significantly higher than the original level? Explain. Assume the $\bar{\alpha}=0.05$ level of significance.
Identify the null and alternative hypotheses for this test.
A.
\[
\begin{array}{l}
H_{0}: p \neq 0.52 \\
H_{1}: p=0.52
\end{array}
\]
E.
\[
\begin{array}{l}
H_{0}: p> 0.52 \\
H_{1}: p=0.52
\end{array}
\]
c. $H_{0}: p=0.52$
D. $H_{0}: p< 0.52$
$E \cdot H_{0}: p=0.52$
$H_{1}: p \neq 0.52$
$H_{1}: p=0.52$
$H_{1}: p< 0.52$
F.
\[
\begin{array}{l}
H_{0}: p=0.52 \\
H_{1}: p> 0.52
\end{array}
\]
Find the test statistic for this hypothesis test.
$z=1.24$ (Round to two decimal places as needed)
Determine the P-value for this hypothesis test.
P-value $=\square$ (Round to three decimal places as needed.)

Step 1 ：Identify the null and alternative hypotheses for this test. The null hypothesis is usually a statement of no effect or no difference and is the hypothesis that is tested. The alternative hypothesis is what we might believe if we find enough evidence against the null hypothesis. In this case, the question is asking if the proportion is significantly higher than a certain value (0.52). So, the null hypothesis should be that the proportion is equal to 0.52 and the alternative hypothesis should be that the proportion is greater than 0.52. The correct hypotheses are: \[H_{0}: p=0.52\] \[H_{1}: p>0.52\]

Step 2 ：Find the test statistic for this hypothesis test. The test statistic is given as z=1.24.

Step 3 ：Determine the P-value for this hypothesis test. The P-value is the probability of observing a test statistic as extreme as the one calculated, assuming the null hypothesis is true. We can calculate the P-value using the cumulative distribution function of the standard normal distribution. The P-value for this hypothesis test is approximately 0.107.

Step 4 ：Final Answer: The correct null and alternative hypotheses are: \[H_{0}: p=0.52\] \[H_{1}: p>0.52\] And the P-value is \(\boxed{0.107}\).