<\/li>\n<li>Conclusion: The mean cost could be &#x1f4b2;1, or less.  At the 5% significance level, there is insufficient evidence to conclude that the mean price of a box of macaroni and cheese is more than &#x1f4b2;1.<\/li>\n<\/ol>\n<\/li>\n<li>(0.8291, 1.241)<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\" id=\"id9402361\">\n<p id=\"id8933019\">&#8220;William Shakespeare: The Tragedy of Hamlet, Prince of Denmark,&#8221; by Jacqueline Ghodsi<\/p>\n<div data-type=\"list\" id=\"element-392\">\n<div data-type=\"title\">THE CHARACTERS (in order of appearance):<\/div>\n<ul>\n<li>HAMLET, Prince of Denmark and student of Statistics<\/li>\n<li>POLONIUS, Hamlet\u2019s tutor<\/li>\n<li>HOROTIO, friend to Hamlet and fellow student<\/li>\n<\/ul>\n<\/div>\n<p>Scene: The great library of the castle, in which Hamlet does his lessons<\/p>\n<p id=\"id10572210\">Act I<\/p>\n<p id=\"id10572215\">(The day is fair, but the face of Hamlet is clouded. He paces the large room. His tutor, Polonius, is reprimanding Hamlet regarding the latter\u2019s recent experience. Horatio is seated at the large table at right stage.)<\/p>\n<p id=\"id10603284\">POLONIUS: My Lord, how cans\u2019t thou admit that thou hast seen a ghost! It is but a figment of your imagination!<\/p>\n<p id=\"id4997547\">HAMLET: I beg to differ; I know of a certainty that five-and-seventy in one hundred of us, condemned to the whips and scorns of time as we are, have gazed upon a spirit of health, or goblin damn\u2019d, be their intents wicked or charitable.<\/p>\n<p id=\"id9031598\">POLONIUS If thou doest insist upon thy wretched vision then let me invest your time; be true to thy work and speak to me through the reason of the null and alternate hypotheses. (He turns to Horatio.) Did not Hamlet himself say, \u201cWhat piece of work is man, how noble in reason, how infinite in faculties? Then let not this foolishness persist. Go, Horatio, make a survey of three-and-sixty and discover what the true proportion be. For my part, I will never succumb to this fantasy, but deem man to be devoid of all reason should thy proposal of at least five-and-seventy in one hundred hold true.<\/p>\n<p id=\"id10821885\">HORATIO (to Hamlet): What should we do, my Lord?<\/p>\n<p id=\"id10821892\">HAMLET: Go to thy purpose, Horatio.<\/p>\n<p id=\"id9142272\">HORATIO: To what end, my Lord?<\/p>\n<p id=\"id9142279\">HAMLET: That you must teach me. But let me conjure you by the rights of our fellowship, by the consonance of our youth, but the obligation of our ever-preserved love, be even and direct with me, whether I am right or no.<\/p>\n<p id=\"id5011881\">(Horatio exits, followed by Polonius, leaving Hamlet to ponder alone.)<\/p>\n<p id=\"id10626816\">Act II<\/p>\n<p id=\"id10626821\">(The next day, Hamlet awaits anxiously the presence of his friend, Horatio. Polonius enters and places some books upon the table just a moment before Horatio enters.)<\/p>\n<p id=\"id11671359\">POLONIUS: So, Horatio, what is it thou didst reveal through thy deliberations?<\/p>\n<p id=\"id11671367\">HORATIO: In a random survey, for which purpose thou thyself sent me forth, I did discover that one-and-forty believe fervently that the spirits of the dead walk with us. Before my God, I might not this believe, without the sensible and true avouch of mine own eyes.<\/p>\n<p id=\"id4945509\">POLONIUS: Give thine own thoughts no tongue, Horatio. (Polonius turns to Hamlet.) But look to\u2019t I charge you, my Lord. Come Horatio, let us go together, for this is not our test. (Horatio and Polonius leave together.)<\/p>\n<p id=\"id4986106\">HAMLET: To reject, or not reject, that is the question: whether \u2018tis nobler in the mind to suffer the slings and arrows of outrageous statistics, or to take arms against a sea of data, and, by opposing, end them. (Hamlet resignedly attends to his task.)<\/p>\n<p id=\"id10766013\">(Curtain falls)<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"exer7\">\n<div data-type=\"problem\" class=\"problem\" id=\"id9402839\">\n<p id=\"id8293070\">&#8220;Untitled,&#8221; by Stephen Chen<\/p>\n<p id=\"id5050494\">I&#8217;ve often wondered how software is released and sold to the public. Ironically, I work for a company that sells products with known problems. Unfortunately, most of the problems are difficult to create, which makes them difficult to fix. I usually use the test program X, which tests the product, to try to create a specific problem. When the test program is run to make an error occur, the likelihood of generating an error is 1%.<\/p>\n<p id=\"id11215303\">So, armed with this knowledge, I wrote a new test program Y that will generate the same error that test program X creates, but more often. To find out if my test program is better than the original, so that I can convince the management that I&#8217;m right, I ran my test program to find out how often I can generate the same error. When I ran my test program 50 times, I generated the error twice. While this may not seem much better, I think that I can convince the management to use my test program instead of the original test program. Am I right?