Chapter Hypothesis Testing with Two Samples 1 of 70 8 © 2012 Pearson Education, Inc. All rights reserved.

Chapter Outline 8.1 Testing the Difference Between Means (Large Independent Samples) 8.2 Testing the Difference Between Means (Small Independent Samples) 8.3 Testing the Difference Between Means (Dependent Samples) 8.4 Testing the Difference Between Proportions © 2012 Pearson Education, Inc. All rights reserved. 2 of 70

Section 8.1 Objectives Determine whether two samples are independent or dependent Perform a two-sample z-test for the difference between two means μ 1 and μ 2 using large independent samples © 2012 Pearson Education, Inc. All rights reserved. 4 of 70

Two Sample Hypothesis Test Compares two parameters from two populations. Sampling methods:  Independent Samples The sample selected from one population is not related to the sample selected from the second population.  Dependent Samples (paired or matched samples) Each member of one sample corresponds to a member of the other sample. © 2012 Pearson Education, Inc. All rights reserved. 5 of 70

Example: Independent and Dependent Samples Classify the pair of samples as independent or dependent. Sample 1: Weights of 65 college students before their freshman year begins. Sample 2: Weights of the same 65 college students after their freshmen year. Solution: Dependent Samples (The samples can be paired with respect to each student) © 2012 Pearson Education, Inc. All rights reserved. 7 of 70

Example: Independent and Dependent Samples Classify the pair of samples as independent or dependent. Sample 1: Scores for 38 adults males on a psycho- logical screening test for attention-deficit hyperactivity disorder. Sample 2: Scores for 50 adult females on a psycho- logical screening test for attention-deficit hyperactivity disorder. Solution: Independent Samples (Not possible to form a pairing between the members of the samples; the sample sizes are different, and the data represent scores for different individuals.) © 2012 Pearson Education, Inc. All rights reserved. 8 of 70

Two Sample Hypothesis Test with Independent Samples 1.Null hypothesis H 0  A statistical hypothesis that usually states there is no difference between the parameters of two populations.  Always contains the symbol ≤, =, or ≥. 2.Alternative hypothesis H a  A statistical hypothesis that is true when H 0 is false.  Always contains the symbol >, ≠, or <. © 2012 Pearson Education, Inc. All rights reserved. 9 of 70

Two Sample Hypothesis Test with Independent Samples H 0 : μ 1 = μ 2 H a : μ 1 ≠ μ 2 H 0 : μ 1 ≤ μ 2 H a : μ 1 > μ 2 H 0 : μ 1 ≥ μ 2 H a : μ 1 < μ 2 Regardless of which hypotheses you use, you always assume there is no difference between the population means, or μ 1 = μ 2. © 2012 Pearson Education, Inc. All rights reserved. 10 of 70

Two Sample z - Test for the Difference Between Means Three conditions are necessary to perform a z-test for the difference between two population means μ 1 and μ 2. 1. The samples must be randomly selected. 2. The samples must be independent. 3. Each sample size must be at least 30, or, if not, each population must have a normal distribution with a known standard deviation. © 2012 Pearson Education, Inc. All rights reserved. 11 of 70

Two Sample z - Test for the Difference Between Means If these requirements are met, the sampling distribution for (the difference of the sample means) is a normal distribution with Sampling distribution for : Mean: Standard error: © 2012 Pearson Education, Inc. All rights reserved. 12 of 70

Two Sample z - Test for the Difference Between Means Test statistic is. The standardized test statistic is When the samples are large, you can use s 1 and s 2 in place of σ 1 and σ 2. If the samples are not large, you can still use a two-sample z-test, provided the populations are normally distributed and the population standard deviations are known. © 2012 Pearson Education, Inc. All rights reserved. 13 of 70

Using a Two - Sample z - Test for the Difference Between Means (Large Independent Samples) 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. 14 of 70

Using a Two - Sample z - Test for the Difference Between Means (Large Independent Samples) 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. 15 of 70

Example: Two - Sample z - Test for the Difference Between Means A credit card watchdog group claims that there is a difference in the mean credit card debts of households in New York and Texas. The results of a random survey of 250 households from each state are shown below. The two samples are independent. Do the results support the group’s claim? Use α = 0.05. (Adapted from PlasticRewards.com) New YorkTexas s 1 = $1045.70s 2 = $1361.95 n 1 = 250n 2 = 250 © 2012 Pearson Education, Inc. All rights reserved. 16 of 70

