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%.