#### Transcript Ch7-Sec7.2

Section 7.2 Hypothesis Testing for the Mean (Large Samples) 1 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 2 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 H0. 2. If P > , then fail to reject H0. 3 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? Solution: Because 0.0237 < 0.05, you should reject the null hypothesis. 2. 0.01? Solution: Because 0.0237 > 0.01, you should fail to reject the null hypothesis. 4 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). 5 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 H0 if the level of significance is α = 0.01. Solution: For a left-tailed test, P = (Area in left tail) P = 0.0129 -2.23 0 z Because 0.0129 > 0.01, you should fail to reject H0 6 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 H0 if the level of significance is α = 0.05. Solution: For a two-tailed test, P = 2(Area in tail of test statistic) 1 – 0.9838 = 0.0162 0.9838 0 2.14 P = 2(0.0162) = 0.0324 z Because 0.0324 < 0.05, you should reject H0 7 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 x The standardized test statistic is z x where standard error z x n n When n 30, the sample standard deviation s can be substituted for . 8 Using P-values for a z-Test for Mean μ In Words 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. 9 In Symbols State H0 and Ha. Identify . z x n Use Table 4 in Appendix B. Using P-values for a z-Test for Mean μ In Words In Symbols 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. 10 Reject H0 if P-value is less than or equal to . Otherwise, fail to reject H0. Example: Hypothesis Testing Using Pvalues 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. 11 Solution: Hypothesis Testing Using Pvalues • • • • H0: μ ≥ 30 min Ha: μ < 30 min = 0.01 Test Statistic: z x n 28.5 30 3.5 36 2.57 12 • P-value 0.0051 -2.57 0 z • Decision: 0.0051 < 0.01 Reject H0 At the 1% level of significance, you have sufficient evidence to conclude the mean delivery time is less than 30 minutes. Example: Hypothesis Testing Using Pvalues 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. 13 Solution: Hypothesis Testing Using Pvalues • • • • H0: μ = $143,260 Ha: μ ≠ $143,260 = 0.05 Test Statistic: z x n 135, 000 143, 260 30, 000 1.51 14 30 • P-value P = 2(0.0655) = 0.1310 0.0655 -1.51 0 z • Decision: 0.1310 > 0.05 Fail to reject H0 At the 5% level of significance, there is not sufficient evidence to conclude the mean franchise investment is different from $143,260. 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 z0 separates the rejection region from the nonrejection region. 15 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) z0. 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). 16 Example: Finding Critical Values Find the critical value and rejection region for a two-tailed test with = 0.05. Solution: 1 – α = 0.95 ½α = 0.025 z0 -z0 = -1.96 ½α = 0.025 0 z0 =z01.96 z The rejection regions are to the left of -z0 = -1.96 and to the right of z0 = 1.96. 17 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 H0. 2. is not in the rejection region, then fail to reject H0. Fail to reject Ho. Fail to reject H0. Reject H0. z < z0 Reject Ho. z0 z 0 Fail to reject H0 Left-Tailed Test Reject H0 z < -z0 z0 18 0 0 z > z0 Right-Tailed Test Reject H0 z z0 z > z0 Two-Tailed Test z0 z Using Rejection Regions for a z-Test for a Mean μ In Words 1. State the claim mathematically and verbally. Identify the null and alternative hypotheses. 2. Specify the level of significance. In Symbols State H0 and Ha. Identify . 3. Sketch the sampling distribution. 4. Determine the critical value(s). 5. Determine the rejection region(s). 19 Use Table 4 in Appendix B. Using Rejection Regions for a z-Test for a Mean μ In Words 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. 20 In Symbols x or if n 30 n use s. z If z is in the rejection region, reject H0. Otherwise, fail to reject H0. 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. 21 Solution: Testing with Rejection Regions • • • • H0: μ ≥ $45,000 Ha: μ < $45,000 = 0.05 Rejection Region: • Test Statistic x 43,500 45, 000 z n 5200 30 0.05 -1.645 0 -1.58 22 z 1.58 • Decision: Fail to reject H0 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. 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) 23 Solution: Testing with Rejection Regions • • • • H0: μ = $10,460 Ha: μ ≠ $10,460 = 0.05 Rejection Region: 0.025 -1.96 -2.24 24 • Test Statistic x 10,345 10, 460 z n 1540 900 0.025 0 1.96 z 2.24 • Decision: Reject H0 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. 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 25