<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-idp10825072\">\n<ol id=\"fs-idp10825200\" data-number-style=\"lower-alpha\">\n<li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> = 0.01<\/li>\n<li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> &gt; 0.01<\/li>\n<li>Let <em data-effect=\"italics\">P\u2032<\/em> = the proportion of errors generated<\/li>\n<li>Normal for a single proportion<\/li>\n<li>2.13<\/li>\n<li>0.0165<\/li>\n<li>Check student\u2019s solution.<\/li>\n<li>\n<ol id=\"fs-idm61739936\" data-number-style=\"lower-roman\">\n<li>Alpha: 0.05<\/li>\n<li>Decision: Reject the null hypothesis<\/li>\n<li>Reason for decision: The <em data-effect=\"italics\">p<\/em>-value is less than 0.05.<\/li>\n<li>Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the proportion of errors generated is more than 0.01.<\/li>\n<\/ol>\n<\/li>\n<li>Confidence interval: (0, 0.094).\n<div data-type=\"newline\"><\/div>\n<p>The\u201cplus-4s\u201d confidence interval is (0.004, 0.144).<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\" id=\"id9402904\">\n<p id=\"id10126776\">&#8220;Japanese Girls\u2019 Names&#8221;<\/p>\n<p id=\"id9977266\">by Kumi Furuichi<\/p>\n<p id=\"id5031080\">It used to be very typical for Japanese girls\u2019 names to end with \u201cko.\u201d (The trend might have started around my grandmothers\u2019 generation and its peak might have been around my mother\u2019s generation.) \u201cKo\u201d means \u201cchild\u201d in Chinese characters. Parents would name their daughters with \u201cko\u201d attaching to other Chinese characters which have meanings that they want their daughters to become, such as Sachiko\u2014happy child, Yoshiko\u2014a good child, Yasuko\u2014a healthy child, and so on.<\/p>\n<p id=\"id4930619\">However, I noticed recently that only two out of nine of my Japanese girlfriends at this school have names which end with \u201cko.\u201d More and more, parents seem to have become creative, modernized, and, sometimes, westernized in naming their children.<\/p>\n<p id=\"id5048749\">I have a feeling that, while 70 percent or more of my mother\u2019s generation would have names with \u201cko\u201d at the end, the proportion has dropped among my peers. I wrote down all my Japanese friends\u2019, ex-classmates\u2019, co-workers, and acquaintances\u2019 names that I could remember. Following are the names. (Some are repeats.) Test to see if the proportion has dropped for this generation.<\/p>\n<p id=\"id12545176\">Ai, Akemi, Akiko, Ayumi, Chiaki, Chie, Eiko, Eri, Eriko, Fumiko, Harumi, Hitomi, Hiroko, Hiroko, Hidemi, Hisako, Hinako, Izumi, Izumi, Junko, Junko, Kana, Kanako, Kanayo, Kayo, Kayoko, Kazumi, Keiko, Keiko, Kei, Kumi, Kumiko, Kyoko, Kyoko, Madoka, Maho, Mai, Maiko, Maki, Miki, Miki, Mikiko, Mina, Minako, Miyako, Momoko, Nana, Naoko, Naoko, Naoko, Noriko, Rieko, Rika, Rika, Rumiko, Rei, Reiko, Reiko, Sachiko, Sachiko, Sachiyo, Saki, Sayaka, Sayoko, Sayuri, Seiko, Shiho, Shizuka, Sumiko, Takako, Takako, Tomoe, Tomoe, Tomoko, Touko, Yasuko, Yasuko, Yasuyo, Yoko, Yoko, Yoko, Yoshiko, Yoshiko, Yoshiko, Yuka, Yuki, Yuki, Yukiko, Yuko, Yuko.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"exerrr\">\n<div data-type=\"problem\" class=\"problem\" id=\"id9403278\">\n<p id=\"id4925098\">&#8220;Phillip\u2019s Wish,&#8221; by Suzanne Osorio<\/p>\n<p id=\"eip-idm86797472\">My nephew likes to play<\/p>\n<p id=\"eip-idm258943856\">Chasing the girls makes his day.<\/p>\n<p id=\"eip-idm258943424\">He asked his mother<\/p>\n<p id=\"eip-idm258943008\">If it is okay<\/p>\n<p id=\"eip-idm258942624\">To get his ear pierced.<\/p>\n<p id=\"eip-idm258942208\">She said, \u201cNo way!\u201d<\/p>\n<p id=\"eip-idm258941776\">To poke a hole through your ear,<\/p>\n<p id=\"eip-idm258941344\">Is not what I want for you, dear.<\/p>\n<p id=\"eip-idm258940912\">He argued his point quite well,<\/p>\n<p id=\"eip-idm258940480\">Says even my macho pal,  Mel,<\/p>\n<p id=\"eip-idm2866384\">Has gotten this done.<\/p>\n<p id=\"eip-idm2866000\">It\u2019s all just for fun.<\/p>\n<p id=\"eip-idm2865536\">C\u2019mon please, mom, please, what the hell.<\/p>\n<p id=\"eip-idm2865040\">Again Phillip complained to his mother,<\/p>\n<p id=\"eip-idm2864608\">Saying half his friends (including their brothers)<\/p>\n<p id=\"eip-idm2864160\">Are piercing their ears<\/p>\n<p id=\"eip-idm2863744\">And they have no fears<\/p>\n<p id=\"eip-idm2863328\">He wants to be like the others.<\/p>\n<p id=\"eip-idm2862896\">She said, \u201cI think it\u2019s much less.<\/p>\n<p id=\"eip-idm2862416\">We must do a hypothesis test.<\/p>\n<p id=\"eip-idm190635664\">And if you are right,<\/p>\n<p id=\"eip-idm190635280\">I won\u2019t put up a fight.<\/p>\n<p id=\"eip-idm190634816\">But, if not, then my case will rest.\u201d<\/p>\n<p id=\"eip-idm190634336\">We proceeded to call fifty guys<\/p>\n<p id=\"eip-idm190633904\">To see whose prediction would fly.