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. © 2012 Pearson Education, Inc. All rights reserved. 1 of 101

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 H0. This region is indicated by the alternative hypothesis. © 2012 Pearson Education, Inc. All rights reserved. 2 of 101

Left-tailed Test The alternative hypothesis Ha contains the less-than inequality symbol (<). H0: μ ≥ k Ha: μ < k P is the area to the left of the standardized test statistic. z 1 2 3 –3 –2 –1 Test statistic © 2012 Pearson Education, Inc. All rights reserved. 3 of 101

Right-tailed Test The alternative hypothesis Ha contains the greater-than inequality symbol (>). H0: μ ≤ k Ha: μ > k P is the area to the right of the standardized test statistic. z 1 2 3 –3 –2 –1 Test statistic © 2012 Pearson Education, Inc. All rights reserved. 4 of 101

Two-tailed Test The alternative hypothesis Ha contains the not-equal-to symbol (≠). Each tail has an area of ½P. H0: μ = k Ha: μ ≠ k P is twice the area to the right of the positive standardized test statistic. P is twice the area to the left of the negative standardized test statistic. z 1 2 3 –3 –2 –1 Test statistic Test statistic © 2012 Pearson Education, Inc. All rights reserved. 5 of 101

Example: Identifying The Nature of a Test For each claim, state H0 and Ha. 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 school publicizes that the proportion of its students who are involved in at least one extracurricular activity is 61%. Solution: H0: Ha: p = 0.61 p ≠ 0.61 Two-tailed test © 2012 Pearson Education, Inc. All rights reserved. 6 of 101

Example: Identifying The Nature of a Test For each claim, state H0 and Ha. 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 car dealership announces that the mean time for an oil change is less than 15 minutes. Solution: z -z P-value area H0: Ha: μ ≥ 15 min μ < 15 min Left-tailed test © 2012 Pearson Education, Inc. All rights reserved. 7 of 101

Example: Identifying The Nature of a Test For each claim, state H0 and Ha. 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 company advertises that the mean life of its furnaces is more than 18 years. Solution: z P-value area H0: Ha: μ ≤ 18 yr μ > 18 yr Right-tailed test © 2012 Pearson Education, Inc. All rights reserved. 8 of 101

Making a Decision Decision Rule Based on P-value Compare the P-value with α. If P ≤ α , then reject H0. If P > α, then fail to reject H0. Claim Decision Claim is H0 Claim is Ha Reject H0 Fail to reject H0 There is enough evidence to reject the claim There is enough evidence to support the claim There is not enough evidence to reject the claim There is not enough evidence to support the claim © 2012 Pearson Education, Inc. All rights reserved. 9 of 101

Example: Interpreting a Decision You perform a hypothesis test for the following claim. How should you interpret your decision if you reject H0? If you fail to reject H0? H0 (Claim): A school publicizes that the proportion of its students who are involved in at least one extracurricular activity is 61%. Solution: The claim is represented by H0. © 2012 Pearson Education, Inc. All rights reserved. 10 of 101

Solution: Interpreting a Decision If you reject H0, then you should conclude “there is enough evidence to reject the school’s claim that the proportion of students who are involved in at least one extracurricular activity is 61%.” If you fail to reject H0, then you should conclude “there is not enough evidence to reject the school’s claim that proportion of students who are involved in at least one extracurricular activity is 61%.” © 2012 Pearson Education, Inc. All rights reserved. 11 of 101

Example: Interpreting a Decision You perform a hypothesis test for the following claim. How should you interpret your decision if you reject H0? If you fail to reject H0? Ha (Claim): A car dealership announces that the mean time for an oil change is less than 15 minutes. Solution: The claim is represented by Ha. H0 is “the mean time for an oil change is greater than or equal to 15 minutes.” © 2012 Pearson Education, Inc. All rights reserved. 12 of 101

Solution: Interpreting a Decision If you reject H0, then you should conclude “there is enough evidence to support the dealership’s claim that the mean time for an oil change is less than 15 minutes.” If you fail to reject H0, then you should conclude “there is not enough evidence to support the dealership’s claim that the mean time for an oil change is less than 15 minutes.” © 2012 Pearson Education, Inc. All rights reserved. 13 of 101

Steps for Hypothesis Testing State the claim mathematically and verbally. Identify the null and alternative hypotheses. H0: ? Ha: ? Specify the level of significance. α = ? Determine the standardized sampling distribution and sketch its graph. Calculate the test statistic and its corresponding standardized test statistic. Add it to your sketch. This sampling distribution is based on the assumption that H0 is true. z z Standardized test statistic © 2012 Pearson Education, Inc. All rights reserved. 14 of 101

Steps for Hypothesis Testing Find the P-value. Use the following decision rule. 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? No Fail to reject H0. Yes Reject H0. © 2012 Pearson Education, Inc. All rights reserved. 15 of 101

Section 7.1 Summary Stated a null hypothesis and an alternative hypothesis Identified type I and type II 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 © 2012 Pearson Education, Inc. All rights reserved. 16 of 101

Hypothesis Testing for the Mean (Large Samples) Section 7.2 Hypothesis Testing for the Mean (Large Samples) © 2012 Pearson Education, Inc. All rights reserved. 17 of 101

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 © 2012 Pearson Education, Inc. All rights reserved. 18 of 101

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 α. If P ≤ α, then reject H0. If P > α, then fail to reject H0. © 2012 Pearson Education, Inc. All rights reserved. 19 of 101

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 α = 0.05? α = 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. 20 of 101

