91影视

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

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

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.

.