Hypothesis Testing with Two Samples

Two Population Means with Unknown Standard Deviations

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[latexpage]

  1. The two independent samples are simple random samples from two distinct populations.
  2. For the two distinct populations:
    • if the sample sizes are small, the distributions are important (should be normal)
    • if the sample sizes are large, the distributions are not important (need not be normal)

The test comparing two independent population means with unknown and possibly unequal population standard deviations is called the Aspin-Welch t-test. The degrees of freedom formula was developed by Aspin-Welch.

The comparison of two population means is very common. A difference between the two samples depends on both the means and the standard deviations. Very different means can occur by chance if there is great variation among the individual samples. In order to account for the variation, we take the difference of the sample means,
\({\overline{X}}_{1}\) – \({\overline{X}}_{2}\), and divide by the standard error in order to standardize the difference. The result is a t-score test statistic.

Because we do not know the population standard deviations, we estimate them using the two sample standard deviations from our independent samples. For the hypothesis test, we calculate the estimated standard deviation, or standard error, of the difference in sample means, \({\overline{X}}_{1}\) – \({\overline{X}}_{2}\).

The standard error is:\(\sqrt{\frac{\left({s}_{1}{\right)}^{2}}{{n}_{1}}+\frac{\left({s}_{2}{\right)}^{2}}{{n}_{2}}}\)

The test statistic (t-score) is calculated as follows:

\(\frac{\text{(}{\overline{x}}_{1}–{\overline{x}}_{2}\text{)}–\text{(}{\mu }_{1}–{\mu }_{2}\text{)}}{\sqrt{\frac{{\text{(}{s}_{1}\text{)}}^{2}}{{n}_{1}}+\frac{{\text{(}{s}_{2}\text{)}}^{2}}{{n}_{2}}}}\)
where:
s1 and s2, the sample standard deviations, are estimates of σ1 and σ2, respectively.
σ1 and σ1 are the unknown population standard deviations.
\({\overline{x}}_{1}\)
and
\({\overline{x}}_{2}\)
are the sample means. μ1 and μ2 are the population means.

The number of degrees of freedom (df) requires a somewhat complicated calculation. However, a computer or calculator calculates it easily. The df are not always a whole number. The test statistic calculated previously is approximated by the Student’s t-distribution with df as follows:

Degrees of freedom\(df=\frac{{\left(\frac{{\left({s}_{1}\right)}^{2}}{{n}_{1}}+\frac{{\left({s}_{2}\right)}^{2}}{{n}_{2}}\right)}^{2}}{\left(\frac{1}{{n}_{1}–1}\right){\left(\frac{{\left({s}_{1}\right)}^{2}}{{n}_{1}}\right)}^{2}+\left(\frac{1}{{n}_{2}–1}\right){\left(\frac{{\left({s}_{2}\right)}^{2}}{{n}_{2}}\right)}^{2}}\)

When both sample sizes n1 and n2 are five or larger, the Student’s t approximation is very good. Notice that the sample variances (s1)2 and (s2)2 are not pooled. (If the question comes up, do not pool the variances.)

It is not necessary to compute this by hand. A calculator or computer easily computes it.
Independent groups

The average amount of time boys and girls aged seven to 11 spend playing sports each day is believed to be the same. A study is done and data are collected, resulting in the data in [link]. Each populations has a normal distribution.

Sample Size Average Number of Hours Playing Sports Per Day Sample Standard Deviation
Girls 9 2 \(0.866\)
Boys 16 3.2 1.00

Is there a difference in the mean amount of time boys and girls aged seven to 11 play sports each day? Test at the 5% level of significance.

The population standard deviations are not known. Let g be the subscript for girls and b be the subscript for boys. Then, μg is the population mean for girls and μb is the population mean for boys. This is a test of two independent groups, two population means.

Random variable:
\({\overline{X}}_{g}-{\overline{X}}_{b}\)
= difference in the sample mean amount of time girls and boys play sports each day.

H0: μg = μb  H0: μg – μb = 0

Ha: μg ≠ μb  Ha: μg – μb ≠ 0

The words “the same” tell you H0 has an “=”. Since there are no other words to indicate Ha, assume it says “is different.”
This is a two-tailed test.

Distribution for the test:
Use tdf where df is calculated using the df formula for independent groups, two population means. Using a calculator, df is approximately 18.8462. Do not pool the variances.

