Goodness-of-Fit Test

OpenStaxCollege

The Chi-Square Distribution

Goodness-of-Fit Test

OpenStaxCollege

[latexpage]

In this type of hypothesis test, you determine whether the data “fit” a particular distribution or not. For example, you may suspect your unknown data fit a binomial distribution. You use a chi-square test (meaning the distribution for the hypothesis test is chi-square) to determine if there is a fit or not. The null and the alternative hypotheses for this test may be written in sentences or may be stated as equations or inequalities.

The test statistic for a goodness-of-fit test is:

\(\underset{k}{\Sigma }\frac{{\left(O-E\right)}^{2}}{E}\)

where:

O = observed values (data)
E = expected values (from theory)
k = the number of different data cells or categories

The observed values are the data values and the expected values are the values you would expect to get if the null hypothesis were true. There are n terms of the form \(\frac{{\left(O-E\right)}^{2}}{E}\).

The number of degrees of freedom is df = (number of categories – 1).

The goodness-of-fit test is almost always right-tailed. If the observed values and the corresponding expected values are not close to each other, then the test statistic can get very large and will be way out in the right tail of the chi-square curve.

Note

The expected value for each cell needs to be at least five in order for you to use this test.

Absenteeism of college students from math classes is a major concern to math instructors because missing class appears to increase the drop rate. Suppose that a study was done to determine if the actual student absenteeism rate follows faculty perception. The faculty expected that a group of 100 students would miss class according to [link].

Number of absences per term	Expected number of students
0–2	50
3–5	30
6–8	12
9–11	6
12+	2

A random survey across all mathematics courses was then done to determine the actual number (observed) of absences in a course. The chart in [link] displays the results of that survey.

Number of absences per term	Actual number of students
0–2	35
3–5	40
6–8	20
9–11	1
12+	4

Determine the null and alternative hypotheses needed to conduct a goodness-of-fit test.

H₀: Student absenteeism fits faculty perception.

The alternative hypothesis is the opposite of the null hypothesis.

H_a: Student absenteeism does not fit faculty perception.

a. Can you use the information as it appears in the charts to conduct the goodness-of-fit test?

a. No. Notice that the expected number of absences for the “12+” entry is less than five (it is two). Combine that group with the “9–11” group to create new tables where the number of students for each entry are at least five. The new results are in [link] and [link].

Number of absences per term	Expected number of students
0–2	50
3–5	30
6–8	12
9+	8

Number of absences per term	Actual number of students
0–2	35
3–5	40
6–8	20
9+	5

b. What is the number of degrees of freedom (df)?

b. There are four “cells” or categories in each of the new tables.

df = number of cells – 1 = 4 – 1 = 3

Try It

A factory manager needs to understand how many products are defective versus how many are produced. The number of expected defects is listed in [link].

Number produced	Number defective
0–100	5
101–200	6
201–300	7
301–400	8
401–500	10

A random sample was taken to determine the actual number of defects. [link] shows the results of the survey.

Number produced	Number defective
0–100	5
101–200	7
201–300	8
301–400	9
401–500	11

State the null and alternative hypotheses needed to conduct a goodness-of-fit test, and state the degrees of freedom.

H₀:The number of defaults fits expectations.

H_a:The number of defaults does not fit expectations.

df = 4

Employers want to know which days of the week employees are absent in a five-day work week. Most employers would like to believe that employees are absent equally during the week. Suppose a random sample of 60 managers were asked on which day of the week they had the highest number of employee absences. The results were distributed as in [link]. For the population of employees, do the days for the highest number of absences occur with equal frequencies during a five-day work week? Test at a 5% significance level.

Day of the Week Employees were Most Absent
	Monday	Tuesday	Wednesday	Thursday	Friday
Number of Absences	15	12	9	9	15

The null and alternative hypotheses are:

H₀: The absent days occur with equal frequencies, that is, they fit a uniform distribution.
H_a: The absent days occur with unequal frequencies, that is, they do not fit a uniform distribution.

