Professors collected data consisting of eye color and gender of statistics students. Among 915 male students, 294 had blue eyes. Among 1062 female students, 349 had blue eyes. Use a 0.05 significance level to test the claim that the proportions of blue eyes are the same for males and females. Complete parts (a) through (c) below. Consider the first sample to be the sample of males and the second sample to be the sample of females.
a. Test the claim rising a hypothesis test.
What are the null and alternative hypotheses for the hypothesis test?
A. $H_{0}: p_{1} \leq p_{2}$
$H_{1}: p_{1} \neq p_{2}$
D. $H_{0}: P_{1} \neq p_{2}$
$\mathrm{H}_{1}: \mathrm{p}_{1}=\mathrm{p}_{2}$
B. $H_{0}: p_{1}=p_{2}$
$H_{1}: p_{1}> p_{2}$
E. $H_{0}: p_{1} \geq p_{2}$
$H_{1}: p_{1} \neq p_{2}$
c.
\[
\begin{array}{l}
H_{0}: p_{1}=p_{2} \\
H_{1}: p_{1} \neq p_{2}
\end{array}
\]
F. $H_{0}: p_{1}=p_{2}$
$H_{1}: p_{1}< p_{2}$
Identify the test statistic.
\[
z=
\]
(Round to two decimal places as needed.)

Step 1 ：Given data: number of male students (n1) = 915, number of male students with blue eyes (x1) = 294, number of female students (n2) = 1062, number of female students with blue eyes (x2) = 349.

Step 2 ：Calculate the sample proportions of blue eyes in males (\(\hat{p}_1\)) and females (\(\hat{p}_2\)) using the formulas \(\hat{p}_1 = \frac{x1}{n1}\) and \(\hat{p}_2 = \frac{x2}{n2}\).

Step 3 ：\(\hat{p}_1 = \frac{294}{915} = 0.32131147540983607\) and \(\hat{p}_2 = \frac{349}{1062} = 0.3286252354048964\).

Step 4 ：Calculate the combined sample proportion of blue eyes (\(\hat{p}\)) using the formula \(\hat{p} = \frac{x1 + x2}{n1 + n2}\).

Step 5 ：\(\hat{p} = \frac{294 + 349}{915 + 1062} = 0.32524026302478504\).

Step 6 ：Calculate the test statistic (z-score) using the formula \(z = \frac{(\hat{p}_1 - \hat{p}_2) - 0}{\sqrt{\hat{p}(1-\hat{p})(\frac{1}{n1} + \frac{1}{n2})}}\).

Step 7 ：Substitute the calculated values into the formula: \(z = \frac{(0.32131147540983607 - 0.3286252354048964) - 0}{\sqrt{0.32524026302478504(1-0.32524026302478504)(\frac{1}{915} + \frac{1}{1062})}} = -0.3461253783175717\).

Step 8 ：The test statistic (z-score) is approximately -0.35. This value represents how many standard deviations the observed difference in proportions is from the expected difference under the null hypothesis (which is 0 in this case). A negative z-score indicates that the observed difference is less than the expected difference.

Step 9 ：Final Answer: The test statistic is \(\boxed{-0.35}\).