Calculate the p-value using a Student’s t-distribution:p-value = 0.0054

Graph:

This is a normal distribution curve representing the difference in the average amount of time girls and boys play sports all day. The mean is equal to zero, and the values -1.2, 0, and 1.2 are labeled on the horizontal axis. Two vertical lines extend from -1.2 and 1.2 to the curve. The region to the left of x = -1.2 and the region to the right of x = 1.2 are shaded to represent the p-value. The area of each region is 0.0028.

\({s}_{g}=0.866\)

\({s}_{b}=1\)

So, \({\overline{x}}_{g}–{\overline{x}}_{b}\) = 2 – 3.2 = –1.2

Half the p-value is below –1.2 and half is above 1.2.

Make a decision: Since α > p-value, reject H0. This means you reject μg = μb. The means are different.

Press STAT. Arrow over to TESTS and press 4:2-SampTTest. Arrow over to Stats and press ENTER. Arrow down and enter 2 for the first sample mean, \(\sqrt{0.866}\) for Sx1, 9 for n1, 3.2 for the
second sample mean, 1 for Sx2, and 16 for n2. Arrow down to μ1: and arrow to does not equal μ2. Press ENTER. Arrow down to Pooled: and No. Press ENTER. Arrow down to Calculate and press ENTER. The p-value is p = 0.0054, the dfs are approximately 18.8462, and the test statistic is -3.14. Do the procedure again but instead of Calculate do Draw.

Conclusion: At the 5% level of significance, the sample data show there is sufficient evidence to conclude that the mean number of hours that girls and boys aged seven to 11 play sports per day is different (mean number of hours boys aged seven to 11 play sports per day is greater than the mean number of hours played by girls OR the mean number of hours girls aged seven to 11 play sports per day is greater than the mean number of hours played by boys).

Try It

Two samples are shown in [link]. Both have normal distributions. The means for the two populations are thought to be the same. Is there a difference in the means? Test at the 5% level of significance.

Sample Size Sample Mean Sample Standard Deviation
Population A 25 5 1
Population B 16 4.7 1.2

The p-value is 0.4125, which is much higher than 0.05, so we decline to reject the null hypothesis. There is not sufficient evidence to conclude that the means of the two populations are not the same.

NOTE

When the sum of the sample sizes is larger than 30 (n1 + n2 > 30) you can use the normal distribution to approximate the Student’s t.

A study is done by a community group in two neighboring colleges to determine which one graduates students with more math classes. College A samples 11 graduates. Their average is four math classes with a standard deviation of 1.5 math classes. College B samples nine graduates. Their average is 3.5 math classes with a standard deviation of one math class. The community group believes that a student who graduates from college A has taken more math classes, on the average. Both populations have a normal distribution. Test at a 1% significance level. Answer the following questions.

a. Is this a test of two means or two proportions?

a. two means

b. Are the populations standard deviations known or unknown?

b. unknown

c. Which distribution do you use to perform the test?

c. Student’s t

d. What is the random variable?

d. \({\overline{X}}_{A}-{\overline{X}}_{B}\)

e. What are the null and alternate hypotheses? Write the null and alternate hypotheses in words and in symbols.

e.

\({H}_{o}:{\mu }_{A}\le {\mu }_{B}\)
\({H}_{a}:{\mu }_{A}>{\mu }_{B}\)

f. Is this test right-, left-, or two-tailed?

f.

This is a normal distribution curve with mean equal to 0. A vertical line near the tail of the curve to the right of zero extends from the axis to the curve. The region under the curve to the right of the line is shaded.

right

g. What is the p-value?

g. 0.1928

h. Do you reject or not reject the null hypothesis?

h. Do not reject.

i. Conclusion:

i. At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that a student who graduates from college A has taken more math classes, on the average, than a student who graduates from college B.

Try It

A study is done to determine if Company A retains its workers longer than Company B. Company A samples 15 workers, and their average time with the company is five years with a standard deviation of 1.2. Company B samples 20 workers, and their average time with the company is 4.5 years with a standard deviation of 0.8. The populations are normally distributed.

  1. Are the population standard deviations known?
  2. Conduct an appropriate hypothesis test. At the 5% significance level, what is your conclusion?
  1. They are unknown.
  2. The p-value = 0.0878. At the 5% level of significance, there is insufficient evidence to conclude that the workers of Company A stay longer with the company.

