According to a survey conducted by the Association for Dressings and Sauces, 80% of American adults eat salad once a week. A nutritionist suspects that this percentage is not accurate. She conducts a survey of 357 American adults and finds that 271 of them eat salad once a week. Use a 0.01 significance level to test the claim that the proportion of American adults who eat salad once a week is different from 80%. Hint: When you calculate p^, round to at least 4 decimals Claim: Select an answer p < 0.8 p 0.8 u = 0.8 p 0.8 p = 0.8 u < 0.8 p 0.8 u 0.8 u > 0.8 p > 0.8 u 0.8 u 0.8 which corresponds to Select an answer H0: u 0.8 H0: p 0.8 H1: p < 0.8 H1: p 0.8 H0: p 0.8 H1: p > 0.8 H0: p = 0.8 Opposite: Select an answer u < 0.8 p 0.8 p > 0.8 p 0.8 u 0.8 u = 0.8 u 0.8 u 0.8 p = 0.8 p 0.8 p < 0.8 u > 0.8 which corresponds to Select an answer H1: p 0.8 H0: p 0.8 H0: p = 0.8 H1: p < 0.8 H0: p 0.8 H1: p > 0.8 H0: u 0.8 The test is: Select an answer right-tailed two-tailed left-tailed The test statistic is: z=Select an answer -1.46 -1.62 -1.76 -1.63 -1.93 The Critical Values are: z22= Select an answer 1.25 1.6 2.58 1.88 1.41 Based on this we: Select an answer Fail to reject the null hypothesis Reject the null hypothesis Conclusion: There Select an answer does not does appear to be enough evidence to support the claim that the proportion of American adults who eat salad once a week is different from 80%.