11.4: Goodness-of-Fit Test
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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:
\[\sum_{k} \frac{(O-E)^{2}}{E}\nonumber\]
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 \(df\) = (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 number of expected values inside each cell needs to be at least five in order to use this test.
Exercise \(\PageIndex{1}\)
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 Table \(\PageIndex{1}\).
| 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 Table \(\PageIndex{2}\) 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.
\(\bf{H_a}\) : Student absenteeism fits faculty perception.
The alternative hypothesis is the opposite of the null hypothesis.
\(\bf{H_a}\) : 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
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Solution 11.4
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 Table \(\PageIndex{3}\) and Table \(\PageIndex{4}\).
Number of absences per term Expected number of students 0–2 50 3–5 30 6–8 12 9+ 8 Table 11.3
Number of absences per term Actual number of students 0–2 35 3–5 40 6–8 20 9+ 5 Table \(\PageIndex{4}\)
b. What is the number of degrees of freedom (\(df\))?
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Solution 11.4
b. There are four "cells" or categories in each of the new tables.
\(d f=\text { number of cells }-1=4-1=3\)
Example \(\PageIndex{1}\)
A factory manager needs to understand how many products are defective versus how many are produced. The number of expected defects is listed in Table \(\PageIndex{5}\).
| 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. Table \(\PageIndex{6}\) 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.
Exercise \(\PageIndex{2}\)
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 Table \(\PageIndex{7}\). 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.
| Monday | Tuesday | Wednesday | Thursday | Friday | |
|---|---|---|---|---|---|
| Number of absences | 15 | 12 | 9 | 9 | 15 |
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Solution 11.5
The null and alternative hypotheses are:
- \(H_0\): 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.
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 \chi2 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 \((15, 12, 9, 9, 15)\)
- \((O – E)\)
- \((O – E)^2\)
- \(\frac{(O-E)^{2}}{E}\)
Now add (sum) the last column. The sum is three. This is the \(\chi^2\) test statistic.
The calculated test statistics is 3 and the critical value of the \(\chi^2\) distribution at 4 degrees of freedom the 0.05 level of confidence is 9.48. This value is found in the \(\chi^2\) table at the 0.05 column on the degrees of freedom row 4.
\(\text{The degrees of freedom are the number of cells }– 1 = 5 – 1 = 4\)
Next, complete a graph like the following one with the proper labeling and shading. (You should shade the right tail.)
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\[\bf{\chi}_{c}^{2}=\sum_{k} \frac{(O-E)^{2}}{E}=3\nonumber\]
The decision is not to reject the null hypothesis because the calculated value of the test statistic is not in the tail of the distribution.
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.
Example \(\PageIndex{2}\)
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 56 students were asked on which night of the week they did the most homework. The results were distributed as in Table \(\PageIndex{8}\).
| 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?
Exercise \(\PageIndex{3}\)
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 Table \(\PageIndex{9}\).
| 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 Table \(\PageIndex{10}\).
| 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?
- Answer
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Solution 11.6
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 Table \(\PageIndex{11}\).
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 Table \(\PageIndex{11}\) Therefore, the expected frequencies are 60, 96, 330, 66, and 48.
\(H_0\): 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.
Distribution for the test: \(\chi_{4}^{2} \text { where } d f=(\text { the number of cells })-1=5-1=4\).
Calculate the test statistic: \(\chi^2 = 29.65\)
Graph:
The graph of the Chi-square shows the distribution and marks the critical value with four degrees of freedom at 99% level of confidence, α = .01, 13.277. The graph also marks the calculated chi squared test statistic of 29.65. Comparing the test statistic with the critical value, as we have done with all other hypothesis tests, we reach the conclusion.
Make a decision: Because the test statistic is in the tail of the distribution we cannot accept 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.
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.
Example \(\PageIndex{3}\)
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 Table \(\PageIndex{12}\).
| 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 Table \(\PageIndex{13}\).
| 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?
Exercise \(\PageIndex{4}\)
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.
- Answer
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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 from the binomial probability 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.
\(\bf{H_0}\): The coins are fair.
\(\bf{H_a}\): The coins are not fair.
Distribution for the test: \(\chi_2^2\) where \(df = 3 – 1 = 2\).
Calculate the test statistic: \(\chi^2 = 2.14\).
Graph:
The graph of the Chi-square shows the distribution and marks the critical value with two degrees of freedom at 95% level of confidence, \(\alpha = 0.05\), 5.991. The graph also marks the calculated \(\chi^2\) test statistic of 2.14. Comparing the test statistic with the critical value, as we have done with all other hypothesis tests, we reach the conclusion.
Conclusion: There is insufficient evidence to conclude that the coins are not fair: we cannot reject the null hypothesis that the coins are fair.