A professor at a large community college wanted to determine whether there is a difference in the means of final exam scores between students who took his statistics course online and the students who took his face-to-face statistics class. He believed that the mean of the final exam scores for the online class would be lower than that of the face-to-face class. Was the professor correct? The randomly selected 30 final exam scores from each group are listed in [link] and [link].

Online Class
67.6 41.2 85.3 55.9 82.4 91.2 73.5 94.1 64.7 64.7
70.6 38.2 61.8 88.2 70.6 58.8 91.2 73.5 82.4 35.5
94.1 88.2 64.7 55.9 88.2 97.1 85.3 61.8 79.4 79.4
Face-to-face Class
77.9 95.3 81.2 74.1 98.8 88.2 85.9 92.9 87.1 88.2
69.4 57.6 69.4 67.1 97.6 85.9 88.2 91.8 78.8 71.8
98.8 61.2 92.9 90.6 97.6 100 95.3 83.5 92.9 89.4

Is the mean of the Final Exam scores of the online class lower than the mean of the Final Exam scores of the face-to-face class? Test at a 5% significance level. Answer the following questions:

  1. Is this a test of two means or two proportions?
  2. Are the population standard deviations known or unknown?
  3. Which distribution do you use to perform the test?
  4. What is the random variable?
  5. What are the null and alternative hypotheses? Write the null and alternative hypotheses in words and in symbols.
  6. Is this test right, left, or two tailed?
  7. What is the p-value?
  8. Do you reject or not reject the null hypothesis?
  9. At the ___ level of significance, from the sample data, there ______ (is/is not) sufficient evidence to conclude that ______.

(See the conclusion in [link], and write yours in a similar fashion)

First put the data for each group into two lists (such as L1 and L2). Press STAT. Arrow over to TESTS and press 4:2SampTTest. Make sure Data is highlighted and press ENTER. Arrow down and enter L1 for the first list and L2 for the second list. Arrow down to μ1: and arrow to ≠ μ2 (does not equal). Press ENTER. Arrow down to Pooled: No. Press ENTER. Arrow down to Calculate and press ENTER.

Note

Be careful not to mix up the information for Group 1 and Group 2!

  1. two means
  2. unknown
  3. Student’s t
  4. \({\overline{X}}_{1}–{\overline{X}}_{2}\)
    1. H0: μ1 = μ2 Null hypothesis: the means of the final exam scores are equal for the online and face-to-face statistics classes.
    2. Ha: μ1 < μ2 Alternative hypothesis: the mean of the final exam scores of the online class is less than the mean of the final exam scores of the face-to-face class.
  6. left-tailed
  7. p-value = 0.0011
    This is a normal distribution curve with mean equal to zero. A vertical line near the tail of the curve to the left of zero extends from the axis to the curve. The region under the curve to the left of the line is shaded representing p-value = 0.0011.
  8. Reject the null hypothesis
  9. The professor was correct. The evidence shows that the mean of the final exam scores for the online class is lower than that of the face-to-face class.

    At the 5% level of significance, from the sample data, there is (is/is not) sufficient evidence to conclude that the mean of the final exam scores for the online class is less than the mean of final exam scores of the face-to-face class.

Cohen’s Standards for Small, Medium, and Large Effect SizesCohen’s d is a measure of effect size based on the differences between two means. Cohen’s d, named for United States statistician Jacob Cohen, measures the relative strength of the differences between the means of two populations based on sample data. The calculated value of effect size is then compared to Cohen’s standards of small, medium, and large effect sizes.

Cohen’s Standard Effect Sizes
Size of effect d
Small 0.2
medium 0.5
Large 0.8

Cohen’s d is the measure of the difference between two means divided by the pooled standard deviation: \(d=\frac{{\overline{x}}_{1}–{\overline{x}}_{2}}{{s}_{pooled}}\) where \({s}_{pooled}=\sqrt{\frac{\left({n}_{1}–1\right){s}_{1}^{2}+\left({n}_{2}–1\right){s}_{2}^{2}}{{n}_{1}+{n}_{2}–2}}\)

Calculate Cohen’s d for [link]. Is the size of the effect small, medium, or large? Explain what the size of the effect means for this problem.

μ1 = 4 s1 = 1.5 n1 = 11

μ2 = 3.5 s2 = 1 n2 = 9

d = 0.384

The effect is small because 0.384 is between Cohen’s value of 0.2 for small effect size and 0.5 for medium effect size. The size of the differences of the means for the two colleges is small indicating that there is not a significant difference between them.