If the absent days occur with equal frequencies, then, out of 60 absent days (the total in the sample: 15 + 12 + 9 + 9 + 15 = 60), there would be 12 absences on Monday, 12 on Tuesday, 12 on Wednesday, 12 on Thursday, and 12 on Friday. These numbers are the expected (E) values. The values in the table are the observed (O) values or data.

This time, calculate the χ² test statistic by hand. Make a chart with the following headings and fill in the columns:

Expected (E) values (12, 12, 12, 12, 12)
Observed (O) values (15, 12, 9, 9, 15)
(O – E)
(O – E)²
\(\frac{{\left(O\text{ – }E\right)}^{2}}{E}\)

Now add (sum) the last column. The sum is three. This is the χ² test statistic.

To find the p-value, calculate P(χ² > 3). This test is right-tailed. (Use a computer or calculator to find the p-value. You should get p-value = 0.5578.)

The dfs are the number of cells – 1 = 5 – 1 = 4

Press 2nd DISTR. Arrow down to χ²cdf. Press ENTER. Enter (3,10^99,4). Rounded to four decimal places, you should see 0.5578, which is the p-value.

Next, complete a graph like the following one with the proper labeling and shading. (You should shade the right tail.)

This is a blank nonsymmetrical chi-square curve for the test statistic of the days of the week absent.

The decision is not to reject the null hypothesis.

Conclusion: At a 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the absent days do not occur with equal frequencies.

TI-83+ and some TI-84 calculators do not have a special program for the test statistic for the goodness-of-fit test. The next example [link] has the calculator instructions. The newer TI-84 calculators have in STAT TESTS the test Chi2 GOF. To run the test, put the observed values (the data) into a first list and the expected values (the values you expect if the null hypothesis is true) into a second list. Press STAT TESTS and Chi2 GOF. Enter the list names for the Observed list and the Expected list. Enter the degrees of freedom and press calculate or draw. Make sure you clear any lists before you start. To Clear Lists in the calculators: Go into STAT EDIT and arrow up to the list name area of the particular list. Press CLEAR and then arrow down. The list will be cleared. Alternatively, you can press STAT and press 4 (for ClrList). Enter the list name and press ENTER.

Try It

Teachers want to know which night each week their students are doing most of their homework. Most teachers think that students do homework equally throughout the week. Suppose a random sample of 49 students were asked on which night of the week they did the most homework. The results were distributed as in [link].

	Sunday	Monday	Tuesday	Wednesday	Thursday	Friday	Saturday
Number of Students	11	8	10	7	10	5	5

From the population of students, do the nights for the highest number of students doing the majority of their homework occur with equal frequencies during a week? What type of hypothesis test should you use?

df = 6

p-value = 0.6093

We decline to reject the null hypothesis. There is not enough evidence to support that students do not do the majority of their homework equally throughout the week.

One study indicates that the number of televisions that American families have is distributed (this is the given distribution for the American population) as in [link].

Number of Televisions	Percent
0	10
1	16
2	55
3	11
4+	8

The table contains expected (E) percents.

A random sample of 600 families in the far western United States resulted in the data in [link].

Number of Televisions	Frequency
	Total = 600
0	66
1	119
2	340
3	60
4+	15

The table contains observed (O) frequency values.

At the 1% significance level, does it appear that the distribution “number of televisions” of far western United States families is different from the distribution for the American population as a whole?

This problem asks you to test whether the far western United States families distribution fits the distribution of the American families. This test is always right-tailed.

The first table contains expected percentages. To get expected (E) frequencies, multiply the percentage by 600. The expected frequencies are shown in [link].

Number of Televisions	Percent	Expected Frequency
0	10	(0.10)(600) = 60
1	16	(0.16)(600) = 96
2	55	(0.55)(600) = 330
3	11	(0.11)(600) = 66
over 3	8	(0.08)(600) = 48

Therefore, the expected frequencies are 60, 96, 330, 66, and 48. In the TI calculators, you can let the calculator do the math. For example, instead of 60, enter 0.10*600.

H₀: The “number of televisions” distribution of far western United States families is the same as the “number of televisions” distribution of the American population.

H_a: The “number of televisions” distribution of far western United States families is different from the “number of televisions” distribution of the American population.