<\/p>\n<p id=\"eip-idm190633472\">Nineteen of the fifty<\/p>\n<p id=\"eip-idm190633056\">Said piercing was nifty<\/p>\n<p id=\"eip-idm190632640\">And earrings they\u2019d occasionally buy.<\/p>\n<p id=\"eip-idm190632192\">Then there\u2019s the other thirty-one,<\/p>\n<p id=\"eip-idm96693360\">Who said they\u2019d never have this done.<\/p>\n<p id=\"eip-idm96692848\">So now this poem\u2019s finished.<\/p>\n<p id=\"eip-idm96692416\">Will his hopes be diminished,<\/p>\n<p id=\"eip-idm96692032\">Or will my nephew have his fun?<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-idm61462416\">\n<ol id=\"fs-idp31538240\" data-number-style=\"lower-alpha\">\n<li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> = 0.50<\/li>\n<li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> &lt; 0.50<\/li>\n<li>Let <em data-effect=\"italics\">P\u2032<\/em> = the proportion of friends that has a pierced ear.<\/li>\n<li>normal for a single proportion<\/li>\n<li>\u20131.70<\/li>\n<li><em data-effect=\"italics\">p<\/em>-value = 0.0448<\/li>\n<li>Check student\u2019s solution.<\/li>\n<li>\n<ol id=\"fs-idm172480592\" data-number-style=\"lower-roman\">\n<li>Alpha: 0.05<\/li>\n<li>Decision: Reject the null hypothesis<\/li>\n<li>Reason for decision: The <em data-effect=\"italics\">p<\/em>-value is less than 0.05. (However, they are very close.)<\/li>\n<li>Conclusion: There is sufficient evidence to support the claim that less than 50% of his friends have pierced ears.<\/li>\n<\/ol>\n<\/li>\n<li>Confidence Interval: (0.245, 0.515): The \u201cplus-4s\u201d confidence interval is (0.259, 0.519).<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"exercise\">\n<div data-type=\"problem\" class=\"problem\" id=\"id9403352\">\n<p id=\"id8450089\">&#8220;The Craven,&#8221; by Mark Salangsang<\/p>\n<p id=\"eip-idm8295648\">Once upon a morning dreary<\/p>\n<p id=\"eip-idm8295216\">In stats class I was weak and weary.<\/p>\n<p id=\"eip-idm8294784\">Pondering over last night\u2019s homework<\/p>\n<p id=\"eip-idp24682128\">Whose answers were now on the board<\/p>\n<p id=\"eip-idp24682560\">This I did and nothing more.<\/p>\n<div data-type=\"newline\" data-count=\"1\"><\/div>\n<p id=\"eip-idp24683360\">While I nodded nearly napping<\/p>\n<p id=\"eip-idp24683792\">Suddenly, there came a tapping.<\/p>\n<p id=\"eip-idp24684224\">As someone gently rapping,<\/p>\n<p id=\"eip-idp24684656\">Rapping my head as I snore.<\/p>\n<p id=\"eip-idp24685088\">Quoth the teacher, \u201cSleep no more.\u201d<\/p>\n<div data-type=\"newline\" data-count=\"1\"><\/div>\n<p id=\"eip-idp24685904\">\u201cIn every class you fall asleep,\u201d<\/p>\n<p id=\"eip-idm21424304\">The teacher said, his voice was deep.<\/p>\n<p id=\"eip-idm21423872\">\u201cSo a tally I\u2019ve begun to keep<\/p>\n<p id=\"eip-idm21423376\">Of every class you nap and snore.<\/p>\n<p id=\"eip-idm21422944\">The percentage being forty-four.\u201d<\/p>\n<div data-type=\"newline\" data-count=\"1\"><\/div>\n<p id=\"eip-idm21422112\">\u201cMy dear teacher I must confess,<\/p>\n<p id=\"eip-idm21421648\">While sleeping is what I do best.  <\/p>\n<p id=\"eip-idm21421216\">The percentage, I think, must be less,<\/p>\n<p id=\"eip-idm21420784\">A percentage less than forty-four.\u201d<\/p>\n<p id=\"eip-idm57340704\">This I said and nothing more.<\/p>\n<div data-type=\"newline\" data-count=\"1\"><\/div>\n<p id=\"eip-idm57339904\">\u201cWe\u2019ll see,\u201d he said and walked away,<\/p>\n<p id=\"eip-idm57339376\">And fifty classes from that day<\/p>\n<p id=\"eip-idm57338944\">He counted till the month of May<\/p>\n<p id=\"eip-idm57338512\">The classes in which I napped and snored.<\/p>\n<p id=\"eip-idm57338128\">The number he found was twenty-four.<\/p>\n<div data-type=\"newline\" data-count=\"1\"><\/div>\n<p id=\"eip-idm57337328\">At a significance level of 0.05,<\/p>\n<p id=\"eip-idm57336896\">Please tell me am I still alive?<\/p>\n<p id=\"eip-idm11006496\">Or did my grade just take a dive<\/p>\n<p id=\"eip-idm11006112\">Plunging down beneath the floor?<\/p>\n<p id=\"eip-idm11005680\">Upon thee I hereby implore.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\" id=\"id9403596\">\n<p>Toastmasters International cites a report by Gallop Poll that 40% of Americans fear public speaking. A student believes that less than 40% of students at her school fear public speaking. She randomly surveys 361 schoolmates and finds that 135 report they fear public speaking. Conduct a hypothesis test to determine if the percent at her school is less than 40%.\n  <\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-idm62899616\">\n<ol id=\"fs-idm62899488\" data-number-style=\"lower-alpha\">\n<li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> = 0.40<\/li>\n<li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> &lt; 0.40<\/li>\n<li>Let <em data-effect=\"italics\">P\u2032<\/em> = the proportion of schoolmates who fear public speaking.<\/li>\n<li>normal for a single proportion<\/li>\n<li>\u20131.01<\/li>\n<li><em data-effect=\"italics\">p<\/em>-value = 0.1563<\/li>\n<li>Check student\u2019s solution.