Calculate Cohen’s d for [link]. Is the size of the effect small, medium or large? Explain what the size of the effect means for this problem.

d = 0.834; Large, because 0.834 is greater than Cohen’s 0.8 for a large effect size. The size of the differences between the means of the Final Exam scores of online students and students in a face-to-face class is large indicating a significant difference.

Try It

Weighted alpha is a measure of risk-adjusted performance of stocks over a period of a year. A high positive weighted alpha signifies a stock whose price has risen while a small positive weighted alpha indicates an unchanged stock price during the time period. Weighted alpha is used to identify companies with strong upward or downward trends. The weighted alpha for the top 30 stocks of banks in the northeast and in the west as identified by Nasdaq on May 24, 2013 are listed in [link] and [link], respectively.

Northeast
94.2 75.2 69.6 52.0 48.0 41.9 36.4 33.4 31.5 27.6
77.3 71.9 67.5 50.6 46.2 38.4 35.2 33.0 28.7 26.5
76.3 71.7 56.3 48.7 43.2 37.6 33.7 31.8 28.5 26.0
West
126.0 70.6 65.2 51.4 45.5 37.0 33.0 29.6 23.7 22.6
116.1 70.6 58.2 51.2 43.2 36.0 31.4 28.7 23.5 21.6
78.2 68.2 55.6 50.3 39.0 34.1 31.0 25.3 23.4 21.5

Is there a difference in the weighted alpha of the top 30 stocks of banks in the northeast and in the west? Test at a 5% significance level. Answer the following questions:

  1. Is this a test of two means or two proportions?
  2. Are the population standard deviations known or unknown?
  3. Which distribution do you use to perform the test?
  4. What is the random variable?
  5. What are the null and alternative hypotheses? Write the null and alternative hypotheses in words and in symbols.
  6. Is this test right, left, or two tailed?
  7. What is the p-value?
  8. Do you reject or not reject the null hypothesis?
  9. At the ___ level of significance, from the sample data, there ______ (is/is not) sufficient evidence to conclude that ______.
  10. Calculate Cohen’s d and interpret it.
  1. two means
  2. unknown
  3. Student’s-t
  4. \({\overline{X}}_{1}\text{ – }{\overline{X}}_{2}\)
    1. H0 : μ1 = μ2 Null hypothesis: the means of the weighted alphas are equal.
    2. Ha : μ1 ≠ μ2 Alternative hypothesis : the means of the weighted alphas are not equal.
  6. two-tailed
  7. p-value = 0.8787
  8. Do not reject the null hypothesis
  9. This indicates that the trends in stocks are about the same in the top 30 banks in each region.
    This is a normal distribution curve with mean equal to zero. Both the right and left tails of the curve are shaded. Each tail represents 1/2(p-value) = 0.4394.

    5% level of significance, from the sample data, there is not sufficient evidence to conclude that the mean weighted alphas for the banks in the northeast and the west are different

  10. d = 0.040, Very small, because 0.040 is less than Cohen’s value of 0.2 for small effect size. The size of the difference of the means of the weighted alphas for the two regions of banks is small indicating that there is not a significant difference between their trends in stocks.

References

Data from Graduating Engineer + Computer Careers. Available online at http://www.graduatingengineer.com

Data from Microsoft Bookshelf.

Data from the United States Senate website, available online at www.Senate.gov (accessed June 17, 2013).

“List of current United States Senators by Age.” Wikipedia. Available online at http://en.wikipedia.org/wiki/List_of_current_United_States_Senators_by_age (accessed June 17, 2013).

“Sectoring by Industry Groups.” Nasdaq. Available online at http://www.nasdaq.com/markets/barchart-sectors.aspx?page=sectors&base=industry (accessed June 17, 2013).

“Strip Clubs: Where Prostitution and Trafficking Happen.” Prostitution Research and Education, 2013. Available online at www.prostitutionresearch.com/ProsViolPosttrauStress.html (accessed June 17, 2013).

“World Series History.” Baseball-Almanac, 2013. Available online at http://www.baseball-almanac.com/ws/wsmenu.shtml (accessed June 17, 2013).

