1. Chapter 7 Hypothesis Testing with One Sample 1 Larson/Farber 4th ed.

2. Chapter Outline 7.1 Introduction to Hypothesis Testing 7.2 Hypothesis Testing for the Mean (Large Samples) 7.3 Hypothesis Testing for the Mean (Small Samples) 7.4 Hypothesis Testing for Proportions 7.5 Hypothesis Testing for Variance and Standard Deviation 2 Larson/Farber 4th ed.

3. Section 7.1 Introduction to Hypothesis Testing 3 Larson/Farber 4th ed.

4. Section 7.1 Objectives State a null hypothesis and an alternative hypothesis Identify type I and type I errors and interpret the level of significance Determine whether to use a one-tailed or two-tailed statistical test and find a p-value Make and interpret a decision based on the results of a statistical test Write a claim for a hypothesis test 4 Larson/Farber 4th ed.

5. Hypothesis Tests Hypothesis test A process that uses sample statistics to test a claim about the value of a population parameter. For example: An automobile manufacturer advertises that its new hybrid car has a mean mileage of 50 miles per gallon. To test this claim, a sample would be taken. If the sample mean differs enough from the advertised mean, you can decide whether to reject the claim or fail to reject the claim. 5 Larson/Farber 4th ed.

6. Hypothesis Tests Statistical hypothesis A statement, or claim, about a population parameter. Need a pair of hypotheses the null hypothesis that represents the claim the other, its alternative hypothesis 6 Larson/Farber 4th ed.

7. Stating a Hypothesis Null hypothesis A statistical hypothesis that contains a statement of equality, =. Denoted H 0 read “H subzero” or “H naught.” Alternative hypothesis A statement of inequality such as >, , or <. Denoted H a read “H sub-a.” complementary statements 7 Larson/Farber 4th ed.

8. Stating a Hypothesis To write the null and alternative hypotheses, translate the claim made about the population parameter from a verbal statement to a mathematical statement. Then write its alternative. H 0 : μ = k H a : μ > k H 0 : μ = k H a : μ ≠ k 8 Larson/Farber 4th ed.

9. Example: Stating the Null and Alternative Hypotheses Write the claim as a mathematical sentence. State the null and alternative hypotheses. A university publicizes that the proportion of its students who graduate in 4 years is 82%. Equality condition Complement of H 0 H0:Ha:H0:Ha: (Claim) p = 0.82 p ≠ 0.82 Solution: 9 Larson/Farber 4th ed.

10. μ = 2.5 gallons per minute Example: Stating the Null and Alternative Hypotheses Write the claim as a mathematical sentence. State the null and alternative hypotheses. A water faucet manufacturer announces that the mean flow rate of a certain type of faucet is 2.5 gallons per minute. Our water pressure is low. Claim H0:Ha:H0:Ha: μ < 2.5 gallons per minute Solution: 10 Larson/Farber 4th ed.

11. μ = 20 ounces Example: Stating the Null and Alternative Hypotheses Write the claim as a mathematical sentence. State the null and alternative hypotheses. A cereal company advertises that the mean weight of the contents of its 20-ounce size cereal boxes is 20 ounces. We think the boxes contain less cereal. Alternative Hypothesis Null Hypothesis (Claim) H0:Ha:H0:Ha: μ < 20 ounces Solution: 11 Larson/Farber 4th ed.

12. Types of Errors Always begin the hypothesis test assuming that the equality condition in the null hypothesis is true. At the end of the test, one of two decisions will be made:  reject the null hypothesis  fail to reject the null hypothesis Because our decision is based on a sample, there is the possibility that a wrong decision was made. Don’t worry, it’s not your fault. 12 Larson/Farber 4th ed.

13. Types of Errors A type I error occurs if the null hypothesis is rejected when it is true. A type II error occurs if the null hypothesis is not rejected when it is false. Actual Truth of H 0 DecisionH 0 is trueH 0 is false Do not reject H 0 Correct DecisionType II Error Reject H 0 Type I ErrorCorrect Decision 13 Larson/Farber 4th ed.

