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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:
where:
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 .
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.
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 |
H 0 : 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
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.
H 0 :The number of defaults fits expectations.
H
a :The number of defaults does not fit expectations.
df = 4
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