A random sample of 852 births included 432 boys. Use a 0.10 significance level to test the claim that $51.1 \%$ of newborn babies are boys. Do the results support the belief that $51.1 \%$ of newborn babies are boys?
C.
\[
\begin{array}{l}
H_{0}: p=0.511 \\
H_{1}: p> 0.511
\end{array}
\]
D.
\[
\begin{array}{l}
H_{0}: p \neq 0.511 \\
H_{1}: p=0.511
\end{array}
\]

Identify the test statistic for this hypothesis test.
The test statistic for this hypothesis test is -0.23 .
(Round to two decimal places as needed.)
Identify the P-value for this hypothesis test.
The P-value for this hypothesis test is $\square$.
(Round to three decimal places as needed.)
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Step 1 ：State the hypotheses. The first hypothesis, the null hypothesis, is that the proportion of newborn babies that are boys is 0.511. The second hypothesis, the alternative hypothesis, is that the proportion is not 0.511. In mathematical terms, these hypotheses are \(H_{0}: p=0.511\) and \(H_{1}: p \neq 0.511\).

Step 2 ：Calculate the sample proportion (\(p_{hat}\)) by dividing the number of boys (432) by the total number of births (852). This gives \(p_{hat} = \frac{432}{852} = 0.507\).

Step 3 ：Calculate the test statistic using the formula \(\frac{p_{hat} - p_{0}}{\sqrt{p_{0}(1-p_{0})/n}}\), where \(p_{0}\) is the hypothesized proportion, \(p_{hat}\) is the sample proportion, and \(n\) is the sample size. Substituting the given values gives a test statistic of -0.23.

Step 4 ：Calculate the P-value. This is the probability of observing a test statistic as extreme as the one calculated, assuming the null hypothesis is true. Since this is a two-tailed test, we need to calculate the two-tailed P-value. The P-value is 0.817.

Step 5 ：Since the P-value is greater than the significance level of 0.10, we do not reject the null hypothesis. This means that the data does not provide strong evidence against the null hypothesis, so we do not have enough evidence to say that the proportion of newborn babies that are boys is not 0.511.

Step 6 ：The final answer is: The test statistic for this hypothesis test is \(-0.23\) and the P-value for this hypothesis test is \(\boxed{0.817}\).