11.3: Goodness-of-Fit Test
- Page ID
- 20094
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:
\[\sum_k \frac{(O - E)^{2}}{E}\]
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 \(\frac{(O - E)^{2}}{E}\).
The number of degrees of freedom is \(df = (\text{number of categories} - 1)\).
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.
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 the table below.
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 the table below 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.
- \(H_{0}\): Student absenteeism fits faculty perception.
- \(H_{a}\): 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?
Answer
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 the table below.
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 (\(df\))?
Answer
b. There are four "cells" or categories in each of the new tables.
\(df = \text{number of cells} - 1 = 4 - 1 = 3\)
A factory manager needs to understand how many products are defective versus how many are produced. The number of expected defects is listed in the table below.
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. The table below 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.
Answer
\(H_{0}\):The number of defects fits expectations.
\(H_{a}\):The number of defects 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 the table below. 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 using the \(p\)-value method.
Monday | Tuesday | Wednesday | Thursday | Friday | |
---|---|---|---|---|---|
Number of Absences | 15 | 12 | 9 | 9 | 15 |
Answer
Determine the Hypothesis:
- \(H_{0}\): 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.
Calculate the Evidence:
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 \(\chi^{2}\) test statistic by hand. Make a chart with the following headings and fill in with the calculations:
- Observed (\(O\)) values \((15, 12, 9, 9, 15)\)
- Expected (\(E\)) values \((12, 12, 12, 12, 12)\)
- \((O – E)\)
- \((O – E)^{2}\)
- \(\frac{(O - E)^{2}}{E}\)
Observed values | Expected values | \((O – E)\) | \((O – E)^{2}\) | \(\frac{(O - E)^{2}}{E}\) |
---|---|---|---|---|
15 | 12 | 3 | 9 | 0.75 |
12 | 12 | 0 | 0 | 0 |
9 | 12 | -3 | 9 | 0.75 |
9 | 12 | -3 | 9 | 0.75 |
15 | 12 | 3 | 9 | 0.75 |
Now add (sum) the last column. The sum is three. This is the \(\chi^{2}\) test statistic, \(\chi^{2}=3\).
All Goodness-of-Fit Tests will be right-tailed. To find the \(p\)-value, use the Excel equation \(=\text{CHISQ.DIST.RT}(x,df)\).
The \(x\) is the test statistic, which is \(\chi^{2}=3\), and the degrees of freedom are calculated as the number of cells \(-1\), \(df=5-1\).
Enter in Excel \(=\text{CHISQ.DIST.RT}(3,4)=0.5578\)
Next, complete a graph like the following one with the proper labeling and shading. (You should shade the right tail.)
Make a Decision:
The decision is not to reject the null hypothesis since the \(p\)-value is larger than \(\alpha=0.05\).
Determine the 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.
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 the table below.
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 at a significance level of \(\alpha=0.10\)?
Answer
- \(H_{0}\): The days students do their homework occur with equal frequencies, that is, they fit a uniform distribution.
- \(H_{a}\): The days students do their homework occur with unequal frequencies, that is, they do not fit a uniform distribution.
\(df = 6\)
\(p\text{-value} = 0.6093\)
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 the table below.
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 the table below.
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? Use the critical value method.
Answer
Determine the Hypothesis:
\(H_{0}\): 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.
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.
Calculate the Evidence:
Since all Chi-Square tests are right-tailed, use the Excel formula \(=\text{CHISQ.INV.RT}(\alpha,df)\) to find the critical value.
The degrees of freedom are the number of cells \(-1\), \(df=5-1=4\). Be careful, on Chi-Square tests, the degrees of freedom are not the sample size \(-1\).
To find the critical value, enter the formula into Excel: \(=\text{CHISQ.INV.RT}(0.01,4)=13.2767\).
Now work to calculate the test statistic.
The first table contains expected percentages. To get expected (E) frequencies, multiply the percentage by 600. The expected frequencies are shown in the table below.
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.
Make a chart with the following headings and fill in with the calculations:
- Observed (\(O\)) values
- Expected (\(E\)) values
- \((O – E)\)
- \((O – E)^{2}\)
- \(\frac{(O - E)^{2}}{E}\)
Observed values | Expected values | \((O – E)\) | \((O – E)^{2}\) | \(\frac{(O - E)^{2}}{E}\) |
---|---|---|---|---|
66 | 60 | 6 | 36 | 0.6 |
119 | 96 | 23 | 529 | 5.51 |
340 | 330 | 10 | 100 | 0.30 |
60 | 66 | -6 | 36 | 0.55 |
15 | 48 | -33 | 1089 | 22.69 |
Now add (sum) the last column. The sum is 22.69. This is the \(\chi^{2}\) test statistic, \(\chi^{2}=22.69\).