Chapter Review

Two population means from independent samples where the population standard deviations are not known

  • Random Variable: \({\overline{X}}_{1}-{\overline{X}}_{2}\) = the difference of the sampling means
  • Distribution: Student’s t-distribution with degrees of freedom (variances not pooled)

Formula Review

Standard error: SE = \(\sqrt{\frac{{\left({s}_{1}\right)}^{2}}{{n}_{1}}+\frac{{\left({s}_{2}\right)}^{2}}{{n}_{2}}}\)

Test statistic (t-score): t = \(\frac{\left({\overline{x}}_{1}-{\overline{x}}_{2}\right)-\left({\mu }_{1}-{\mu }_{2}\right)}{\sqrt{\frac{{\left({s}_{1}\right)}^{2}}{{n}_{1}}+\frac{{\left({s}_{2}\right)}^{2}}{{n}_{2}}}}\)

Degrees of freedom:

\(df= \frac{{\left(\frac{{\left({s}_{1}\right)}^{2}}{{n}_{1}}+ \frac{{\left({s}_{2}\right)}^{2}}{{n}_{2}}\right)}^{2}}{\left(\frac{1}{{n}_{1}-1}\right){\left(\frac{{\left({s}_{1}\right)}^{2}}{{n}_{1}}\right)}^{2}+\left(\frac{1}{{n}_{2}-1}\right){\left(\frac{{\left({s}_{2}\right)}^{2}}{{n}_{2}}\right)}^{2}}\)

where:

s1 and s2 are the sample standard deviations, and n1 and n2 are the sample sizes.

\({\overline{x}}_{1}\) and \({\overline{x}}_{2}\) are the sample means.

Cohen’s d is the measure of effect size:

\(d=\frac{{\overline{x}}_{1}-{\overline{x}}_{2}}{{s}_{pooled}}\)

where \({s}_{pooled}=\sqrt{\frac{\left({n}_{1}-1\right){s}_{1}^{2}+\left({n}_{2}-1\right){s}_{2}^{2}}{{n}_{1}+{n}_{2}-2}}\)

Use the following information to answer the next 15 exercises: Indicate if the hypothesis test is for

  1. independent group means, population standard deviations, and/or variances known
  2. independent group means, population standard deviations, and/or variances unknown
  3. matched or paired samples
  4. single mean
  5. two proportions
  6. single proportion

It is believed that 70% of males pass their drivers test in the first attempt, while 65% of females pass the test in the first attempt. Of interest is whether the proportions are in fact equal.

two proportions

A new laundry detergent is tested on consumers. Of interest is the proportion of consumers who prefer the new brand over the leading competitor. A study is done to test this.

A new windshield treatment claims to repel water more effectively. Ten windshields are tested by simulating rain without the new treatment. The same windshields are then treated, and the experiment is run again. A hypothesis test is conducted.

matched or paired samples

The known standard deviation in salary for all mid-level professionals in the financial industry is 💲11,000. Company A and Company B are in the financial industry. Suppose samples are taken of mid-level professionals from Company A and from Company B. The sample mean salary for mid-level professionals in Company A is 💲80,000. The sample mean salary for mid-level professionals in Company B is 💲96,000. Company A and Company B management want to know if their mid-level professionals are paid differently, on average.

The average worker in Germany gets eight weeks of paid vacation.

single mean

According to a television commercial, 80% of dentists agree that Ultrafresh toothpaste is the best on the market.

It is believed that the average grade on an English essay in a particular school system for females is higher than for males. A random sample of 31 females had a mean score of 82 with a standard deviation of three, and a random sample of 25 males had a mean score of 76 with a standard deviation of four.

independent group means, population standard deviations and/or variances unknown

The league mean batting average is 0.280 with a known standard deviation of 0.06. The Rattlers and the Vikings belong to the league. The mean batting average for a sample of eight Rattlers is 0.210, and the mean batting average for a sample of eight Vikings is 0.260. There are 24 players on the Rattlers and 19 players on the Vikings. Are the batting averages of the Rattlers and Vikings statistically different?

In a random sample of 100 forests in the United States, 56 were coniferous or contained conifers. In a random sample of 80 forests in Mexico, 40 were coniferous or contained conifers. Is the proportion of conifers in the United States statistically more than the proportion of conifers in Mexico?

two proportions

A new medicine is said to help improve sleep. Eight subjects are picked at random and given the medicine. The means hours slept for each person were recorded before starting the medication and after.

It is thought that teenagers sleep more than adults on average. A study is done to verify this. A sample of 16 teenagers has a mean of 8.9 hours slept and a standard deviation of 1.2. A sample of 12 adults has a mean of 6.9 hours slept and a standard deviation of 0.6.

independent group means, population standard deviations and/or variances unknown

Varsity athletes practice five times a week, on average.