<\/li>\n<li>\n<ol id=\"fs-idm92461088\" data-number-style=\"lower-roman\">\n<li>Alpha: 0.05<\/li>\n<li>Decision: Do not reject the null hypothesis.<\/li>\n<li>Reason for decision: The <em data-effect=\"italics\">p<\/em>-value is greater than 0.05.<\/li>\n<li>Conclusion: There is insufficient evidence to support the claim that less than 40% of students at the school fear public speaking.<\/li>\n<\/ol>\n<\/li>\n<li>Confidence Interval: (0.3241, 0.4240): The \u201cplus-4s\u201d confidence interval is (0.3257, 0.4250).<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\" id=\"id9403657\">\n<p>Sixty-eight percent of online courses taught at community colleges nationwide were taught by full-time faculty. To test if 68% also represents California\u2019s percent for full-time faculty teaching the online classes, Long Beach City College (LBCC) in California, was randomly selected for comparison. In the same year, 34 of the 44 online courses LBCC offered were taught by full-time faculty. Conduct a hypothesis test to determine if 68% represents California. NOTE: For more accurate results, use more California community colleges and this past year&#8217;s data.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\" id=\"id9403858\">\n<p>According to an article in <em data-effect=\"italics\">Bloomberg Businessweek<\/em>, New York City&#8217;s most recent adult smoking rate is 14%. Suppose that a survey is conducted to determine this year\u2019s rate. Nine out of 70 randomly chosen N.Y. City residents reply that they smoke. Conduct a hypothesis test to determine if the rate is still 14% or if it has decreased.<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-idp21365712\">\n<ol id=\"fs-idp21365968\" data-number-style=\"lower-alpha\">\n<li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> = 0.14<\/li>\n<li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> &lt; 0.14<\/li>\n<li>Let <em data-effect=\"italics\">P\u2032<\/em> = the proportion of NYC residents that smoke.<\/li>\n<li>normal for a single proportion<\/li>\n<li>\u20130.2756<\/li>\n<li><em data-effect=\"italics\">p<\/em>-value = 0.3914<\/li>\n<li>Check student\u2019s solution.<\/li>\n<li>\n<ol id=\"fs-idm186262784\" data-number-style=\"lower-roman\">\n<li>alpha: 0.05<\/li>\n<li>Decision: Do not reject the null hypothesis.<\/li>\n<li>Reason for decision: The <em data-effect=\"italics\">p<\/em>-value is greater than 0.05.<\/li>\n<li>At the 5% significance level, there is insufficient evidence to conclude that the proportion of NYC residents who smoke is less than 0.14.<\/li>\n<\/ol>\n<\/li>\n<li>Confidence Interval: (0.0502, 0.2070): The \u201cplus-4s\u201d confidence interval (see chapter 8) is (0.0676, 0.2297).<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\" id=\"id9403905\">\n<p>The mean age of De Anza College students in a previous term was 26.6 years old. An instructor thinks the mean age for online students is older than 26.6. She randomly surveys 56 online students and finds that the sample mean is 29.4 with a standard deviation of 2.1. Conduct a hypothesis test.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\" id=\"id9404099\">\n<p>Registered nurses earned an average annual salary of &#x1f4b2;69,110. For that same year, a survey was conducted of 41 California registered nurses to determine if the annual salary is higher than &#x1f4b2;69,110 for California nurses. The sample average was &#x1f4b2;71,121 with a sample standard deviation of &#x1f4b2;7,489. Conduct a hypothesis test.\n  <\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-idm11266464\">\n<ol id=\"fs-idm126211024\" data-number-style=\"lower-alpha\">\n<li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> = 69,110<\/li>\n<li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &gt; 69,110<\/li>\n<li>Let \\(\\overline{X}\\) = the mean salary in dollars for California registered nurses.<\/li>\n<li>Student&#8217;s <em data-effect=\"italics\">t<\/em>-distribution<\/li>\n<li><em data-effect=\"italics\">t<\/em> = 1.719<\/li>\n<li><em data-effect=\"italics\">p<\/em>-value: 0.0466<\/li>\n<li>Check student\u2019s solution.<\/li>\n<li>\n<ol id=\"fs-idm11928480\" data-number-style=\"lower-roman\">\n<li>Alpha: 0.05<\/li>\n<li>Decision: Reject the null hypothesis.<\/li>\n<li>Reason for decision: The <em data-effect=\"italics\">p<\/em>-value is less than 0.05.<\/li>\n<li>Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean salary of California registered nurses exceeds &#x1f4b2;69,110.<\/li>\n<\/ol>\n<\/li>\n<li>(&#x1f4b2;68,757, &#x1f4b2;73,485) <\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\" id=\"id9404155\">\n<p>La Leche League International reports that the mean age of weaning a child from breastfeeding is age four to five worldwide. In America, most nursing mothers wean their children much earlier. Suppose a random survey is conducted of 21 U.S. mothers who recently weaned their children. The mean weaning age was nine months (3\/4 year) with a standard deviation of 4 months. Conduct a hypothesis test to determine if the mean weaning age in the U.S. is less than four years old.