Chapter 7 Hypothesis Testing with One Sample LarsonFarber

Chapter 7 Hypothesis Testing with One Sample Larson/Farber 4 th ed. 1

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 Larson/Farber 4 th ed. 2

Section 7. 1 Introduction to Hypothesis Testing Larson/Farber 4 th ed. 3

Section 7. 1 Objectives • State a null hypothesis and an alternative hypothesis • Identify type I and type I 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 Larson/Farber 4 th ed. 4

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 whether to reject the claim or fail to reject the claim. Larson/Farber 4 th ed. 5

Hypothesis Tests Statistical hypothesis • A statement, or claim, about a population parameter. • Need a pair of hypotheses • the null hypothesis that represents the claim • the other, its alternative hypothesis Larson/Farber 4 th ed. 6

Stating a Hypothesis Null hypothesis • A statistical hypothesis that contains a statement of equality, =. • Denoted H 0 read “H subzero” or “H naught. ” Alternative hypothesis • A statement of inequality such as >, , or <. • Denoted Ha read “H sub -a. ” complementary statements Larson/Farber 4 th ed. 7

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 alternative. H 0: μ = k Ha: μ > k H 0: μ = k Ha: μ ≠ k Larson/Farber 4 th ed. 8

Example: Stating the Null and Alternative Hypotheses Write the claim as a mathematical sentence. State the null and alternative hypotheses. A university publicizes that the proportion of its students who graduate in 4 years is 82%. Solution: H 0: p = 0. 82 Ha: p ≠ 0. 82 Equality condition (Claim) Complement of H 0 Larson/Farber 4 th ed. 9

Example: Stating the Null and Alternative Hypotheses Write the claim as a mathematical sentence. State the null and alternative hypotheses. A water faucet manufacturer announces that the mean flow rate of a certain type of faucet is 2. 5 gallons per minute. Our water pressure is low. Solution: H 0: μ = 2. 5 gallons per minute Claim Ha: μ < 2. 5 gallons per minute Larson/Farber 4 th ed. 10

Example: Stating the Null and Alternative Hypotheses Write the claim as a mathematical sentence. State the null and alternative hypotheses. A cereal company advertises that the mean weight of the contents of its 20 -ounce size cereal boxes is 20 ounces. We think the boxes contain less cereal. Solution: H 0: μ = 20 ounces Null Hypothesis (Claim) Ha: μ < 20 ounces Alternative Hypothesis Larson/Farber 4 th ed. 11

Types of Errors • 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 our decision is based on a sample, there is the possibility that a wrong decision was made. • Don’t worry, it’s not your fault. Larson/Farber 4 th ed. 12

Types of Errors Decision Do not reject H 0 Reject H 0 Actual Truth of H 0 is true H 0 is false Correct Decision Type II Error Type I Error Correct Decision • 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. Larson/Farber 4 th ed. 13

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) Larson/Farber 4 th ed. 14

Solution: Identifying Type I and Type II Errors Let p represent the proportion of chicken that is contaminated. Hypotheses: H 0: p = 0. 2 Ha: p > 0. 2 Larson/Farber 4 th ed. 15

Solution: Identifying Type I and Type II Errors Hypotheses: H 0: p = 0. 2 (Claim) Ha: p > 0. 2 A type I error is rejecting H 0 when it is true. (RT 1) The actual proportion of contaminated chicken has not exceeded 0. 2, but you decide to reject H 0. A type II error is failing to reject H 0 when it is false. (FF 2) The actual proportion of contaminated chicken is greater than 0. 2, but you do not reject H 0. Larson/Farber 4 th ed. 16

Solution: Identifying Type I and Type II Errors Hypotheses: H 0: p = 0. 2 Ha: p > 0. 2 • 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. Larson/Farber 4 th ed. 17

Level of Significance Level of significance • Your maximum allowable probability of making a type I error. Known as the alpha level. § Denoted by , 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: § = 0. 10 = 0. 05 = 0. 01 Larson/Farber 4 th ed. 18

