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

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

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 Larson/Farber 4th ed.

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. Larson/Farber 4th ed.

Example: Interpreting a P-value The P-value for a hypothesis test is P = 0.0132. What is your decision if the level of significance is Solution: Because 0.0132 < 0.05, you should reject the null hypothesis. Solution: Because 0.0132 > 0.01, you should fail to reject the null hypothesis. Larson/Farber 4th ed. 2.0.01? 1.0.05?

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). Larson/Farber 4th ed.

Example: Finding the P-value Find the P-value for a left-tailed hypothesis test with a test statistic of z = -1.99. Decide whether to reject H 0 if the level of significance is α = 0.01. z 0-1.99 P = 0.0233 Solution: For a left-tailed test, P = (Area in left tail) Because 0.0233 > 0.01, you should fail to reject H 0 Larson/Farber 4th ed.

z 01.82 Example: Finding the P-value Find the P-value for a two-tailed hypothesis test with a test statistic of z = 1.82. 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 0.0688 > 0.05, you should fail to reject H 0 0.9838 1 – 0.9656 = 0.0344 P = 2(0.0344) = 0.0688 Larson/Farber 4th ed.

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 . Larson/Farber 4th ed.

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. Larson/Farber 4th ed. In WordsIn Symbols

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. Larson/Farber 4th ed. In WordsIn Symbols

Example: Hypothesis Testing Using P- values A manufacturer of sprinkler systems designed for fire protection claims that the average activating temperature is at least 135˚F. To test the claim, you randomly select a sample of 32 systems and find the mean activation temperature to be 133˚F with a standard deviation of 3.3˚F. Is there enough evidence to support the claim at  = 0.01? Use a P-value. Larson/Farber 4th ed.

Solution: Hypothesis Testing Using P- values H 0 : H a :  = Test Statistic: μ ≥ 135 (Claim) μ < 135 0.01 Decision: At the 10% level of significance, you have sufficient evidence to reject the manufacturer’s claim that the average activating temperature is at least 135 degrees. 0-3.43 z 0.003 P-value 0.0003 < 0.01 Reject H 0 Larson/Farber 4th ed.

Example: Hypothesis Testing Using P- values An Alabama politician claims that the mean annual salary for engineering managers in Alabama is more than the national mean of \$100,800. We take a random sample of 34 engineering managers salaries in Alabama. At α = 0.03, is there enough evidence to support the politician’s claim? Larson/Farber 4th ed.

Solution: Hypothesis Testing Using P- values H 0 : H a :  = Test Statistic: μ <= \$100,800 μ > \$100,800, (Claim) 0.03 (0.9700), z = 1.88 Decision: At the 3% level of significance, there is not enough evidence to support the politician’s claim! P-value P ≈ 1 Larson/Farber 4th ed. => ~ 0 Fail to reject H 0 z 01.88 0.03 -6.58 1 > 0.03

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. Larson/Farber 4th ed.

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). Larson/Farber 4th ed.

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 0 = -1.96 and to the right of z 0 = 1.96. z 0 = 1.96-z 0 = -1.96 Solution: Larson/Farber 4th ed.

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 Larson/Farber 4th ed.

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. Larson/Farber 4th ed. In WordsIn Symbols

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

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. Larson/Farber 4th ed.

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 Larson/Farber 4th ed.

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) Larson/Farber 4th ed.

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 Larson/Farber 4th ed.

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: 1 - 45 EO Larson/Farber 4th ed.

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