Question

Researchers sent 5000 resumes in response to job ads that appeared in the Boston Globe and Chicago Tribune. The resumes were identical except that 2500 of them had "white sounding" first names, such as Brett and Emily, whereas the other 2500 had "black sounding" names such as Tamika and Rasheed. The resumes of the first type elicited 250 responses and the resumes of the second type only 167 responses (these numbers are very consistent with information that appeared in a Jan. 15. 2003, report by the Associated Press). Does this data strongly suggest that a resume with a "black" name is less likely to result in a response than is a resume with a "white" name?

Short answer

\({\rm{ Reject null hypothesis}}{\rm{. }}\)

Step-by-step solution

To obtain a large- sample test procedure for the hypothesis

A Large-Sample Test Procedure:

The two proportion z statistic - Consider test in which\({H_0}:{p_1} - {p_2} = 0\). The test statistic value for the large samples is

\(z = \frac{{{{\hat p}_1} - {{\hat p}_2}}}{{\sqrt {\hat p\hat q \cdot \frac{1}{m} + \frac{1}{n}} }}\)

where

\(\hat p = \frac{m}{{m + n}} \cdot {\hat p_1} + \frac{n}{{m + n}} \cdot {\hat p_2}\)

Depending on alternative hypothesis, the P value can be determined as the corresponding area under the standard normal curve. The test should be used when\(m \cdot {\hat p_1},m \cdot {\hat q_1},n \cdot {\hat p_1}\), and \(n \cdot {\hat q_1}\) are at least 10.

The hypotheses of interest are \({H_0}:{p_1} - {p_2} = 0\)versus\({H_a}:{p_1} - {p_2} > 0\). From the given values, the estimates are

\(\begin{array}{l}{{\hat p}_1} = \frac{{250}}{{2500}} = 0.1\\{{\hat p}_2} = \frac{{167}}{{2500}} = 0.0668\end{array}\)

Final proof

Also, the \(\hat p\)is

\(\begin{array}{l}\hat p = \frac{m}{{m + n}} \cdot {{\hat p}_1} + \frac{n}{{m + n}} \cdot {{\hat p}_2} = \frac{{250}}{{250 + 167}} \cdot \frac{{250}}{{2500}} + \frac{{167}}{{250 + 167}} \cdot \frac{{167}}{{2500}}\\ = 0.0834\end{array}\)

The z statistic value is

\(\begin{array}{l} = \frac{{{{\hat p}_1} - {{\hat p}_2}}}{{\sqrt {\hat p\hat q \cdot \frac{1}{m} + \frac{1}{n}} }} = \frac{{0.1 - 0.0668}}{{\sqrt {0.0834 \cdot (1 - 0.0834) \cdot (1/250 + 1/167)} }}\\ = 4.2.\end{array}\)

The P values is two times the area of the z curve to the right of the |z| value; thus

\(P = 2 \cdot P(Z > 4.2) = 2 \cdot (1 - P(Z \le 4.2)) = 2 \cdot 0 = 0 < \alpha \)

so

Reject null hypothesis

in any reasonable level. Based on this experiment, it is more likely for a white name to get a response than for a black name.

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