1. Section 7.2 Hypothesis Testing for the Mean (Large Samples) © 2012 Pearson Education, Inc. All rights reserved. 1 of 31

2. 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 (start on slide 15) Use rejection regions for a z-test © 2012 Pearson Education, Inc. All rights reserved. 2 of 31

3. Using P-values to Make a Decision Decision Rule Based on P-value To use a P-value to make a conclusion in a hypothesis test, compare the P-value with . 1.If P  , then reject H 0. 2.If P > , then fail to reject H 0. © 2012 Pearson Education, Inc. All rights reserved. 3 of 31

4. 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. © 2012 Pearson Education, Inc. All rights reserved. 4 of 31

5. 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). © 2012 Pearson Education, Inc. All rights reserved. 5 of 31

6. 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 © 2012 Pearson Education, Inc. All rights reserved. 6 of 31

7. 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 © 2012 Pearson Education, Inc. All rights reserved. 7 of 31

8. 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 . © 2012 Pearson Education, Inc. All rights reserved. 8 of 31

9. 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. In WordsIn Symbols © 2012 Pearson Education, Inc. All rights reserved. 9 of 31

10. 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. In WordsIn Symbols © 2012 Pearson Education, Inc. All rights reserved. 10 of 31

11. Example: Hypothesis Testing Using P- values In auto racing, a pit crew claims that its mean pit stop time (for 4 new tires and fuel) is less than 13 seconds. A random selection of 32 pit stop times has a sample mean of 12.9 seconds and a standard deviation of 0.19 seconds. Is there enough evidence to support the claim at  = 0.01? Use a P-value. © 2012 Pearson Education, Inc. All rights reserved. 11 of 31

12. Solution: Hypothesis Testing Using P- values H 0 : H a :  = Test Statistic: μ ≥ 13 sec μ < 13 sec (claim) 0.01 Decision: At the 1% level of significance, you have sufficient evidence to conclude the mean pit stop time is less than 13 seconds. P-value 0.0014 < 0.01 Reject H 0 © 2012 Pearson Education, Inc. All rights reserved. 12 of 31

13. Example: Hypothesis Testing Using P- values The National Institute of Diabetes and Digestive and Kidney Diseases reports that the average cost of bariatric (weight loss) surgery is $22,500. You think this information is incorrect. You randomly select 30 bariatric surgery patients and find that the average cost for their surgeries is $21,545 with a standard deviation of $3015. Is there enough evidence to support your claim at  = 0.05? Use a P-value. © 2012 Pearson Education, Inc. All rights reserved. 13 of 31

14. Solution: Hypothesis Testing Using P- values H 0 : H a :  = Test Statistic: μ = $22,500 μ ≠ 22,500 (claim) 0.05 Decision: At the 5% level of significance, there is not sufficient evidence to support the claim that the mean cost of bariatric surgery is different from $22,500. P-value 0.0836 > 0.05 Fail to reject H 0 © 2012 Pearson Education, Inc. All rights reserved. 14 of 31

15. Rejection Regions and Critical Values Rejection region (or critical region) The range of values for which the null hypothesis is not probable. If a test statistic falls in this region, the null hypothesis is rejected. A critical value z 0 separates the rejection region from the nonrejection region. © 2012 Pearson Education, Inc. All rights reserved. 15 of 31

16. Rejection Regions and Critical Values Finding Critical Values in a Normal Distribution 1.Specify the level of significance . 2.Decide whether the test is left-, right-, or two-tailed. 3.Find the critical value(s) z 0. If the hypothesis test is a.left-tailed, find the z-score that corresponds to an area of , b.right-tailed, find the z-score that corresponds to an area of 1 – , c.two-tailed, find the z-score that corresponds to ½  and 1 – ½ . 4.Sketch the standard normal distribution. Draw a vertical line at each critical value and shade the rejection region(s). © 2012 Pearson Education, Inc. All rights reserved. 16 of 31

17. Example: Finding Critical Values Find the critical value and rejection region for a Left-Tailed Test with  = 0.05. z 0z0z0 α = 0.05 1 – α = 0.95 The rejection region are to the left of -z 0 = -1.645. -z 0 = -1.645 Solution: © 2012 Pearson Education, Inc. All rights reserved. 17 of 31

18. Example: Finding Critical Values Find the critical value and rejection region for a Right-Tailed Test with  = 0.05. z 0z0z0 α = 0.05 1 – α = 0.95 The rejection region are to the right of z 0 = 1.645. z 0 = 1.645 Solution: © 2012 Pearson Education, Inc. All rights reserved. 18 of 31

19. Example: Finding Critical Values Find the critical value and rejection region for a two-tailed test with  = 0.05. z 0z0z0 z0z0 ½ α = 0.025 1 – α = 0.95 The rejection regions are to the left of -z 0 = -1.96 and to the right of z 0 = 1.96. z 0 = 1.96-z 0 = -1.96 Solution: © 2012 Pearson Education, Inc. All rights reserved. 19 of 31

20. Page 377 Try It Yourself 7 and 8 Example: Finding Critical Values 20 of 31

21. 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 © 2012 Pearson Education, Inc. All rights reserved. 21 of 31

22. 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. In WordsIn Symbols © 2012 Pearson Education, Inc. All rights reserved. 22 of 31

23. 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. In WordsIn Symbols © 2012 Pearson Education, Inc. All rights reserved. 23 of 31

24. Example: Testing with Rejection Regions Employees at a construction and mining company claim that the mean salary of the company’s mechanical engineers is less than that of the one of its competitors, which is $68,000. A random sample of 30 of the company’s mechanical engineers has a mean salary of $66,900 with a standard deviation of $5500. At α = 0.05, test the employees’ claim. © 2012 Pearson Education, Inc. All rights reserved. 24 of 31

25. Solution: Testing with Rejection Regions H 0 : H a :  = Rejection Region: μ ≥ $68,000 μ < $68,000 (claim) 0.05 Decision: At the 5% level of significance, there is not sufficient evidence to support the employees’ claim that the mean salary is less than $68,000. Test Statistic Fail to reject H 0 © 2012 Pearson Education, Inc. All rights reserved. 25 of 31

26. Making a Decision Decision Rule Claim DecisionClaim is H 0 Claim is H a 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 © 2012 Pearson Education, Inc. All rights reserved. 26 of 31

27. Testing with Rejection Regions © 2012 Pearson Education, Inc. All rights reserved. 27 of 31 Page 379 Try It Yourself 9

28. Example: Testing with Rejection Regions The U.S. Department of Agriculture claims that the mean cost of raising a child from birth to age 2 by husband-wife families in the U.S. is $13,120. A random sample of 500 children (age 2) has a mean cost of $12,925 with a standard deviation of $1745. At α = 0.10, is there enough evidence to reject the claim? (Adapted from U.S. Department of Agriculture Center for Nutrition Policy and Promotion) © 2012 Pearson Education, Inc. All rights reserved. 28 of 31

29. Solution: Testing with Rejection Regions H 0 : H a :  = Rejection Region: μ = $13,120 (claim) μ ≠ $13,120 0.10 Decision: At the 10% level of significance, you do not have enough evidence to conclude the mean cost of raising a child from birth to age 2 by husband- wife families in the U.S. is $13,120. Test Statistic Reject H 0 © 2012 Pearson Education, Inc. All rights reserved. 29 of 31

30. Testing with Rejection Regions © 2012 Pearson Education, Inc. All rights reserved. 30 of 31 Page 380 Try It Yourself 10

31. 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 HW: pg 382 17-31 odd © 2012 Pearson Education, Inc. All rights reserved. 31 of 31