Chapter Hypothesis Testing with One Sample 1 of 101 7 © 2012 Pearson Education, Inc. All rights reserved.

Chapter Outline 7.1 Introduction to Hypothesis Testing 7.2 Hypothesis Testing for the Mean (Large Samples) 7.3 Hypothesis Testing for the Mean (Small Samples) 7.4 Hypothesis Testing for Proportions 7.5 Hypothesis Testing for Variance and Standard Deviation © 2012 Pearson Education, Inc. All rights reserved. 2 of 101

Section 7.1 Objectives State a null hypothesis and an alternative hypothesis Identify type I and type II errors and interpret the level of significance Determine whether to use a one-tailed or two-tailed statistical test and find a p-value Make and interpret a decision based on the results of a statistical test Write a claim for a hypothesis test © 2012 Pearson Education, Inc. All rights reserved. 4 of 101

Hypothesis Tests Hypothesis test A process that uses sample statistics to test a claim about the value of a population parameter. For example: An automobile manufacturer advertises that its new hybrid car has a mean mileage of 50 miles per gallon. To test this claim, a sample would be taken. If the sample mean differs enough from the advertised mean, you can decide the advertisement is wrong. © 2012 Pearson Education, Inc. All rights reserved. 5 of 101

Hypothesis Tests Statistical hypothesis A statement, or claim, about a population parameter. Need a pair of hypotheses one that represents the claim the other, its complement When one of these hypotheses is false, the other must be true. © 2012 Pearson Education, Inc. All rights reserved. 6 of 101

Stating a Hypothesis Null hypothesis A statistical hypothesis that contains a statement of equality such as ≤, =, or ≥. Denoted H 0 read “H sub-zero” or “H naught.” Alternative hypothesis A statement of strict inequality such as >, ≠, or <. Must be true if H 0 is false. Denoted H a read “H sub-a.” complementary statements © 2012 Pearson Education, Inc. All rights reserved. 7 of 101

Stating a Hypothesis To write the null and alternative hypotheses, translate the claim made about the population parameter from a verbal statement to a mathematical statement. Then write its complement. H 0 : μ ≤ k H a : μ > k H 0 : μ ≥ k H a : μ < k H 0 : μ = k H a : μ ≠ k Regardless of which pair of hypotheses you use, you always assume μ = k and examine the sampling distribution on the basis of this assumption. © 2012 Pearson Education, Inc. All rights reserved. 8 of 101

Example: Stating the Null and Alternative Hypotheses Write the claim as a mathematical sentence. State the null and alternative hypotheses and identify which represents the claim. 1. A school publicizes that the proportion of its students who are involved in at least one extracurricular activity is 61%. Equality condition Complement of H 0 H0:Ha:H0:Ha: (Claim) p = 0.61 p ≠ 0.61 Solution: © 2012 Pearson Education, Inc. All rights reserved. 9 of 101

μ ≥ 15 minutes Example: Stating the Null and Alternative Hypotheses Write the claim as a mathematical sentence. State the null and alternative hypotheses and identify which represents the claim. 2. A car dealership announces that the mean time for an oil change is less than 15 minutes. Inequality condition Complement of H a H0:Ha:H0:Ha: (Claim) μ < 15 minutes Solution: © 2012 Pearson Education, Inc. All rights reserved. 10 of 101

μ ≤ 18 years Example: Stating the Null and Alternative Hypotheses Write the claim as a mathematical sentence. State the null and alternative hypotheses and identify which represents the claim. 3. A company advertises that the mean life of its furnaces is more than 18 years Inequality condition Complement of H a H0:Ha:H0:Ha: (Claim) μ > 18 years Solution: © 2012 Pearson Education, Inc. All rights reserved. 11 of 101