Statistical Tests • After stating the null and alternative hypotheses and specifying the level of significance (alpha level), 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. Population parameter Test statistic μ p σ2 Larson/Farber 4 th ed. s 2 Standardized test statistic z (Section 7. 2 n 30) t (Section 7. 3 n < 30) z (Section 7. 4) χ2 (Section 7. 5) 19

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. Larson/Farber 4 th ed. 20

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. Larson/Farber 4 th ed. 21

Left-tailed Test • The alternative hypothesis Ha contains the less-than inequality symbol (<). H 0: μ = k Ha: μ < k P is the area to the left of the test statistic. z -3 -2 -1 0 1 2 3 Test statistic Larson/Farber 4 th ed. 22

Right-tailed Test • The alternative hypothesis Ha contains the greaterthan inequality symbol (>). H 0: μ = k P is the area Ha: μ > k to the right of the test statistic. z -3 -2 -1 0 1 2 3 Test statistic Larson/Farber 4 th ed. 23

Two-tailed Test • The alternative hypothesis Ha contains the not equal inequality symbol (≠). Each tail has an area of ½P. H 0: μ = k Ha: μ k P is twice the area to the right of the positive test statistic. P is twice the area to the left of the negative test statistic. z -3 Larson/Farber 4 th ed. -2 -1 Test statistic 0 1 2 Test statistic 3 24

Example: Identifying The Nature of a Test For each claim, state H 0 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 university publicizes that the proportion of its students who graduate in 4 years is 82%. Solution: H 0: p = 0. 82 ½ P-value Ha: p ≠ 0. 82 area Two-tailed test Larson/Farber 4 th ed. -z 0 z z 25

Example: Identifying The Nature of a Test For each claim, state H 0 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 water faucet manufacturer announces that the mean flow rate of a certain type of faucet is less than 2. 5 gallons per minute. Solution: P-value H 0: μ = 2. 5 gpm area μ < 2. 5 gpm Ha : Left-tailed test Larson/Farber 4 th ed. -z 0 z 26

Example: Identifying The Nature of a Test For each claim, state H 0 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 cereal company advertises that the mean weight of the contents of its 20 -ounce size cereal boxes is more than 20 ounces. Solution: P-value H 0: μ = 20 oz area Ha: μ > 20 oz Right-tailed test Larson/Farber 4 th ed. 0 z z 27

Making a Decision Rule Based on P-value • Compare the P-value with . § If P < , then reject H 0. § If P > , then fail to reject H 0. Hypotheses Decision Rule Reject H 0 There is enough evidence to reject the claim Fail to reject H 0 There is not enough evidence to reject the claim to support the claim Larson/Farber 4 th ed. There is enough evidence to support the claim 28

When the p-value is lower than the alpha level, we reject the null hypothesis and use the alternative hypothesis “When it’s low, let it go…reject the null. ” When the p-value is higher than the alpha level, we fail to reject the null hypothesis. “When it’s high, let it fly…we fail to reject the null hypothesis. ” Larson/Farber 4 th ed. 29

Example: Interpreting a Decision • Larson/Farber 4 th ed. 30

Solution: Interpreting a Decision • The claim is represented by H 0. If you reject H 0 you should conclude “There is sufficient evidence to reject the university’s claim that the proportion of its students who graduate in 4 years is 82% and therefore conclude that the proportion of graduates in 4 years is higher or lower than the university’s claim. ” If you fail to reject H 0, you should conclude “There is sufficient evidence to support the university’s claim that its four-year graduation rate is 82%. ” Larson/Farber 4 th ed. 31

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? Honda Automobile Company states that the mean stopping distance (on a dry surface) for a Honda Civic is 136 feet. Consumer Reports believes that the stopping distance is greater than 136 feet. Larson/Farber 4 th ed. 32

Solution: Interpreting a Decision • If you reject H 0 you should conclude “There is sufficient evidence to support Consumer Reports’ claim that the stopping distance for a Honda Civic is greater than 136 feet. ” • If you fail to reject H 0, you should conclude “There is not enough evidence to conclude that the stopping distance for a Honda Civic is greater than 136 feet. ” Larson/Farber 4 th ed. 33