14. Example: Identifying Type I and Type II Errors The USDA limit for salmonella contamination for chicken is 20%. A meat inspector reports that the chicken produced by a company exceeds the USDA limit. You perform a hypothesis test to determine whether the meat inspector’s claim is true. When will a type I or type II error occur? Which is more serious? (Source: United States Department of Agriculture) 14 Larson/Farber 4th ed.

15. Let p represent the proportion of chicken that is contaminated. Solution: Identifying Type I and Type II Errors H0:Ha:H0:Ha: p = 0.2 p > 0.2 Hypotheses: 15 Larson/Farber 4th ed.

16. Solution: Identifying Type I and Type II Errors A type I error is rejecting H 0 when it is true. (RT1) The actual proportion of contaminated chicken has not exceeded 0.2, but you decide to reject H 0. A type II error is failing to reject H 0 when it is false. (FF2) The actual proportion of contaminated chicken is greater than 0.2, but you do not reject H 0. H0:Ha:H0:Ha: p = 0.2 p > 0.2 Hypotheses: (Claim) 16 Larson/Farber 4th ed.

17. Solution: Identifying Type I and Type II Errors H0:Ha:H0:Ha: p = 0.2 p > 0.2 Hypotheses: With a type I error, you might create a health scare and hurt the sales of chicken producers who were actually meeting the USDA limits. With a type II error, you could be allowing chicken that exceeded the USDA contamination limit to be sold to consumers. A type II error could result in sickness or even death. 17 Larson/Farber 4th ed.

18. Level of Significance Level of significance Your maximum allowable probability of making a type I error. Known as the alpha level.  Denoted by , the lowercase Greek letter alpha. By setting the level of significance at a small value, you are saying that you want the probability of rejecting a true null hypothesis to be small. Commonly used levels of significance:   = 0.10  = 0.05  = 0.01  Use  = 0.05 when the level of significance is not given. 18 Larson/Farber 4th ed.

19. Statistical Tests After stating the null and alternative hypotheses and specifying the level of significance (alpha level), a random sample is taken from the population and sample statistics are calculated. The statistic that is compared with the parameter in the null hypothesis is called the test statistic. σ2σ2 χ 2 (Section 7.5)s2s2 z (Section 7.4) p t (Section 7.3 n < 30) z (Section 7.2 n  30) μ Standardized test statistic Test statisticPopulation parameter 19 Larson/Farber 4th ed.

20. P-values P-value (or probability value) The probability, if the null hypothesis is true, of obtaining a sample statistic with a value as extreme or more extreme than the one determined from the sample data. Depends on the nature of the test. 20 Larson/Farber 4th ed.

21. Nature of the Test Three types of hypothesis tests  left-tailed test  right-tailed test  two-tailed test The type of test depends on the region of the sampling distribution that favors a rejection of H 0. This region is indicated by the alternative hypothesis. 21 Larson/Farber 4th ed.

22. Left-tailed Test The alternative hypothesis H a contains the less-than inequality symbol (<). z 0123-3-2 Test statistic H 0 : μ = k H a : μ < k P is the area to the left of the test statistic. 22 Larson/Farber 4th ed.

23. The alternative hypothesis H a contains the greater- than inequality symbol (>). z 0123-3-2 Right-tailed Test H 0 : μ = k H a : μ > k Test statistic P is the area to the right of the test statistic. 23 Larson/Farber 4th ed.

24. Two-tailed Test The alternative hypothesis H a contains the not equal inequality symbol (≠). Each tail has an area of ½P. z 0123-3-2 Test statistic H 0 : μ = k H a : μ  k P is twice the area to the left of the negative test statistic. P is twice the area to the right of the positive test statistic. 24 Larson/Farber 4th ed.

25. Example: Identifying The Nature of a Test For each claim, state H 0 and H a. Then determine whether the hypothesis test is a left-tailed, right-tailed, or two-tailed test. Sketch a normal sampling distribution and shade the area for the P-value. A university publicizes that the proportion of its students who graduate in 4 years is 82%. H0:Ha:H0:Ha: p = 0.82 p ≠ 0.82 Two-tailed test z 0-zz ½ P-value area Solution: 25 Larson/Farber 4th ed.