Types of Errors No matter which hypothesis represents the claim, always begin the hypothesis test assuming that the equality condition in the null hypothesis is true. At the end of the test, one of two decisions will be made:  reject the null hypothesis  fail to reject the null hypothesis Because your decision is based on a sample, there is the possibility of making the wrong decision. © 2012 Pearson Education, Inc. All rights reserved. 12 of 101

Types of Errors A type I error occurs if the null hypothesis is rejected when it is true. A type II error occurs if the null hypothesis is not rejected when it is false. Actual Truth of H 0 DecisionH 0 is trueH 0 is false Do not reject H 0 Correct DecisionType II Error Reject H 0 Type I ErrorCorrect Decision © 2012 Pearson Education, Inc. All rights reserved. 13 of 101

Example: Identifying Type I and Type II Errors The USDA limit for salmonella contamination for chicken is 20%. A meat inspector reports that the chicken produced by a company exceeds the USDA limit. You perform a hypothesis test to determine whether the meat inspector’s claim is true. When will a type I or type II error occur? Which is more serious? (Source: United States Department of Agriculture) © 2012 Pearson Education, Inc. All rights reserved. 14 of 101

Let p represent the proportion of chicken that is contaminated. Solution: Identifying Type I and Type II Errors H0:Ha:H0:Ha: p ≤ 0.2 p > 0.2 Hypotheses: (Claim) 0.160.180.200.220.24 p H 0 : p ≤ 0.20H 0 : p > 0.20 Chicken meets USDA limits. Chicken exceeds USDA limits. © 2012 Pearson Education, Inc. All rights reserved. 15 of 101

Solution: Identifying Type I and Type II Errors A type I error is rejecting H 0 when it is true. The actual proportion of contaminated chicken is less than or equal to 0.2, but you decide to reject H 0. A type II error is failing to reject H 0 when it is false. The actual proportion of contaminated chicken is greater than 0.2, but you do not reject H 0. H0:Ha:H0:Ha: p ≤ 0.2 p > 0.2 Hypotheses: (Claim) © 2012 Pearson Education, Inc. All rights reserved. 16 of 101

Solution: Identifying Type I and Type II Errors H0:Ha:H0:Ha: p ≤ 0.2 p > 0.2 Hypotheses: (Claim) With a type I error, you might create a health scare and hurt the sales of chicken producers who were actually meeting the USDA limits. With a type II error, you could be allowing chicken that exceeded the USDA contamination limit to be sold to consumers. A type II error could result in sickness or even death. © 2012 Pearson Education, Inc. All rights reserved. 17 of 101

Level of Significance Level of significance Your maximum allowable probability of making a type I error.  Denoted by α, the lowercase Greek letter alpha. By setting the level of significance at a small value, you are saying that you want the probability of rejecting a true null hypothesis to be small. Commonly used levels of significance:  α = 0.10α = 0.05α = 0.01 P(type II error) = β (beta) © 2012 Pearson Education, Inc. All rights reserved. 18 of 101

Statistical Tests After stating the null and alternative hypotheses and specifying the level of significance, a random sample is taken from the population and sample statistics are calculated. The statistic that is compared with the parameter in the null hypothesis is called the test statistic. σ2σ2 χ 2 (Section 7.5)s2s2 z (Section 7.4) p t (Section 7.3 n < 30) z (Section 7.2 n ≥ 30) μ Standardized test statistic Test statisticPopulation parameter © 2012 Pearson Education, Inc. All rights reserved. 19 of 101

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. 20 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 H 0. This region is indicated by the alternative hypothesis. © 2012 Pearson Education, Inc. All rights reserved. 21 of 101

Left-tailed Test The alternative hypothesis H a contains the less-than inequality symbol (<). z 0123–3–2–1 Test statistic H 0 : μ ≥ k H a : μ < k P is the area to the left of the standardized test statistic. © 2012 Pearson Education, Inc. All rights reserved. 22 of 101

The alternative hypothesis H a contains the greater- than inequality symbol (>). Right-tailed Test H 0 : μ ≤ k H a : μ > k Test statistic © 2012 Pearson Education, Inc. All rights reserved. 23 of 101 z 0123–3–2–1 P is the area to the right of the standardized test statistic.