Solution: Two - Sample z - Test for the Difference Between Means H 0 : H a :  n 1 =, n 2 = Rejection Region: Test Statistic: 0.05 250 μ 1 = μ 2 μ 1 ≠ μ 2 (claim) Decision: At the 5% level of significance, there is not enough evidence to support the group’s claim that there is a difference in the mean credit card debts of households in New York and Texas. Fail to Reject H 0. © 2012 Pearson Education, Inc. All rights reserved. 17 of 70 z = –1.11

Example: Using Technology to Perform a Two-Sample z-Test A travel agency claims that the average daily cost of meals and lodging for vacationing in Texas is less than the same average costs for vacationing in Virginia. The table shows the results of a random survey of vacationers in each state. The two samples are independent. At α = 0.01, is there enough evidence to support the claim? (Adapted from American Automobile Association) Texas (1)Virginia (2) s 1 = $18s 2 = $24 n 1 = 50n 2 = 35 © 2012 Pearson Education, Inc. All rights reserved. 18 of 70

z 0 0.01 Solution: Using Technology to Perform a Two-Sample z-Test Decision: At the 1% level of significance, there is not enough evidence to support the travel agency’s claim. Fail to Reject H 0. Rejection Region: –1.25 –2.33 © 2012 Pearson Education, Inc. All rights reserved. 20 of 70

Section 8.1 Summary Determined whether two samples are independent or dependent Performed a two-sample z-test for the difference between two means μ 1 and μ 2 using large independent samples © 2012 Pearson Education, Inc. All rights reserved. 21 of 70

Assignment Page 434 5-20, 22-32 even Larson/Farber 5th ed 22

Chapter 8 Quiz 1 Do problem 23 on page 436 Answer part a), part c), part d), and part e) 20 points -5 points each answer Larson/Farber 5th ed 23

Group Project (needs work, not useable…) Many students drive to school I claim that the mean number of students who drive to school is greater than or equal to the mean number of students who do NOT drive to school Using a simple random sample, gather at least 30 observations of students who drive, and at least 30 observations of students who do not drive Assume ơ= 5.2 Assume the proportion of student drivers is 65%. Create a group project with the following information: Names of members in the group 1.Sample 1: n= _____ Sample 2: n= 2.Identify H 0 and H a and identify which is my claim 3.Create a 95% confidence interval for the sample means for each sample(chapter 5) (Z- Interval –if your sample is more than 30) 4.Using a.05 confidence level (α), evaluate my claim that the sample mean of drivers is greater than or equal to the sample mean of non-drivers (chapter 6). Do you FTR or reject my claim? (chapter 6) 5.Using a.05 confidence level (α), evaluate my claim that the sample means of drivers and non-drivers is the same (chapter 7) (Z Test) Larson/Farber 5th ed 24

Two Sample t - Test for the Difference Between Means If samples of size less than 30 are taken from normally- distributed populations, a t-test may be used to test the difference between the population means μ 1 and μ 2. Three conditions are necessary to use a t-test for small independent samples. 1.The samples must be randomly selected. 2.The samples must be independent. 3.Each population must have a normal distribution. © 2012 Pearson Education, Inc. All rights reserved. 27 of 70

Two Sample t - Test for the Difference Between Means The test statistic is. The standardized test statistic is The standard error and the degrees of freedom of the sampling distribution depend on whether the population variances and are equal. © 2012 Pearson Education, Inc. All rights reserved. 28 of 70

 The standard error for the sampling distribution of is Two Sample t - Test for the Difference Between Means Variances are equal  Information from the two samples is combined to calculate a pooled estimate of the standard deviation.  d.f.= n 1 + n 2 – 2 © 2012 Pearson Education, Inc. All rights reserved. 29 of 70

Variances are not equal  If the population variances are not equal, then the standard error is  d.f = smaller of n 1 – 1 and n 2 – 1 Two Sample t - Test for the Difference Between Means © 2012 Pearson Education, Inc. All rights reserved. 30 of 70

Normal or t-Distribution? Are both sample sizes at least 30? Are both populations normally distributed? You cannot use the z-test or the t-test. No Yes Are both population standard deviations known? Use the z-test. Yes No Are the population variances equal? Use the z-test. Use the t-test with d.f = smaller of n 1 – 1 or n 2 – 1. Use the t-test with Yes No Yes d.f = n 1 + n 2 – 2. © 2012 Pearson Education, Inc. All rights reserved. 31 of 70