26. Example: Identifying The Nature of a Test For each claim, state H 0 and H a. Then determine whether the hypothesis test is a left-tailed, right-tailed, or two-tailed test. Sketch a normal sampling distribution and shade the area for the P-value. A water faucet manufacturer announces that the mean flow rate of a certain type of faucet is less than 2.5 gallons per minute. H0:Ha:H0:Ha: Left-tailed test z 0-z P-value area μ = 2.5 gpm μ < 2.5 gpm Solution: 26 Larson/Farber 4th ed.

27. Example: Identifying The Nature of a Test For each claim, state H 0 and H a. Then determine whether the hypothesis test is a left-tailed, right-tailed, or two-tailed test. Sketch a normal sampling distribution and shade the area for the P-value. A cereal company advertises that the mean weight of the contents of its 20-ounce size cereal boxes is more than 20 ounces. H0:Ha:H0:Ha: Right-tailed test z 0 z P-value area μ = 20 oz μ > 20 oz Solution: 27 Larson/Farber 4th ed.

28. Making a Decision Decision Rule Based on P-value Compare the P-value with .  If P < , then reject H 0.  If P > , then fail to reject H 0. Hypotheses DecisionRule Reject H 0 Fail to reject H 0 There is enough evidence to reject the claim There is not enough evidence to reject the claim There is enough evidence to support the claim There is not enough evidence to support the claim 28 Larson/Farber 4th ed.

29. 29 When the p-value is lower than the alpha level, we reject the null hypothesis and use the alternative hypothesis “When it’s low, let it go…reject the null.” When the p-value is higher than the alpha level, we fail to reject the null hypothesis. “When it’s high, let it fly…we fail to reject the null hypothesis.”

30. Example: Interpreting a Decision 30 Larson/Farber 4th ed.

31. Solution: Interpreting a Decision The claim is represented by H 0. If you reject H 0 you should conclude “There is sufficient evidence to reject the university’s claim that the proportion of its students who graduate in 4 years is 82% and therefore conclude that the proportion of graduates in 4 years is higher or lower than the university’s claim.” 31 Larson/Farber 4th ed. If you fail to reject H 0, you should conclude “There is sufficient evidence to support the university’s claim that its four-year graduation rate is 82%.”

32. Example: Interpreting a Decision You perform a hypothesis test for the following claim. How should you interpret your decision if you reject H 0 ? If you fail to reject H 0 ? Honda Automobile Company states that the mean stopping distance (on a dry surface) for a Honda Civic is 136 feet. Consumer Reports believes that the stopping distance is greater than 136 feet. 32 Larson/Farber 4th ed.

33. Solution: Interpreting a Decision If you reject H 0 you should conclude “There is sufficient evidence to support Consumer Reports’ claim that the stopping distance for a Honda Civic is greater than 136 feet.” If you fail to reject H 0, you should conclude “There is not enough evidence to conclude that the stopping distance for a Honda Civic is greater than 136 feet.” 33 Larson/Farber 4th ed.

34. z 0 Steps for Hypothesis Testing 1.State the claim mathematically or verbally. Identify the null and alternative hypotheses. H 0 : ? H a : ? 2.Specify the level of significance. α = ? 3.Determine the standardized sampling distribution and draw its graph. 4.Calculate the test statistic and its standardized value. Add it to your sketch. z 0 Test statistic This sampling distribution is based on the assumption that H 0 is true. 34 Larson/Farber 4th ed.

35. Steps for Hypothesis Testing 5.Find the P-value. 6.Use the following decision rule. 7.Write a statement to interpret the decision in the context of the original claim. Is the P-value less than or equal to the level of significance? Fail to reject H 0. Yes Reject H 0. No 35 Larson/Farber 4th ed.

