Contents

Goodness-of-Fit Test

In this type of hypothesis test, you determine whether the data fit a particular distribution. 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. 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{k}{Σ} \frac{{(O - E)}^{2}}{E} \underset{k}{Σ} \frac{{(O - E)}^{2}}{E}

where

O = observed values (data),
E = expected values (from theory), and
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 $\frac{{(O - E)}^{2}}{E} \frac{{(O - E)}^{2}}{E}$ .

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 expected value for each cell needs to be at least five for you to use this test.

Example 11.1

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 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 number of observed absences in a course. 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?

Show solution

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 is 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)?

Show solution

Solution

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

df = number of cells – 1 = 4 – 1 = 3.

Try It 11.1

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 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 11.2

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 percent 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

Show solution

Solution

The null and alternative hypotheses are as follows:

H₀: 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 χ² 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
$\frac{{(O – E)}^{2}}{E} \frac{{(O – E)}^{2}}{E}$

Now add (sum) the last column. The sum is three. This is the χ² test statistic.

To find the p-value, calculate P(χ² > 3). This test is right-tailed. Use a computer or calculator to find the p-value. You should get p-value = 0.5578.

The dfs are the number of cells – 1 = 5 – 1 = 4.

Using the TI-83, 83+, 84, 84+ Calculator

Press 2nd DISTR. Arrow down to χ²cdf. Press ENTER. Enter (3,10^99,4). Rounded to four decimal places, you should see .5578, which is the p-value.

Next, complete a graph like the following one with the proper labeling and shading. You should shade the right tail.

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

Figure 11.4

The decision is not to reject the null hypothesis.

Conclusion: At a 5 percent level of significance, from the sample data, there is not sufficient evidence to conclude that the absent days do not occur with equal frequencies.

Using the TI-83, 83+, 84, 84+ Calculator

TI-83+ and some TI-84 calculators do not have a special program for the test statistic for the goodness-of-fit test. The next example, Example 11.3, has the calculator instructions. The newer TI-84 calculators have in STAT TESTS the test Chi2 GOF. To run the test, put the observed values—the data—into a first list and the expected values—the values you expect if the null hypothesis is true—into a second list. Press STAT TESTS and Chi2 GOF. Enter the list names for the Observed list and the Expected list. Enter the degrees of freedom and press Calculate or Draw. Make sure you clear any lists before you start. To Clear Lists in the calculators: Go into STAT EDIT and arrow up to the list name area of the particular list. Press CLEAR and then arrow down. The list will be cleared. Alternatively, you can press STAT and press 4 for ClrList. Enter the list name and press ENTER.

Try It 11.2

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 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 11.3

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 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 U.S. 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 percent significance level, does it appear that the distribution number of televisions of far western U.S. families is different from the distribution for the American population as a whole?

Show solution

Solution

This problem asks you to test whether the far western U.S. 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 11.11.

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
more than 3	8	(0.08)(600) = 48

Table 11.11

Therefore, the expected frequencies are 60, 96, 330, 66, and 48. In the TI calculators, you can let the calculator do the math. For example, instead of 60, enter 0.10 * 600.

H₀: The number of televisions distribution of far western U.S. families is the same as the number of televisions distribution of the American population.

H_a: The number of televisions distribution of far western U.S. families is different from the number of televisions distribution of the American population.

Distribution for the test: $χ_{4}^{2} χ_{4}^{2}$ where df = (the number of cells) – 1 = 5 – 1 = 4.

Note

df ≠ 600 – 1

Calculate the test statistic: χ² = 29.65

Graph

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.

Figure 11.5

Probability statement: p-value = P(χ² > 29.65) = .000006

Compare α and the p-value:

α = .01
p-value = 0.000006

So, α > p-value.

Make a decision: Since α > p-value, reject H_o.

This means you reject the hypothesis that the distribution for the far western states is the same as that of the American population as a whole.

Conclusion: At the 1 percent 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.

Using the TI-83, 83+, 84, 84+ Calculator

Press STAT and ENTER. Make sure to clear lists L1, L2, and L3 if they have data in them—see the note at the end of Example 11.2. Into L1, put the observed frequencies 66, 119, 349, 60, 15. Into L2, put the expected frequencies .10*600, .16*600, .55*600, .11*600, .08*600. Arrow over to list L3 and up to the name area L3. Enter (L1-L2)^2/L2 and ENTER. Press 2nd QUIT. Press 2nd LIST and arrow over to MATH. Press 5. You should see sum (Enter L3). Rounded to two decimal places, you should see 29.65. Press 2nd DISTR. Press 7 or Arrow down to 7:χ2cdf and press ENTER. Enter (29.65,1E99,4). Rounded to four places, you should see 5.77E-6 = .000006 (rounded to six decimal places), which is the p-value.

The newer TI-84 calculators have in STAT TESTS the test Chi2 GOF. To run the test, put the observed values (the data) into a first list and the expected values—the values you expect if the null hypothesis is true—into a second list. Press STAT TESTS and Chi2 GOF. Enter the list names for the Observed list and the Expected list. Enter the degrees of freedom and press Calculate or Draw. Make sure you clear any lists before you start.

Try It 11.3

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 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 percent 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? What is the p-value?

Example 11.4

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 percent significance level.

Show solution

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. 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: $χ_{2}^{2} χ_{2}^{2}$ 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 11.6

Probability statement: p-value = P(χ² > 2.14) = 0.3430.

Compare α and the p-value:

α = .05
p-value = 0.3430

α < p-value.

Make a decision: Since α < p-value, do not reject H₀.

Conclusion: There is insufficient evidence to conclude that the coins are not fair.

Using the TI-83, 83+, 84, 84+ Calculator

Press STAT and ENTER. Make sure you clear lists L1, L2, and L3 if they have data in them. Into L1, put the observed frequencies 20, 57, 23. Into L2, put the expected frequencies 25, 50, 25. Arrow over to list L3 and up to the name area L3. Enter (L1-L2)^2/L2 and ENTER. Press 2nd QUIT. Press 2nd LIST and arrow over to MATH. Press 5. You should see sum. Enter L3. Rounded to two decimal places, you should see 2.14. Press 2nd DISTR. Arrow down to 7:χ2cdf—or press 7. Press ENTER. Enter 2.14,1E99,2). Rounded to four places, you should see .3430, which is the p-value.

The newer TI-84 calculators have in STAT TESTS the test Chi2 GOF. To run the test, put the observed values—the data—into a first list and the expected values—the values you expect if the null hypothesis is true—into a second list. Press STAT TESTS and Chi2 GOF. Enter the list names for the Observed list and the Expected list. Enter the degrees of freedom and press Calculate or Draw. Make sure you clear any lists before you start.

Try It 11.4

Students in a social studies class hypothesize that the literacy rates around the world for every region are 82 percent. Table 11.14 shows the actual literacy rates around the world broken down by region. What are the test statistic and the degrees of freedom?

MDG Region	Adult Literacy Rate (%)
Developed regions	99
Commonwealth of Independent States	99.5
Northern Africa	67.3
Sub-Saharan Africa	62.5
Latin America and the Caribbean	91
Eastern Asia	93.8
Southern Asia	61.9
Southeastern Asia	91.9
Western Asia	84.5
Oceania	66.4

Table 11.14