Chapter

1. Chapter 10 Hypothesis Tests Regarding a Parameter

2. Section 10.3 Hypothesis Tests for a Population Mean

3. Objectives Test hypotheses about a mean Understand the difference between statistical significance and practical significance.

4. Objective 1 Test Hypotheses about a Mean

5. To test hypotheses regarding the population mean assuming the population standard deviation is unknown, we use the t-distribution rather than the Z-distribution. When we replace σ with s, follows Student’s t-distribution with n –1 degrees of freedom.

6. The t-distribution is different for different degrees of freedom. Properties of the t-Distribution

7. The t-distribution is different for different degrees of freedom. The t-distribution is centered at 0 and is symmetric about 0. Properties of the t-Distribution

8. The t-distribution is different for different degrees of freedom. The t-distribution is centered at 0 and is symmetric about 0. The area under the curve is 1. Because of the symmetry, the area under the curve to the right of 0 equals the area under the curve to the left of 0 equals 1/2. Properties of the t-Distribution

9. 4. As t increases (or decreases) without bound, the graph approaches, but never equals, 0. Properties of the t-Distribution

10. As t increases (or decreases) without bound, the graph approaches, but never equals, 0. The area in the tails of thet-distribution is a little greater than the area in the tails of the standard normal distribution because using s as an estimate of σ introduces more variability to the t-statistic. Properties of the t-Distribution

11. 6. As the sample size n increases, the density curve of t gets closer to the standard normal density curve. This result occurs because as the sample size increases, the values of s get closer to the values of σby the Law of Large Numbers. Properties of the t-Distribution

12. To test hypotheses regarding the population mean, we use the following steps, provided that: The sample is obtained using simple random sampling. The sample has no outliers, and the population from which the sample is drawn is normally distributed or the sample size is large (n ≥ 30). The sampled values are independent of each other. Testing Hypotheses Regarding aPopulation Mean

13. Step 1:Determine the null and alternative hypotheses. Again, the hypotheses can be structured in one of three ways:

14. Step 2:Select a level of significance, α, based on the seriousness of making a Type I error.

15. Classical Approach Step 3:Compute the test statistic which follows the Student’s t-distribution with n – 1 degrees of freedom.

16. Classical Approach Use Table VI to determine the critical value.

17. Classical Approach Two-Tailed

18. Classical Approach Left-Tailed

19. Classical Approach Right-Tailed

20. Classical Approach Step 4:Compare the critical value with the test statistic:

21. P-Value Approach By Hand Step 3:Compute the test statistic which follows the Student’s t-distribution with n – 1 degrees of freedom.

22. P-Value Approach Use Table VI to approximate the P-value.

23. P-Value Approach Two-Tailed

24. P-Value Approach Left-Tailed

25. P-Value Approach Right-Tailed

26. P-Value Approach Technology Step 3:Use a statistical spreadsheet or calculator with statistical capabilities to obtain the P-value. The directions for obtaining the P-value using the TI-83/84 Plus graphing calculator, MINITAB, Excel, and StatCrunch, are in the Technology Step-by-Step in the text.

27. P-Value Approach Step 4:If the P-value < α, reject the null hypothesis.

28. Step 5:State the conclusion.

29. The procedure is robust, which means that minor departures from normality will not adversely affect the results of the test. However, for small samples, if the data have outliers, the procedure should not be used.

30. Parallel Example 1: Testing a Hypothesis about a Population Mean, Large Sample Assume the resting metabolic rate (RMR) of healthy males in complete silence is 5710 kJ/day. Researchers measured the RMR of 45 healthy males who were listening to calm classical music and found their mean RMR to be 5708.07 with a standard deviation of 992.05. At the α= 0.05 level of significance, is there evidence to conclude that the mean RMR of males listening to calm classical music is different than 5710 kJ/day?

31. Solution We assume that the RMR of healthy males is 5710 kJ/day. This is a two-tailed test since we are interested in determining whether the RMR differs from 5710 kJ/day. Since the sample size is large, we follow the steps for testing hypotheses about a population mean for large samples.

32. Solution Step 1:H0: μ=5710 versus H1: μ ≠5710 Step 2: The level of significance is α =0.05. Step 3: The sample mean is = 5708.07 and the sample standard deviation is s=992.05. The test statistic is

33. Solution: Classical Approach Step 4: Since this is a two-tailed test, we determine the critical values at the α =0.05 level of significance with n –1 = 45 – 1 = 44 degrees of freedomto be approximately –t0.025 = –2.021 and t0.025 = 2.021. Step 5: Since the test statistic, t0 = –0.013, is between the critical values, we fail to reject the null hypothesis.

