Question

The Gallup Poll asked 1015 U.S. adults whether they believed that people should pay sales tax on items purchased over the internet. Of these, 437 said they supported such a tax. Does the survey provide convincing evidence that less than 45% of U.S. adults favor an internet sales tax? Use the 0.05 level of significance.

Prompt 1: The null hypothesis Ho

Prompt 1 Answer

Prompt 2: The alternative hypothesis Ha

Prompt 2 Answer

Prompt 3: The test statistic

Prompt 3: Answer

Prompt 4: p-value

Prompt 4: Answer

Prompt 5: Reject or fail to reject the null hypothesis

Prompt 5: Answer

Prompt 6: Is the claim supported at the 0.05 level of significance

prompt 6: answer

Answer 1

Solution-

Prompt 1: The null hypothesis Ho

Ho : P = 0.45

Prompt 2: The alternative hypothesis Ha

H1 : P < 0.45

Prompt 3: The test statistic

Z = -2.202

Prompt 4: p-value

P = 0.0138

Prompt 5: Reject or fail to reject the null hypothesis

Reject Ho

Prompt 6: Is the claim supported at the 0.05 level of significance

Yes, there is enough evidence to support the claim

◆ test calculation and results-

The following information is provided: The sample size is N = 1050, the number of favorable cases is X = 437, and the sample proportion is p = X 0.4162, and the significance level is a = 0.05 437 1050 — N (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho:p= 0.45 Ha:p < 0.45 This corresponds to a left-tailed test, for which a z-test for one population proportion needs to be used. (2) Rejection Region Based on the information provided, the significance level is a = -0.05, and the critical value for a left- tailed test is ze = -1.64 The rejection region for this left-tailed test is R = {z: z < -1.64} (3) Test Statistics The z-statistic is computed as follows: 2 = p - po V po(1 – po) /n 0.4162 – 0.45 /0.45(1 – 0.45)/1050 -2.202 (4) Decision about the null hypothesis Since it is observed that z = -2.202 < ze = -1.64, it is then concluded that the null hypothesis is rejected. 0.0138 < 0.05, it is Using the P-value approach: The p-value is p = 0.0138, and since p = concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population proportion p is less than po, at the a = 0.05 significance level. Confidence Interval The 95% confidence interval for p is: 0.386 < p < 0.446.

Answer 2

p is 0.386 < p<0.446