Steps for Hypothesis Testing 1. State the claim mathematically or verbally. Identify the null and alternative hypotheses. H 0: ? Ha: ? 2. Specify the level of significance. This sampling distribution is based on the assumption α = ? that H 0 is true. 3. Determine the standardized sampling distribution and draw its graph. z 0 4. Calculate the test statistic and its standardized value. Add it to your sketch. z 0 Test statistic Larson/Farber 4 th ed. 34

Steps for Hypothesis Testing 5. Find the P-value. 6. Use the following decision rule. Is the P-value less than or equal to the level of significance? No Fail to reject H 0. Yes Reject H 0. 7. Write a statement to interpret the decision in the context of the original claim. Larson/Farber 4 th ed. 35

Section 7. 1 Summary • Stated a null hypothesis and an alternative hypothesis • Identified type I and type I 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 Larson/Farber 4 th ed. 36

Section 7. 2 Hypothesis Testing for the Mean (Large Samples) Larson/Farber 4 th ed. 37

Section 7. 2 Objectives • Find P-values and use them to test a mean μ • Use P-values for a z-test • Find critical values and rejection regions in a normal distribution • Use rejection regions for a z-test Larson/Farber 4 th ed. 38

Using P-values to Make a Decision • Larson/Farber 4 th ed. 39

Example: Interpreting a P-value The P-value for a hypothesis test is P = 0. 0237. What is your decision if the level of significance is 1. 0. 05? Solution: Because 0. 0237 < 0. 05, you should reject the null hypothesis. 2. 0. 01? Solution: Because 0. 0237 > 0. 01, you should fail to reject the null hypothesis. Larson/Farber 4 th ed. 40

Finding the P-value After determining the hypothesis test’s standardized test statistic and the test statistic’s corresponding area, do one of the following to find the P-value. a. For a left-tailed test, P = (Area in left tail). b. For a right-tailed test, P = (Area in right tail). c. For a two-tailed test, P = 2(Area in tail of test statistic). Larson/Farber 4 th ed. 41

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. Solution: For a left-tailed test, P = (Area in left tail) P = 0. 0129 -2. 23 0 z Because 0. 0129 > 0. 01, you should fail to reject H 0 Larson/Farber 4 th ed. 42

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) 1 – 0. 9838 = 0. 0162 0. 9838 0 2. 14 P = 2(0. 0162) = 0. 0324 z Because 0. 0324 < 0. 05, you should reject H 0 Larson/Farber 4 th ed. 43

Z-Test for a Mean μ • Can be used when the population is normal and is known, or for any population when the sample size n is at least 30. • The test statistic is the sample mean • The standardized test statistic is z • When n 30, the sample standard deviation s can be substituted for . Larson/Farber 4 th ed. 44

Using P-values for a z-Test for Mean μ In Words 1. State the claim mathematically and verbally. Identify the null and alternative hypotheses. 2. Specify the level of significance. In Symbols State H 0 and Ha. Identify . 3. Determine the standardized test statistic. 4. Find the area that corresponds to z. Larson/Farber 4 th ed. Use Table 4 in Appendix B. 45

Using P-values for a z-Test for Mean μ In Words In Symbols 5. Find the P-value. a. For a left-tailed test, P = (Area in left tail). b. For a right-tailed test, P = (Area in right tail). c. For a two-tailed test, P = 2(Area in tail of test statistic). Reject H 0 if P-value 6. Make a decision to reject or is less than or equal fail to reject the null hypothesis. to . Otherwise, fail to reject H 0. 7. Interpret the decision in the context of the original claim. Larson/Farber 4 th ed. 46

Example: Hypothesis Testing Using Pvalues In an advertisement, a pizza shop claims that its mean delivery time is 30 minutes. A random selection of 36 delivery times has a sample mean of 28. 5 minutes and a standard deviation of 3. 5 minutes. Is there enough evidence to support the claim at = 0. 01? Use a Pvalue. Larson/Farber 4 th ed. 47