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. For a left-tailed test, P = (Area in left tail). For a right-tailed test, P = (Area in right tail). For a two-tailed test, P = 2(Area in tail of test statistic). © 2012 Pearson Education, Inc. All rights reserved. 21 of 101

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) z -2.23 P = 0.0129 Because 0.0129 > 0.01, you should fail to reject H0. © 2012 Pearson Education, Inc. All rights reserved. 22 of 101

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) z 2.14 1 – 0.9838 = 0.0162 P = 2(0.0162) = 0.0324 0.9838 Because 0.0324 < 0.05, you should reject H0. © 2012 Pearson Education, Inc. All rights reserved. 23 of 101

Z-Test for a Mean μ Can be used when the population is normal and σ 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 σ. © 2012 Pearson Education, Inc. All rights reserved. 24 of 101

Using P-values for a z-Test for Mean μ In Words In Symbols State the claim mathematically and verbally. Identify the null and alternative hypotheses. Specify the level of significance. Determine the standardized test statistic. Find the area that corresponds to z. State H0 and Ha. Identify α. Use Table 4 in Appendix B. © 2012 Pearson Education, Inc. All rights reserved. 25 of 101

Using P-values for a z-Test for Mean μ In Words In Symbols Find the P-value. For a left-tailed test, P = (Area in left tail). For a right-tailed test, P = (Area in right tail). For a two-tailed test, P = 2(Area in tail of test statistic). Make a decision to reject or fail to reject the null hypothesis. Interpret the decision in the context of the original claim. Reject H0 if P-value is less than or equal to α. Otherwise, fail to reject H0. © 2012 Pearson Education, Inc. All rights reserved. 26 of 101

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 second. Is there enough evidence to support the claim at α = 0.01? Use a P-value. © 2012 Pearson Education, Inc. All rights reserved. 27 of 101

Solution: Hypothesis Testing Using P-values Ha: α = Test Statistic: μ ≥ 13 sec μ < 13 sec (Claim) 0.01 0.0014 < 0.01 Reject H0 . Decision: At the 1% level of significance, you have sufficient evidence to support the claim that the mean pit stop time is less than 13 seconds. © 2012 Pearson Education, Inc. All rights reserved. 28 of 101

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 α = 0.05? Use a P-value. (Adapted from National Institute of Diabetes and Digestive and Kidney Diseases) © 2012 Pearson Education, Inc. All rights reserved. 29 of 101

Solution: Hypothesis Testing Using P-values Ha: α = Test Statistic: μ = $22,500 μ ≠ 22,500 (Claim) 0.05 Decision: 0.0836 > 0.05 Fail to reject H0 . 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. © 2012 Pearson Education, Inc. All rights reserved. 30 of 101

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. © 2012 Pearson Education, Inc. All rights reserved. 31 of 101

Rejection Regions and Critical Values Finding Critical Values in a Normal Distribution Specify the level of significance α. Decide whether the test is left-, right-, or two-tailed. Find the critical value(s) z0. If the hypothesis test is left-tailed, find the z-score that corresponds to an area of α, right-tailed, find the z-score that corresponds to an area of 1 – α, two-tailed, find the z-score that corresponds to ½α and 1 – ½α. 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. 32 of 101

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 is in the rejection region, then reject H0. is not in the rejection region, then fail to reject H0. z z0 Fail to reject H0. Reject H0. Left-Tailed Test z < z0 z z0 Reject Ho. Fail to reject Ho. z > z0 Right-Tailed Test z –z0 Two-Tailed Test z0 z < –z0 z > z0 Reject H0 Fail to reject H0 © 2012 Pearson Education, Inc. All rights reserved. 33 of 101

Example: Finding Critical Values Find the critical value and rejection region for a two-tailed test with α = 0.05. Solution: 1 – α = 0.95 z z0 ½α = 0.025 ½α = 0.025 –z0 = –1.96 z0 = 1.96 The rejection regions are to the left of –z0 = –1.96 and to the right of z0 = 1.96. © 2012 Pearson Education, Inc. All rights reserved. 34 of 101

Using Rejection Regions for a z-Test for a Mean μ In Words In Symbols State the claim mathematically and verbally. Identify the null and alternative hypotheses. Specify the level of significance. Determine the critical value(s). Determine the rejection region(s). State H0 and Ha. Identify α. Use Table 4 in Appendix B. © 2012 Pearson Education, Inc. All rights reserved. 35 of 101

Using Rejection Regions for a z-Test for a Mean μ In Words In Symbols Find the standardized test statistic. Make a decision to reject or fail to reject the null hypothesis. Interpret the decision in the context of the original claim. If z is in the rejection region, reject H0. Otherwise, fail to reject H0. © 2012 Pearson Education, Inc. All rights reserved. 36 of 101

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. 37 of 101

Solution: Testing with Rejection Regions Ha: α = Rejection Region: μ ≥ $68,000 μ < $68,000 (Claim) Test Statistic 0.05 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 $68,000. © 2012 Pearson Education, Inc. All rights reserved. 38 of 101

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. 39 of 101

Solution: Testing with Rejection Regions Ha: α = Rejection Region: μ = $13,120 (Claim) μ ≠ $13,120 Test Statistic 0.10 Decision: Reject H0 . At the 10% level of significance, you have enough evidence to reject the claim that the mean cost of raising a child from birth to age 2 by husband-wife families in the U.S. is $13,120. © 2012 Pearson Education, Inc. All rights reserved. 40 of 101

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 © 2012 Pearson Education, Inc. All rights reserved. 41 of 101