A sample of 12 in-state graduate school programs at school A has a mean tuition of 💲64,000 with a standard deviation of 💲8,000. At school B, a sample of 16 in-state graduate programs has a mean of 💲80,000 with a standard deviation of 💲6,000. On average, are the mean tuitions different?

independent group means, population standard deviations and/or variances unknown

A new WiFi range booster is being offered to consumers. A researcher tests the native range of 12 different routers under the same conditions. The ranges are recorded. Then the researcher uses the new WiFi range booster and records the new ranges. Does the new WiFi range booster do a better job?

A high school principal claims that 30% of student athletes drive themselves to school, while 4% of non-athletes drive themselves to school. In a sample of 20 student athletes, 45% drive themselves to school. In a sample of 35 non-athlete students, 6% drive themselves to school. Is the percent of student athletes who drive themselves to school more than the percent of nonathletes?

two proportions

Use the following information to answer the next three exercises: A study is done to determine which of two soft drinks has more sugar. There are 13 cans of Beverage A in a sample and six cans of Beverage B. The mean amount of sugar in Beverage A is 36 grams with a standard deviation of 0.6 grams. The mean amount of sugar in Beverage B is 38 grams with a standard deviation of 0.8 grams. The researchers believe that Beverage B has more sugar than Beverage A, on average. Both populations have normal distributions.

Are standard deviations known or unknown?

What is the random variable?

The random variable is the difference between the mean amounts of sugar in the two soft drinks.

Is this a one-tailed or two-tailed test?

Use the following information to answer the next 12 exercises: The U.S. Center for Disease Control reports that the mean life expectancy was 47.6 years for whites born in 1900 and 33.0 years for nonwhites. Suppose that you randomly survey death records for people born in 1900 in a certain county. Of the 124 whites, the mean life span was 45.3 years with a standard deviation of 12.7 years. Of the 82 nonwhites, the mean life span was 34.1 years with a standard deviation of 15.6 years. Conduct a hypothesis test to see if the mean life spans in the county were the same for whites and nonwhites.

Is this a test of means or proportions?

means

State the null and alternative hypotheses.

  1. H0: __________
  2. Ha: __________

Is this a right-tailed, left-tailed, or two-tailed test?

two-tailed

In symbols, what is the random variable of interest for this test?

In words, define the random variable of interest for this test.

the difference between the mean life spans of whites and nonwhites

Which distribution (normal or Student’s t) would you use for this hypothesis test?

Explain why you chose the distribution you did for [link].

This is a comparison of two population means with unknown population standard deviations.

Calculate the test statistic and p-value.

Sketch a graph of the situation. Label the horizontal axis. Mark the hypothesized difference and the sample difference. Shade the area corresponding to the p-value.

Check student’s solution.

Find the p-value.

At a pre-conceived α = 0.05, what is your:

Decision:
Reason for the decision:
Conclusion (write out in a complete sentence):
  1. Reject the null hypothesis
  2. p-value < 0.05
  3. There is not enough evidence at the 5% level of significance to support the claim that life expectancy in the 1900s is different between whites and nonwhites.

Does it appear that the means are the same? Why or why not?

Homework

DIRECTIONS: For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in Appendix E. 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.

NOTE

If you are using a Student’s t-distribution for a homework problem in what follows, including for paired data, you may assume that the underlying population is normally distributed. (When using these tests in a real situation, you must first prove that assumption, however.)

The mean number of English courses taken in a two–year time period by male and female college students is believed to be about the same. An experiment is conducted and data are collected from 29 males and 16 females. The males took an average of three English courses with a standard deviation of 0.8. The females took an average of four English courses with a standard deviation of 1.0. Are the means statistically the same?

A student at a four-year college claims that mean enrollment at four–year colleges is higher than at two–year colleges in the United States. Two surveys are conducted. Of the 35 two–year colleges surveyed, the mean enrollment was 5,068 with a standard deviation of 4,777. Of the 35 four-year colleges surveyed, the mean enrollment was 5,466 with a standard deviation of 8,191.