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"exer12\">\n<div data-type=\"problem\" class=\"problem\" id=\"id8184194\">\n<p id=\"id10291861\">Over the past few decades, public health officials have examined the link between weight concerns and teen girls&#8217; smoking. Researchers surveyed a group of 273 randomly selected teen girls living in Massachusetts (between 12 and 15 years old). After four years the girls were surveyed again. Sixty-three said they smoked to stay thin. Is there good evidence that more than thirty percent of the teen girls smoke to stay thin?<\/p>\n<div data-type=\"newline\"><\/div>\n<p>After conducting the test, your decision and conclusion are<\/p>\n<ol id=\"id10291869\" data-number-style=\"lower-alpha\">\n<li>Reject <em data-effect=\"italics\">H<sub>0<\/sub><\/em>: There is sufficient evidence to conclude that more than 30% of teen girls smoke to stay thin.<\/li>\n<li>Do not reject <em data-effect=\"italics\">H<sub>0<\/sub><\/em>: There is not sufficient evidence to conclude that less than 30% of teen girls smoke to stay thin.<\/li>\n<li>Do not reject <em data-effect=\"italics\">H<sub>0<\/sub><\/em>: There is not sufficient evidence to conclude that more than 30% of teen girls smoke to stay thin.<\/li>\n<li>Reject <em data-effect=\"italics\">H<sub>0<\/sub><\/em>: There is sufficient evidence to conclude that less than 30% of teen girls smoke to stay thin.<\/li>\n<\/ol>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"id8184383\">\n<p id=\"sulution\">c<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"exer14\">\n<div data-type=\"problem\" class=\"problem\" id=\"id8184753\">\n<p id=\"eip-idp66540992\">A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening night midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 of them attended the midnight showing.\n<\/p>\n<div data-type=\"newline\"><\/div>\n<p>At a 1% level of significance, an appropriate conclusion is:<\/p>\n<ol id=\"id6241780\" data-number-style=\"lower-alpha\">\n<li>There is insufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is less than 20%.<\/li>\n<li>There is sufficient evidence to conclude that the  percent of EVC students who attended the midnight showing of Harry Potter is more than 20%.<\/li>\n<li>There is sufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is less than 20%.<\/li>\n<li>There is insufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is at least 20%.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"exer19\">\n<div data-type=\"problem\" class=\"problem\" id=\"id8186196\">\n<p id=\"id11679199\">Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test.\n<\/p>\n<div data-type=\"newline\"><\/div>\n<div data-type=\"newline\"><\/div>\n<p>At a significance level of<br \/>\n<em data-effect=\"italics\">a<\/em> = 0.05, what is the correct conclusion?<\/p>\n<ol id=\"id4986877\" data-number-style=\"lower-alpha\">\n<li>There is enough evidence to conclude that the mean number of hours is more than 4.75<\/li>\n<li>There is enough evidence to conclude that the mean number of hours is more than 4.5<\/li>\n<li>There is not enough evidence to conclude that the mean number of hours is more than 4.5<\/li>\n<li>There is not enough evidence to conclude that the mean number of hours is more than 4.75<\/li>\n<\/ol>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"id8186317\">\n<p id=\"paragraph123\">c<\/p>\n<\/div>\n<\/div>\n<p id=\"eip-964\">Instructions: For the following ten exercises,\n<\/p>\n<div data-type=\"newline\"><\/div>\n<p>Hypothesis testing: For the following ten exercises, answer each question.<\/p>\n<ol id=\"eip-76\" data-number-style=\"lower-alpha\">\n<li>State the null and alternate hypothesis.<\/li>\n<li>State the <em data-effect=\"italics\">p<\/em>-value.<\/li>\n<li>State alpha.<\/li>\n<li>What is your decision?<\/li>\n<li>Write a conclusion.<\/li>\n<li>Answer any other questions asked in the problem.<\/li>\n<\/ol>\n<div data-type=\"exercise\" class=\"exercise\" id=\"eip-69\">\n<div data-type=\"problem\" class=\"problem\">\n<p id=\"eip-971\">According to the Center for Disease Control website, in 2011 at least 18% of high school students have smoked a cigarette. An Introduction to Statistics class in Davies County, KY conducted a hypothesis test at the local high school (a medium sized\u2013approximately 1,200 students\u2013small city demographic) to determine if the local high school\u2019s percentage was lower. One hundred fifty students were chosen at random and surveyed. Of the 150 students surveyed, 82 have smoked. Use a significance level of 0.05 and using appropriate statistical evidence, conduct a hypothesis test and state the conclusions.