Test the hypothesis using the P-value approach. Be sure to verify the requirements of the test.
\[
\begin{array}{l}
H_{0}: p=0.76 \text { versus } H_{1}: p \neq 0.76 \\
n=500, x=370, \alpha=0.1
\end{array}
\]
A. Yes, because $n p_{0}\left(1-p_{0}\right)=91.2$
B. No, because $n p_{0}\left(1-p_{0}\right)=$

Now find $\hat{p}$.
$\hat{p}=0.74$ (Type an integer or a decimal. Do not round)
Find the test statistic $z_{0}$
$\mathrm{z}_{0}=\square($ Round to two decimal places as needed.)

Step 1 ：Given the null hypothesis \(H_{0}: p=0.76\) and the alternative hypothesis \(H_{1}: p \neq 0.76\). The sample size is \(n=500\), and the number of successes in the sample is \(x=370\). The significance level is \(\alpha=0.1\).

Step 2 ：First, check the requirements of the test. The sample size should be large enough such that both \(np\) and \(n(1-p)\) are greater than or equal to 10. Here, \(np_0(1-p_0) = 500 \times 0.76 \times (1-0.76) = 91.2\), which is greater than 10. So, the requirements of the test are met.

Step 3 ：Next, calculate the sample proportion \(\hat{p}\), which is the number of successes divided by the sample size. Here, \(\hat{p} = \frac{x}{n} = \frac{370}{500} = 0.74\).

Step 4 ：Then, calculate the test statistic \(z_{0}\), which is \((\hat{p} - p_{0}) / \sqrt{(p_{0} \times (1 - p_{0})) / n}\), where \(p_{0}\) is the hypothesized population proportion under the null hypothesis. Here, \(z_{0} = \frac{0.74 - 0.76}{\sqrt{(0.76 \times (1 - 0.76)) / 500}} \approx -1.05\).

Step 5 ：Final Answer: The requirements of the test are met, so the answer to the first part of the question is A. Yes, because \(n p_{0}(1-p_{0})=91.2\). The sample proportion \(\hat{p}\) is \(\boxed{0.74}\). The test statistic \(z_{0}\) is \(\boxed{-1.05}\).