1. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. Chapter Hypothesis Tests Regarding a Parameter 10

2. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. Section Hypothesis Tests for a Population Proportion 10.2

3. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-3 Objectives 1.Explain the logic of hypothesis testing 2.Test the hypotheses about a population proportion 3.Test hypotheses about a population proportion using the binomial probability distribution.

4. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-4 Objective 1 Explain the Logic of Hypothesis Testing

5. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-5 A researcher obtains a random sample of 1000 people and finds that 534 are in favor of the banning cell phone use while driving, so = 534/1000. Does this suggest that more than 50% of people favor the policy? Or is it possible that the true proportion of registered voters who favor the policy is some proportion less than 0.5 and we just happened to survey a majority in favor of the policy? In other words, would it be unusual to obtain a sample proportion of 0.534 or higher from a population whose proportion is 0.5? What is convincing, or statistically significant, evidence?

6. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-6 When observed results are unlikely under the assumption that the null hypothesis is true, we say the result is statistically significant. When results are found to be statistically significant, we reject the null hypothesis.

7. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-7 To determine if a sample proportion of 0.534 is statistically significant, we build a probability model. Since np(1 – p) = 100(0.5)(1 – 0.5) = 250 ≥ 10 and the sample size (n = 1000) is sufficiently smaller than the population size, we can use the normal model to describe the variability in. The mean of the distribution of is and the standard deviation is

8. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-8 Sampling distribution of the sample proportion

9. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-9 We may consider the sample evidence to be statistically significant (or sufficient) if the sample proportion is too many standard deviations, say 2, above the assumed population proportion of 0.5. The Logic of the Classical Approach

10. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-10 Recall that our simple random sample yielded a sample proportion of 0.534, so standard deviations above the hypothesized proportion of 0.5. Therefore, using our criterion, we would reject the null hypothesis.

11. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-11 Why does it make sense to reject the null hypothesis if the sample proportion is more than 2 standard deviations away from the hypothesized proportion? The area under the standard normal curve to the right of z = 2 is 0.0228.

12. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-12 If the null hypothesis were true (population proportion is 0.5), then 1 – 0.0228 = 0.9772 = 97.72% of all sample proportions will be less than.5 + 2(0.016) = 0.532 and only 2.28% of the sample proportions will be more than 0.532.

13. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-13 If the sample proportion is too many standard deviations from the proportion stated in the null hypothesis, we reject the null hypothesis. Hypothesis Testing Using the Classical Approach

14. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-14 Objective 2 Test hypotheses about a population proportion.

15. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-15 Recall: The best point estimate of p, the proportion of the population with a certain characteristic, is given by where x is the number of individuals in the sample with the specified characteristic and n is the sample size.

16. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-16 Recall: The sampling distribution of is approximately normal, with mean and standard deviation provided that the following requirements are satisfied: 1.The sample is a simple random sample. 2. np(1-p) ≥ 10. 3.The sampled values are independent of each other.

17. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-17 Testing Hypotheses Regarding a Population Proportion, p To test hypotheses regarding the population proportion, we can use the steps that follow, provided that: The sample is obtained by simple random sampling. np 0 (1 – p 0 ) ≥ 10. The sampled values are independent of each other.

18. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-18 Step 1: Determine the null and alternative hypotheses. The hypotheses can be structured in one of three ways:

19. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-19 Step 2: Select a level of significance, α, based on the seriousness of making a Type I error.

20. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-20 Step 3: Compute the test statistic Note: We use p 0 in computing the standard error rather than. This is because, when we test a hypothesis, the null hypothesis is always assumed true. Classical Approach

21. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-21 Classical Approach Two-Tailed (critical value) Use Table V to determine the critical value.

22. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-22 Classical Approach Left-Tailed (critical value)

23. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-23 Classical Approach Right-Tailed (critical value)

24. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-24 Step 4: Compare the critical value with the test statistic: Classical Approach

25. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-25 Step 5: State the conclusion.

26. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-26 Parallel Example 1: Testing a Hypothesis about a Population Proportion: Large Sample Size In 1997, 46% of Americans said they did not trust the media “when it comes to reporting the news fully, accurately and fairly”. In a 2007 poll of 1010 adults nationwide, 525 stated they did not trust the media. At the α = 0.05 level of significance, is there evidence to support the claim that the percentage of Americans that do not trust the media to report fully and accurately has increased since 1997? Source: Gallup Poll

27. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-27 Solution We want to know if p > 0.46. First, we must verify the requirements to perform the hypothesis test: 1.This is a simple random sample. 2.np 0 (1 – p 0 ) = 1010(0.46)(1 – 0.46) = 250.8 > 10 3.Since the sample size is less than 5% of the population size, the assumption of independence is met.

28. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-28 Solution Step 1: H 0 : p = 0.46 versus H 1 : p > 0.46 Step 2: The level of significance is α = 0.05. Step 3: The sample proportion is. The test statistic is then

29. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-29 Solution: Classical Approach Step 4: Since this is a right-tailed test, we determine the critical value at the α = 0.05 level of significance to be z 0.05 = 1.645. Step 5: Since the test statistic, z 0 = 3.83, is greater than the critical value 1.645, we reject the null hypothesis.

30. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-30 Solution Step 6: There is sufficient evidence at the α = 0.05 level of significance to conclude that the percentage of Americans that do not trust the media to report fully and accurately has increased since 1997.

31. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-31 Objective 3 Test hypotheses about a population proportion using the binomial probability distribution.

32. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-32 Parallel Example 4: Hypothesis Test for a Population Proportion A recent survey found that 68.6% of the population own their homes. In a random sample of 150 heads of households, 92 responded that they owned their homes. At the α = 0.01level of significance, does that suggest a difference from the national proportion(different than 68.6%)?

33. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-33 Parallel Example 4: Hypothesis Test for a Population Proportion: Approach: Step 1: Determine the null and alternative hypotheses Step 2: Check whether np 0 (1–p 0 ) is greater than or equal to 10, where p 0 is the proportion stated in the null hypothesis. If it is, then the sampling distribution of is approximately normal and we can use the steps for a large sample size. Otherwise we use the following Steps 3 and 4.

34. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-34 Solution Check the requirements: From the null hypothesis, we have p 0 = 0.686. n=150, so np 0 (1– p 0 )=32.31>10. And the sample size is less than 5% of the population size.Thus, the sampling distribution of is approximately normal. Step 1: H 0 : p = 0.686 versus H 1 : p # 0.686 Step 2: The level of significance is α = 0.01.

35. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-35 Solution Step 3: The sample proportion is. The test statistic is then … Step 4: Since this is a two-tailed test, we determine the critical value at the α = 0.01 level of significance to be z 0.005 = 2.58 and -z 0.005 =- 2.58. Step 5: Since the test statistic, z 0 = -1.93 does not fall within the critical region, we do not reject the null hypothesis.

36. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-36 Step 6: There is insufficient evidence at the α = 0.01 level of significance to conclude that the proportion of … is different than 0.686

37. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. Section Hypothesis Tests for a Population Mean 10.3

38. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-38 Objectives 1.Test hypotheses about a mean 2.Understand the difference between statistical significance and practical significance.

39. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-39 Objective 1 Test Hypotheses about a Mean

40. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-40 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.

41. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-41 1.The t-distribution is different for different degrees of freedom. Properties of the t-Distribution

42. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-42 1.The t-distribution is different for different degrees of freedom. 2.The t-distribution is centered at 0 and is symmetric about 0. Properties of the t-Distribution

43. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-43 1.The t-distribution is different for different degrees of freedom. 2.The t-distribution is centered at 0 and is symmetric about 0. 3.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

44. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-44 4.As t increases (or decreases) without bound, the graph approaches, but never equals, 0. Properties of the t-Distribution

45. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-45 4.As t increases (or decreases) without bound, the graph approaches, but never equals, 0. 5.The area in the tails of the t-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

46. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-46 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 σ by the Law of Large Numbers. Properties of the t-Distribution

47. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-47 Testing Hypotheses Regarding a Population Mean 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.

48. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-48 Step 1: Determine the null and alternative hypotheses. Again, the hypotheses can be structured in one of three ways:

49. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-49 Step 2: Select a level of significance, α, based on the seriousness of making a Type I error.

50. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-50 Classical Approach Step 3: Compute the test statistic which follows the Student’s t-distribution with n – 1 degrees of freedom.

51. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-51 Classical Approach Use Table VI to determine the critical value.

52. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-52 Classical Approach Two-Tailed

53. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-53 Classical Approach Left-Tailed

54. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-54 Classical Approach Right-Tailed

55. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-55 Classical Approach Step 4: Compare the critical value with the test statistic:

56. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-56 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.

57. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-57 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?

58. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-58 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.

59. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-59 Solution Step 1: H 0 : μ = 5710 versus H 1 : 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

60. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-60 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 freedom to be approximately –t 0.025 = –2.021 and t 0.025 = 2.021. Step 5: Since the test statistic, t 0 = –0.013, is between the critical values(not in the critical region), we fail to reject the null hypothesis.

61. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-61 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.

62. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-62 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?

63. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-63 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.

64. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-64 Assumption of normality appears reasonable.

65. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-65 No outliers.

66. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-66 Solution Step 1: H 0 : μ = 5.67 versus H 1 : 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

67. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-67 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 freedom to be –t 0.025 = –2.11 and t 0.025 = 2.11. Step 5: Since the test statistic, t 0 = 2.75, is greater than the critical value 2.11(falls within the critical region), we reject the null hypothesis.

68. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-68 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.

69. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-69 Objective 2 Understand the Difference between Statistical Significance and Practical Significance

70. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-70 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.

71. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-71 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.

72. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-72 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.

73. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-73 Solution Step 1: To determine whether the mean age has increased, this is a right-tailed test with H 0 : μ = 25.2 versus H 1 : 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

74. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-74 Solution Step 4: Since this is a right-tailed test, we determine the critical value at the α = 0.05 level of significance with n – 1 = 1199 degrees of freedom to be t 0.05 = 1.646. Step 5: Since the test statistic, t 0 = 2.17, is greater than the critical value 1.646(falls within the critical region), we reject the null hypothesis.

75. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-75 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).

76. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 10-76 Large Sample Sizes Large sample sizes can lead to statistically significant results while the difference between the statistic and parameter is not enough to be considered practically significant.