Chapter 7. Hypothesis Testing with One Sample

Chapter 7: Hypothesis Testing with One Sample 20324. H 0 : p = 0.103, which makes q = 0.897, n = 800, and p ˆ = 0.12. The test statistic is thenz = p ˆ − p 0.12 − 0.103= = 0.017pq 0.103⋅ 0.897 0.0107 = 1.582n 800In Exercises 25-32, use the given information to find the P-value. (Hint: See Figure 7-6.)25. It is a right tailed test, so the P-value is the area to the right of the test statistic, z = 0.55. Using the methods of5-2, the P-value is 1-0.7088 = 0.2912.26. It is a left tailed test, so the P-value is the area to the left of the test statistic,z = -1.72. Using the methods of 5-2, the P-value is 0.0427.27. It is a two-tailed test, and the test statistic z = 1.95, is to the right of center so the p-value is the twice the areato the right of the test statistic. Using the methods of 5-2, the P-value is 2×(1–0.9744) = 0.0512.28. It is a two-tailed test, and the test statistic z = -1.63, is to the left of center so the p-value is the twice the area tothe left of the test statistic. Using the methods of 5-2, the P-value is 2×(0.0516) = 0.1032.29. Since H 1 : p > 0.29, it is a right tailed test, so the P-value is the area to the right of the test statistic, z = 1.97.Using the methods of 5-2, the P-value is 1-0.9756 = 0.0244.30. Since H 1 : p ≠ 0.30, it is a two-tailed test, and the test statistic z = 2.44, is to the right of center so the p-value isthe twice the area to the right of the test statistic. Using the methods of 5-2, the P-value is 2×(1–0.9927) =0.0146.31. Since H 1 : p ≠ 0.31, it is a two-tailed test, and the test statistic z = 0.77, is to the right of center so the p-value isthe twice the area to the right of the test statistic. Using the methods of 5-2, the P-value is 2×(1–0.7794) =0.4412.32. Since H 1 : p < 0.32, it is a left tailed test, so the p-value is the area to the left of the test statistic, z = -1.90.Using the methods of 5-2, the P-value is 0.0287.In Exercises 33-36, state the final conclusion in simple, non-technical terms. Be sure to address the originalclaim. (Hint: See Figure 7-7)33. The original claim did not include equality, H 0 was rejected, so the conclusion is:The sample data support the claim that the proportion of men who are married is greater than 0.5.34. The original claim did not include equality, H 0 was rejected, so the conclusion is:The sample data support the claim that the proportion of college graduates who smoke is less than 0.2735. The original claim did not include equality, H 0 was not rejected, so the conclusion is:There is no sufficient sample evidence to support the claim that the proportion of fatal commercial aviationcrashes is different from 0.038.36. The original claim did include equality, H 0 was rejected, so the conclusion is:There is sufficient evidence to warrant the rejection of the claim that the proportion of M&Ms that are blue isequal to 0.10.In exercises 37-40, identify the type I error and the type II error that correspond to the given hypothesis.37. Type I: Conclude there is sufficient evidence to support that the proportion of married women is greater than0.5 when in actuality p = 0.5Type II: Fail to reject that the proportion of married women is 0.5 when in actuality p > 0.5.