Two-Sample t-Test for the Difference Between Means (Small Independent Samples) 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 5 in Appendix B. d.f. = n 1 + n 2 – 2 or d.f. = smaller of n 1 – 1 or n 2 – 1. In WordsIn Symbols © 2012 Pearson Education, Inc. All rights reserved. 32 of 70

Two-Sample t-Test for the Difference Between Means (Small Independent Samples) 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. 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. 33 of 70

Example: Two - Sample t - Test for the Difference Between Means The results of a state mathematics test for random samples of students taught by two different teachers at the same school are shown below. Can you conclude that there is a difference in the mean mathematics test scores for the students of the two teachers? Use α = 0.01. Assume the populations are normally distributed and the population variances are not equal. Teacher 1Teacher 2 s 1 = 39.7s 2 = 24.5 n 1 = 8n 2 = 18 © 2012 Pearson Education, Inc. All rights reserved. 34 of 70

Solution: Two - Sample t - Test for the Difference Between Means H 0 : H a : α  d.f. = Rejection Region: Test Statistic: 0.10 8 – 1 = 7 μ 1 = μ 2 μ 1 ≠ μ 2 (claim) Decision: At the 10% level of significance, there is not enough evidence to support the claim that the mean mathematics test scores for the students of the two teachers are different. Fail to Reject H 0. © 2012 Pearson Education, Inc. All rights reserved. 35 of 70 t ≈ 0.922

On the Calculator Use Stats  Test  2-SampTTest Enter data When it asks “Pooled” choose no Compare the p-value to the alpha Larson/Farber 5th ed 36

Example: Two - Sample t - Test for the Difference Between Means A manufacturer claims that the mean calling range (in feet) of its 2.4-GHz cordless telephone is greater than that of its leading competitor. You perform a study using 14 randomly selected phones from the manufacturer and 16 randomly selected similar phones from its competitor. The results are shown below. At α = 0.05, can you support the manufacturer’s claim? Assume the populations are normally distributed and the population variances are equal. ManufacturerCompetitor s 1 = 45 fts 2 = 30 ft n 1 = 14 n 2 = 16 © 2012 Pearson Education, Inc. All rights reserved. 37 of 70

Solution: Two - Sample t - Test for the Difference Between Means H 0 : H a : α = d.f. = Rejection Region: 0.05 14 + 16 – 2 = 28 μ 1 ≤ μ 2 μ 1 > μ 2 (claim) t 01.701 0.05 © 2012 Pearson Education, Inc. All rights reserved. 38 of 70

Solution: Two - Sample t - Test for the Difference Between Means H 0 : H a : α = d.f. = Rejection Region: Test Statistic: 0.05 14 + 16 – 2 = 28 μ 1 ≤ μ 2 μ 1 > μ 2 (claim) 1.811 Decision: At the 5% level of significance, there is enough evidence to support the manufacturer’s claim that its phone has a greater calling range than its competitors. Reject H 0. t 01.701 0.05 © 2012 Pearson Education, Inc. All rights reserved. 40 of 70

On the Calculator Use same steps as previously except choose yes when it asks “pooled?” This is because it tells us to assume the population variances are the same Therefore you add both “n” values, and the D.F. is n+n-2 Larson/Farber 5th ed 41

Assignment Page 446 3-18 Larson/Farber 5th ed 43

Chapter 8 Quiz 2 Problem 20 on page 448 1.Ho= and Ha= (and identify claim) 2.T = 3.P = _____. Compare to Alpha. Reject or FTR? 4.Claim is ____________ Larson/Farber 5th ed 44

The test statistic is the mean of these differences.  t - Test for the Difference Between Means To perform a two-sample hypothesis test with dependent samples, the difference between each data pair is first found:  d = x 1 – x 2 Difference between entries for a data pair Mean of the differences between paired data entries in the dependent samples © 2012 Pearson Education, Inc. All rights reserved. 47 of 70

t - Test for the Difference Between Means Three conditions are required to conduct the test. 1.The samples must be randomly selected. 2.The samples must be dependent (paired). 3.Both populations must be normally distributed. If these requirements are met, then the sampling distribution for is approximated by a t-distribution with n – 1 degrees of freedom, where n is the number of data pairs. -t0-t0 t0t0 μdμd © 2012 Pearson Education, Inc. All rights reserved. 48 of 70

