According to a survey, $85 \%$ of American adults eat salad at least once a week. A nutritionist suspects that the percentage is higher than this. She conducts a survey of 200 American adults and finds that 171 of them eat salad at least once a week. Conduct the appropriate test that addresses the nutritionist's suspicion. Use an $\alpha=0.05$ level of significance. State the null and alternative hypothesis: Assumptions: $S R S, n<0.05 N, n p_{0}\left(1-p_{0}\right)=$

Step 1 ：State the null and alternative hypothesis: The null hypothesis is that the proportion of American adults who eat salad at least once a week is 0.85 (or 85%). The alternative hypothesis, which is what the nutritionist suspects, is that the proportion is greater than 0.85.

Step 2 ：We can use a one-sample z-test for proportions to test these hypotheses. The test statistic is calculated as follows: \[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \] where \(\hat{p}\) is the sample proportion, \(p_0\) is the proportion under the null hypothesis, and \(n\) is the sample size.

Step 3 ：The p-value is then calculated as the probability of observing a z-score as extreme as the one calculated, under the null hypothesis. If the p-value is less than the significance level \(\alpha\), we reject the null hypothesis.

Step 4 ：Given that \(\hat{p} = 0.855\), \(p_0 = 0.85\), and \(n = 200\), we calculate the z-score to be approximately 0.198.

Step 5 ：The p-value is approximately 0.422, which is greater than the significance level of 0.05, so we do not reject the null hypothesis.

Step 6 ：This means that the data does not provide strong evidence to support the nutritionist's suspicion that the proportion of American adults who eat salad at least once a week is greater than 85%.

Step 7 ：\(\boxed{\text{Final Answer: We do not reject the null hypothesis at the 0.05 level of significance. The data does not provide strong evidence to support the claim that the proportion of American adults who eat salad at least once a week is greater than 85%.}}\)