Solution: Hypothesis Testing Using Pvalues • • H 0: Ha: μ < 30 min = 0. 01 Test Statistic: • P-value 0. 0051 -2. 57 0 z • Decision: 0. 0051 < 0. 01 Reject H 0 Larson/Farber 4 th ed. 48

Example: Hypothesis Testing Using Pvalues You think that the average franchise investment information shown in the graph is incorrect, so you randomly select 30 franchises and determine the necessary investment for each. The sample mean investment is $135, 000 with a standard deviation of $30, 000. Is there enough evidence to support your claim at = 0. 05? Use a P-value. Larson/Farber 4 th ed. 49

Solution: Hypothesis Testing Using Pvalues • • H 0: μ = $143, 260 Ha: μ ≠ $143, 260 = 0. 05 Test Statistic: • P-value 0. 0655 -1. 51 Larson/Farber 4 th ed. P = 2(0. 0655) = 0. 1310 0 z • Decision: 0. 1310 > 0. 05 Fail to reject H 0 50

Rejection Regions and Critical Values • Larson/Farber 4 th ed. 51

Rejection Regions and Critical Values Larson/Farber 4 th ed. 52

Example: Finding Critical Values Find the critical value and rejection region for a twotailed test with = 0. 05. Solution: 1 – α = 0. 95 ½α = 0. 025 z 0 -z = -1. 96 ½α = 0. 025 0 z = 1. 96 z 0 z The rejection regions are to the left of -z = -1. 96 and to the right of z = 1. 96. Larson/Farber 4 th ed. 53

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. Fail to reject Ho. Fail to reject H 0. Reject H 0. z < z 0 Reject Ho. z 0 Fail to reject H 0 Left-Tailed Test Reject H 0 z < -z 0 0 0 z z > z 0 Right-Tailed Test Reject H 0 z z 0 z > z 0 Two-Tailed Test Larson/Farber 4 th ed. 54

Using Rejection Regions for a z-Test for a Mean μ In Words In Symbols 1. State the claim mathematically and verbally. Identify the null and alternative hypotheses. State H 0 and Ha. 2. Specify the level of significance. Identify . 3. Sketch the sampling distribution. 4. Determine the critical value(s). 5. Determine the rejection region(s). Larson/Farber 4 th ed. Use Table 4 in Appendix B. 55

Using Rejection Regions for a z-Test for a Mean μ In Words In Symbols 6. Find the standardized test statistic. 7. Make a decision to reject or fail to reject the null hypothesis. 8. Interpret the decision in the context of the original claim. Larson/Farber 4 th ed. If z is in the rejection region, reject H 0. Otherwise, fail to reject H 0. 56

Example: Testing with Rejection Regions Employees in a large accounting firm claim that the mean salary of the firm’s accountants is the same as that of its competitor’s, which is $45, 000. A random sample of 30 of the firm’s accountants has a mean salary of $43, 500 with a standard deviation of $5200. At α = 0. 05, test the employees’ claim. Larson/Farber 4 th ed. 57

Solution: Testing with Rejection Regions • • H 0: Ha: μ < $45, 000 = 0. 05 Rejection Region: • Test Statistic 0. 05 -1. 645 0 -1. 58 Larson/Farber 4 th ed. z • Decision: Fail to reject H 0 Since the p-value falls outside the rejection region, we fail to reject the null hypothesis and can say that there is not sufficient evidence to support the employees’ claim that the mean salary is less than $45, 000. 58

Example: Testing with Rejection Regions The U. S. Department of Agriculture reports that the mean cost of raising a child from birth to age 2 in a rural area is $10, 460. You believe this value is incorrect, so you select a random sample of 900 children (age 2) and find that the mean cost is $10, 345 with a standard deviation of $1540. At α = 0. 05, is there enough evidence to conclude that the mean cost is different from $10, 460? (Adapted from U. S. Department of Agriculture Center for Nutrition Policy and Promotion) Larson/Farber 4 th ed. 59