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\">\n<p>A recent survey in the <em data-effect=\"italics\">N.Y. Times Almanac<\/em> indicated that 48.8% of families own stock. A broker wanted to determine if this survey could be valid. He surveyed a random sample of 250 families and found that 142 owned some type of stock. At the 0.05 significance level, can the survey be considered to be accurate?<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\">\n<ol id=\"eip-idm39194080\" data-number-style=\"lower-alpha\">\n<li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> = 0.488 <em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2260 0.488<\/li>\n<li><em data-effect=\"italics\">p<\/em>-value = 0.0114<\/li>\n<li>alpha = 0.05<\/li>\n<li>Reject the null hypothesis.<\/li>\n<li>At the 5% level of significance, there is enough evidence to conclude that 48.8% of families own stocks.<\/li>\n<li>The survey does not appear to be accurate.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"eip-59\">\n<div data-type=\"problem\" class=\"problem\" id=\"eip-900\">\n<p>Driver error can be listed as the cause of approximately 54% of all fatal auto accidents, according to the American Automobile Association. Thirty randomly selected fatal accidents are examined, and it is determined that 14 were caused by driver error. Using \u03b1 = 0.05, is the AAA proportion accurate?<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\">\n<p>The US Department of Energy reported that 51.7% of homes were heated by natural gas. A random sample of 221 homes in Kentucky found that 115 were heated by natural gas. Does the evidence support the claim for Kentucky at the \u03b1 = 0.05 level in Kentucky? Are the results applicable across the country? Why?<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\">\n<ol id=\"eip-idp117547184\" data-number-style=\"lower-alpha\">\n<li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> = 0.517 <em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2260 0.517<\/li>\n<li><em data-effect=\"italics\">p<\/em>-value = 0.9203.<\/li>\n<li>alpha = 0.05.<\/li>\n<li>Do not reject the null hypothesis.<\/li>\n<li>At the 5% significance level, there is not enough evidence to conclude that the proportion of homes in Kentucky that are heated by natural gas is 0.517.<\/li>\n<li>However, we cannot generalize this result to the entire nation. First, the sample\u2019s population is only the state of Kentucky. Second, it is reasonable to assume that homes in the extreme north and south will have extreme high usage and low usage, respectively. We would need to expand our sample base to include these possibilities if we wanted to generalize this claim to the entire nation.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"eip-835\">\n<div data-type=\"problem\" class=\"problem\">\n<p>For Americans using library services, the American Library Association claims that at most 67% of patrons borrow books. The library director in Owensboro, Kentucky feels this is not true, so she asked a local college statistic class to conduct a survey. The class randomly selected 100 patrons and found that 82 borrowed books. Did the class demonstrate that the percentage was higher in Owensboro, KY?  Use \u03b1 = 0.01 level of significance. What is the possible proportion of patrons that do borrow books from the Owensboro Library?<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\">\n<p>The Weather Underground reported that the mean amount of summer rainfall for the northeastern US is at least 11.52 inches. Ten cities in the northeast are randomly selected and the mean rainfall amount is calculated to be 7.42 inches with a standard deviation of 1.3 inches. At the \u03b1 = 0.05 level, can it be concluded that the mean rainfall was below the reported average? What if \u03b1 = 0.01? Assume the amount of summer rainfall follows a normal distribution.<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\">\n<ol id=\"eip-idp46967856\" data-number-style=\"lower-alpha\">\n<li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u00b5<\/em> \u2265 11.52  <em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u00b5<\/em> &lt; 11.52<\/li>\n<li><em data-effect=\"italics\">p<\/em>-value = 0.000002 which is almost 0.<\/li>\n<li>alpha = 0.05.<\/li>\n<li>Reject the null hypothesis.<\/li>\n<li>At the 5% significance level, there is enough evidence to conclude that the mean amount of summer rain in the northeaster US is less than 11.52 inches, on average.<\/li>\n<li>We would make the same conclusion if alpha was 1% because the <em data-effect=\"italics\">p<\/em>-value is almost 0.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\" id=\"eip-444\">\n<p>A survey in the <em data-effect=\"italics\">N.Y. Times Almanac<\/em> finds the mean commute time (one way) is 25.4 minutes for the 15 largest US cities. The Austin, TX chamber of commerce feels that Austin\u2019s commute time is less and wants to publicize this fact.  The mean for 25 randomly selected commuters is 22.1 minutes with a standard deviation of 5.3 minutes. At the \u03b1 = 0.10 level, is the Austin, TX commute significantly less than the mean commute time for the 15 largest US cities?