Make a decision:
Since this is a right-tailed test, everything larger than the critical value is in the rejection region. The test statistic, \(\chi^{2}=22.69\), is larger than the critical value, \(13.2767\), so we Reject the null hypothesis.
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.
Determine the 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.
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 the table below.
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 the table below.
Number of Pets | Frequency |
---|---|
0 | 210 |
1 | 240 |
2 | 320 |
3 | 140 |
4+ | 90 |
At the 2% 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 \(p\text{-value}\)?
Answer
Determine the Hypothesis:
\(H_{0}\): The "number of pets" distribution among students in the United States is the same as the "number of pets" distribution among students in the Eastern United States.
\(H_{a}\): The "number of pets" distribution among students in the United States is different from the "number of pets" distribution among students in the Eastern United States.
This problem asks you to test whether the Eastern United States distribution among students fits the given distribution of the given distribution of the entire United States among students. This test is always right-tailed.
Calculate the Evidence:
Since all Chi-Square tests are right-tailed, use the Excel formula \(=\text{CHISQ.INV.RT}(\alpha,df)\) to find the critical value.
The degrees of freedom are the number of cells \(-1\), \(df=5-1=4\). Be careful, on Chi-Square tests, the degrees of freedom are not the sample size \(-1\).
To find the critical value, enter the formula into Excel: \(=\text{CHISQ.INV.RT}(0.02,4)=11.6678\).
Now work to calculate the test statistic.
The first table contains expected percentages. To get expected (E) frequencies, multiply the percentage by 600. The expected frequencies are shown in the table below.
Make a chart with the following headings and fill in with the calculations:
- Observed (\(O\)) values
- Expected (\(E\)) values
- \((O – E)\)
- \((O – E)^{2}\)
- \(\frac{(O - E)^{2}}{E}\)
Observed values | Expected values | \((O – E)\) | \((O – E)^{2}\) | \(\frac{(O - E)^{2}}{E}\) |
---|---|---|---|---|
210 | (0.18)(1000) = 180 | 30 | 900 | 5 |
240 | (0.25)(1000) = 250 | -10 | 100 | 0.4 |
320 | (0.30)(1000) = 300 | 20 | 400 | 1.33 |
140 | (0.18)(1000) = 180 | -40 | 1600 | 8.89 |
90 | (0.09)(1000) = 90 | 0 | 0 | 0 |
Now add (sum) the last column. The sum is 15.62. This is the \(\chi^{2}\) test statistic, \(\chi^{2}=15.62\).
Make a decision:
Since this is a right-tailed test, everything larger than the critical value is in the rejection region. The test statistic, \(\chi^{2}=15.62\), is larger than the critical value, \(11.6678\). This means the test statistic is in the rejection region, so we Reject the null hypothesis.
Determine the Conclusion:
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 23 pair of flips resulting in no heads, 57 pair of flips resulting in one head, and 20 pair of flips resulting in two heads. Are the coins fair? Test at a 7% significance level.
Answer
Determine the Hypothesis:
\(H_{0}\): The coins are fair.
\(H_{a}\): The coins are not fair.
The question, "Are the coins fair?" is the same as saying, "Does the distribution with 23 pair of flips resulting in no heads, 57 pair of flips resulting in one head, and 20 pair of flips resulting in two heads fit the expected distribution?" This test is always right-tailed.
Calculate the Evidence:
Work to calculate the test statistic.
Make a chart with the following headings and fill in with the calculations:
- Observed (\(O\)) values
- Expected (\(E\)) values
- \((O – E)\)
- \((O – E)^{2}\)
- \(\frac{(O - E)^{2}}{E}\)
Determine the Expected values by using probability.
Observed values | Expected values | \((O – E)\) | \((O – E)^{2}\) | \(\frac{(O - E)^{2}}{E}\) |
---|---|---|---|---|
23 | (0.25)(100) = 25 | -2 | 4 | 0.16 |
57 | (0.50)(100) = 50 | 7 | 49 | 0.98 |
20 | (0.25)(100) = 25 | -5 | 25 | 1 |
Now add (sum) the last column. The sum is 2.14. This is the \(\chi^{2}\) test statistic, \(\chi^{2}=2.14\).
All Goodness-of-Fit Tests will be right-tailed. To find the \(p\)-value, use the Excel equation \(=\text{CHISQ.DIST.RT}(x,df)\).
The \(x\) is the test statistic, which is \(\chi^{2}=2.14\), and the degrees of freedom are calculated as the number of cells \(-1\), \(df=3-1=2\).