Solution: Testing with Rejection Regions • • H 0: μ = $10, 460 Ha: μ ≠ $10, 460 = 0. 05 Rejection Region: 0. 025 -1. 96 -2. 24 Larson/Farber 4 th ed. • Test Statistic 0. 025 0 1. 96 z • Decision: Reject H 0 At the 5% level of significance, you have enough evidence to conclude the mean cost of raising a child from birth to age 2 in a rural area is significantly different from $10, 460. 60

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 Larson/Farber 4 th ed. 61

Section 7. 3 Hypothesis Testing for the Mean (Small Samples) Larson/Farber 4 th ed. 62

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 Larson/Farber 4 th ed. 63

Finding Critical Values in a t-Distribution 1. Identify the level of significance . 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, ” column with a negative sign, b. right-tailed, use “One Tail, ” column with a positive sign, c. two-tailed, use “Two Tails, ” column with a negative and a positive sign. Larson/Farber 4 th ed. 64

Example: Finding Critical Values for t • Solution: • The degrees of freedom are d. f. = n – 1 = 21 – 1 = 20. • Look at α = 0. 05 in the “One Tail, ” column. • Because the test is lefttailed, the critical value is negative. Larson/Farber 4 th ed. 0. 05 -1. 725 0 t 65

Example: Finding Critical Values for t Find the critical values t and -t for a two-tailed test given = 0. 05 and n = 26. Solution: • The degrees of freedom are d. f. = n – 1 = 26 – 1 = 25. • Look at α = 0. 05 in the “Two Tail, ” column. • Because the test is twotailed, one critical value is negative and one is positive. Larson/Farber 4 th ed. 0. 025 -2. 060 0 0. 025 2. 060 t 66

t-Test for a Mean μ (n < 30, 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, 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. Larson/Farber 4 th ed. 67

Using the t-Test for a Mean μ (Small Sample) In Words In Symbols 1. State the claim mathematically and verbally. Identify the null and alternative hypotheses. State H 0 and Ha. 2. Specify the level of significance. Identify . 3. Identify the degrees of freedom and sketch the sampling distribution. d. f. = n – 1. 4. Determine any critical value(s). Use Table 5 in Appendix B. Larson/Farber 4 th ed. 68

Using the t-Test for a Mean μ (Small Sample) In Words In Symbols 5. Determine any rejection region(s). 6. Find the standardized test statistic. 7. Make a decision to reject or fail to reject the null hypothesis. 8. Interpret the decision in the context of the original claim. Larson/Farber 4 th ed. If t is in the rejection region, reject H 0. Otherwise, fail to reject H 0. 69

Example: Testing μ with a Small Sample A used car dealer says that the mean price of a 2005 Honda Pilot LX is $23, 900. You suspect this claim is incorrect and find that a random sample of 14 similar vehicles has a mean price of $23, 000 and a standard deviation of $1113. 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) Larson/Farber 4 th ed. 70

Solution: Testing μ with a Small Sample • • • H 0: Ha: μ < $23, 900 α = 0. 05 df = 14 – 1 = 13 Rejection Region: • Test Statistic: • Decision: Reject H 0 0. 05 -1. 771 0 t At the 0. 05 level of significance, there is enough evidence to reject the claim that the mean price of a 2005 Honda Pilot LX is $23, 900 -3. 026 Larson/Farber 4 th ed. 71

Example: Testing μ with a Small Sample An industrial company claims that the mean p. H level of the water in a nearby river is 6. 8. You randomly select 19 water samples and measure the p. H 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. Larson/Farber 4 th ed. 72

Solution: Testing μ with a Small Sample • • • H 0: μ = 6. 8 Ha: μ ≠ 6. 8 α = 0. 05 df = 19 – 1 = 18 Rejection Region: 0. 025 -2. 101 0 0. 025 2. 101 • Test Statistic: • Decision: Fail to reject H 0 At the 0. 05 level of significance, there is not enough evidence to reject the claim that the mean p. H is 6. 8. t -1. 816 Larson/Farber 4 th ed. 73

