1. SECTION 7.2 Hypothesis Testing for the Mean (Large Samples) 1 Larson/Farber 4th ed.

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  Use rejection regions for a z-test 2Larson/Farber 4th ed.

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. 3Larson/Farber 4th ed.

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. 4Larson/Farber 4th ed.

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). 5Larson/Farber 4th ed.

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 6Larson/Farber 4th ed.

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 0.983 8 1 – 0.9838 = 0.0162 P = 2(0.0162) = 0.0324 7Larson/Farber 4th ed.

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 . 8Larson/Farber 4th ed.

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. 9Larson/Farber 4th ed. In WordsIn Symbols

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. 10Larson/Farber 4th ed. In WordsIn Symbols

11. Example: Hypothesis Testing Using P- values In an advertisement, a pizza shop claims that its mean delivery time is less than 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. 11Larson/Farber 4th ed.

12. Solution: Hypothesis Testing Using P- values H 0 : H a :  = Test Statistic: μ ≥ 30 min μ < 30 min 0.01 Decision: At the 1% level of significance, you have sufficient evidence to conclude the mean delivery time is less than 30 minutes. z 0-2.57 0.0051 P-value 0.0051 < 0.01 Reject H 0 12Larson/Farber 4th ed.

13. 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. 13Larson/Farber 4th ed.

14. Solution: Hypothesis Testing Using P- values H 0 : H a :  = Test Statistic: μ =$143,260 μ ≠ $143,260 0.05 Decision: At the 5% level of significance, there is not sufficient evidence to conclude the mean franchise investment is different from $143,260. P-value P = 2(0.0655) = 0.1310 0.1310 > 0.05 Fail to reject H 0 z 0-1.51 0.0655 14Larson/Farber 4th ed.

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. 15Larson/Farber 4th ed.

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). 16Larson/Farber 4th ed.

17. 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: 17Larson/Farber 4th ed.

18. 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 18Larson/Farber 4th ed.

19. 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. 19Larson/Farber 4th ed. In WordsIn Symbols

20. 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. 20Larson/Farber 4th ed. In WordsIn Symbols

21. Example: Testing with Rejection Regions Employees in a large accounting firm claim that the mean salary of the firm’s accountants is less than 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. 21Larson/Farber 4th ed.

22. Solution: Testing with Rejection Regions H 0 : H a :  = Rejection Region: μ ≥ $45,000 μ < $45,000 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 $45,000. Test Statistic z 0-1.645 0.05 -1.58 -1.645 Fail to reject H 0 22Larson/Farber 4th ed.

23. 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) 23Larson/Farber 4th ed.

24. 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 24Larson/Farber 4th ed.

25. 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 25Larson/Farber 4th ed.