Distribution for the test: \({\chi }_{4}^{2}\) where df = (the number of cells) – 1 = 5 – 1 = 4.

Note

df ≠ 600 – 1

Calculate the test statistic:χ2 = 29.65

Graph:

This is a nonsymmetric chi-square curve with values of 0, 4, and 29.65 labeled on the horizontal axis. The value 4 coincides with the peak of the curve. A vertical upward line extends from 29.65 to the curve, and the region to the right of this line is shaded. The shaded area is equal to the p-value.

Probability statement:p-value = P(χ² > 29.65) = 0.000006

Compare α and the p-value:

α = 0.01

p-value = 0.000006

So, α > p-value.

Make a decision: Since α > p-value, reject H_o.

This means you reject the belief that the distribution for the far western states is the same as that of the American population as a whole.

Conclusion: At the 1% significance level, from the data, there is sufficient evidence to conclude that the “number of televisions” distribution for the far western United States is different from the “number of televisions” distribution for the American population as a whole.

Press STAT and ENTER. Make sure to clear lists L1, L2, and L3 if they have data in them (see the note at the end of [link]). Into L1, put the observed frequencies 66, 119, 349, 60, 15. Into L2, put the expected frequencies .10*600, .16*600, .55*600, .11*600, .08*600. Arrow over to list L3 and up to the name area "L3". Enter (L1-L2)^2/L2 and ENTER. Press 2nd QUIT. Press 2nd LIST and arrow over to MATH. Press 5. You should see "sum" (Enter L3). Rounded to 2 decimal places, you should see 29.65. Press 2nd DISTR. Press 7 or Arrow down to 7:χ2cdf and press ENTER. Enter (29.65,1E99,4). Rounded to four places, you should see 5.77E-6 = .000006 (rounded to six decimal places), which is the p-value.

The newer TI-84 calculators have in STAT TESTS the test Chi2 GOF. To run the test, put the observed values (the data) into a first list and the expected values (the values you expect if the null hypothesis is true) into a second list. Press STAT TESTS and Chi2 GOF. Enter the list names for the Observed list and the Expected list. Enter the degrees of freedom and press calculate or draw. Make sure you clear any lists before you start.

Try It

The expected percentage of the number of pets students have in their homes is distributed (this is the given distribution for the student population of the United States) as in [link].

Number of Pets	Percent
0	18
1	25
2	30
3	18
4+	9

A random sample of 1,000 students from the Eastern United States resulted in the data in [link].

Number of Pets	Frequency
0	210
1	240
2	320
3	140
4+	90

At the 1% significance level, does it appear that the distribution “number of pets” of students in the Eastern United States is different from the distribution for the United States student population as a whole? What is the p-value?

p-value = 0.0036

We reject the null hypothesis that the distributions are the same. There is sufficient evidence to conclude that the distribution “number of pets” of students in the Eastern United States is different from the distribution for the United States student population as a whole.

Suppose you flip two coins 100 times. The results are 20 HH, 27 HT, 30 TH, and 23 TT. Are the coins fair? Test at a 5% significance level.

This problem can be set up as a goodness-of-fit problem. The sample space for flipping two fair coins is {HH, HT, TH, TT}. Out of 100 flips, you would expect 25 HH, 25 HT, 25 TH, and 25 TT. This is the expected distribution. The question, “Are the coins fair?” is the same as saying, “Does the distribution of the coins (20 HH, 27 HT, 30 TH, 23 TT) fit the expected distribution?”

Random Variable: Let X = the number of heads in one flip of the two coins. X takes on the values 0, 1, 2. (There are 0, 1, or 2 heads in the flip of two coins.) Therefore, the number of cells is three. Since X = the number of heads, the observed frequencies are 20 (for two heads), 57 (for one head), and 23 (for zero heads or both tails). The expected frequencies are 25 (for two heads), 50 (for one head), and 25 (for zero heads or both tails). This test is right-tailed.

H₀: The coins are fair.

H_a: The coins are not fair.

Distribution for the test:\({\chi }_{2}^{2}\)
where
df = 3 – 1 = 2.