Example: Using P-values with t-Tests The American Automobile Association claims that the mean daily meal cost for a family of four traveling on vacation in Florida is $118. A random sample of 11 such families has a mean daily meal cost of $128 with a standard deviation of $20. Is there enough evidence to reject the claim at α = 0. 10? Assume the population is normally distributed. (Adapted from American Automobile Association) Larson/Farber 4 th ed. 74

Solution: Using P-values with t-Tests • H 0: μ = $118 • Ha: μ ≠ $118 TI-83/84 set up: Calculate: Draw: • Decision: 0. 1664 > 0. 10 Fail to reject H 0. At the 0. 10 level of significance, there is not enough evidence to reject the claim that the mean daily meal cost for a family of four traveling on vacation in Florida is $118. Larson/Farber 4 th ed. 75

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 Larson/Farber 4 th ed. 76

Section 7. 4 Hypothesis Testing for Proportions Larson/Farber 4 th ed. 77

Section 7. 4 Objectives • Use the z-test to test a population proportion p Larson/Farber 4 th ed. 78

z-Test for a Population Proportion • A statistical test for a population proportion. • 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. Larson/Farber 4 th ed. 79

Using a z-Test for a Proportion p Verify that np ≥ 5 and nq ≥ 5 In Words In Symbols 1. State the claim mathematically and verbally. Identify the null and alternative hypotheses. State H 0 and Ha. 2. Specify the level of significance. Identify . 3. Sketch the sampling distribution. 4. Determine any critical value(s). Larson/Farber 4 th ed. Use Table 5 in Appendix B. 80

Using a z-Test for a Proportion p In Words In Symbols 5. Determine any rejection region(s). 6. Find the standardized test statistic. 7. Make a decision to reject or fail to reject the null hypothesis. 8. Interpret the decision in the context of the original claim. Larson/Farber 4 th ed. If z is in the rejection region, reject H 0. Otherwise, fail to reject H 0. 81

Example: Hypothesis Test for Proportions Zogby International claims that 45% of people in the United States support making cigarettes illegal within the next 5 to 10 years. You decide to test this claim and ask a random sample of 200 people in the United States whether they support making cigarettes illegal within the next 5 to 10 years. Of the 200 people, 49% support this law. At α = 0. 05 is there enough evidence to reject the claim? Solution: • Verify that np ≥ 5 and nq ≥ 5. np = 200(0. 45) = 90 and nq = 200(0. 55) = 110 Larson/Farber 4 th ed. 82

Solution: Hypothesis Test for Proportions • • H 0: p = 0. 45 Ha: p ≠ 0. 45 = 0. 05 Rejection Region: 0. 025 -1. 96 0. 025 0 1. 96 1. 14 Larson/Farber 4 th ed. • Test Statistic z • Decision: Fail to reject H 0 At the 5% level of significance, there is not enough evidence to reject the claim that 45% of people in the U. S. support making cigarettes illegal within the next 5 to 10 years. 83

Example: Hypothesis Test for Proportions The Pew Research Center claims that 55% of U. S. adults regularly watch their local television news. You decide to test this claim and ask a random sample of 425 adults in the United States whether they regularly watch their local television news. Of the 425 adults, 255 respond yes. At α = 0. 05 is there enough evidence to support the claim? Solution: • Verify that np ≥ 5 and nq ≥ 5. np = 425(0. 55) ≈ 234 and nq = 425 (0. 45) ≈ 191 Larson/Farber 4 th ed. 84

Solution: Hypothesis Test for Proportions • • H 0: Ha: p > 0. 55 = 0. 05 Rejection Region: • Test Statistic 0. 05 0 1. 645 2. 07 Larson/Farber 4 th ed. z • Decision: Reject H 0 At the 5% level of significance, there is enough evidence to support the claim that more than 55% of U. S. adults regularly watch their local television news. 85

Section 7. 4 Summary • Used the z-test to test a population proportion p Larson/Farber 4 th ed. 86