Assume that you plan to use a significance level of $\alpha = 0.05$ test the claim that p $\mathrm { p } _ { 1 } = \mathrm { p } _ { 2 }$ , Use the given sample sizes and
numbers of successes to find the pooled estimate $\overline { \mathrm { p } }$ . Round your answer to the nearest thousandth.
-$$\begin{array} { l l }
\mathrm { n } _ { 1 } = 255 & \mathrm { n } _ { 2 } = 270 \
\mathrm { x } _ { 1 } = 82 & \mathrm { x } _ { 2 } = 88
\end{array}$$
A) 0.292
B) 0.324
C) 0.227
D) 0.162

Question

Assume that you plan to use a significance level of $\alpha = 0.05$ test the claim that p $\mathrm { p } _ { 1 } = \mathrm { p } _ { 2 }$ , Use the given sample sizes and
numbers of successes to find the pooled estimate $\overline { \mathrm { p } }$ . Round your answer to the nearest thousandth.
-$$\begin{array} { l l } 
\mathrm { n } _ { 1 } = 255 & \mathrm { n } _ { 2 } = 270 \
\mathrm { x } _ { 1 } = 82 & \mathrm { x } _ { 2 } = 88
\end{array}$$
A) 0.292
B) 0.324
C) 0.227
D) 0.162

Quizplus · Accepted Answer

The pooled estimate of the proportion is calculated using the formula:

$$\overline { \mathrm { p } } = \frac { x _ { 1 } + x _ { 2 } } { n _ { 1 } + n _ { 2 } } = \frac {82 + 88} {255 + 270} = 0.324.$$

To test the claim that $\mathrm { p } _ { 1 } = \mathrm { p } _ { 2 }$ at a significance level of $\alpha = 0.05$, we would use a two-sample z-test for proportions, with the test statistic:

$$z = \frac{(\hat{p}_1 - \hat{p}_2) - 0}{\sqrt{\overline{p}(1-\overline{p})\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}$$

where $\hat{p}_1$ and $\hat{p}_2$ are the sample proportions, and $\overline{p}$ is the pooled estimate of the proportion. We have already calculated $\overline{p}=0.324$, and we can calculate:

$$\hat{p}_1 = \frac{x_1}{n_1} = \frac{82}{255} \approx 0.322$$
$$\hat{p}_2 = \frac{x_2}{n_2} = \frac{88}{270} \approx 0.326$$

Substituting these values into the formula for the test statistic, we get:

$$z = \frac{(0.322 - 0.326) - 0}{\sqrt{0.324(1-0.324)\left(\frac{1}{255}+\frac{1}{270}\right)}} \approx -0.507.$$

Using a standard normal distribution table, we can find the p-value for this test statistic to be approximately 0.612. Since this p-value is greater than our significance level of 0.05, we fail to reject the null hypothesis and do not have enough evidence to conclude that the proportions are different.

Assume That You Plan to Use a Significance Level Of \alpha