Two-tailed Test The alternative hypothesis H a contains the not-equal- to symbol (≠). Each tail has an area of ½P. z 0123–3–2–1 Test statistic H 0 : μ = k H a : μ ≠ k P is twice the area to the left of the negative standardized test statistic. P is twice the area to the right of the positive standardized test statistic. © 2012 Pearson Education, Inc. All rights reserved. 24 of 101

Example: Identifying The Nature of a Test For each claim, state H 0 and H a. 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. 1. A school publicizes that the proportion of its students who are involved in at least one extracurricular activity is 61%. H0:Ha:H0:Ha: p = 0.61 p ≠ 0.61 Two-tailed test Solution: © 2012 Pearson Education, Inc. All rights reserved. 25 of 101

Example: Identifying The Nature of a Test For each claim, state H 0 and H a. 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. 2. A car dealership announces that the mean time for an oil change is less than 15 minutes. H0:Ha:H0:Ha: Left-tailed test z 0-z P-value area μ ≥ 15 min μ < 15 min Solution: © 2012 Pearson Education, Inc. All rights reserved. 26 of 101

Example: Identifying The Nature of a Test For each claim, state H 0 and H a. 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. 3. A company advertises that the mean life of its furnaces is more than 18 years. H0:Ha:H0:Ha: Right-tailed test z 0 z P-value area μ ≤ 18 yr μ > 18 yr Solution: © 2012 Pearson Education, Inc. All rights reserved. 27 of 101

Making a Decision Decision Rule Based on P-value Compare the P-value with α.  If P ≤ α, then reject H 0.  If P > α, then fail to reject H 0. 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. 28 of 101

Example: Interpreting a Decision You perform a hypothesis test for the following claim. How should you interpret your decision if you reject H 0 ? If you fail to reject H 0 ? 1. H 0 (Claim): A school publicizes that the proportion of its students who are involved in at least one extracurricular activity is 61%. © 2012 Pearson Education, Inc. All rights reserved. 29 of 101 Solution: The claim is represented by H 0.

Solution: Interpreting a Decision If you reject H 0, 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 H 0, 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. 30 of 101

Example: Interpreting a Decision You perform a hypothesis test for the following claim. How should you interpret your decision if you reject H 0 ? If you fail to reject H 0 ? 2. H a (Claim): A car dealership announces that the mean time for an oil change is less than 15 minutes. Solution: The claim is represented by H a. H 0 is “the mean time for an oil change is greater than or equal to 15 minutes.” © 2012 Pearson Education, Inc. All rights reserved. 31 of 101

Solution: Interpreting a Decision If you reject H 0, 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 H 0, 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. 32 of 101

z 0 Steps for Hypothesis Testing 1.State the claim mathematically and verbally. Identify the null and alternative hypotheses. H 0 : ? H a : ? 2.Specify the level of significance. α = ? 3.Determine the standardized sampling distribution and sketch its graph. 4.Calculate the test statistic and its corresponding standardized test statistic. Add it to your sketch. z 0 Standardized test statistic This sampling distribution is based on the assumption that H 0 is true. © 2012 Pearson Education, Inc. All rights reserved. 33 of 101

Steps for Hypothesis Testing 5.Find the P-value. 6.Use the following decision rule. 7.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? Fail to reject H 0. Yes Reject H 0. No © 2012 Pearson Education, Inc. All rights reserved. 34 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. 35 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. 37 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. 1.If P ≤ α, then reject H 0. 2.If P > α, then fail to reject H 0. © 2012 Pearson Education, Inc. All rights reserved. 38 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 1. α = 0.05? 2. 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. 39 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. 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. 40 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 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. 41 of 101

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. 42 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. 43 of 101