<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\" id=\"eip-426\">\n<p>A report by the Gallup Poll found that a woman visits her doctor, on average, at most 5.8 times each year. A random sample of 20 women results in these yearly visit totals<\/p>\n<p id=\"eip-idm37849296\"><span data-type=\"list\" data-list-type=\"labeled-item\" data-display=\"inline\" data-mark-suffix=\"\"><span data-type=\"item\">3<\/span><span data-type=\"item\">2<\/span><span data-type=\"item\">1<\/span><span data-type=\"item\">3<\/span><span data-type=\"item\">7<\/span><span data-type=\"item\">2<\/span><span data-type=\"item\">9<\/span><span data-type=\"item\">4<\/span><span data-type=\"item\">6<\/span><span data-type=\"item\">6<\/span><span data-type=\"item\">8<\/span><span data-type=\"item\">0<\/span><span data-type=\"item\">5<\/span><span data-type=\"item\">6<\/span><span data-type=\"item\">4<\/span><span data-type=\"item\">2<\/span><span data-type=\"item\">1<\/span><span data-type=\"item\">3<\/span><span data-type=\"item\">4<\/span><span data-type=\"item\">1<\/span><\/span><\/p>\n<div data-type=\"newline\"><\/div>\n<p>At the <em data-effect=\"italics\">\u03b1<\/em> = 0.05 level can it be concluded that the sample mean is higher than 5.8 visits per year?\n<\/div>\n<div data-type=\"solution\" class=\"solution\">\n<ol id=\"eip-idp114932976\" data-number-style=\"lower-alpha\">\n<li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u00b5<\/em> \u2264 5.8\t<em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u00b5<\/em> &gt; 5.8<\/li>\n<li><em data-effect=\"italics\">p<\/em>-value =  0.9987<\/li>\n<li>alpha = 0.05<\/li>\n<li>Do not reject the null hypothesis.<\/li>\n<li>At the 5% level of significance, there is not enough evidence to conclude that a woman visits her doctor, on average, more than 5.8 times a year.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\" id=\"eip-484\">\n<p>According to the <em data-effect=\"italics\">N.Y. Times Almanac<\/em> the mean family size in the U.S. is 3.18. A sample of a college math class resulted in the following family sizes:\n<\/p>\n<div data-type=\"newline\"><\/div>\n<p><span data-type=\"list\" data-list-type=\"labeled-item\" data-display=\"inline\"><span data-type=\"item\">5<\/span><span data-type=\"item\">4<\/span><span data-type=\"item\">5<\/span><span data-type=\"item\">4<\/span><span data-type=\"item\">4<\/span><span data-type=\"item\">3<\/span><span data-type=\"item\">6<\/span><span data-type=\"item\">4<\/span><span data-type=\"item\">3<\/span><span data-type=\"item\">3<\/span><span data-type=\"item\">5<\/span><span data-type=\"item\">5<\/span><span data-type=\"item\">6<\/span><span data-type=\"item\">3<\/span><span data-type=\"item\">3<\/span><span data-type=\"item\">2<\/span><span data-type=\"item\">7<\/span><span data-type=\"item\">4<\/span><span data-type=\"item\">5<\/span><span data-type=\"item\">2<\/span><span data-type=\"item\">2<\/span><span data-type=\"item\">2<\/span><span data-type=\"item\">3<\/span><span data-type=\"item\">2<\/span><\/span><\/p>\n<div data-type=\"newline\"><\/div>\n<p>At <em data-effect=\"italics\">\u03b1<\/em> = 0.05 level, is the class\u2019 mean family size greater than the national average? Does the Almanac result remain valid? Why?\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\">\n<div data-type=\"problem\" class=\"problem\">\n<p>The student academic group on a college campus claims that freshman students study at least 2.5 hours per day, on average. One Introduction to Statistics class was skeptical. The class took a random sample of 30 freshman students and found a mean study time of 137 minutes with a standard deviation of 45 minutes. At <em data-effect=\"italics\">\u03b1<\/em> = 0.01 level, is the student academic group\u2019s claim correct?<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\">\n<ol id=\"eip-idp64977040\" data-number-style=\"lower-alpha\">\n<li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u00b5<\/em> \u2265 150\t<em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u00b5<\/em> &lt; 150<\/li>\n<li><em data-effect=\"italics\">p<\/em>-value = 0.0622<\/li>\n<li>alpha = 0.01<\/li>\n<li>Do not reject the null hypothesis.<\/li>\n<li>At the 1% significance level, there is not enough evidence to conclude that freshmen students study less than 2.5 hours per day, on average.<\/li>\n<li>The student academic group\u2019s claim appears to be correct.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"references\" data-depth=\"1\" id=\"fs-idp118682672\">\n<h1 data-type=\"title\">References<\/h1>\n<p>Data from Amit Schitai. Director of Instructional Technology and Distance Learning. LBCC.<\/p>\n<p>Data from <em data-effect=\"italics\">Bloomberg Businessweek<\/em>. Available online at http:\/\/www.businessweek.com\/news\/2011-<br \/>\n09-15\/nyc-smoking-rate-falls-to-record-low-of-14-bloomberg-says.html.<\/p>\n<p>Data from energy.gov. Available online at http:\/\/energy.gov (accessed June 27. 2013).