11.5 The chi-square distribution: test for homogeneity

$H_o$ : The distribution of living conditions for male college students is the same as the distribution of living conditions for female college students.

$H_a$ : The distribution of living conditions for male college students is not the same as the distribution of living conditions for female college students.

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

Distribution for the test: ${\chi }_3^2$

Calculate the test statistic: ${\chi }^2=10.1287$ (calculator or computer)

Probability statement: $\text{p-value}=P({\chi }^2>10.1287)=0.0175$

TI-83+ and TI-84 calculator: Press the MATRX key and arrow over to EDIT . Press 1:[A] . Press 2 ENTER 4 ENTER . Enter the table values by row. Press ENTER after each. Press 2nd QUIT . Press STAT and arrow over to TESTS . Arrow down to C:χ2-TEST . Press ENTER . You should see Observed:[A] and Expected:[B] . Arrow down to Calculate . Press ENTER . The test statistic is 10.1287 and the $\text{p-value}=0.0175$ . Do the procedure a second time but arrow down to Draw instead of calculate .

Compare $\alpha$ and the $\text{p-value}$ : Since no $\alpha$ is given, assume $\alpha =0.05$ . $\text{p-value}=0.0175$ . $\alpha >\text{p-value}$ .

Make a decision: Since $\alpha >\text{p-value}$ , reject $H_o$ . This means that the distributions are not the same.

Conclusion: At a 5% level of significance, from the data, there is sufficient evidence toconclude that the distributions of living conditions 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 make any conclusions about how they differ.