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 α. Use Table 4 in Appendix B. In WordsIn Symbols © 2012 Pearson Education, Inc. All rights reserved. 44 of 101

Using P-values for a z-Test for Mean μ Reject H 0 if P-value is less than or equal to α. 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. 45 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. 46 of 101

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 support the claim that 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. 47 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. 48 of 101

Solution: Hypothesis Testing Using P- values H 0 : H a : α = 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. 49 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 z 0 separates the rejection region from the nonrejection region. © 2012 Pearson Education, Inc. All rights reserved. 50 of 101

Rejection Regions and Critical Values Finding Critical Values in a Normal Distribution 1.Specify the level of significance α. 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 α, b.right-tailed, find the z-score that corresponds to an area of 1 – α, c.two-tailed, find the z-score that corresponds to ½ α and 1 – ½ α. 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. 51 of 101

Example: Finding Critical Values Find the critical value and rejection region for a two- tailed test with α = 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: © 2012 Pearson Education, Inc. All rights reserved. 52 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 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. 53 of 101

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.Determine the critical value(s). 4.Determine the rejection region(s). State H 0 and H a. Identify α. Use Table 4 in Appendix B. In WordsIn Symbols © 2012 Pearson Education, Inc. All rights reserved. 54 of 101

Using Rejection Regions for a z-Test for a Mean μ 5.Find the standardized 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. 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. 55 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. 56 of 101

Solution: Testing with Rejection Regions H 0 : H a : α = 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. 57 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. 58 of 101

Solution: Testing with Rejection Regions H 0 : H a : α = Rejection Region: μ = $13,120 (Claim) μ ≠ $13,120 0.10 Decision: 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. Test Statistic Reject H 0. © 2012 Pearson Education, Inc. All rights reserved. 59 of 101

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 © 2012 Pearson Education, Inc. All rights reserved. 60 of 101

Section 7.3 Objectives Find critical values in a t-distribution Use the t-test to test a mean μ Use technology to find P-values and use them with a t-test to test a mean μ © 2012 Pearson Education, Inc. All rights reserved. 62 of 101

Finding Critical Values in a t-Distribution 1.Identify the level of significance α. 2.Identify the degrees of freedom d.f. = n – 1. 3.Find the critical value(s) using Table 5 in Appendix B in the row with n – 1 degrees of freedom. If the hypothesis test is a.left-tailed, use “One Tail, α ” column with a negative sign, b.right-tailed, use “One Tail, α ” column with a positive sign, c.two-tailed, use “Two Tails, α ” column with a negative and a positive sign. © 2012 Pearson Education, Inc. All rights reserved. 63 of 101

Example: Finding Critical Values for t Find the critical value t 0 for a left-tailed test given α = 0.05 and n = 21. Solution: The degrees of freedom are d.f. = n – 1 = 21 – 1 = 20. Look at α = 0.05 in the “One Tail, α” column. Because the test is left- tailed, the critical value is negative. t 0 t 0 = –1.725 0.05 © 2012 Pearson Education, Inc. All rights reserved. 64 of 101

Example: Finding Critical Values for t Find the critical values –t 0 and t 0 for a two-tailed test given α = 0.10 and n = 26. Solution: The degrees of freedom are d.f. = n – 1 = 26 – 1 = 25. Look at α = 0.10 in the “Two Tail, α” column. Because the test is two- tailed, one critical value is negative and one is positive. © 2012 Pearson Education, Inc. All rights reserved. 65 of 101

t-Test for a Mean μ (n < 30, σ Unknown) t-Test for a Mean A statistical test for a population mean. The t-test can be used when the population is normal or nearly normal, σ is unknown, and n < 30. The test statistic is the sample mean The standardized test statistic is t. The degrees of freedom are d.f. = n – 1. © 2012 Pearson Education, Inc. All rights reserved. 66 of 101