36. Section 7.1 Summary Stated a null hypothesis and an alternative hypothesis Identified type I and type I errors and interpreted the level of significance Determined whether to use a one-tailed or two-tailed statistical test and found a p-value Made and interpreted a decision based on the results of a statistical test Wrote a claim for a hypothesis test 36 Larson/Farber 4th ed.

37. Section 7.2 Hypothesis Testing for the Mean (Large Samples) 37 Larson/Farber 4th ed.

38. Section 7.2 Objectives Find P-values and use them to test a mean μ Use P-values for a z-test Find critical values and rejection regions in a normal distribution Use rejection regions for a z-test 38 Larson/Farber 4th ed.

39. Using P-values to Make a Decision 39 Larson/Farber 4th ed.

40. Example: Interpreting a P-value The P-value for a hypothesis test is P = 0.0237. What is your decision if the level of significance is 1.0.05? 2.0.01? Solution: Because 0.0237 < 0.05, you should reject the null hypothesis. Solution: Because 0.0237 > 0.01, you should fail to reject the null hypothesis. 40 Larson/Farber 4th ed.

41. Finding the P-value After determining the hypothesis test’s standardized test statistic and the test statistic’s corresponding area, do one of the following to find the P-value. a.For a left-tailed test, P = (Area in left tail). b.For a right-tailed test, P = (Area in right tail). c.For a two-tailed test, P = 2(Area in tail of test statistic). 41 Larson/Farber 4th ed.

42. Example: Finding the P-value Find the P-value for a left-tailed hypothesis test with a test statistic of z = -2.23. Decide whether to reject H 0 if the level of significance is α = 0.01. z 0-2.23 P = 0.0129 Solution: For a left-tailed test, P = (Area in left tail) Because 0.0129 > 0.01, you should fail to reject H 0 42 Larson/Farber 4th ed.

43. z 02.14 Example: Finding the P-value Find the P-value for a two-tailed hypothesis test with a test statistic of z = 2.14. Decide whether to reject H 0 if the level of significance is α = 0.05. Solution: For a two-tailed test, P = 2(Area in tail of test statistic) Because 0.0324 < 0.05, you should reject H 0 0.9838 1 – 0.9838 = 0.0162 P = 2(0.0162) = 0.0324 43 Larson/Farber 4th ed.

44. Z-Test for a Mean μ Can be used when the population is normal and  is known, or for any population when the sample size n is at least 30. The test statistic is the sample mean The standardized test statistic is z When n  30, the sample standard deviation s can be substituted for . 44 Larson/Farber 4th ed.

45. Using P-values for a z-Test for Mean μ 1.State the claim mathematically and verbally. Identify the null and alternative hypotheses. 2.Specify the level of significance. 3.Determine the standardized test statistic. 4.Find the area that corresponds to z. State H 0 and H a. Identify . Use Table 4 in Appendix B. 45 Larson/Farber 4th ed. In WordsIn Symbols

46. Using P-values for a z-Test for Mean μ Reject H 0 if P-value is less than or equal to . Otherwise, fail to reject H 0. 5.Find the P-value. a.For a left-tailed test, P = (Area in left tail). b.For a right-tailed test, P = (Area in right tail). c.For a two-tailed test, P = 2(Area in tail of test statistic). 6.Make a decision to reject or fail to reject the null hypothesis. 7.Interpret the decision in the context of the original claim. 46 Larson/Farber 4th ed. In WordsIn Symbols

47. Example: Hypothesis Testing Using P- values In an advertisement, a pizza shop claims that its mean delivery time is 30 minutes. A random selection of 36 delivery times has a sample mean of 28.5 minutes and a standard deviation of 3.5 minutes. Is there enough evidence to support the claim at  = 0.01? Use a P- value. 47 Larson/Farber 4th ed.

48. Solution: Hypothesis Testing Using P- values H 0 : H a :  = Test Statistic: μ < 30 min 0.01 Decision: z 0-2.57 0.0051 P-value 0.0051 < 0.01 Reject H 0 48 Larson/Farber 4th ed.