34. Solution: P-Value Approach Step 4: Since this is a two-tailed test, the P-value is the area under the t-distribution with n –1 = 45 – 1 = 44 degrees of freedomto the left of –t0.025= –0.013 and to the right of t0.025 = 0.013. That is, P-value = P(t < -0.013) + P(t > 0.013) = 2 P(t > 0.013). 0.50 < P-value. Step 5: Since the P-value is greater than the level of significance (0.05<0.5), we fail to reject the null hypothesis.

35. Solution Step 6: There is insufficient evidence at the α= 0.05 level of significance to conclude that the mean RMR of males listening to calm classical music differs from 5710 kJ/day.

36. Parallel Example 3: Testing a Hypothesis about a Population Mean, Small Sample According to the United States Mint, quarters weigh 5.67 grams. A researcher is interested in determining whether the “state” quarters have a weight that is different from 5.67 grams. He randomly selects 18 “state” quarters, weighs them and obtains the following data. 5.70 5.67 5.73 5.61 5.70 5.67 5.65 5.62 5.73 5.65 5.79 5.73 5.77 5.71 5.70 5.76 5.73 5.72 At the α =0.05 level of significance, is there evidence to conclude that state quarters have a weight different than 5.67 grams?

37. Solution We assume that the weight of the state quarters is 5.67 grams. This is a two-tailed test since we are interested in determining whether the weight differs from 5.67 grams. Since the sample size is small, we must verify that the data come from a population that is normally distributed with no outliers before proceeding to Steps 1-6.

38. Assumption of normality appears reasonable.

39. No outliers.

40. Solution Step 1:H0: μ =5.67 versus H1: μ ≠5.67 Step 2: The level of significance is α =0.05. Step 3: From the data, the sample mean is calculated to be 5.7022 and the sample standard deviation is s=0.0497. The test statistic is

41. Solution: Classical Approach Step 4: Since this is a two-tailed test, we determine the critical values at the α =0.05 level of significance with n – 1 = 18 – 1 = 17 degrees of freedomto be –t0.025 = –2.11 and t0.025 = 2.11. Step 5: Since the test statistic, t0 = 2.75, is greater than the critical value 2.11, we reject the null hypothesis.

42. Solution: P-Value Approach Step 4: Since this is a two-tailed test, the P-value is the area under the t-distribution with n – 1 = 18 – 1 = 17 degrees of freedomto the left of –t0.025 = –2.75 and to the right of t0.025 = 2.75. That is, P-value = P(t < -2.75) + P(t > 2.75) = 2 P(t > 2.75). 0.01 < P-value < 0.02. Step 5: Since the P-value is less than the level of significance (0.02<0.05), we reject the null hypothesis.

43. Solution Step 6: There is sufficient evidence at the α= 0.05 level of significance to conclude that the mean weight of the state quarters differs from 5.67 grams.

44. Objective 5 Distinguish between Statistical Significance and Practical Significance

45. When a large sample size is used in a hypothesis test, the results could be statistically significant even though the difference between the sample statistic and mean stated in the null hypothesis may have no practical significance.

46. Practical significance refers to the idea that, while small differences between the statistic and parameter stated in the null hypothesis are statistically significant, the difference may not be large enough to cause concern or be considered important.

47. Parallel Example 7: Statistical versus Practical Significance In 2003, the average age of a mother at the time of her first childbirth was 25.2. To determine if the average age has increased, a random sample of 1200 mothers is taken and is found to have a sample mean age of 25.5 with a standard deviation of 4.8, determine whether the mean age has increased using a significance level of α=0.05.

48. Solution Step 1: To determine whether the mean age has increased, this is a right-tailed test with H0: μ =25.2 versus H1: μ >25.2. Step 2: The level of significance is α =0.05. Step 3: Recall that the sample mean is 25.5. The test statistic is then

49. Solution Step 4: Since this is a right-tailed test, the critical value with α = 0.05 and 1200 – 1 = 1199 degrees of freedom is P-value = P(t > 2.17). From Table VI: .01 < P-value < .02 Step 5: Because the P-value is less than the level of significance, 0.05, we reject the null hypothesis.

50. Solution Step 6: There is sufficient evidence at the 0.05 significance level to conclude that the mean age of a mother at the time of her first childbirth is greater than 25.2. Although we found the difference in age to be significant, there is really no practical significance in the age difference (25.2 versus 25.5).