10.4: Test for Homogeneity

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Test for Homogeneity

jkesler

[latexpage]

The goodness–of–fit test can be used to decide whether a population fits a given distribution, but it will not suffice to decide whether two populations follow the same unknown distribution. A different test, called the test for homogeneity, can be used to draw a conclusion about whether two populations have the same distribution. To calculate the test statistic for a test for homogeneity, follow the same procedure as with the test of independence.

Note

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

Hypotheses

H₀: The distributions of the two populations are the same.
H₁: The distributions of the two populations are not the same.

Test Statistic

Use a $\chi^2$ test statistic. It is computed in the same way as the test for independence.

Degrees of Freedom (df)

df = number of columns – 1

Requirements

All values in the table must be greater than or equal to five.

Common Uses

Comparing two populations. For example: men vs. women, before vs. after, east vs. west. The variable is categorical with more than two possible response values.

Example 10.8

Do male and female college students have the same distribution of living arrangements? Use a level of significance of 0.05. Suppose that 250 randomly selected male college students and 300 randomly selected female college students were asked about their living arrangements: dormitory, apartment, with parents, other. The results are shown in Table 10.19. Do male and female college students have the same distribution of living arrangements?

	Dormitory	Apartment	With Parents	Other
Males	72	84	49	45
Females	91	86	88	35

Table 10.19Distribution of Living Arragements for College Males and College Females

Solution 10.8

H₀: The distribution of living arrangements for male college students is the same as the distribution of living arrangements for female college students.


H₁: The distribution of living arrangements for male college students is not the same as the distribution of living arrangements for female college students.


Degrees of Freedom (df):
df = number of columns – 1 = 4 – 1 = 3

Distribution for the test:

$\chi_3^2$

Create the supporting tables

1. Expand the given observed table to include the row totals and column totals.

   	Dormitory	Apartment	With Parents	Other	Row Totals
   Males	72	84	49	45	250
   Females	91	86	88	35	300
   Col Totals	163	170	137	80	550

2. Create the table of expected values using the same formula for each value as we did with the test for independence in the previous section.
   $$E = \frac{(\text{row total})(\text{col total})}{\text{total num surveyed}}$$

   74.091	77.273	62.273	36.364
   88.909	92.727	74.727	43.636

3. Create the $O-E$ table, subtracting the values in the second table from the values in the first table.
   

   -2.091	6.727	-13.273	8.636
   2.091	-6.727	13.273	-8.636

4. Create the residuals $\frac{(O-E)^2}{E}$ table by squaring each value in the previous $O-E$ table (step 3) and dividing each value by the values in the expected value $E$ table (step 2).
   

   0.059	0.586	2.829	2.051
   0.049	0.488	2.358	1.709

Calculate the test statistic:

Add the values together in the $\frac{(O-E)^2}{E}$ table to get the test statistic.

χ² = 10.129

Probability statement:

In the same way we did for a Test for Independence, you can use the Chi Square Distribution table to get a range for the p-value, or use the Google Sheets function CHISQ.DIST.RT, along with the test statistic χ² = 10.129 and the degrees of freedom df = 3, to find the exact value.

=CHISQ.DIST.RT(10.129,3)

p-value = P(χ² >10.129) = 0.0175

Compare α and the p-value: Since no α is given, assume α = 0.05. p-value = 0.0175. α > p-value.

Make a decision: Since α > p-value, reject H₀. This means that the distributions are not the same.

Conclusion: At a 5% level of significance, from the data, there is sufficient evidence to conclude that the distributions of living arrangements for male and female college students are not the same.

Notice that the conclusion is only that the distributions are not the same. We cannot use the test for homogeneity to draw any conclusions about how they differ.

Just like with a Test for Independence, you can find a critical value instead of a p-value. Then you compare the critical value to the test statistic to decide whether or not to reject H₀.

 

Try It 10.8

Do families and singles have the same distribution of cars? Use a level of significance of 0.05. Suppose that 100 randomly selected families and 200 randomly selected singles were asked what type of car they drove: sport, sedan, hatchback, truck, van/SUV. The results are shown in Table 10.20. Do families and singles have the same distribution of cars? Test at a level of significance of 0.05.

	Sport	Sedan	Hatchback	Truck	Van/SUV
Family	5	15	35	17	28
Single	45	65	37	46	7

Table 10.20

Example 10.9

Both before and after a recent earthquake, surveys were conducted asking voters which of the three candidates they planned on voting for in the upcoming city council election. Has there been a change since the earthquake? Use a level of significance of 0.05. Table 10.21 shows the results of the survey. Has there been a change in the distribution of voter preferences since the earthquake?

	Perez	Chung	Stevens
Before	167	128	135
After	214	197	225

Table 10.21

Solution 10.9

H₀: The distribution of voter preferences was the same before and after the earthquake.

H₁: The distribution of voter preferences was not the same before and after the earthquake.

Degrees of Freedom (df):
df = number of columns – 1 = 3 – 1 = 2

Distribution for the test:

$\chi_2^2$

Create the supporting tables

1. Expand the given observed table to include the row totals and column totals.

   	Perez	Chung	Stevens	Row Totals
   Before	167	128	135	430
   After	214	197	225	636
   Col Totals	381	325	360	1066

2. Create the table of expected values using the same formula for each value as we did with the test for independence in the previous section.
   $$E = \frac{(\text{row total})(\text{col total})}{\text{total num surveyed}}$$

   153.687	131.098	145.216
   227.313	193.902	214.784

3. Create the $O-E$ table, subtracting the values in the second table from the values in the first table.
   

   13.313	-3.098	-10.216
   -13.313	3.098	10.216

4. Create the residuals $\frac{(O-E)^2}{E}$ table by squaring each value in the previous $O-E$ table (step 3) and dividing each value by the values in the expected value $E$ table (step 2).
   

   1.153	0.073	0.719
   0.78	0.049	0.486

Calculate the test statistic:

χ² = 3.26

Critical Value:

Look in the α = 0.05 column and the df = 2 row.

The critical value is 5.991

Compare the test statistic and the critical value:

3.26 < 5.991

Make a decision: Since the test statistic is less than the critical value, do not reject H_o.

Conclusion: At a 5% level of significance, from the data, there is insufficient evidence to conclude that the distribution of voter preferences was not the same before and after the earthquake.

Try It 10.9

Ivy League schools receive many applications, but only some can be accepted. At the schools listed in Table 10.22, two types of applications are accepted: regular and early decision.

Application Type Accepted	Brown	Columbia	Cornell	Dartmouth	Penn	Yale
Regular	2,115	1,792	5,306	1,734	2,685	1,245
Early Decision	577	627	1,228	444	1,195	761

Table 10.22

We want to know if the number of regular applications accepted follows the same distribution as the number of early applications accepted. State the null and alternative hypotheses, the degrees of freedom and the test statistic, sketch the graph of the p-value, and draw a conclusion about the test of homogeneity.