49. Example: Hypothesis Testing Using P- values You think that the average franchise investment information shown in the graph is incorrect, so you randomly select 30 franchises and determine the necessary investment for each. The sample mean investment is $135,000 with a standard deviation of $30,000. Is there enough evidence to support your claim at  = 0.05? Use a P-value. 49 Larson/Farber 4th ed.

50. Solution: Hypothesis Testing Using P- values H 0 : H a :  = Test Statistic: μ = $143,260 μ ≠ $143,260 0.05 Decision: P-value P = 2(0.0655) = 0.1310 0.1310 > 0.05 Fail to reject H 0 z 0-1.51 0.0655 50 Larson/Farber 4th ed.

51. Rejection Regions and Critical Values 51 Larson/Farber 4th ed.

52. Rejection Regions and Critical Values 52 Larson/Farber 4th ed.

53. Example: Finding Critical Values Find the critical value and rejection region for a two- tailed test with  = 0.05. z 0z0z0 z0z0 ½α = 0.025 1 – α = 0.95 The rejection regions are to the left of -z = -1.96 and to the right of z = 1.96. z = 1.96-z = -1.96 Solution: 53 Larson/Farber 4th ed.

54. Decision Rule Based on Rejection Region To use a rejection region to conduct a hypothesis test, calculate the standardized test statistic, z. If the standardized test statistic 1. is in the rejection region, then reject H 0. 2. is not in the rejection region, then fail to reject H 0. z 0 z0z0 Fail to reject H 0. Reject H 0. Left-Tailed Test z < z 0 z 0 z0z0 Reject H o. Fail to reject H o. z > z 0 Right-Tailed Test z 0 z0z0 Two-Tailed Test z0z0 z < -z 0 z > z 0 Reject H 0 Fail to reject H 0 Reject H 0 54 Larson/Farber 4th ed.

55. Using Rejection Regions for a z-Test for a Mean μ 1.State the claim mathematically and verbally. Identify the null and alternative hypotheses. 2.Specify the level of significance. 3.Sketch the sampling distribution. 4.Determine the critical value(s). 5.Determine the rejection region(s). State H 0 and H a. Identify . Use Table 4 in Appendix B. 55 Larson/Farber 4th ed. In WordsIn Symbols

56. Using Rejection Regions for a z-Test for a Mean μ 6.Find the standardized test statistic. 7.Make a decision to reject or fail to reject the null hypothesis. 8.Interpret the decision in the context of the original claim. If z is in the rejection region, reject H 0. Otherwise, fail to reject H 0. 56 Larson/Farber 4th ed. In WordsIn Symbols

57. Example: Testing with Rejection Regions Employees in a large accounting firm claim that the mean salary of the firm’s accountants is the same as that of its competitor’s, which is $45,000. A random sample of 30 of the firm’s accountants has a mean salary of $43,500 with a standard deviation of $5200. At α = 0.05, test the employees’ claim. 57 Larson/Farber 4th ed.

58. Solution: Testing with Rejection Regions H 0 : H a :  = Rejection Region: μ < $45,000 0.05 Decision: Since the p-value falls outside the rejection region, we fail to reject the null hypothesis and can say that there is not sufficient evidence to support the employees’ claim that the mean salary is less than $45,000. Test Statistic z 0-1.645 0.05 -1.58 -1.645 Fail to reject H 0 58 Larson/Farber 4th ed.

59. Example: Testing with Rejection Regions The U.S. Department of Agriculture reports that the mean cost of raising a child from birth to age 2 in a rural area is $10,460. You believe this value is incorrect, so you select a random sample of 900 children (age 2) and find that the mean cost is $10,345 with a standard deviation of $1540. At α = 0.05, is there enough evidence to conclude that the mean cost is different from $10,460? (Adapted from U.S. Department of Agriculture Center for Nutrition Policy and Promotion) 59 Larson/Farber 4th ed.

60. Solution: Testing with Rejection Regions H 0 : H a :  = Rejection Region: μ = $10,460 μ ≠ $10,460 0.05 Decision: At the 5% level of significance, you have enough evidence to conclude the mean cost of raising a child from birth to age 2 in a rural area is significantly different from $10,460. Test Statistic z 0-1.96 0.025 1.96 0.025 -1.961.96 -2.24 Reject H 0 60 Larson/Farber 4th ed.

