SESSION:11 The Chi-Square Distribution11.3 Goodness-of-Fit Test

Session:11 The Chi-Square Distribution

11.3 Goodness-of-Fit Test

Introductory Business Statistics | Leadership Development – Micro-Learning Session

Rice University 2020 | Michael Laverty, Colorado State University Global Chris Littel, North Carolina State University| https://openstax.org/details/books/introductory-business-statistics

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:

\underset{�}{�} \frac{{(� - �)}^{2}}{�}

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 $\frac{{(O - E)}^{2}}{E}$

$\frac{{(� - �)}^{2}}{�}$ .

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.

EXAMPLE 11.4

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 11.1.

Number of absences per term	Expected number of students
0–2	50
3–5	30
6–8	12
9–11	6
12+	2

Table 11.1

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 11.2 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

Table 11.2

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.

H_a: Student absenteeism does not fit faculty perception.

Problem

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

Solution

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 11.3 and Table 11.4.

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 11.4

Problem

b. What is the number of degrees of freedom (df)?

Solution

b. There are four “cells” or categories in each of the new tables.

df = number of cells – 1 = 4 – 1 = 3

TRY IT 11.4

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 11.5.

Number produced	Number defective
0–100	5
101–200	6
201–300	7
301–400	8
401–500	10

Table 11.5

A random sample was taken to determine the actual number of defects. Table 11.6 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

Table 11.6

State the null and alternative hypotheses needed to conduct a goodness-of-fit test, and state the degrees of freedom.

EXAMPLE 11.5

Problem

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 11.7. 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

Table 11.7 Day of the Week Employees were Most Absent

This is a blank nonsymmetrical chi-square curve for the test statistic of the days of the week absent.

TRY IT 11.5

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 11.8.

	Sunday	Monday	Tuesday	Wednesday	Thursday	Friday	Saturday
Number of students	11	8	10	7	10	5	5

Table 11.8

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?

EXAMPLE 11.6

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 11.9.

Number of Televisions	Percent
0	10
1	16
2	55
3	11
4+	8

Table 11.9

The table contains expected (E) percents.

A random sample of 600 families in the far western United States resulted in the data in Table 11.10.

Number of Televisions	Frequency
0	66
1	119
2	340
3	60
4+	15
	Total = 600

Table 11.10

The table contains observed (O) frequency values.

Problem

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?

This is a nonsymmetric chi-square curve with values of 0, 4, and 29.65 labeled on the horizontal axis. The value 4 coincides with the peak of the curve. A vertical upward line extends from 29.65 to the curve, and the region to the right of this line is shaded. The shaded area is equal to the p-value.

TRY IT 11.6

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 11.12.

Number of pets	Percent
0	18
1	25
2	30
3	18
4+	9

Table 11.12

A random sample of 1,000 students from the Eastern United States resulted in the data in Table 11.13.

Number of pets	Frequency
0	210
1	240
2	320
3	140
4+	90

Table 11.13

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?

EXAMPLE 11.7

Problem

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.

Solution

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.

H₀: The coins are fair.

H_a: The coins are not fair.

Distribution for the test: $χ_{2}^{2}$

$�_{2}^{2}$ where df = 3 – 1 = 2.

Calculate the test statistic: χ² = 2.14

Graph:

This is a nonsymmetrical chi-square curve with values of 0 and 2.14 labeled on the horizontal axis. A vertical upward line extends from 2.14 to the curve and the region to the right of this line is shaded. The shaded area is equal to the p-value.

Figure 11.7

The graph of the Chi-square shows the distribution and marks the critical value with two degrees of freedom at 95% level of confidence, α = 0.05, 5.991. The graph also marks the calculated χ² 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.