Subscripts: 1: two-year colleges; 2: four-year colleges

  1. H0: μ1 ≥ μ2
  2. Ha: μ1 < μ2
  3. \({\overline{X}}_{1}–{\overline{X}}_{2}\) is the difference between the mean enrollments of the two-year colleges and the four-year colleges.
  4. Student’s-t
  5. test statistic: -0.2480
  6. p-value: 0.4019
  7. Check student’s solution.
    1. Alpha: 0.05
    2. Decision: Do not reject
    3. Reason for Decision: p-value > alpha
    4. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean enrollment at four-year colleges is higher than at two-year colleges.

At Rachel’s 11th birthday party, eight girls were timed to see how long (in seconds) they could hold their breath in a relaxed position. After a two-minute rest, they timed themselves while jumping. The girls thought that the mean difference between their jumping and relaxed times would be zero. Test their hypothesis.

Relaxed time (seconds) Jumping time (seconds)
26 21
47 40
30 28
22 21
23 25
45 43
37 35
29 32

Mean entry-level salaries for college graduates with mechanical engineering degrees and electrical engineering degrees are believed to be approximately the same. A recruiting office thinks that the mean mechanical engineering salary is actually lower than the mean electrical engineering salary. The recruiting office randomly surveys 50 entry level mechanical engineers and 60 entry level electrical engineers. Their mean salaries were 💲46,100 and 💲46,700, respectively. Their standard deviations were 💲3,450 and 💲4,210, respectively. Conduct a hypothesis test to determine if you agree that the mean entry-level mechanical engineering salary is lower than the mean entry-level electrical engineering salary.

Subscripts: 1: mechanical engineering; 2: electrical engineering

  1. H0: µ1 ≥ µ2
  2. Ha: µ1 < µ2
  3. \({\overline{X}}_{1}-{\overline{X}}_{2}\) is the difference between the mean entry level salaries of mechanical engineers and electrical engineers.
  4. t108
  5. test statistic: t = –0.82
  6. p-value: 0.2061
  7. Check student’s solution.
    1. Alpha: 0.05
    2. Decision: Do not reject the null hypothesis.
    3. Reason for Decision: p-value > alpha
    4. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the mean entry-level salaries of mechanical engineers is lower than that of electrical engineers.

Marketing companies have collected data implying that teenage girls use more ring tones on their cellular phones than teenage boys do. In one particular study of 40 randomly chosen teenage girls and boys (20 of each) with cellular phones, the mean number of ring tones for the girls was 3.2 with a standard deviation of 1.5. The mean for the boys was 1.7 with a standard deviation of 0.8. Conduct a hypothesis test to determine if the means are approximately the same or if the girls’ mean is higher than the boys’ mean.

Use the information from [link] to answer the next four exercises.

Using the data from Lap 1 only, conduct a hypothesis test to determine if the mean time for completing a lap in races is the same as it is in practices.

  1. H0: µ1 = µ2
  2. Ha: µ1 ≠ µ2
  3. \({\overline{X}}_{1}-{\overline{X}}_{2}\) is the difference between the mean times for completing a lap in races and in practices.
  4. t20.32
  5. test statistic: –4.70
  6. p-value: 0.0001
  7. Check student’s solution.
    1. Alpha: 0.05
    2. Decision: Reject the null hypothesis.
    3. Reason for Decision: p-value < alpha
    4. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean time for completing a lap in races is different from that in practices.

Repeat the test in Exercise 10.83, but use Lap 5 data this time.

Repeat the test in Exercise 10.83, but this time combine the data from Laps 1 and 5.

  1. H0: µ1 = µ2
  2. Ha: µ1 ≠ µ2
  3. is the difference between the mean times for completing a lap in races and in practices.
  4. t40.94
  5. test statistic: –5.08
  6. p-value: zero
  7. Check student’s solution.
    1. Alpha: 0.05
    2. Decision: Reject the null hypothesis.
    3. Reason for Decision: p-value < alpha
    4. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean time for completing a lap in races is different from that in practices.

In two to three complete sentences, explain in detail how you might use Terri Vogel’s data to answer the following question. “Does Terri Vogel drive faster in races than she does in practices?”

Use the following information to answer the next two exercises. The Eastern and Western Major League Soccer conferences have a new Reserve Division that allows new players to develop their skills. Data for a randomly picked date showed the following annual goals.

Western Eastern
Los Angeles 9 D.C. United 9
FC Dallas 3 Chicago 8
Chivas USA 4 Columbus 7
Real Salt Lake 3 New England 6
Colorado 4 MetroStars 5
San Jose 4 Kansas City 3

Conduct a hypothesis test to answer the next two exercises.

