Question

In the discrimination case Connecticut v. Teal, the following data were given concerning a CT state agency's record of employees rejected or selected for promotion.

	Selected	Rejected
Blacks	26	22
Whites	206	53

In discrimination cases, sometimes the blacks and whites described in such a table are viewed as samples from theoretical populations that might result if large numbers of blacks and whites were considered for promotion by the agency. Test the claim that the population of whites selected for promotion is larger than that of blacks at 2.5% level of significance.

Answer 1

Solution :

The null and alternative hypotheses are as follows :

i.e. The population proportion of whites selected for promotion is equal to that of blacks.

i.e. The population proportion of whites selected for promotion is greater than that of blacks.

To test the hypothesis we shall use z-test for testing the equality of two population proportions. The test statistic is given as follows :

Where, are the sample proportion of selected whites and blacks respectively, are the sample sizes for whites and blacks respectively.

Q = 1 - P

Sample proportions of whites who are selected for promotion is,

Sample proportions of blacks who are selected for promotion is,

Q = 1 - 0.7557 = 0.2443

The value of the test statistic is 3.7574.

Since, our test is right-tailed test, therefore we shall obtain right-tailed p-value for the test statistic. The p-value is given as follows :

p-value = P(Z > value of the test statistic)

p-value = P(Z > 3.7574)

p-value = 0.0001

The p-value is 0.0001.

Significance level = 2.5% = 0.025

(0.0001 < 0.025)

Since, p-value is less than the significance level of 2.5%, therefore we shall reject the null hypothesis (H​​​​​​0) at 2.5% significance level.

Conclusion : At 2.5% significance level, there is sufficient evidence to support the claim that the population of whites selected for promotion is larger than that of blacks.