<\/p>\n<p>Data from Gallup\u00ae. Available online at www.gallup.com (accessed June 27, 2013).<\/p>\n<p>Data from <em data-effect=\"italics\">Growing by Degrees<\/em> by Allen and Seaman.<\/p>\n<p>Data from La Leche League International. Available online at http:\/\/www.lalecheleague.org\/Law\/BAFeb01.html.<\/p>\n<p>Data from the American Automobile Association. Available online at www.aaa.com (accessed June 27, 2013).<\/p>\n<p>Data from the American Library Association. Available online at www.ala.org (accessed June 27, 2013).<\/p>\n<p>Data from the Bureau of Labor Statistics. Available online at http:\/\/www.bls.gov\/oes\/current\/oes291111.htm.<\/p>\n<p id=\"fs-idp1\">Data from the Centers for Disease Control and Prevention. Available online at www.cdc.gov (accessed June 27, 2013)<\/p>\n<p>Data from the U.S. Census Bureau, available online at http:\/\/quickfacts.census.gov\/qfd\/states\/00000.html (accessed June 27, 2013).<\/p>\n<p>Data from the United States Census Bureau. Available online at http:\/\/www.census.gov\/hhes\/socdemo\/language\/.<\/p>\n<p id=\"eip-79\">Data from Toastmasters International. Available online at http:\/\/toastmasters.org\/artisan\/detail.asp?CategoryID=1&amp;SubCategoryID=10&amp;ArticleID=429&amp;Page=1.<\/p>\n<p>Data from Weather Underground. Available online at www.wunderground.com (accessed June 27, 2013).<\/p>\n<p>Federal Bureau of Investigations. \u201cUniform Crime Reports and Index of Crime in Daviess in the State of Kentucky enforced by Daviess County from 1985 to 2005.\u201d Available online at http:\/\/www.disastercenter.com\/kentucky\/crime\/3868.htm (accessed June 27, 2013).<\/p>\n<p>\u201cFoothill-De Anza Community College District.\u201d De Anza College, Winter 2006. Available online at http:\/\/research.fhda.edu\/factbook\/DAdemofs\/Fact_sheet_da_2006w.pdf.<\/p>\n<p>Johansen, C., J. Boice, Jr., J. McLaughlin, J. Olsen. \u201cCellular Telephones and Cancer\u2014a Nationwide Cohort Study in Denmark.\u201d Institute of Cancer Epidemiology and the Danish Cancer Society, 93(3):203-7. Available online at http:\/\/www.ncbi.nlm.nih.gov\/pubmed\/11158188 (accessed June 27, 2013).<\/p>\n<p>Rape, Abuse &amp; Incest National Network. \u201cHow often does sexual assault occur?\u201d RAINN, 2009. Available online at http:\/\/www.rainn.org\/get-information\/statistics\/frequency-of-sexual-assault (accessed June 27, 2013).<\/p>\n<\/div>\n<div data-type=\"glossary\" class=\"textbox shaded\">\n<h2 data-type=\"glossary-title\">Glossary<\/h2>\n<dl class=\"definition\">\n<dt>Central Limit Theorem<\/dt>\n<dd>\n     Given a random variable (RV) with known mean \\(\\mu \\) and known standard deviation \u03c3. We are sampling with size <em data-effect=\"italics\">n<\/em> and we are interested in two new RVs &#8211; the sample mean, \\(\\overline{X}\\), and the sample sum, \\(\\Sigma X\\).<br \/>\nIf the size <em data-effect=\"italics\">n<\/em> of the sample is sufficiently large, then \\(\\overline{X}~N\\left(\\mu \\text{,}\\frac{\\sigma }{\\sqrt{n}}\\right)\\) and \\(\\Sigma X~N\\left(n\\mu ,\\sqrt{n}\\sigma \\right)\\). If the size <em data-effect=\"italics\">n<\/em> of the sample is sufficiently large, then the distribution of the sample means and the distribution of the sample sums will approximate a normal distribution regardless of the shape of the population. The mean of the sample  means will equal the population mean and the mean of the sample sums will equal <em data-effect=\"italics\">n<\/em> times the population mean. The standard deviation of the distribution of the sample means, \\(\\frac{\\sigma }{\\sqrt{n}}\\), is called the standard error of the mean.<\/dd>\n<\/dl>\n<\/div>\n","protected":false},"author":3,"menu_order":6,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":["openstaxcollege"],"pb_section_license":"cc-by"},"chapter-type":[],"contributor":[57],"license":[50],"class_list":["post-339","chapter","type-chapter","status-publish","hentry","contributor-openstaxcollege","license-cc-by"],"part":306,"_links":{"self":[{"href":"https:\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-json\/pressbooks\/v2\/chapters\/339","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-json\/wp\/v2\/users\/3"}],"version-history":[{"count":0,"href":"https:\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-json\/pressbooks\/v2\/chapters\/339\/revisions"}],"part":[{"href":"https:\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-json\/pressbooks\/v2\/parts\/306"}],"metadata":[{"href":"https:\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-json\/pressbooks\/v2\/chapters\/339\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-json\/wp\/v2\/media?parent=339"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-json\/pressbooks\/v2\/chapter-type?post=339"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-json\/wp\/v2\/contributor?post=339"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks-dev.oer.hawaii.edu\/introductorystatistics\/wp-json\/wp\/v2\/license?post=339"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}