Question

ccording to a survey conducted by the Association for Dressings and Sauces (Links to an external site.), 30% of American adults eat salad once each week. A nutritionist suspects that this percentage is not accurate. She conducts a survey of 121 American adults and finds that 30 of them eat salad once per week. Use a 0.1 significance level to test the claim that the proportion of American adults who eat salad once per week is different from 30%.

The test statistic is:                             [ Select ]                       ["-4.17", "-1.69", "-4.22", "-1.25", "-2.88", "-3.45"]      

The p-value is:                            [ Select ]                       ["0.2112", "0.4224", "0.2995", "0.1056", "0.3321"]      

Based on this we:                            [ Select ]                       ["Fail to reject the null hypothesis", "Reject the null hypothesis"]      

Conclusion There                              [ Select ]                       ["does", "does not"]         appear to be enough evidence to support the claim that the proportion of American adults who eat salad once per week is different from 30%.

Answer 1

The answers are:

• Z = -1.25
• p = 0.2112
• Fail to reject the null hypothesis
• There does not appear to be enough evidence to support the claim that the proportion of American adults who eat salad once per week is different from 30%.

The test:

One-Proportion Z test

The following information is provided: The sample size is N = 121, the number of favorable cases is X = 30 and the sample proportion is pˉ​=X/N​=30/121​=0.2479, and the significance level is α=0.1 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: p =0.3 Ha: p ≠0.3 This corresponds to a Two-tailed test, for which a z-test for one population proportion needs to be used. (2a) Critical Value Based on the information provided, the significance level is α=0.1, therefore the critical value for this Two-tailed test is Zc​=1.6449. This can be found by either using excel or the Z distribution table. (2b) Rejection Region The rejection region for this Two-tailed test is |Z|>1.6449 i.e. Z>1.6449 or Z<-1.6449 (3) Test Statistics The z-statistic is computed as follows: (4) The p-value The p-value is the probability of obtaining sample results as extreme or more extreme than the sample results obtained, under the assumption that the null hypothesis is true. In this case, the p-value is p =P(|Z|>1.2498)=0.2114 (5) The Decision about the null hypothesis (a) Using traditional method Since it is observed that |Z|=1.2498 < Zc​=1.6449, it is then concluded that the null hypothesis is Not rejected. (b) Using p-value method Using the P-value approach: The p-value is p=0.2114, and since p=0.2114>0.1, it is concluded that the null hypothesis is Not rejected. (6) Conclusion It is concluded that the null hypothesis Ho is Not rejected. Therefore, there is Not enough evidence to claim that the population proportion p is different than 0.3, at the 0.1 significance level.