# 11.3 Goodness-of-fit test

This module describes how the chi-square distribution is used to conduct goodness-of-fit test.

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 binomialdistribution. 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 alternate hypotheses for this test may be written in sentences or may be stated as equations orinequalities.

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 degrees of freedom are $\text{df = (number of categories - 1)}$ .

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 statisticcan 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 5 in order 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 follows faculty perception. The faculty expected that a group of 100 students would miss class according to the following chart.

 Number 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 next chart displays the result of that survey.

 Number 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 alternate hypotheses needed to conduct a goodness-of-fit test.

${H}_{o}$ : Student absenteeism fits faculty perception.

The alternate hypothesis is the opposite of the null hypothesis.

${H}_{a}$ : Student absenteeism does not fit faculty perception.

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

No. Notice that the expected number of absences for the "12+" entry is less than 5 (it is 2). Combine that group with the "9 - 11" group to create new tables where the number of students for each entry are at least 5. The new tables are below.

 Number absences per term Expected number of students 0 - 2 50 3 - 5 30 6 - 8 12 9+ 8
 Number absences per term Actual number of students 0 - 2 35 3 - 5 40 6 - 8 20 9+ 5

What are the degrees of freedom ( $\mathrm{df}$ )?

There are 4 "cells" or categories in each of the new tables.

$\mathrm{df = number of cells - 1 = 4 - 1 = 3}$

Employers particularly want to know which days of the week employees are absent in a five day work week. Most employers wouldlike 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 did they have the highest number of employee absences. The results were distributed as follows:

Day of the week employees were most absent
Monday Tuesday Wednesday Thursday Friday
Number of Absences 15 12 9 9 15

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Ali
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I think
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1 It is estimated that 30% of all drivers have some kind of medical aid in South Africa. What is the probability that in a sample of 10 drivers: 3.1.1 Exactly 4 will have a medical aid. (8) 3.1.2 At least 2 will have a medical aid. (8) 3.1.3 More than 9 will have a medical aid.