Let $: p _ { 1 } - p _ { 2 }$ denote two population proportions, and let $\hat { p } _ { 1 } = .39 , \text { and } \hat { p } _ { 2 } = .49$ be the sample proportions of samples of sizes 150 and 200, respectively. Then a large sample confidence interval for $: p _ { 1 } - p _ { 2 }$ with a confidence level of approximately 99% is determined by
A) $- .10 \pm 2.58 ( .0532 )$
B) $- .10 \pm 2.58 ( .0028 )$
C) $- .10 \pm 1.96 ( .0051 )$
D) $- .10 \pm 1.96 ( .0711 )$
E) $- .10 \pm 1.65 ( .0501 )$

Question

Quizplus · Accepted Answer

The correct confidence interval is calculated using the formula for the difference between two proportions, $\hat{p}_1 - \hat{p}_2 \pm Z * \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}}$, where $Z$ is the Z-score corresponding to the desired confidence level. For a 99% confidence level, $Z = 2.58$. Substituting the given values, the standard error is calculated as $\sqrt{\frac{0.39(1-0.39)}{150} + \frac{0.49(1-0.49)}{200}} = 0.0532$, making the correct choice A.

Let :p1−p2: p _ { 1 } - p _ { 2 }:p1​−p2​

Let $: p _ { 1 } - p _ { 2 }$