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:
\[\sum_k \dfrac{(O-E)^2}{E}\]
where:
- \(\mathrm{O}=\) 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 \(\dfrac{(O-E)^2}{E}\).
The number of degrees of freedom is \(d f=\) (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 number of expected values inside each cell needs to be at least five in order to use this test.
Absenteeism in health education and physical activity courses is a major concern for public health instructors because missing practical sessions appears to decrease the mastery of essential clinical skills. Suppose that a study was done to determine if the actual student absenteeism rate follows faculty perception in a Kinesiology department. 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 health education and physical activity 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.
H0: Student absenteeism fits faculty perception.
The alternative hypothesis is the opposite of the null hypothesis.
Ha: 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?
b. What is the number of degrees of freedom (df)?
- Answer
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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 \(\PageIndex{4}\).
Table \(\PageIndex{3}\) Number of absences per term Expected number of students 0–2 50 3–5 30 6–8 12 9+ 8 Table \(\PageIndex{4}\) Number of absences per term Actual number of students 0–2 35 3–5 40 6–8 20 9+ 5 b. There are four "cells" or categories in each of the new tables.
df = number of cells – 1 = 4 – 1 = 3
Public health researchers want to know if patients are more likely to forget their daily medication on specific days of a week. For clinical consistency, researchers would like to believe that missed doses occur equally across the week. Suppose a random sample of 60 patients with chronic hypertension were monitored. Each patient reported the specific day of the week they were most likely to miss their scheduled dose. The results for the primary five-day work week (Monday–Friday) were distributed as follows:
| Monday | Tuesday | Wednesday | Thursday | Friday | |
|---|---|---|---|---|---|
| Number of missed abscenses | 15 | 12 | 9 | 9 | 15 |
Answer-
The null and alternative hypotheses are:
- \(H_0\) : The missed days occur with equal frequencies, that is, they fit a uniform distribution.
- \(H_a\) : The missed days occur with unequal frequencies, that is, they do not fit a uniform distribution.
If the missed medications occur with equal frequencies, then, out of 60 missed medications (the total in the sample: \(15+12+9+9+15=60\) ), there would be 12 missed medications 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 the columns:
- Expected \((E)\) values \((12,12,12,12,12)\)
- Observed (O) values (15, 12, 9, 9, 15)
- \((O-E)\)
- \((O-E)^2\)
- \(\dfrac{(O-E)^2}{E}\)
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.
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.)
Figure \(\PageIndex{1}\) 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 the 5% level of significance, there is not sufficient evidence to conclude that missed medication days occur with unequal frequencies. The distribution appears to be uniform.
In Public Health and Lifestyle Medicine, researchers often track the sedentary behavior of families by looking at their home entertainment habits. Excessive screen time is a significant correlate of physical inactivity and metabolic health risks. One study indicates that the number of home fitnesss and exercise streaming services that American families have is distributed (this is the given distribution for the general population) as follows: Table \(\PageIndex{9}\).
| Number of Streaming Services | 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 Streaming Services | Frequency |
|---|---|
| 0 | 66 |
| 1 | 119 |
| 2 | 340 |
| 3 | 60 |
| 4+ | 15 |
| Total = 600 |
The table contains observed (O) frequency values.
Problem
At the 1% significance level, does it appear that the distribution "number of fitness and exercise streaming services" of far western United States families is different from the distribution for the American population as a whole?
- Answer
-
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 \(\PageIndex{11}\).
Table \(\PageIndex{11}\) Number of Streaming Services 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.
\(H_0\) : The "number of streaming services" distribution of far western United States families is the same as the "number of streaming services" distribution of the American population.
\(H_a\) : The "number of streaming services" distribution of far western United States families is different from the "number of streaming services" distribution of the American population.
Distribution for the test: \(\chi_4^2\) where \(d f=(\) the number of cells \()-1=5-1=4\).
Calculate the test statistic: \(\chi_2=29.65\)
Figure \(\PageIndex{2}\) The graph of the Chi-square shows the distribution and marks the critical value with four degrees of freedom at \(99 \%\) level of confidence, \(\alpha=.01,13.277\). 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 fitness and exercise streaming services" distribution for the far western United States is different from the "number of streaming services" distribution for the American population as a whole.