Question

According to a survey conducted by the Association for Dressings and Sauces, 70% of American adults eat salad once a week. A nutritionist suspects that this percentage is not accurate. She conducts a survey of 234 American adults and finds that 150 of them eat salad once a week. Use a 0.1 significance level to test the claim that the proportion of American adults who eat salad once a week is different from 70%.
Hint: When you calculate ˆpp^, round to at least 4 decimals

Claim: Select an answer u < 0.7 u = 0.7 u ≤ 0.7 p < 0.7 p = 0.7 u > 0.7 u ≠ 0.7 p ≠ 0.7 u ≥ 0.7 p ≥ 0.7 p ≤ 0.7 p > 0.7  which corresponds to Select an answer H1: p < 0.7 H0: p ≠ 0.7 H0: u ≥ 0.7 H0: p = 0.7 H1: p ≠ 0.7 H1: p > 0.7 H0: p ≤ 0.7

Opposite: Select an answer u = 0.7 p ≤ 0.7 u ≠ 0.7 p = 0.7 p < 0.7 p ≥ 0.7 u ≤ 0.7 p ≠ 0.7 p > 0.7 u ≥ 0.7 u < 0.7 u > 0.7  which corresponds to Select an answer H1: p ≠ 0.7 H1: p > 0.7 H0: p ≤ 0.7 H0: p ≠ 0.7 H0: u ≥ 0.7 H1: p < 0.7 H0: p = 0.7


The test is: Select an answer right-tailed two-tailed left-tailed

The test statistic is: zz=Select an answer -1.74 -1.97 -1.84 -1.47 -2.34

The Critical Values are: zα2zα2= Select an answer ± 1.04 ± 1.64 ± 1.15 ± 1.28 ± 1.44

Based on this we: Select an answer Fail to reject the null hypothesis Reject the null hypothesis

Conclusion: There Select an answer does does not  appear to be enough evidence to support the claim that the proportion of American adults who eat salad once a week is different from 70%.

Answer 1

The statistical software output for this problem is :

Claim : p ≠ 0.7

H0: p = 0.7

H1: p ≠ 0.7

Test statistics = -1.97

z_α/2 = ± 1.64

Reject the null hypothesis