Using the t-Test for a Mean μ (Small Sample) 1.State the claim mathematically and verbally. Identify the null and alternative hypotheses. 2.Specify the level of significance. 3.Identify the degrees of freedom. 4.Determine the critical value(s). 5.Determine the rejection region(s). State H 0 and H a. Identify α. Use Table 5 in Appendix B. d.f. = n – 1. In WordsIn Symbols © 2012 Pearson Education, Inc. All rights reserved. 67 of 101

Using the t-Test for a Mean μ (Small Sample) 6.Find the standardized test statistic and sketch the sampling distribution 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 t is in the rejection region, reject H 0. Otherwise, fail to reject H 0. In WordsIn Symbols © 2012 Pearson Education, Inc. All rights reserved. 68 of 101

Example: Testing μ with a Small Sample A used car dealer says that the mean price of a 2008 Honda CR-V is at least $20,500. You suspect this claim is incorrect and find that a random sample of 14 similar vehicles has a mean price of $19,850 and a standard deviation of $1084. Is there enough evidence to reject the dealer’s claim at α = 0.05? Assume the population is normally distributed. (Adapted from Kelley Blue Book) © 2012 Pearson Education, Inc. All rights reserved. 69 of 101

Solution: Testing μ with a Small Sample H 0 : H a : α = df = Rejection Region: Test Statistic: Decision: μ ≥ $20,500 (Claim) μ < $20,500 0.05 14 – 1 = 13 At the 5% level of significance, there is enough evidence to reject the claim that the mean price of a 2008 Honda CR-V is at least $20,500. Reject H 0. © 2012 Pearson Education, Inc. All rights reserved. 70 of 101 t ≈ –2.244

Example: Testing μ with a Small Sample An industrial company claims that the mean pH level of the water in a nearby river is 6.8. You randomly select 19 water samples and measure the pH of each. The sample mean and standard deviation are 6.7 and 0.24, respectively. Is there enough evidence to reject the company’s claim at α = 0.05? Assume the population is normally distributed. © 2012 Pearson Education, Inc. All rights reserved. 71 of 101

Solution: Testing μ with a Small Sample H 0 : H a : α = df = Rejection Region: Test Statistic: Decision: μ = 6.8 (Claim) μ ≠ 6.8 0.05 19 – 1 = 18 At the 5% level of significance, there is not enough evidence to reject the claim that the mean pH is 6.8. t 0–2.101 0.025 2.101 0.025 –2.1012.101 –1.816 Fail to reject H 0. © 2012 Pearson Education, Inc. All rights reserved. 72 of 101

Example: Using P-values with t-Tests A Department of Motor Vehicles office claims that the mean wait time is less than 14 minutes. A random sample of 10 people has a mean wait time of 13 minutes with a standard deviation of 3.5 minutes. At α = 0.10, test the office’s claim. Assume the population is normally distributed. © 2012 Pearson Education, Inc. All rights reserved. 73 of 101

Solution: Using P-values with t-Tests © 2012 Pearson Education, Inc. All rights reserved. 74 of 101 H 0 : H a : Decision: μ ≥ 14 min μ < 14 min (Claim) TI-83/84 setup:Calculate:Draw: At the 10% level of significance, there is not enough evidence to support the office’s claim that the mean wait time is less than 14 minutes. P ≈ 0.1949 Since 0.1949 > 0.10, fail to reject H 0.

Section 7.3 Summary Found critical values in a t-distribution Used the t-test to test a mean μ Used technology to find P-values and used them with a t-test to test a mean μ © 2012 Pearson Education, Inc. All rights reserved. 75 of 101

z-Test for a Population Proportion z-Test for a Population Proportion p A statistical test for a population proportion p. Can be used when a binomial distribution is given such that np ≥ 5 and nq ≥ 5. The test statistic is the sample proportion. The standardized test statistic is z. © 2012 Pearson Education, Inc. All rights reserved. 78 of 101