Enter in Excel \(=\text{CHISQ.DIST.RT}(2.14,2)=0.3430\)
Draw the graph.
Make a decision:
\(\alpha = 0.07\)
\(p\text{-value} = 0.3430\)
\(\alpha < p\text{-value}\).Since \(\alpha < p\text{-value}\), do not reject \(H_{0}\).
Conclusion: There is insufficient evidence to conclude that the coins are not fair.
Students in a social studies class hypothesize that the literacy rates across the world for every region are 82%. The table below shows the actual literacy rates across the world broken down by region. Conduct a Chi-Square Goodness-of-Fit test to see if the actual literacy rates differ from the hypothesized literacy rates with \(\alpha=0.02\). Use the Critical Value method.
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 |
Answer
Determine the Hypothesis:
\(H_{0}\): The adult literacy rates across the world is 82% for every region.
\(H_{a}\): The adult literacy rates across the world is not 82% for every region.
This test is always right-tailed.
Calculate the Evidence:
Since all Chi-Square tests are right-tailed, use the Excel formula \(=\text{CHISQ.INV.RT}(\alpha,df)\) to find the critical value.
The degrees of freedom are the number of cells \(-1\), \(df=10-1=9\). Be careful, on Chi-Square tests, the degrees of freedom are not the sample size \(-1\).
To find the critical value, enter the formula into Excel: \(=\text{CHISQ.INV.RT}(0.02,9)=19.6790\).
Now work to calculate the test statistic.
Make a chart with the following headings and fill in with the calculations:
- Observed (\(O\)) values
- Expected (\(E\)) values
- \((O – E)\)
- \((O – E)^{2}\)
- \(\frac{(O - E)^{2}}{E}\)
The expected values will be 82 for all regions.
Observed values | Expected values | \((O – E)\) | \((O – E)^{2}\) | \(\frac{(O - E)^{2}}{E}\) |
---|---|---|---|---|
99.0 | 82 | 17 | 289 | 3.52 |
99.5 | 82 | 17.5 | 306.25 | 3.73 |
67.3 | 82 | -14.7 | 216.09 | 2.63 |
62.5 | 82 | -19.5 | 380.25 | 4.63 |
91.0 | 82 | 9 | 81 | 0.99 |
93.8 | 82 | 11.8 | 139.24 | 1.70 |
61.9 | 82 | -20.1 | 404.01 | 4.93 |
91.9 | 82 | 9.9 | 98.01 | 1.20 |
84.5 | 82 | 2.5 | 6.25 | 0.08 |
66.4 | 82 | -15.6 | 243.36 | 2.97 |
Now add (sum) the last column. The sum is 26.38. This is the \(\chi^{2}\) test statistic, \(\chi^{2}=26.38\).
Make a decision:
Since this is a right-tailed test, everything larger than the critical value is in the rejection region. The test statistic, \(\chi^{2}=26.38\), is larger than the critical value, \(19.6790\), so we Reject the null hypothesis.
Determine the Conclusion: With \(\alpha=0.02\), from the data, there is sufficient evidence to conclude that the adult literacy rates across the world is not 82% for every region.
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 www.arlingtonva.us/department.../file84429.pdf (accessed May 24,2013).
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
\(\sum_k \frac{(O - E)^{2}}{E}\) goodness-of-fit test statistic where:
\(O\): observed values
\(E\): expected value
\(k\): 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.
Answer
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 the table below.
Grade | Proportion |
---|---|
A | 0.25 |
B | 0.30 |
C | 0.35 |
D | 0.10 |
The actual distribution for a class of 20 is in the table below.
Grade | Frequency |
---|---|
A | 7 |
B | 7 |
C | 5 |
D | 1 |
\(df =\) ______
Answer
3
State the null and alternative hypotheses.
\(\chi^{2} \text{test statistic} =\) ______
Answer
2.04
\(p\text{-value} =\) ______
At the 5% significance level, what can you conclude?
Answer
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 the table below.
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 the table below.
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.
\(H_{0}\): _______
Answer
\(H_{0}\): 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?
Answer
right-tailed
degrees of freedom = _______
\(\chi^{2} \text{test statistic}\) = _______
Answer
88,621
\(p\text{-value} =\) _______
Graph the situation. Label and scale the horizontal axis. Mark the mean and test statistic. Shade in the region corresponding to the \(p\text{-value}\).
Let \(\alpha = 0.05\)
Decision: ________________
Reason for the Decision: ________________
Conclusion (write out in complete sentences): ________________
Answer
Graph: Check student’s solution.
Decision: Reject the null hypothesis.
Reason for the Decision: \(p\text{-value} < \alpha\)
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?