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

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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

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Section 7.1 Introduction to Hypothesis Testing © 2012 Pearson Education, Inc. All rights reserved. 3 of 101

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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

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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

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Hypothesis Tests Statistical hypothesis A statement, or claim, about a population parameter (such as the mean). 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

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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.” (or H –o) 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.” (or H – a) complementary statements © 2012 Pearson Education, Inc. All rights reserved. 7 of 101

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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

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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

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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

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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

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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. (NOTE: This may or may not be the claim) At the end of the test, one of two decisions will be made: reject the null hypothesis fail to reject the null hypothesis (FTR) 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

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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. Hint: 1 comes before 2, True comes before 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 KNOW THESE

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Types of Errors Another way of looking at a Type 1 and Type 2 error: Type 1 Error –Rejecting a true “claim” Type 2 Error –Accepting (Failing To Reject, or FTR) a false “claim” Larson/Farber 5th ed. 14

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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. 15 of 101

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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. 16 of 101

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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. 17 of 101

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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. 18 of 101

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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) We won’t cover this much, it’s higher level statistics. © 2012 Pearson Education, Inc. All rights reserved. 19 of 101

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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. 20 of 101

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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. 21 of 101

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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. In other words, we’re testing the alternative hypothesis © 2012 Pearson Education, Inc. All rights reserved. 22 of 101

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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. 23 of 101

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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. 24 of 101 z 0123–3–2–1 P is the area to the right of the standardized test statistic.

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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. 25 of 101

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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. 26 of 101

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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. 27 of 101

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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. 28 of 101

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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. 29 of 101 Remember, α is your maximum allowable probability of making a type I error.

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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. 30 of 101 Solution: The claim is represented by H 0.

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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. 31 of 101

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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. 32 of 101

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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. 33 of 101

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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. 34 of 101

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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. 35 of 101 Remember: P-Value is t he 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. So, if my sample is more extreme than my level of significance, I have to reject the null hypothesis What you evaluating is the AREA in the tail

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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. 36 of 101

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Assignment Page 367 5-9,11-16,21-24 Larson/Farber 5th ed. 37

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Section 7.2 Hypothesis Testing for the Mean (Large Samples) © 2012 Pearson Education, Inc. All rights reserved. 38 of 101

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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. 39 of 101

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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. 40 of 101

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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. –This would be outside the limits of the tail Solution: Because 0.0237 > 0.01, you should fail to reject the null hypothesis. -This would be inside our limits © 2012 Pearson Education, Inc. All rights reserved. 41 of 101

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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. 42 of 101

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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. 43 of 101

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On a Calculator Calculate P-Value 2 nd Vars normalcdf For a left tail test enter (–EE99,Zscore) For a right tail test enter (Zscore, EE99) For a 2 tail test, do one or the other and multiply your answer times 2 (for 2 tails) Larson/Farber 5th ed. 44

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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. 45 of 101

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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. 46 of 101

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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. 47 of 101

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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. 48 of 101

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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. 49 of 101

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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. 50 of 101

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On a Calculator Stats Tests Z-Test Input Stats Enter the mean, the standard deviation, the sample mean, and n Here were our 2 equations: We will test the claim, so chose < μ 0 We get a p-value of.00145, which is less than.01 Larson/Farber 5th ed. 51 H 0 : H a : = μ ≥ 13 sec μ < 13 sec (Claim) 0.01 (alpha) Therefore, we should reject the null hypothesis, because our sample value is outside the limit we set for ourselves

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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. 52 of 101

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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. 53 of 101

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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. 54 of 101

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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. 55 of 101

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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. 56 of 101 Calculator: Again, we can use invNorm(.025) and select either or both tails

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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. 57 of 101

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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. 58 of 101

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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. 59 of 101

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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. 60 of 101

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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. 61 of 101

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On a Calculator Stats Tests Z-Test Input Stats Enter the mean, the standard deviation, the sample mean, and n Here were our 2 equations: We will test the claim, so chose < μ 0 We get a z-score of -1.09, which is inside our desired area 62 H 0 : H a : = Therefore, we should not reject the null hypothesis, because our sample value is inside the limit we set for ourselves μ ≥ $68,000 μ < $68,000 (Claim) 0.05 Note: You could also compare P- Value to alpha. The p-value is not less than alpha, so we do not reject the null hypothesis

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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. 63 of 101

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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. 64 of 101

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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. 65 of 101

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Assignment P. 381 3-12 P. 382 15-22, 25-34 Larson/Farber 5th ed. 66

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Chapter 7 Quiz 1 Using data from question 40 on page 384: 1.Write a null hypothesis and an alternate hypothesis (identify which is the claim) 2.Calculate the critical Z-value 3.Calculate the sample standard deviation 4.Calculate the p-value 5.Using a confidence level α =.09, is there enough information to reject the company’s claim? 4 points each question Larson/Farber 5th ed. 67

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Section 7.3 Hypothesis Testing for the Mean (Small Samples) © 2012 Pearson Education, Inc. All rights reserved. 68 of 101

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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. 69 of 101

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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. 70 of 101

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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. 71 of 101

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On the Calculator 2 nd Vars InvT(area,D.F.) Area = α = 0.05 n = 21 so D.F. = 20 Solve InvT(.05,20) and get -1.7247 or -1.725 NOTE: This is a left tailed test. For a right tailed test, use the “Positive” answer 1.725 Larson/Farber 5th ed. 72

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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. 73 of 101