Using a z-Test for a Proportion p 1.State the claim mathematically and verbally. Identify the null and alternative hypotheses. 2.Specify the level of significance. 3.Determine the critical value(s). 4.Determine the rejection region(s). State H 0 and H a. Identify α. Use Table 4 in Appendix B. Verify that np ≥ 5 and nq ≥ 5. In WordsIn Symbols © 2012 Pearson Education, Inc. All rights reserved. 79 of 101

Using a z-Test for a Proportion p 5.Find the standardized test statistic and sketch the sampling distribution. 6.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. In WordsIn Symbols © 2012 Pearson Education, Inc. All rights reserved. 80 of 101

Example: Hypothesis Test for Proportions A research center claims that less than 50% of U.S. adults have accessed the Internet over a wireless network with a laptop computer. In a random sample of 100 adults, 39% say they have accessed the Internet over a wireless network with a laptop computer. At α = 0.01, is there enough evidence to support the researcher’s claim? (Adopted from Pew Research Center) Solution: Verify that np ≥ 5 and nq ≥ 5. np = 100(0.50) = 50 and nq = 100(0.50) = 50 © 2012 Pearson Education, Inc. All rights reserved. 81 of 101

Solution: Hypothesis Test for Proportions H 0 : H a : α = Rejection Region: p ≥ 0.5 p < 0.5 Claim 0.01 Decision: At the 1% level of significance, there is not enough evidence to support the claim that less than 50% of U.S. adults have accessed the Internet over a wireless network with a laptop computer. Test Statistic Fail to reject H 0. © 2012 Pearson Education, Inc. All rights reserved. 82 of 101 z = –2.2

Example: Hypothesis Test for Proportions A research center claims that 25% of college graduates think a college degree is not worth the cost. You decide to test this claim and ask a random sample of 200 college graduates whether they think a college degree is not worth the cost. Of those surveyed, 21% reply yes. At α = 0.10 is there enough evidence to reject the claim? Solution: Verify that np ≥ 5 and nq ≥ 5. np = 200(0.25) = 50 and nq = 200 (0.75) = 150 © 2012 Pearson Education, Inc. All rights reserved. 83 of 101

Solution: Hypothesis Test for Proportions H 0 : H a : α = Rejection Region: p = 0.25 (Claim) p ≠ 0.25 0.10 Decision: At the 10% level of significance, there is not enough evidence to reject the claim that 25% of college graduates think a college degree is not worth the cost. Test Statistic Fail to reject H 0. © 2012 Pearson Education, Inc. All rights reserved. 84 of 101 z ≈ –1.31

Finding Critical Values for the χ 2 -Test 1.Specify the level of significance α. 2.Determine the degrees of freedom d.f. = n – 1. 3.The critical values for the χ 2 -distribution are found in Table 6 in Appendix B. To find the critical value(s) for a a.right-tailed test, use the value that corresponds to d.f. and α. b.left-tailed test, use the value that corresponds to d.f. and 1 – α. c.two-tailed test, use the values that corresponds to d.f. and ½α, and d.f. and 1 – ½α. © 2012 Pearson Education, Inc. All rights reserved. 88 of 101

Example: Finding Critical Values for χ 2 Find the critical χ 2 -value for a left-tailed test when n = 11 and α = 0.01. Solution: Degrees of freedom: n – 1 = 11 – 1 = 10 d.f. The area to the right of the critical value is 1 – α = 1 – 0.01 = 0.99. From Table 6, the critical value is. © 2012 Pearson Education, Inc. All rights reserved. 90 of 101 χ 0 = 2.558

Example: Finding Critical Values for χ 2 Find the critical χ 2 -value for a two-tailed test when n = 9 and α = 0.05. Solution: Degrees of freedom: n – 1 = 9 – 1 = 8 d.f. The areas to the right of the critical values are From Table 6, the critical values are and. © 2012 Pearson Education, Inc. All rights reserved. 91 of 101

