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%.
Get Answers For Free
Most questions answered within 1 hours.