Question

calculating probabilities from the sampling distribution of $hat{p}_1 - hat{p}_2$
at university a, 46% of the undergraduate students are female. at university b, 52% of the undergraduate students are female. suppose a random sample of 100 students from both schools was selected. assume the sampling distribution of $hat{p}_a - hat{p}_b$ is approximately normal.
what is the probability that the proportion of females selected from university a is greater than the proportion of females selected from university b?
if needed, use the z-table to answer the question.
report your answer to 3 decimal places.

Explanation:

Step1: Define given parameters

Let $p_A = 0.46$, $n_A = 100$, $p_B = 0.52$, $n_B = 100$. We need $P(\hat{p}_A > \hat{p}_B) = P(\hat{p}_A - \hat{p}_B > 0)$.

Step2: Calculate mean of $\hat{p}_A - \hat{p}_B$

$\mu_{\hat{p}_A - \hat{p}_B} = p_A - p_B$
$\mu_{\hat{p}_A - \hat{p}_B} = 0.46 - 0.52 = -0.06$

Step3: Calculate standard error of $\hat{p}_A - \hat{p}_B$

$\sigma_{\hat{p}_A - \hat{p}_B} = \sqrt{\frac{p_A(1-p_A)}{n_A} + \frac{p_B(1-p_B)}{n_B}}$
$\sigma_{\hat{p}_A - \hat{p}_B} = \sqrt{\frac{0.46(1-0.46)}{100} + \frac{0.52(1-0.52)}{100}} = \sqrt{\frac{0.46\times0.54}{100} + \frac{0.52\times0.48}{100}}$
$\sigma_{\hat{p}_A - \hat{p}_B} = \sqrt{\frac{0.2484}{100} + \frac{0.2496}{100}} = \sqrt{0.002484 + 0.002496} = \sqrt{0.00498} \approx 0.07057$

Step4: Calculate z-score for 0

$z = \frac{0 - \mu_{\hat{p}_A - \hat{p}_B}}{\sigma_{\hat{p}_A - \hat{p}_B}}$
$z = \frac{0 - (-0.06)}{0.07057} = \frac{0.06}{0.07057} \approx 0.850$

Step5: Find corresponding probability

$P(Z > 0.85) = 1 - P(Z \leq 0.85)$. From z-table, $P(Z \leq 0.85) = 0.8023$, so $1 - 0.8023 = 0.1977$.

Answer:

0.198