Symbols used for the t - Test for μ d The number of pairs of data The difference between entries for a data pair, d = x 1 – x 2 The hypothesized mean of the differences of paired data in the population n d Symbol Description © 2012 Pearson Education, Inc. All rights reserved. 49 of 70

Symbols used for the t - Test for μ d Symbol Description The mean of the differences between the paired data entries in the dependent samples The standard deviation of the differences between the paired data entries in the dependent samples sdsd © 2012 Pearson Education, Inc. All rights reserved. 50 of 70

t - Test for the Difference Between Means (Dependent Samples) 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 5 in Appendix B if n > 29 use the last row (∞). d.f. = n – 1 In WordsIn Symbols © 2012 Pearson Education, Inc. All rights reserved. 52 of 70

t - Test for the Difference Between Means (Dependent Samples) 5.Determine the rejection region(s). 6.Calculate and 5.Find the standardized test statistic. In WordsIn Symbols © 2012 Pearson Education, Inc. All rights reserved. 53 of 70

t - Test for the Difference Between Means (Dependent Samples) 8.Make a decision to reject or fail to reject the null hypothesis. 9.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. 54 of 70

Example: t - Test for the Difference Between Means Athletes12345678 Height (old)2422252835323027 Height (new)2625 2933343530 A shoe manufacturer claims that athletes can increase their vertical jump heights using the manufacturer’s new Strength Shoes ®. The vertical jump heights of eight randomly selected athletes are measured. After the athletes have used the Strength Shoes ® for 8 months, their vertical jump heights are measured again. The vertical jump heights (in inches) for each athlete are shown in the table. Assuming the vertical jump heights are normally distributed, is there enough evidence to support the manufacturer’s claim at α = 0.10? (Adapted from Coaches Sports Publishing) © 2012 Pearson Education, Inc. All rights reserved. 55 of 70

How to write the H 0 : / H a : The claim is always the alternate You compare the sample mean to what you want For example, these shoes are claimed to increase height. Therefore your current height is too low μ d < 0 (claim) The “0” is the 0 used to represent population mean on a standardized normal graph (a z score of zero is the mean) So the mean difference is less than the expected population mean Larson/Farber 5th ed 56

Solution: Two - Sample t - Test for the Difference Between Means H 0 : H a : α = d.f. = Rejection Region: 0.10 8 – 1 = 7 μ d ≥ 0 μ d < 0 (claim) d = (jump height before shoes) – (jump height after shoes) © 2012 Pearson Education, Inc. All rights reserved. 57 of 70

Solution: Two - Sample t - Test for the Difference Between Means BeforeAfterdd2d2 2426–24 2225–39 25 00 2829–11 3533 24 3234–24 3035–525 2730–39 Σ = –14Σ = 56 d = (jump height before shoes) – (jump height after shoes) © 2012 Pearson Education, Inc. All rights reserved. 58 of 70

Solution: Two - Sample t - Test for the Difference Between Means H 0 : H a : α = d.f. = Rejection Region: Test Statistic: 0.10 8 – 1 = 7 µ d ≥ 0 μ d < 0 (claim) Decision: d = (jump height before shoes) – (jump height after shoes) At the 10% level of significance, there is enough evidence to support the shoe manufacturer’s claim that athletes can increase their vertical jump heights using the new Strength Shoes ®. Reject H 0. © 2012 Pearson Education, Inc. All rights reserved. 59 of 70 t ≈ –2.333

On the Calculator What you are really testing is the difference between the means Enter the “before” data in List 1 Enter the “after” data in List 2 Go to the header of List 3, and enter List 1 minus List 2 Now you have the difference between the means Go to stats  test  t-test Run program using data from List 3 Enter “0” for the population mean µ Test the Alternate Hypothesis Larson/Farber 5th ed 60 Do “Try It Yourself 2 on page 455

Chapter 8 Quiz 3 Problem 20 on page 459. Use an alpha of.05 1.Ho= and Ha= (and identify claim) 2.T = 3.P = _____. Compare to Alpha. Reject or FTR? 4.Claim is ____________ Larson/Farber 5th ed 63