Calculate the test statistic:χ² = 2.14

Graph:

This is a nonsymmetrical chi-square curve with values of 0 and 2.14 labeled on the horizontal axis. A vertical upward line extends from 2.14 to the curve and the region to the right of this line is shaded. The shaded area is equal to the p-value.

Probability statement:p-value = P(χ² > 2.14) = 0.3430

Compare α and the p-value:

α = 0.05

p-value = 0.3430

α < p-value.

Make a decision: Since α < p-value, do not reject H₀.

Conclusion: There is insufficient evidence to conclude that the coins are not fair.

Press STAT and ENTER. Make sure you
clear lists L1, L2, and L3 if they have data in them. Into L1, put the observed
frequencies 20, 57, 23. Into L2, put the expected frequencies 25, 50, 25. Arrow
over to list L3 and up to the name area "L3". Enter (L1-L2)^2/L2 and
ENTER. Press 2nd QUIT. Press 2nd LIST and arrow over to MATH. Press
5. You should see "sum".Enter L3. Rounded to two decimal places, you
should see 2.14. Press 2nd DISTR. Arrow down to 7:χ2cdf (or press 7). Press
ENTER. Enter 2.14,1E99,2). Rounded to four places, you should see .3430, which
is the p-value.

The newer TI-84 calculators have in STAT TESTS the test Chi2 GOF. To run the test, put the observed values (the data) into a first list and the expected values (the values you expect if the null hypothesis is true) into a second list. Press STAT TESTS and Chi2 GOF. Enter the list names for the Observed list and the Expected list. Enter the degrees of freedom and press calculate or draw. Make
sure you clear any lists before you start.

Try It

Students in a social studies class hypothesize that the literacy rates across the world for every region are 82%. [link] shows the actual literacy rates across the world broken down by region. What are the test statistic and the degrees of freedom?

MDG Region	Adult Literacy Rate (%)
Developed Regions	99.0
Commonwealth of Independent States	99.5
Northern Africa	67.3
Sub-Saharan Africa	62.5
Latin America and the Caribbean	91.0
Eastern Asia	93.8
Southern Asia	61.9
South-Eastern Asia	91.9
Western Asia	84.5
Oceania	66.4

degrees of freedom = 9

chi² test statistic = 26.38

This is a nonsymmetric chi-square curve with df = 9. The values 0, 9, and 26.38 are labeled on the horizontal axis. The value 9 coincides with the peak of the curve. A vertical upward line extends from 26.38 to the curve, and the region to the right of this line is shaded. The shaded area is equal to the p-value.

Press STAT and ENTER. Make sure you clear lists L1, L2, and L3 if they have data in them. Into L1, put the observed frequencies 99, 99.5, 67.3, 62.5, 91, 93.8, 61.9, 91.9, 84.5, 66.4. Into L2, put the expected frequencies 82, 82, 82, 82, 82, 82, 82, 82, 82, 82. Arrow over to list L3 and up to the name area "L3". Enter (L1-L2)^2/L2 and ENTER. Press 2nd QUIT. Press 2nd LIST and arrow over to MATH. Press 5. You should see "sum". Enter L3. Rounded to two decimal places, you should see 26.38. Press 2nd DISTR. Arrow down to 7:χ2cdf (or press 7). Press ENTER. Enter 26.38,1E99,9). Rounded to four places, you should see .0018, which is the p-value.

The newer TI-84 calculators have in STAT TESTS the test Chi2 GOF. To run the test, put the observed values (the data) into a first list and the expected values (the values you expect if the null hypothesis is true) into a second list. Press STAT TESTS and Chi2 GOF. Enter the list names for the Observed list and the Expected list. Enter the degrees of freedom and press calculate or draw. Make sure you clear any lists before you start.

References

Data from the U.S. Census Bureau

Data from the College Board. Available online at http://www.collegeboard.com.

Data from the U.S. Census Bureau, Current Population Reports.

Ma, Y., E.R. Bertone, E.J. Stanek III, G.W. Reed, J.R. Hebert, N.L. Cohen, P.A. Merriam, I.S. Ockene, “Association between Eating Patterns and Obesity in a Free-living US Adult Population.” American Journal of Epidemiology volume 158, no. 1, pages 85-92.