61. Section 7.2 Summary Found P-values and used them to test a mean μ Used P-values for a z-test Found critical values and rejection regions in a normal distribution Used rejection regions for a z-test 61 Larson/Farber 4th ed.

62. Section 7.3 Hypothesis Testing for the Mean (Small Samples) 62 Larson/Farber 4th ed.

63. Section 7.3 Objectives Find critical values in a t-distribution Use the t-test to test a mean μ Use technology to find P-values and use them with a t-test to test a mean μ 63 Larson/Farber 4th ed.

64. Finding Critical Values in a t-Distribution 1.Identify the level of significance . 2.Identify the degrees of freedom d.f. = n – 1. 3.Find the critical value(s) using Table 5 in Appendix B in the row with n – 1 degrees of freedom. If the hypothesis test is a.left-tailed, use “One Tail,  ” column with a negative sign, b.right-tailed, use “One Tail,  ” column with a positive sign, c.two-tailed, use “Two Tails,  ” column with a negative and a positive sign. 64 Larson/Farber 4th ed.

65. Example: Finding Critical Values for t Solution: The degrees of freedom are d.f. = n – 1 = 21 – 1 = 20. Look at α = 0.05 in the “One Tail,  ” column. Because the test is left- tailed, the critical value is negative. t 0-1.725 0.05 65 Larson/Farber 4th ed.

66. Example: Finding Critical Values for t Find the critical values t and -t for a two-tailed test given  = 0.05 and n = 26. Solution: The degrees of freedom are d.f. = n – 1 = 26 – 1 = 25. Look at α = 0.05 in the “Two Tail,  ” column. Because the test is two- tailed, one critical value is negative and one is positive. t 0-2.060 0.025 2.060 0.025 66 Larson/Farber 4th ed.

67. t-Test for a Mean μ (n < 30,  Unknown) t-Test for a Mean A statistical test for a population mean. The t-test can be used when the population is normal or nearly normal,  is unknown, and n < 30. The test statistic is the sample mean The standardized test statistic is t. The degrees of freedom are d.f. = n – 1. 67 Larson/Farber 4th ed.

68. Using the t-Test for a Mean μ (Small Sample) 1.State the claim mathematically and verbally. Identify the null and alternative hypotheses. 2.Specify the level of significance. 3.Identify the degrees of freedom and sketch the sampling distribution. 4.Determine any critical value(s). State H 0 and H a. Identify . Use Table 5 in Appendix B. d.f. = n – 1. 68 Larson/Farber 4th ed. In WordsIn Symbols

69. Using the t-Test for a Mean μ (Small Sample) 5.Determine any rejection region(s). 6.Find the standardized test statistic. 7.Make a decision to reject or fail to reject the null hypothesis. 8.Interpret the decision in the context of the original claim. If t is in the rejection region, reject H 0. Otherwise, fail to reject H 0. 69 Larson/Farber 4th ed. In WordsIn Symbols

70. Example: Testing μ with a Small Sample A used car dealer says that the mean price of a 2005 Honda Pilot LX is $23,900. You suspect this claim is incorrect and find that a random sample of 14 similar vehicles has a mean price of $23,000 and a standard deviation of $1113. Is there enough evidence to reject the dealer’s claim at α = 0.05? Assume the population is normally distributed. (Adapted from Kelley Blue Book) 70 Larson/Farber 4th ed.

71. Solution: Testing μ with a Small Sample H 0 : H a : α = df = Rejection Region: μ < $23,900 0.05 14 – 1 = 13 t 0-1.771 0.05 -3.026 -1.771 At the 0.05 level of significance, there is enough evidence to reject the claim that the mean price of a 2005 Honda Pilot LX is $23,900 Reject H 0 71 Larson/Farber 4th ed.

