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: $S R S, n< 0.05 N, n p_{0}\left(1-p_{0}\right)=$
Test statistic. Round to 4 decimals:
P-value. Round to 4 decimals:
Conclusion: There is evidence at the $\alpha=0.05$ level of significance to conclude that the percentage of adult Americans that salad at least once a week is ("equal to" or "higher than" or "less than" or "not equal to") $85 \%$.

Step 1 ：State the null and alternative hypothesis: The null hypothesis is that the percentage of American adults who eat salad at least once a week is 85%. The alternative hypothesis, based on the nutritionist's suspicion, is that the percentage is higher than 85%.

Step 2 ：Calculate the test statistic: This is done by subtracting the hypothesized proportion (0.85) from the sample proportion (171/200), and dividing the result by the standard error of the proportion. The standard error is calculated as the square root of (p(1-p)/n), where p is the hypothesized proportion and n is the sample size.

Step 3 ：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. Since this is a one-tailed test (we are only interested in whether the true proportion is greater than 0.85), we find the p-value by looking up the test statistic in a Z-table and subtracting the result from 1.

Step 4 ：Compare the p-value to the level of significance (0.05): If the p-value is less than 0.05, we reject the null hypothesis and conclude that the percentage of American adults who eat salad at least once a week is higher than 85%. If the p-value is greater than 0.05, we do not reject the null hypothesis and conclude that the percentage is not significantly different from 85%.

Step 5 ：Final Answer: The null hypothesis is that the percentage of American adults who eat salad at least once a week is 85%. The alternative hypothesis is that the percentage is higher than 85%. The test statistic is approximately 0.198 and the p-value is approximately 0.4215. Since the p-value is greater than the level of significance (0.05), we do not reject the null hypothesis. Therefore, there is not enough evidence at the 0.05 level of significance to conclude that the percentage of adult Americans that eat salad at least once a week is higher than 85%. Thus, the conclusion is that the percentage is \(\boxed{\text{'equal to' 85 \%}}\).