The exact distribution for the hypothesis test is:

  1. the normal distribution
  2. the Student’s t-distribution
  3. the uniform distribution
  4. the exponential distribution

If the level of significance is 0.05, the conclusion is:

  1. There is sufficient evidence to conclude that the W Division teams score fewer goals, on average, than the E teams
  2. There is insufficient evidence to conclude that the W Division teams score more goals, on average, than the E teams.
  3. There is insufficient evidence to conclude that the W teams score fewer goals, on average, than the E teams score.
  4. Unable to determine

c

Suppose a statistics instructor believes that there is no significant difference between the mean class scores of statistics day students on Exam 2 and statistics night students on Exam 2. She takes random samples from each of the populations. The mean and standard deviation for 35 statistics day students were 75.86 and 16.91. The mean and standard deviation for 37 statistics night students were 75.41 and 19.73. The “day” subscript refers to the statistics day students. The “night” subscript refers to the statistics night students. A concluding statement is:

  1. There is sufficient evidence to conclude that statistics night students’ mean on Exam 2 is better than the statistics day students’ mean on Exam 2.
  2. There is insufficient evidence to conclude that the statistics day students’ mean on Exam 2 is better than the statistics night students’ mean on Exam 2.
  3. There is insufficient evidence to conclude that there is a significant difference between the means of the statistics day students and night students on Exam 2.
  4. There is sufficient evidence to conclude that there is a significant difference between the means of the statistics day students and night students on Exam 2.

Researchers interviewed street prostitutes in Canada and the United States. The mean age of the 100 Canadian prostitutes upon entering prostitution was 18 with a standard deviation of six. The mean age of the 130 United States prostitutes upon entering prostitution was 20 with a standard deviation of eight. Is the mean age of entering prostitution in Canada lower than the mean age in the United States? Test at a 1% significance level.

Test: two independent sample means, population standard deviations unknown.

Random variable: \({\overline{X}}_{1}-{\overline{X}}_{2}\)

Distribution: H0: μ1 = μ2Ha: μ1 < μ2 The mean age of entering prostitution in Canada is lower than the mean age in the United States.

This is a normal distribution curve with mean equal to zero. A vertical line near the tail of the curve to the left of zero extends from the axis to the curve. The region under the curve to the left of the line is shaded representing p-value = 0.0157.

Graph: left-tailed

p-value : 0.0151

Decision: Do not reject H0.

Conclusion: At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that the mean age of entering prostitution in Canada is lower than the mean age in the United States.

A powder diet is tested on 49 people, and a liquid diet is tested on 36 different people. Of interest is whether the liquid diet yields a higher mean weight loss than the powder diet. The powder diet group had a mean weight loss of 42 pounds with a standard deviation of 12 pounds. The liquid diet group had a mean weight loss of 45 pounds with a standard deviation of 14 pounds.

Suppose a statistics instructor believes that there is no significant difference between the mean class scores of statistics day students on Exam 2 and statistics night students on Exam 2. She takes random samples from each of the populations. The mean and standard deviation for 35 statistics day students were 75.86 and 16.91, respectively. The mean and standard deviation for 37 statistics night students were 75.41 and 19.73. The “day” subscript refers to the statistics day students. The “night” subscript refers to the statistics night students. An appropriate alternative hypothesis for the hypothesis test is:

  1. μday > μnight
  2. μday < μnight
  3. μday = μnight
  4. μday ≠ μnight

d

Glossary

Degrees of Freedom (df)
the number of objects in a sample that are free to vary.
Standard Deviation
A number that is equal to the square root of the variance and measures how far data values are from their mean; notation: s for sample standard deviation and σ for population standard deviation.
Variable (Random Variable)
a characteristic of interest in a population being studied. Common notation for variables are upper-case Latin letters X, Y, Z,… Common notation for a specific value from the domain (set of all possible values of a variable) are lower-case Latin letters x, y, z,…. For example, if X is the number of children in a family, then x represents a specific integer 0, 1, 2, 3, …. Variables in statistics differ from variables in intermediate algebra in two following ways.

  • The domain of the random variable (RV) is not necessarily a numerical set; the domain may be expressed in words; for example, if X = hair color, then the domain is {black, blond, gray, green, orange}.
  • We can tell what specific value x of the random variable X takes only after performing the experiment.

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Two Population Means with Unknown Standard Deviations Copyright © 2013 by OpenStaxCollege is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.