According to a survey conducted by the Association for Dressings and SaucesOpens externally, 40% of American adults eat salad once each week. A nutritionist suspects that this percentage is not accurate. She conducts a survey of 545 American adults and finds that 202 of them eat salad once per week. Use a 0.01 significance level to test the claim that the proportion of American adults who eat salad once per week is smaller than 40%. Hint: When you calculate ˆpp^, round to at least 4 decimals
Claim : p ≥ 0.4 / p ≠ 0.4 / p = 0.4 / u ≠ 0.4 / u ≤ 0.4 / u < 0.4 / u > 0.4 / p ≤ 0.4 / u ≥ 0.4 / p > 0.4 / u = 0.4 which corresponds to H1: p ≠ 0.4 / H0: p = 0.4 / H0: p ≤ 0.4 / H1: p > 0.4 / H1 : p < 0.4 / H0 : p ≠ 0.4 / H0: u ≥ 0.4
Opposite : p ≥ 0.4 / u ≤ 0.4 / p < 0.4 / p ≤ 0.4 / u ≥ 0.4 / u ≠ 0.4 / u = 0.4 / u < 0.4 / p = 0.4 / p ≠ 0.4 / u > 0.4 / p > 0.4 which corresponds to H0: p ≤ 0.4 / H0 : u ≥ 0.4 / H0: p =0.4 / H0: p ≠ 0.4 / H1: p ≠ 0.4 / H1 : p < 0.4 / H1 p > 0.4
The test is: Select an answer : left-tailed / two-tailed right-tailed /right tailed
The test statistic is: z = 2.46 / 2.25 / 2.13 / 2.74 / 1.91
The Critical Values are: z a/2 : ± 1.64 / ± 1.04 / ± 1.15 / ± 1.44 / ± 1.28
Based on this we: Select an answer : Reject the null hypothesis / Fail to reject the null hypothesis
Conclusion: There Select an answer (does or does not)