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On a Calculator 2 nd Vars InvT(area,D.F.) Area = α = 0.10 HOWEVER, this is the total area. So we need to divide this in half since InvT solves for one tail n = 26 so D.F. = 25 Solve InvT(.05,25) and get -1.708 NOTE: This is a two tailed test. Therefore, we must have 2 rejection regions -1.708 and 1.708 This will give us our total area of.10 Larson/Farber 5th ed. 74

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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. 75 of 101

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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. 76 of 101

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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. 77 of 101

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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. 78 of 101

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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. 79 of 101 t ≈ –2.244

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On the Calculator Stats Test Ttest You must test the ALTERNATE Hypothesis Which means you have to know what it is Enter data Test μ < $20,500 You will get a t-test value of -2.244 If you had calculated the t- critical value, you could compare, and see that the test value is less than the critical value, and thus reject the null hypothesis You get a p-value of.02 This is less than the alpha of.05 Thus, the probability of occurrence is in the rejection region, so we reject the null hypothesis Larson/Farber 5th ed. 80

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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. 81 of 101

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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. 82 of 101

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On the Calculator Stats Test Ttest You must test the ALTERNATE Hypothesis Which means you have to know what it is Enter data Test μ ≠ 6.8 You will get a t-test value of -1.816 Remember, this is a 2 tail test, so you also have a t-test value of 1.816 If you had calculated the T-critical value you could compare these You get a p-value of.09 This is not less than the alpha of.05 Thus, the probability of occurrence is in not in the rejection region, so we fail to reject the null hypothesis Larson/Farber 5th ed. 83

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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. 84 of 101

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Solution: Using P-values with t-Tests © 2012 Pearson Education, Inc. All rights reserved. 85 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.

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Chapter 7 Quiz 2 Using data from question 39 on page 384: 1.Write a null hypothesis and an alternate hypothesis (identify which is the claim) 2.Calculate the critical Z-value (based on α =.06) 3.Calculate the sample standard deviation 4.Calculate the p-value 5.Using a confidence level α =.06, is there enough information to support the scientist’s claim? 4 points each question Larson/Farber 5th ed. 86

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Assignment P. 393 4-30 even Larson/Farber 5th ed. 87

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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. 88 of 101

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Chapter 7 Quiz 3 Using data from question 29 on page 395: 1.Write a null hypothesis and an alternate hypothesis (identify which is the claim) 2.Calculate the p-value 3.Compare the p-value to α =.05 and decide whether to reject or fail to reject the null hypothesis 4.Using a confidence level α =.05, is there enough information to support the university’s claim? (Either the claim is false, or the claim is true). 5 points each question Larson/Farber 5th ed. 89

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Section 7.4 Hypothesis Testing for Proportions © 2012 Pearson Education, Inc. All rights reserved. 90 of 101

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Section 7.4 Objectives Use the z-test to test a population proportion p © 2012 Pearson Education, Inc. All rights reserved. 91 of 101

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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. 92 of 101 Remember: Binomial Distribution means 2 outcomes

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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. 93 of 101

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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. 94 of 101

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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. 95 of 101

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Solution: Hypothesis Test for Proportions H 0 : H a : α = Rejection Region: p ≥ 0.5 p ≠ 0.45 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. 96 of 101 z = –2.2

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On a Calculator Choose Stats Tests 1-PropZTest P 0 =.50 (null hypothesis) X = number of successes. They tell us success happened 39% of the time. There were 100 sampled, so x =.39 x 100 = 39 n = 100 Prop = “not equal” remember, we are testing the Alternative Hypothesis Z test score = -2.2 –we could compare this to a Z critical score if we calculated it P =.02 This is not less than.01, so it is not in the tail, so we do not reject the null hypothesis Larson/Farber 5th ed. 97

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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. 98 of 101

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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 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. 99 of 101 z ≈ –1.31

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On the Calculator 100 Choose Stats Tests 1-PropZTest P 0 =.25 (null hypothesis X = number of successes. They tell us success happened 21% of the time. There were 200 sampled, so x =.21 x 200 = 42 NOTE: You can actually type in.21 x 200 for this entry n = 200 Prop = “not equal” remember, we are testing the Alternative Hypothesis (Binomial is ALWAYS not equal) Z test score = -1.30 –we could compare this to a Z critical score if we calculated it P =.19 This is not less than.10, so it is not in the tail(s), so we do not reject the null hypothesis

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Section 7.4 Summary Used the z-test to test a population proportion p © 2012 Pearson Education, Inc. All rights reserved. 101 of 101

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Assignment Page 401 4-16 Even Larson/Farber 5th ed. 102

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Larson/Farber 5th ed. 103

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Section 7.5 Hypothesis Testing for Variance and Standard Deviation © 2012 Pearson Education, Inc. All rights reserved. 104 of 101

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Section 7.5 Objectives Find critical values for a χ 2 -test Use the χ 2 -test to test a variance or a standard deviation © 2012 Pearson Education, Inc. All rights reserved. 105 of 101

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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. 106 of 101

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Finding Critical Values for the χ 2 -Test © 2012 Pearson Education, Inc. All rights reserved. 107 of 101 1 – Right-tailed χ2χ2 Two-tailed 1 – Left-tailed

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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. 108 of 101 χ 0 = 2.558

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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. 109 of 101

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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. 110 of 101

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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. 111 of 101

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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. 112 of 101

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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. 113 of 101

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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. 114 of 101

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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. 115 of 101

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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. 116 of 101

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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. 117 of 101

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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. 118 of 101

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Section 7.5 Summary Found critical values for a χ 2 -test Used the χ 2 -test to test a variance or a standard deviation © 2012 Pearson Education, Inc. All rights reserved. 119 of 101

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