Based on their records, a hospital claims that the proportion, p, of full-term babies born in the community that weigh more than 7 pounds is $35 \%$. A pediatrician who works with several hospitals in the community would like to verify the hospital's claim. In a random sample of 160 babies born in the community, 39 weighed over 7 pounds. Is there enough evidence to reject the hospital's claim at the 0.10 level of significance?
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Perform a two-tailed test. Then complete the parts below.
Carry your intermediate computations to three or more decimal places. (If necessary, consult a list of formulas.)
(a) State the null hypothesis $H_{0}$ and the alternative hypothesis $H_{1}$.
\[
\begin{array}{l}
H_{0}: \\
H_{1}:
\end{array}
\]
(b) Determine the type of test statistic to use.
(Choose one)
(c) Find the value of the test statistic. (Round to threaior more decimal places.)
(d) Find the $p$-value. (Round to three or more decimal places.)
(e) Can we reject the claim that the proportion of full-term babies born in the community that weigh more than 7 pounds is $35 \%$ ?
Yes ONo

Step 1 ：State the null hypothesis $H_{0}$ and the alternative hypothesis $H_{1}$. The null hypothesis is that the proportion of full-term babies born in the community that weigh more than 7 pounds is 35%. The alternative hypothesis is that the proportion of full-term babies born in the community that weigh more than 7 pounds is not 35%. So, we have: $H_{0}: p = 0.35$ $H_{1}: p \neq 0.35$

Step 2 ：Determine the type of test statistic to use. In this case, we are testing a claim about a population proportion, so we use a z-test for proportions.

Step 3 ：Find the value of the test statistic. We use the formula for the z-test for proportions: $z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}$ where $\hat{p}$ is the sample proportion, $p_0$ is the hypothesized population proportion, and $n$ is the sample size. Substituting the given values, we find that the test statistic is approximately \(-2.818\).

Step 4 ：Find the p-value. The p-value is the probability of observing a test statistic as extreme as, or more extreme than, the observed test statistic, under the null hypothesis. Using a standard normal distribution table or a statistical software, we find that the p-value is approximately \(0.005\).

Step 5 ：Compare the p-value to the level of significance to decide whether to reject the null hypothesis. The level of significance is 0.10. Since the p-value is less than the level of significance, we reject the null hypothesis. Therefore, we can reject the claim that the proportion of full-term babies born in the community that weigh more than 7 pounds is 35%.