The Chi-Square Test χ 2 -Test for a Variance or Standard Deviation A statistical test for a population variance or standard deviation. Can be used when the population is normal. The test statistic is s 2. The standardized test statistic follows a chi-square distribution with degrees of freedom d.f. = n – 1. © 2012 Pearson Education, Inc. All rights reserved. 92 of 101

Using the χ 2 -Test for a Variance or Standard Deviation 1.State the claim mathematically and verbally. Identify the null and alternative hypotheses. 2.Specify the level of significance. 3.Determine the degrees of freedom. 4.Determine the critical value(s). State H 0 and H a. Identify α. Use Table 6 in Appendix B. d.f. = n – 1 In WordsIn Symbols © 2012 Pearson Education, Inc. All rights reserved. 93 of 101

Using the χ 2 -Test for a Variance or Standard Deviation If χ 2 is in the rejection region, reject H 0. Otherwise, fail to reject H 0. 5.Determine the rejection region(s). 6.Find the standardized test statistic and sketch the sampling distribution. 7.Make a decision to reject or fail to reject the null hypothesis. 8.Interpret the decision in the context of the original claim. In WordsIn Symbols © 2012 Pearson Education, Inc. All rights reserved. 94 of 101

Example: Hypothesis Test for the Population Variance A dairy processing company claims that the variance of the amount of fat in the whole milk processed by the company is no more than 0.25. You suspect this is wrong and find that a random sample of 41 milk containers has a variance of 0.27. At α = 0.05, is there enough evidence to reject the company’s claim? Assume the population is normally distributed. © 2012 Pearson Education, Inc. All rights reserved. 95 of 101

Solution: Hypothesis Test for the Population Variance H 0 : H a : α = df = Rejection Region: Test Statistic: Decision: σ 2 ≤ 0.25 (Claim) σ 2 > 0.25 0.05 41 – 1 = 40 Fail to Reject H 0. At the 5% level of significance, there is not enough evidence to reject the company’s claim that the variance of the amount of fat in the whole milk is no more than 0.25. © 2012 Pearson Education, Inc. All rights reserved. 96 of 101

Example: Hypothesis Test for the Standard Deviation A company claims that the standard deviation of the lengths of time it takes an incoming telephone call to be transferred to the correct office is less than 1.4 minutes. A random sample of 25 incoming telephone calls has a standard deviation of 1.1 minutes. At α = 0.10, is there enough evidence to support the company’s claim? Assume the population is normally distributed. © 2012 Pearson Education, Inc. All rights reserved. 97 of 101

Solution: Hypothesis Test for the Standard Deviation H 0 : H a : α = df = Rejection Region: Test Statistic: Decision: σ ≥ 1.4 min. σ < 1.4 min. (Claim) 0.10 25 – 1 = 24 Reject H 0. At the 10% level of significance, there is enough evidence to support the claim that the standard deviation of the lengths of time it takes an incoming telephone call to be transferred to the correct office is less than 1.4 minutes. © 2012 Pearson Education, Inc. All rights reserved. 98 of 101

Example: Hypothesis Test for the Population Variance A sporting goods manufacturer claims that the variance of the strengths of a certain fishing line is 15.9. A random sample of 15 fishing line spools has a variance of 21.8. At α = 0.05, is there enough evidence to reject the manufacturer’s claim? Assume the population is normally distributed. © 2012 Pearson Education, Inc. All rights reserved. 99 of 101

Solution: Hypothesis Test for the Population Variance H 0 : H a : α = df = Rejection Region: Test Statistic: Decision: σ 2 = 15.9 (Claim) σ 2 ≠ 15.9 0.05 15 – 1 = 14 Fail to Reject H 0 At the 5% level of significance, there is not enough evidence to reject the claim that the variance in the strengths of the fishing line is 15.9. © 2012 Pearson Education, Inc. All rights reserved. 100 of 101

Chapter Hypothesis Testing with One Sample 1 of 101 7 © 2012 Pearson Education, Inc. All rights reserved.

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