Compute the test statistic used to test the null hypothesis that p 1 = p 2
-$$\begin{array} { l l }
\mathrm { n } _ { 1 } = 719 & \mathrm { n } _ { 2 } = 3852 \
\mathrm { x } _ { 1 } = 101 & \mathrm { x } _ { 2 } = 612
\end{array}$$
A)0.125
B)0.395
C)0.312
D)0.156

Question

Compute the test statistic used to test the null hypothesis that p 1 = p 2 
-$$\begin{array} { l l } 
\mathrm { n } _ { 1 } = 719 & \mathrm { n } _ { 2 } = 3852 \
\mathrm { x } _ { 1 } = 101 & \mathrm { x } _ { 2 } = 612
\end{array}$$
A)0.125
B)0.395
C)0.312
D)0.156

Quizplus · Accepted Answer

The test statistic for comparing two proportions is calculated using the formula: $$z = \frac{\hat{p}_1 - \hat{p}_2}{\sqrt{\hat{p}(1-\hat{p})(\frac{1}{n_1} + \frac{1}{n_2})}}$$ where $\hat{p}_1 = \frac{x_1}{n_1}$, $\hat{p}_2 = \frac{x_2}{n_2}$, and $\hat{p} = \frac{x_1 + x_2}{n_1 + n_2}$. Plugging in the given values: $\hat{p}_1 = \frac{101}{719}$, $\hat{p}_2 = \frac{612}{3852}$, and $\hat{p} = \frac{101 + 612}{719 + 3852}$, and performing the calculations will yield a test statistic of approximately 0.156.

Compute the Test Statistic Used to Test the Null Hypothesis