Ogden, Cynthia L., Margaret D. Carroll, Brian K. Kit, Katherine M. Flegal, “Prevalence of Obesity in the United States, 2009–2010.” NCHS Data Brief no. 82, January 2012. Available online at http://www.cdc.gov/nchs/data/databriefs/db82.pdf (accessed May 24, 2013).

Stevens, Barbara J., “Multi-family and Commercial Solid Waste and Recycling Survey.” Arlington Count, VA. Available online at http://www.arlingtonva.us/departments/EnvironmentalServices/SW/file84429.pdf (accessed May 24,2013).

Chapter Review

To assess whether a data set fits a specific distribution, you can apply the goodness-of-fit hypothesis test that uses the chi-square distribution. The null hypothesis for this test states that the data come from the assumed distribution. The test compares observed values against the values you would expect to have if your data followed the assumed distribution. The test is almost always right-tailed. Each observation or cell category must have an expected value of at least five.

Formula Review

\(\sum _{k}\frac{{\left(O-E\right)}^{2}}{E}\)
goodness-of-fit test statistic where:

O: observed values

E: expected values

k: number of different data cells or categories

df = k − 1 degrees of freedom

Determine the appropriate test to be used in the next three exercises.

An archeologist is calculating the distribution of the frequency of the number of artifacts she finds in a dig site. Based on previous digs, the archeologist creates an expected distribution broken down by grid sections in the dig site. Once the site has been fully excavated, she compares the actual number of artifacts found in each grid section to see if her expectation was accurate.

An economist is deriving a model to predict outcomes on the stock market. He creates a list of expected points on the stock market index for the next two weeks. At the close of each day’s trading, he records the actual points on the index. He wants to see how well his model matched what actually happened.

a goodness-of-fit test

A personal trainer is putting together a weight-lifting program for her clients. For a 90-day program, she expects each client to lift a specific maximum weight each week. As she goes along, she records the actual maximum weights her clients lifted. She wants to know how well her expectations met with what was observed.

Use the following information to answer the next five exercises: A teacher predicts that the distribution of grades on the final exam will be and they are recorded in [link].

Grade	Proportion
A	0.25
B	0.30
C	0.35
D	0.10

The actual distribution for a class of 20 is in [link].

Grade	Frequency
A	7
B	7
C	5
D	1

\(df=\) ______

3

State the null and alternative hypotheses.

χ² test statistic = ______

2.04

p-value = ______

At the 5% significance level, what can you conclude?

We decline to reject the null hypothesis. There is not enough evidence to suggest that the observed test scores are significantly different from the expected test scores.

Use the following information to answer the next nine exercises: The following data are real. The cumulative number of AIDS cases reported for Santa Clara County is broken down by ethnicity as in [link].

Ethnicity	Number of Cases
White	2,229
Hispanic	1,157
Black/African-American	457
Asian, Pacific Islander	232
	Total = 4,075

The percentage of each ethnic group in Santa Clara County is as in [link].

Ethnicity	Percentage of total county population	Number expected (round to two decimal places)
White	42.9%	1748.18
Hispanic	26.7%
Black/African-American	2.6%
Asian, Pacific Islander	27.8%
	Total = 100%

If the ethnicities of AIDS victims followed the ethnicities of the total county population, fill in the expected number of cases per ethnic group.

Perform a goodness-of-fit test to determine whether the occurrence of AIDS cases follows the ethnicities of the general population of Santa Clara County.

H₀: _______

H₀: the distribution of AIDS cases follows the ethnicities of the general population of Santa Clara County.

H_a: _______

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

right-tailed

degrees of freedom = _______

χ² test statistic = _______

88,621

p-value = _______

Graph the situation. Label and scale the horizontal axis. Mark the mean and test statistic. Shade in the region corresponding to the p-value.

This is a blank graph template. The vertical and horizontal axes are unlabeled.

Let α = 0.05

Decision: ________________

Reason for the Decision: ________________

Conclusion (write out in complete sentences): ________________

Graph: Check student’s solution.