72. Example: Testing μ with a Small Sample An industrial company claims that the mean pH level of the water in a nearby river is 6.8. You randomly select 19 water samples and measure the pH of each. The sample mean and standard deviation are 6.7 and 0.24, respectively. Is there enough evidence to reject the company’s claim at α = 0.05? Assume the population is normally distributed. 72 Larson/Farber 4th ed.

73. Solution: Testing μ with a Small Sample H 0 : H a : α = df = Rejection Region: Test Statistic: Decision: μ = 6.8 μ ≠ 6.8 0.05 19 – 1 = 18 At the 0.05 level of significance, there is not enough evidence to reject the claim that the mean pH is 6.8. t 0-2.101 0.025 2.101 0.025 -2.1012.101 -1.816 Fail to reject H 0 73 Larson/Farber 4th ed.

74. Section 7.3 Summary Found critical values in a t-distribution Used the t-test to test a mean μ Used technology to find P-values and used them with a t-test to test a mean μ 74 Larson/Farber 4th ed.

75. Section 7.4 Hypothesis Testing for Proportions 75 Larson/Farber 4th ed.

76. Section 7.4 Objectives Use the z-test to test a population proportion p 76 Larson/Farber 4th ed.

77. z-Test for a Population Proportion A statistical test for a population proportion. Can be used when a binomial distribution is given such that np  5 and nq  5. The test statistic is the sample proportion. The standardized test statistic is z. 77 Larson/Farber 4th ed.

78. Using a z-Test for a Proportion p 1.State the claim mathematically and verbally. Identify the null and alternative hypotheses. 2.Specify the level of significance. 3.Sketch the sampling distribution. 4.Determine any critical value(s). State H 0 and H a. Identify . Use Table 5 in Appendix B. Verify that np ≥ 5 and nq ≥ 5 78 Larson/Farber 4th ed. In WordsIn Symbols

79. Using a z-Test for a Proportion p 5.Determine any rejection region(s). 6.Find the standardized test statistic. 7.Make a decision to reject or fail to reject the null hypothesis. 8.Interpret the decision in the context of the original claim. If z is in the rejection region, reject H 0. Otherwise, fail to reject H 0. 79 Larson/Farber 4th ed. In WordsIn Symbols

80. Example: Hypothesis Test for Proportions Zogby International claims that 45% of people in the United States support making cigarettes illegal within the next 5 to 10 years. You decide to test this claim and ask a random sample of 200 people in the United States whether they support making cigarettes illegal within the next 5 to 10 years. Of the 200 people, 49% support this law. At α = 0.05 is there enough evidence to reject the claim? Solution: Verify that np ≥ 5 and nq ≥ 5. np = 200(0.45) = 90 and nq = 200(0.55) = 110 80 Larson/Farber 4th ed.

81. Solution: Hypothesis Test for Proportions H 0 : H a :  = Rejection Region: p = 0.45 p ≠ 0.45 0.05 Decision: At the 5% level of significance, there is not enough evidence to reject the claim that 45% of people in the U.S. support making cigarettes illegal within the next 5 to 10 years. Test Statistic z 0-1.96 0.025 1.96 0.025 -1.961.96 1.14 Fail to reject H 0 81 Larson/Farber 4th ed.

82. Example: Hypothesis Test for Proportions The Pew Research Center claims that 55% of U.S. adults regularly watch their local television news. You decide to test this claim and ask a random sample of 425 adults in the United States whether they regularly watch their local television news. Of the 425 adults, 255 respond yes. At α = 0.05 is there enough evidence to support the claim? Solution: Verify that np ≥ 5 and nq ≥ 5. np = 425(0.55) ≈ 234 and nq = 425 (0.45) ≈ 191 82 Larson/Farber 4th ed.

83. Solution: Hypothesis Test for Proportions H 0 : H a :  = Rejection Region: p > 0.55 0.05 Decision: At the 5% level of significance, there is enough evidence to support the claim that more than 55% of U.S. adults regularly watch their local television news. Test Statistic z 01.645 0.05 2.07 1.645 Reject H 0 83 Larson/Farber 4th ed.

84. Section 7.4 Summary Used the z-test to test a population proportion p 84 Larson/Farber 4th ed.