Two-Sample z-Test for Proportions Used to test the difference between two population proportions, p 1 and p 2. Three conditions are required to conduct the test. 1.The samples must be randomly selected. 2.The samples must be independent. 3.The samples must be large enough to use a normal sampling distribution. That is, n 1 p 1 ≥ 5, n 1 q 1 ≥ 5, and n 2 p 2 ≥ 5, n 2 q 2 ≥ 5. © 2012 Pearson Education, Inc. All rights reserved. 66 of 70

Two-Sample z-Test for the Difference Between Proportions If these conditions are met, then the sampling distribution for is a normal distribution. Mean: A weighted estimate of p 1 and p 2 can be found by using Standard error: © 2012 Pearson Education, Inc. All rights reserved. 67 of 70

Two-Sample z-Test for the Difference Between Proportions 1.State the claim. Identify the null and alternative hypotheses. 2.Specify the level of significance. 3.Determine the critical value(s). 4.Determine the rejection region(s). 5.Find the weighted estimate of p 1 and p 2. Verify that State H 0 and H a. Identify α. Use Table 4 in Appendix B. In WordsIn Symbols © 2012 Pearson Education, Inc. All rights reserved. 69 of 70

Two-Sample z-Test for the Difference Between Proportions 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 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. 70 of 70

Example: Two - Sample z - Test for the Difference Between Proportions In a study of 150 randomly selected occupants in passenger cars and 200 randomly selected occupants in pickup trucks, 86% of occupants in passenger cars and 74% of occupants in pickup trucks wear seat belts. At α = 0.10 can you reject the claim that the proportion of occupants who wear seat belts is the same for passenger cars and pickup trucks? (Adapted from National Highway Traffic Safety Administration) Solution: 1 = Passenger cars2 = Pickup trucks © 2012 Pearson Education, Inc. All rights reserved. 71 of 70

Solution: Two - Sample z - Test for the Difference Between Means H 0 : H a : α = n 1 =, n 2 = Rejection Region: 0.10 150 200 p 1 = p 2 (claim) p 1 ≠ p 2 © 2012 Pearson Education, Inc. All rights reserved. 72 of 70

Solution: Two - Sample z - Test for the Difference Between Means H 0 : H a : α  n 1 =, n 2 = Rejection Region: Test Statistic: 0.10 150 200 p 1 = p 2 p 1 ≠ p 2 Decision: At the 10% level of significance, there is enough evidence to reject the claim that the proportion of occupants who wear seat belts is the same for passenger cars and pickup trucks. Reject H 0 © 2012 Pearson Education, Inc. All rights reserved. 75 of 70

On the Calculator First remember to check np>5 and nq>5 Go to Stats  Calc  2-PropZTest To enter xbar 1, enter this: You can enter.86 x 150 and.74 x 200 and let it calculate for you In this case, test that the proportions are not equal Larson/Farber 5th ed 76

Example: Two - Sample z - Test for the Difference Between Proportions A medical research team conducted a study to test the effect of a cholesterol-reducing medication. At the end of the study, the researchers found that of the 4700 randomly selected subjects who took the medication, 301 died of heart disease. Of the 4300 randomly selected subjects who took a placebo, 357 died of heart disease. At α = 0.01 can you conclude that the death rate due to heart disease is lower for those who took the medication than for those who took the placebo? (Adapted from New England Journal of Medicine) Solution: 1 = Medication2 = Placebo © 2012 Pearson Education, Inc. All rights reserved. 77 of 70

z 0–2.33 0.01 Solution: Two - Sample z - Test for the Difference Between Means H 0 : H a : α  n 1 =, n 2 = Rejection Region: 0.01 4700 4300 p 1 ≥ p 2 p 1 < p 2 (claim) © 2012 Pearson Education, Inc. All rights reserved. 78 of 70

z 0–2.33 0.01 Solution: Two - Sample z - Test for the Difference Between Means H 0 : H a : α  n 1 =, n 2 = Rejection Region: Test Statistic: 0.01 4700 4300 p 1 ≥ p 2 p 1 < p 2 (claim) –2.33 –3.46 Decision: At the 1% level of significance, there is enough evidence to conclude that the death rate due to heart disease is lower for those who took the medication than for those who took the placebo. Reject H 0 © 2012 Pearson Education, Inc. All rights reserved. 81 of 70

Chapter Hypothesis Testing with Two Samples 1 of 70 8 © 2012 Pearson Education, Inc. All rights reserved.

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