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.
H0:_=_
H1:_> _
Answer
\(\boxed{\text{Final Answer: We fail to reject the null hypothesis at a 0.05 level of significance. Therefore, we do not have enough evidence to support the claim that the proportion of American adults who eat salad at least once a week is greater than 85%. The test statistic is approximately 0.20, which is less than the critical value of approximately 1.64. Therefore, the result is not statistically significant at the 0.05 level.}}\)
Step 1 ：State the hypotheses. The null hypothesis H0 is that the proportion of American adults who eat salad at least once a week is 85% or 0.85. The alternative hypothesis H1 is that the proportion is greater than 85%.
Step 2 ：Identify a sample statistic. Since we are trying to estimate a population proportion, we choose the sample proportion (171 out of 200) as the sample statistic.
Step 3 ：Select the significance level, α. In this analysis, the significance level is defined as α = 0.05.
Step 4 ：Standardize the sample statistic. Subtract the population proportion from the sample proportion and divide the result by the standard error.
Step 5 ：Find the critical value. The critical value for a one-tailed test is found by looking up the significance level (0.05) in a standard normal distribution table, giving a critical value of 1.6448536269514722.
Step 6 ：Compare the test statistic to the critical value. The calculated z-score of 0.19802950859533505 is less than the critical z-score of 1.6448536269514722.
Step 7 ：Since the test statistic does not exceed the critical value, we fail to reject the null hypothesis. We do not have enough evidence to support the claim that the proportion of American adults who eat salad at least once a week is greater than 85%.
Step 8 ：\(\boxed{\text{Final Answer: We fail to reject the null hypothesis at a 0.05 level of significance. Therefore, we do not have enough evidence to support the claim that the proportion of American adults who eat salad at least once a week is greater than 85%. The test statistic is approximately 0.20, which is less than the critical value of approximately 1.64. Therefore, the result is not statistically significant at the 0.05 level.}}\)
