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

2. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Section The Language of Hypothesis Testing 10.1

3. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-3 1.Determine the null and alternative hypotheses 2.Explain Type I and Type II errors 3.State conclusions to hypothesis tests Objectives

4. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-4 A hypothesis is a statement regarding a characteristic of one or more populations. In this chapter, we look at hypotheses regarding a single population parameter.

5. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-5 Examples of Claims Regarding a Characteristic of a Single Population In 2008, 62% of American adults regularly volunteered their time for charity work. A researcher believes that this percentage is different today. Source: ReadersDigest.com poll created on 2008/05/02

6. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-6 In 2008, 62% of American adults regularly volunteered their time for charity work. A researcher believes that this percentage is different today. According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher believes that the mean length of a call has increased since then. Examples of Claims Regarding a Characteristic of a Single Population

7. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-7 In 2008, 62% of American adults regularly volunteered their time for charity work. A researcher believes that this percentage is different today. According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher believes that the mean length of a call has increased since then. Using an old manufacturing process, the standard deviation of the amount of wine put in a bottle was 0.23 ounces. With new equipment, the quality control manager believes the standard deviation has decreased. Examples of Claims Regarding a Characteristic of a Single Population

8. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-8 CAUTION! We test these types of statements using sample data because it is usually impossible or impractical to gain access to the entire population. If population data are available, there is no need for inferential statistics.

9. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-9 Hypothesis testing is a procedure, based on sample evidence and probability, used to test statements regarding a characteristic of one or more populations.

10. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-10 1.Make a statement regarding the nature of the population. 2.Collect evidence (sample data) to test the statement. 3.Analyze the data to assess the plausibility of the statement. Steps in Hypothesis Testing

11. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-11 The null hypothesis, denoted H 0, is a statement to be tested. The null hypothesis is a statement of no change, no effect or no difference and is assumed true until evidence indicates otherwise.

12. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-12 The alternative hypothesis, denoted H 1, is a statement that we are trying to find evidence to support.

13. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-13 In this chapter, there are three ways to set up the null and alternative hypotheses: 1.Equal versus not equal hypothesis (two-tailed test) H 0 : parameter = some value H 1 : parameter ≠ some value

14. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-14 In this chapter, there are three ways to set up the null and alternative hypotheses: 1.Equal versus not equal hypothesis (two-tailed test) H 0 : parameter = some value H 1 : parameter ≠ some value 2.Equal versus less than (left-tailed test) H 0 : parameter = some value H 1 : parameter < some value

15. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-15 In this chapter, there are three ways to set up the null and alternative hypotheses: 1.Equal versus not equal hypothesis (two-tailed test) H 0 : parameter = some value H 1 : parameter ≠ some value 2.Equal versus less than (left-tailed test) H 0 : parameter = some value H 1 : parameter < some value 3.Equal versus greater than (right-tailed test) H 0 : parameter = some value H 1 : parameter > some value

16. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-16 “In Other Words” The null hypothesis is a statement of status quo or no difference and always contains a statement of equality. The null hypothesis is assumed to be true until we have evidence to the contrary. We seek evidence that supports the statement in the alternative hypothesis.

17. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-17 For each of the following claims, determine the null and alternative hypotheses. State whether the test is two-tailed, left-tailed or right- tailed. a)In 2008, 62% of American adults regularly volunteered their time for charity work. A researcher believes that this percentage is different today. b)According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher believes that the mean length of a call has increased since then. c)Using an old manufacturing process, the standard deviation of the amount of wine put in a bottle was 0.23 ounces. With new equipment, the quality control manager believes the standard deviation has decreased. Parallel Example 2: Forming Hypotheses

18. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-18 a)In 2008, 62% of American adults regularly volunteered their time for charity work. A researcher believes that this percentage is different today. The hypothesis deals with a population proportion, p. If the percentage participating in charity work is no different than in 2008, it will be 0.62 so the null hypothesis is H 0 : p=0.62. Since the researcher believes that the percentage is different today, the alternative hypothesis is a two-tailed hypothesis: H 1 : p≠0.62. Solution

19. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-19 b)According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher believes that the mean length of a call has increased since then. The hypothesis deals with a population mean, μ. If the mean call length on a cellular phone is no different than in 2006, it will be 3.25 minutes so the null hypothesis is H 0 : μ = 3.25. Since the researcher believes that the mean call length has increased, the alternative hypothesis is: H 1 : μ > 3.25, a right-tailed test. Solution

20. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-20 c)Using an old manufacturing process, the standard deviation of the amount of wine put in a bottle was 0.23 ounces. With new equipment, the quality control manager believes the standard deviation has decreased. The hypothesis deals with a population standard deviation, σ. If the standard deviation with the new equipment has not changed, it will be 0.23 ounces so the null hypothesis is H 0 : σ = 0.23. Since the quality control manager believes that the standard deviation has decreased, the alternative hypothesis is: H 1 : σ < 0.23, a left-tailed test. Solution

21. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-21 Objective 2 Explain Type I and Type II Errors

22. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-22 1.Reject the null hypothesis when the alternative hypothesis is true. This decision would be correct. Four Outcomes from Hypothesis Testing

23. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-23 1.Reject the null hypothesis when the alternative hypothesis is true. This decision would be correct. 2.Do not reject the null hypothesis when the null hypothesis is true. This decision would be correct. Four Outcomes from Hypothesis Testing

24. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-24 1.Reject the null hypothesis when the alternative hypothesis is true. This decision would be correct. 2.Do not reject the null hypothesis when the null hypothesis is true. This decision would be correct. 3.Reject the null hypothesis when the null hypothesis is true. This decision would be incorrect. This type of error is called a Type I error. Four Outcomes from Hypothesis Testing

25. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-25 1.Reject the null hypothesis when the alternative hypothesis is true. This decision would be correct. 2.Do not reject the null hypothesis when the null hypothesis is true. This decision would be correct. 3.Reject the null hypothesis when the null hypothesis is true. This decision would be incorrect. This type of error is called a Type I error. 4.Do not reject the null hypothesis when the alternative hypothesis is true. This decision would be incorrect. This type of error is called a Type II error. Four Outcomes from Hypothesis Testing

26. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-26 For each of the following claims, explain what it would mean to make a Type I error. What would it mean to make a Type II error? a)In 2008, 62% of American adults regularly volunteered their time for charity work. A researcher believes that this percentage is different today. b)According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher believes that the mean length of a call has increased since then. Parallel Example 3: Type I and Type II Errors

27. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-27 a)In 2008, 62% of American adults regularly volunteered their time for charity work. A researcher believes that this percentage is different today. A Type I error is made if the researcher concludes that p ≠ 0.62 when the true proportion of Americans 18 years or older who participated in some form of charity work is currently 62%. A Type II error is made if the sample evidence leads the researcher to believe that the current percentage of Americans 18 years or older who participated in some form of charity work is still 62% when, in fact, this percentage differs from 62%. Solution

28. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-28 b) According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher believes that the mean length of a call has increased since then. A Type I error occurs if the sample evidence leads the researcher to conclude that μ > 3.25 when, in fact, the actual mean call length on a cellular phone is still 3.25 minutes. A Type II error occurs if the researcher fails to reject the hypothesis that the mean length of a phone call on a cellular phone is 3.25 minutes when, in fact, it is longer than 3.25 minutes. Solution

29. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-29 α = P(Type I Error) = P(rejecting H 0 when H 0 is true) β =P(Type II Error) = P(not rejecting H 0 when H 1 is true)

30. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-30 The probability of making a Type I error, α, is chosen by the researcher before the sample data is collected. The level of significance, α, is the probability of making a Type I error.

31. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-31 “In Other Words” As the probability of a Type I error increases, the probability of a Type II error decreases, and vice-versa.

32. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-32 Objective 3 State Conclusions to Hypothesis Tests

33. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-33 CAUTION! We never “accept” the null hypothesis, because, without having access to the entire population, we don’t know the exact value of the parameter stated in the null. Rather, we say that we do not reject the null hypothesis. This is just like the court system. We never declare a defendant “innocent”, but rather say the defendant is “not guilty”.

34. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-34 According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher believes that the mean length of a call has increased since then. a)Suppose the sample evidence indicates that the null hypothesis should be rejected. State the wording of the conclusion. b)Suppose the sample evidence indicates that the null hypothesis should not be rejected. State the wording of the conclusion. Parallel Example 4: Stating the Conclusion

35. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-35 a)Suppose the sample evidence indicates that the null hypothesis should be rejected. State the wording of the conclusion. The statement in the alternative hypothesis is that the mean call length is greater than 3.25 minutes. Since the null hypothesis (μ = 3.25) is rejected, there is sufficient evidence to conclude that the mean length of a phone call on a cell phone is greater than 3.25 minutes. Solution

36. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-36 b)Suppose the sample evidence indicates that the null hypothesis should not be rejected. State the wording of the conclusion. Since the null hypothesis (μ = 3.25) is not rejected, there is insufficient evidence to conclude that the mean length of a phone call on a cell phone is greater than 3.25 minutes. In other words, the sample evidence is consistent with the mean call length equaling 3.25 minutes. Solution

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

38. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-38 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.

39. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-39 Objective 1 Explain the Logic of Hypothesis Testing

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

41. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-41 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.

42. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-42 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

43. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-43 Sampling distribution of the sample proportion

44. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-44 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

45. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-45 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.

46. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-46 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.

47. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-47 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.

48. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-48 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

49. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-49 A second criterion we may use for testing hypotheses is to determine how likely it is to obtain a sample proportion of 0.534 or higher from a population whose proportion is 0.5. If a sample proportion of 0.534 or higher is unlikely (or unusual), we have evidence against the statement in the null hypothesis. Otherwise, we do not have sufficient evidence against the statement in the null hypothesis. The Logic of the P-Value Approach

50. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-50 We can compute the probability of obtaining a sample proportion of 0.534 or higher from a population whose proportion is 0.5 using the normal model.

51. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-51 Recall So, we compute The value 0.0158 is called the P-value, which means about 2 samples in 100 will give a sample proportion as high or higher than the one we obtained if the population proportion really is 0.5. Because these results are unusual, we take this as evidence against the statement in the null hypothesis.

52. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-52 If the probability of getting a sample proportion as extreme or more extreme than the one obtained is small under the assumption the statement in the null hypothesis is true, reject the null hypothesis. Hypothesis Testing Using the P-value Approach

53. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-53 Objective 2 Test hypotheses about a population proportion.

54. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-54 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.

55. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-55 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.

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

57. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-57 Step 1: Determine the null and alternative hypotheses. The hypotheses can be structured in one of three ways:

58. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-58 Step 2: Select a level of significance, α, based on the seriousness of making a Type I error.

59. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-59 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

60. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-60 Classical Approach Two-Tailed (critical value) Use Table V to determine the critical value.

61. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-61 Classical Approach Left-Tailed (critical value)

62. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-62 Classical Approach Right-Tailed (critical value)

63. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-63 Step 4: Compare the critical value with the test statistic: Classical Approach

64. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-64 By Hand Step 3: Compute the test statistic. P-Value Approach

65. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-65 P-Value Approach Two-Tailed Use Table V to estimate the P-value.

66. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-66 P-Value Approach Left-Tailed

67. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-67 P-Value Approach Right-Tailed

68. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-68 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. P-Value Approach

69. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-69 Step 4: If the P-value < α, reject the null hypothesis. P-Value Approach

70. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-70 Step 5: State the conclusion.

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

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

73. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-73 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

74. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-74 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.

75. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-75 Solution: P-Value Approach Step 4: Since this is a right-tailed test, the P- value is the area under the standard normal distribution to the right of the test statistic z 0 =3.83. That is, P-value = P(Z > 3.83) ≈ 0. Step 5: Since the P-value is less than the level of significance, we reject the null hypothesis.

76. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-76 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.

77. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-77 Objective 3 Test hypotheses about a population proportion using the binomial probability distribution.

78. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-78 For the sampling distribution of to be approximately normal, we require np(1– p) be at least 10. What if this requirement is not met? We stated that an event was unusual if the probability of observing the event was less than 0.05. This criterion is based on the P-value approach to testing hypotheses; the probability that we computed was the P-value. We use this same approach to test hypotheses regarding a population proportion for small samples.

79. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-79 Parallel Example 4: Hypothesis Test for a Population Proportion: Small Sample Size In 2006, 10.5% of all live births in the United States were to mothers under 20 years of age. A sociologist claims that births to mothers under 20 years of age is decreasing. She conducts a simple random sample of 34 births and finds that 3 of them were to mothers under 20 years of age. Test the sociologist’s claim at the α = 0.01 level of significance.

80. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-80 Parallel Example 4: Hypothesis Test for a Population Proportion: Small Sample Size 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.

81. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-81 Parallel Example 4: Hypothesis Test for a Population Proportion: Small Sample Size Approach: Step 3: Compute the P-value. For right-tailed tests, the P-value is the probability of obtaining x or more successes. For left-tailed tests, the P- value is the probability of obtaining x or fewer successes. The P-value is always computed with the proportion given in the null hypothesis. Step 4: If the P-value is less than the level of significance, α, we reject the null hypothesis.

82. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-82 Solution Step 1: H 0 : p = 0.105 versus H 1 : p < 0.105 Step 2: From the null hypothesis, we have p 0 = 0.105. There were 34 mothers sampled, so np 0 (1– p 0 )=3.57 < 10. Thus, the sampling distribution of is not approximately normal.

83. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-83 Solution Step 3: Let X represent the number of live births in the United States to mothers under 20 years of age. We have x = 3 successes in n = 34 trials so = 3/34= 0.088. We want to determine whether this result is unusual if the population mean is truly 0.105. Thus, P-value = P(X ≤ 3 assuming p=0.105 ) = P(X = 0) + P(X =1) + P(X =2) + P(X = 3) = 0.51

84. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-84 Solution Step 4: The P-value = 0.51 is greater than the level of significance so we do not reject H 0. There is insufficient evidence to conclude that the percentage of live births in the United States to mothers under the age of 20 has decreased below the 2006 level of 10.5%.

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

86. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-86 Objectives 1.Test hypotheses about a mean 2.Understand the difference between statistical significance and practical significance.

87. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-87 Objective 1 Test Hypotheses about a Mean

88. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-88 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.

89. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-89 1.The t-distribution is different for different degrees of freedom. Properties of the t-Distribution

90. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-90 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

91. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-91 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

92. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-92 4.As t increases (or decreases) without bound, the graph approaches, but never equals, 0. Properties of the t-Distribution

93. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-93 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

94. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-94 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

95. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-95 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.

96. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-96 Step 1: Determine the null and alternative hypotheses. Again, the hypotheses can be structured in one of three ways:

97. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-97 Step 2: Select a level of significance, α, based on the seriousness of making a Type I error.

98. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-98 Classical Approach Step 3: Compute the test statistic which follows the Student’s t-distribution with n – 1 degrees of freedom.

99. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-99 Classical Approach Use Table VI to determine the critical value.

100. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-100 Classical Approach Two-Tailed

101. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-101 Classical Approach Left-Tailed

102. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-102 Classical Approach Right-Tailed

103. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-103 Classical Approach Step 4: Compare the critical value with the test statistic:

104. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-104 P-Value Approach By Hand Step 3: Compute the test statistic which follows the Student’s t-distribution with n – 1 degrees of freedom.

105. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-105 P-Value Approach Use Table VI to approximate the P-value.

106. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-106 P-Value Approach Two-Tailed

107. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-107 P-Value Approach Left-Tailed

108. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-108 P-Value Approach Right-Tailed

109. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-109 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.

110. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-110 P-Value Approach Step 4: If the P-value < α, reject the null hypothesis.

111. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-111 Step 5: State the conclusion.

112. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-112 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.

113. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-113 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?

114. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-114 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.

115. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-115 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

116. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-116 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, we fail to reject the null hypothesis.

117. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-117 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 freedom to the left of –t 0.025 = –0.013 and to the right of t 0.025 = 0.013. That is, P- value = 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.

118. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-118 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.

119. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-119 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?

120. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-120 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.

121. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-121 Assumption of normality appears reasonable.

122. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-122 No outliers.

123. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-123 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

124. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-124 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, we reject the null hypothesis.

125. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-125 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 freedom to the left of –t 0.025 = –2.75 and to the right of t 0.025 = 2.75. That is, P-value = 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.

126. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-126 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.

127. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-127 Objective 5 Distinguish between Statistical Significance and Practical Significance

128. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-128 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.

129. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-129 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.

130. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-130 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.

131. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-131 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

132. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-132 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:.001 < P-value <.005 Step 5: Because the P-value is less than the level of significance, 0.05, we reject the null hypothesis.

133. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-133 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).

134. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-134 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.

135. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Section Hypothesis Tests for a Population Standard Deviation 10.4

136. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-136 Objective 1.Test hypotheses about a population standard deviation

137. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-137 Chi-Square Distribution If a simple random sample of size n is obtained from a normally distributed population with mean μ and standard deviation σ, then where s 2 is a sample variance has a chi-square distribution with n – 1 degrees of freedom.

138. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-138 Characteristics of the Chi-Square Distribution 1.It is not symmetric. 2.The shape of the chi-square distribution depends on the degrees of freedom, just as with Student’s t-distribution. 3.As the number of degrees of freedom increases, the chi-square distribution becomes more nearly symmetric. 4.The values of χ 2 are nonnegative (greater than or equal to 0).

139. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-139

140. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-140 Objective 1 Test hypotheses about a population standard deviation.

141. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-141 Testing Hypotheses about a Population Variance or Standard Deviation To test hypotheses about the population variance or standard deviation, we can use the following steps, provided that The sample is obtained using simple random sampling. The population is normally distributed.

142. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-142 Step 1: Determine the null and alternative hypotheses. The hypotheses can be structured in one of three ways:

143. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-143 Step 2: Select a level of significance, α, based on the seriousness of making a Type I error.

144. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-144 Step 3: Compute the test statistic

145. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-145 Use Table VII to determine the critical value using n – 1 degrees of freedom. Classical Approach

146. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-146 Classical Approach Two-Tailed

147. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-147 Classical Approach Left -Tailed

148. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-148 Classical Approach Right-Tailed

149. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-149 Step 4: Compare the critical value with the test statistic: Classical Approach

150. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-150 By Hand Step 3: Compute the test statistic P-Value Approach

151. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-151 Use Table VII to approximate the P-value for a left- or right-tailed test by determining the area under the chi-square distribution with n – 1 degrees of freedom to the left (for a left- tailed test) or right (for a right-tailed test) of the test statistic. For two-tailed tests, it is recommended that technology be used to find the P-value or obtain a confidence interval. P-Value Approach

152. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-152 P-Value Approach Left-Tailed

153. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-153 P-Value Approach Right-Tailed

154. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-154 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 MINITAB and StatCrunch are in the Technology Step-by-Step in the text. P-Value Approach

155. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-155 Step 4: If the P-value < α, reject the null hypothesis. P-Value Approach

156. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-156 Step 5: State the conclusion.

157. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-157 CAUTION! The procedures in this section are not robust. Therefore, if analysis of the data indicates that the variable does not come from a population that is normally distributed, the procedures presented in this section are not valid.

158. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-158 Parallel Example 1: Testing a Hypothesis about a Population Standard Deviation A can of 7-Up states that the contents of the can are 355 ml. A quality control engineer is worried that the filling machine is miscalibrated. In other words, she wants to make sure the machine is not under- or over-filling the cans. She randomly selects 9 cans of 7-Up and measures the contents. She obtains the following data. 351360358356359 358355361352 In section 9.2, we assumed the population standard deviation was 3.2. Test the claim that the population standard deviation, σ, is greater than 3.2 ml at the α = 0.05 level of significance.

159. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-159 Solution Step 1: H 0 : σ = 3.2 versus H 1 : σ > 3.2. This is a right-tailed test. Step 2: The level of significance is α = 0.05. Step 3: From the data, the sample standard deviation is computed to be s = 3.464. The test statistic is

160. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-160 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 with 9 – 1 = 8 degrees of freedom to be χ 2 0.05 = 15.507. Step 5: Since the test statistic,, is less than the critical value, 15.507, we fail to reject the null hypothesis.

161. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-161 Solution: P-Value Approach Step 4: Since this is a right-tailed test, the P-value is the area under the χ 2 -distribution with 9-1=8 to the right of the test statistic. That is, P-value = P(χ 2 > 9.37)>0.10. Step 5: Since the P-value is greater than the level of significance, we fail to reject the null hypothesis.

162. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-162 Solution: Step 6: There is insufficient evidence at the α = 0.05 level of significance to conclude that the standard deviation of the can content of 7-Up is greater than 3.2 ml.

163. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Section Putting It Together: Which Method Do I Use? 10.5

164. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-164 Objective 1.Determine the appropriate hypothesis test to perform

165. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-165 Objective 1 Determine the Appropriate Hypothesis Test to Perform

166. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-166

167. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Section The Probability of a Type II Error and the Power of the Test 10.6

168. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-168 Objectives 1.Determine the probability of making a Type II error 2.Compute the power of the test

169. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-169 Objective 1 Determine the Probability of Making a Type II Error

170. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-170 The Probability of a Type II Error Step 1: Determine the sample mean that separates the rejection region from the nonrejection region.

171. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-171 Step 1 Left-Tailed Test (H 1 : p < p 0 ) Note: z α is the critical z-value corresponding to the level of significance, α.

172. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-172 Step 1 Right-Tailed Test (H 1 : p > p 0 ) Note: z α is the critical z-value corresponding to the level of significance, α.

173. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-173 Step 1 Two-Tailed Test (H 1 : p ≠ p 0 ) Note: z α is the critical z-value corresponding to the level of significance, α.

174. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-174 Step 2: Left-Tailed Test (H 1 : p < p 0 ) Find the area under the normal curve to the right of the sample proportion found in Step 1 assuming a particular value of the population proportion in the alternative hypothesis.

175. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-175 Step 2: Right-Tailed Test (H 1 : p > p 0 ) Find the area under the normal curve to the left of the sample proportion found in Step 1 assuming a particular value of the population proportion in the alternative hypothesis.

176. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-176 Step 2: Two-Tailed Test (H 1 : p ≠ p 0 ) Find the area under the normal curve between and found in Step 1 assuming a particular value of the population proportion in the alternative hypothesis.

177. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-177 Parallel Example 1: Computing the Probability of a Type II Error Earlier, we tested the hypothesis that the mean trade volume of Apple stock was greater than 35.14 million shares, H 0 : μ = 35.14 versus H 1 : μ > 35.14, based upon a random sample of size n = 40 with the population standard deviation, σ, assumed to be 15.07 million shares at the α = 0.1 level of significance. Compute the probability of a type II error given that the population mean is μ = 40.62.

178. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-178 Solution Step 1: Since z 0.9 =1.28, we let z = 1.28, μ 0 = 35.14, σ = 15.07, and n = 40 to find the sample mean that separates the rejection region from the nonrejection region: For any sample mean less than 38.19, we do not reject the null hypothesis.

179. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-179 Solution Step 2:

180. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-180 Solution Step 2:

181. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-181 Solution Step 3: P(Type II error) = P(do not reject H 0 given H 1 is true) = P( < 38.19 given that μ =40.62) = = 0.1539

182. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-182 Objective 2 Compute the Power of the Test

183. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-183 Power of the Test The probability of rejecting the null hypothesis when the alternative hypothesis is true is 1 – β. The value of 1 – β is referred to as the power of the test. The higher the power of the test, the more likely the test will reject the null when the alternative hypothesis is true.

184. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-184 Parallel Example 2: Computing the Power of the Test Compute the power for the Apple stock example.

185. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 10-185 Solution Recall that β = 0.1539. The power of the test is 1 – β = 1 – 0.1539 = 0.8461. There is a 84.61% chance of rejecting the null hypothesis when the true population mean is 40.62.