Decision: Reject the null hypothesis.

Reason for the Decision: p-value < alpha

Conclusion (write out in complete sentences): The make-up of AIDS cases does not fit the ethnicities of the general population of Santa Clara County.

Does it appear that the pattern of AIDS cases in Santa Clara County corresponds to the distribution of ethnic groups in this county? Why or why not?

Homework

For each problem, use a solution sheet to solve the hypothesis test problem. Go to [link] for the chi-square solution sheet. Round expected frequency to two decimal places.

A six-sided die is rolled 120 times. Fill in the expected frequency column. Then, conduct a hypothesis test to determine if the die is fair. The data in [link] are the result of the 120 rolls.

Face Value	Frequency	Expected Frequency
1	15
2	29
3	16
4	15
5	30
6	15

The marital status distribution of the U.S. male population, ages 15 and older, is as shown in [link].

Marital Status	Percent	Expected Frequency
never married	31.3
married	56.1
widowed	2.5
divorced/separated	10.1

Suppose that a random sample of 400 U.S. young adult males, 18 to 24 years old, yielded the following frequency distribution. We are interested in whether this age group of males fits the distribution of the U.S. adult population. Calculate the frequency one would expect when surveying 400 people. Fill in [link], rounding to two decimal places.

Marital Status	Frequency
never married	140
married	238
widowed	2
divorced/separated	20

Marital Status	Percent	Expected Frequency
never married	31.3	125.2
married	56.1	224.4
widowed	2.5	10
divorced/separated	10.1	40.4

The data fits the distribution.
The data does not fit the distribution.
3
chi-square distribution with df = 3
19.27
0.0002
Check student’s solution.
1. Alpha = 0.05
2. Decision: Reject null
3. Reason for decision: p-value < alpha
4. Conclusion: Data does not fit the distribution.

Use the following information to answer the next two exercises: The columns in [link] contain the Race/Ethnicity of U.S. Public Schools for a recent year, the percentages for the Advanced Placement Examinee Population for that class, and the Overall Student Population. Suppose the right column contains the result of a survey of 1,000 local students from that year who took an AP Exam.

Race/Ethnicity	AP Examinee Population	Overall Student Population	Survey Frequency
Asian, Asian American, or Pacific Islander	10.2%	5.4%	113
Black or African-American	8.2%	14.5%	94
Hispanic or Latino	15.5%	15.9%	136
American Indian or Alaska Native	0.6%	1.2%	10
White	59.4%	61.6%	604
Not reported/other	6.1%	1.4%	43

Perform a goodness-of-fit test to determine whether the local results follow the distribution of the U.S. overall student population based on ethnicity.

Perform a goodness-of-fit test to determine whether the local results follow the distribution of U.S. AP examinee population, based on ethnicity.

H₀: The local results follow the distribution of the U.S. AP examinee population
H_a: The local results do not follow the distribution of the U.S. AP examinee population
df = 5
chi-square distribution with df = 5
chi-square test statistic = 13.4
p-value = 0.0199
Check student’s solution.
Alpha = 0.05

Decision: Reject null when a = 0.05

Reason for Decision: p-value < alpha

Conclusion: Local data do not fit the AP Examinee Distribution.

Decision: Do not reject null when a = 0.01

Conclusion: There is insufficient evidence to conclude that local data do not follow the distribution of the U.S. AP examinee distribution.

The City of South Lake Tahoe, CA, has an Asian population of 1,419 people, out of a total population of 23,609. Suppose that a survey of 1,419 self-reported Asians in the Manhattan, NY, area yielded the data in [link]. Conduct a goodness-of-fit test to determine if the self-reported sub-groups of Asians in the Manhattan area fit that of the Lake Tahoe area.

Race	Lake Tahoe Frequency	Manhattan Frequency
Asian Indian	131	174
Chinese	118	557
Filipino	1,045	518
Japanese	80	54
Korean	12	29
Vietnamese	9	21
Other	24	66

Use the following information to answer the next two exercises: UCLA conducted a survey of more than 263,000 college freshmen from 385 colleges in fall 2005. The results of students’ expected majors by gender were reported in The Chronicle of Higher Education (2/2/2006). Suppose a survey of 5,000 graduating females and 5,000 graduating males was done as a follow-up last year to determine what their actual majors were. The results are shown in the tables for [link] and [link]. The second column in each table does not add to 100% because of rounding.

