Chapter

1. Chapter 10 Hypothesis Tests Regarding a Parameter

2. Section 10.3 Hypothesis Tests for a Population Mean- Population Standard Deviation Unknown

3. Objectives • Test the hypotheses about a population mean with  unknown

4. 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  with s, follows Student’s t-distribution with n-1 degrees of freedom.

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

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

7. Properties of the t-Distribution • 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.

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

9. Properties of the t-Distribution • 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  introduces more variability to the t-statistic.

10. Properties of the t-Distribution 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  by the Law of Large Numbers.

11. Objective 1 • Test hypotheses about a population mean with  unknown

12. Testing Hypotheses Regarding a Population Mean with σ Unknown To test hypotheses regarding the population mean with  unknown, 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).

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

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

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

16. Classical Approach Step 4:Use Table VI to determine the critical value using n-1 degrees of freedom.

17. Classical Approach Two-Tailed

18. Classical Approach Left-Tailed

19. Classical Approach Right-Tailed

20. Classical Approach Step 5:Compare the critical value with the test statistic:

21. P-Value Approach Step 4:Use Table VI to estimate the P-value using n-1 degrees of freedom.

22. P-Value Approach Two-Tailed

23. P-Value Approach Left-Tailed

24. P-Value Approach Right-Tailed

25. P-Value Approach Step 5:If the P-value < , reject the null hypothesis.

26. Step 6:State the conclusion.

27. 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 =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?

28. 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.

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

30. Solution: Classical Approach Step 4: Since this is a two-tailed test, we determine the critical values at the =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.

31. 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.

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

33. 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 =0.05 level of significance, is there evidence to conclude that state quarters have a weight different than 5.67 grams?

34. 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.

35. Assumption of normality appears reasonable.

36. No outliers.

37. Solution Step 1:H0: =5.67 versus H1: ≠5.67 Step 2: The level of significance is =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

38. Solution: Classical Approach Step 4: Since this is a two-tailed test, we determine the critical values at the =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.

39. 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.

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

41. Summary • Which test to use? • Provided that the population from which the sample is drawn is normal or that the sample size is large, • if  is known, use the Z-test procedures from Section 10.2 • if  is unknown, use the t-test procedures from this Section.