Conduct a goodness-of-fit test to determine if the actual college majors of graduating females fit the distribution of their expected majors.

Major	Women – Expected Major	Women – Actual Major
Arts & Humanities	14.0%	670
Biological Sciences	8.4%	410
Business	13.1%	685
Education	13.0%	650
Engineering	2.6%	145
Physical Sciences	2.6%	125
Professional	18.9%	975
Social Sciences	13.0%	605
Technical	0.4%	15
Other	5.8%	300
Undecided	8.0%	420

H₀: The actual college majors of graduating females fit the distribution of their expected majors
H_a: The actual college majors of graduating females do not fit the distribution of their expected majors
df = 10
chi-square distribution with df = 10
test statistic = 11.48
p-value = 0.3211
Check student’s solution.
1. Alpha = 0.05
2. Decision: Do not reject null when a = 0.05 and a = 0.01
3. Reason for decision: p-value > alpha
4. Conclusion: There is insufficient evidence to conclude that the distribution of actual college majors of graduating females fits the distribution of their expected majors.

Conduct a goodness-of-fit test to determine if the actual college majors of graduating males fit the distribution of their expected majors.

Major	Men – Expected Major	Men – Actual Major
Arts & Humanities	11.0%	600
Biological Sciences	6.7%	330
Business	22.7%	1130
Education	5.8%	305
Engineering	15.6%	800
Physical Sciences	3.6%	175
Professional	9.3%	460
Social Sciences	7.6%	370
Technical	1.8%	90
Other	8.2%	400
Undecided	6.6%	340

Read the statement and decide whether it is true or false.

In a goodness-of-fit test, the expected values are the values we would expect if the null hypothesis were true.

true

In general, if the observed values and expected values of a goodness-of-fit test are not close together, then the test statistic can get very large and on a graph will be way out in the right tail.

Use a goodness-of-fit test to determine if high school principals believe that students are absent equally during the week or not.

true

The test to use to determine if a six-sided die is fair is a goodness-of-fit test.

In a goodness-of fit test, if the p-value is 0.0113, in general, do not reject the null hypothesis.

false

A sample of 212 commercial businesses was surveyed for recycling one commodity; a commodity here means any one type of recyclable material such as plastic or aluminum. [link] shows the business categories in the survey, the sample size of each category, and the number of businesses in each category that recycle one commodity. Based on the study, on average half of the businesses were expected to be recycling one commodity. As a result, the last column shows the expected number of businesses in each category that recycle one commodity. At the 5% significance level, perform a hypothesis test to determine if the observed number of businesses that recycle one commodity follows the uniform distribution of the expected values.

Business Type	Number in class	Observed Number that recycle one commodity	Expected number that recycle one commodity
Office	35	19	17.5
Retail/Wholesale	48	27	24
Food/Restaurants	53	35	26.5
Manufacturing/Medical	52	21	26
Hotel/Mixed	24	9	12

[link] contains information from a survey among 499 participants classified according to their age groups. The second column shows the percentage of obese people per age class among the study participants. The last column comes from a different study at the national level that shows the corresponding percentages of obese people in the same age classes in the USA. Perform a hypothesis test at the 5% significance level to determine whether the survey participants are a representative sample of the USA obese population.

Age Class (Years)	Obese (Percentage)	Expected USA average (Percentage)
20–30	75.0	32.6
31–40	26.5	32.6
41–50	13.6	36.6
51–60	21.9	36.6
61–70	21.0	39.7

H₀: Surveyed obese fit the distribution of expected obese
H_a: Surveyed obese do not fit the distribution of expected obese
df = 4
chi-square distribution with df = 4
test statistic = 54.01
p-value = 0
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% level of significance, from the data, there is sufficient evidence to conclude that the surveyed obese do not fit the distribution of expected obese.

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