1. + The Practice of Statistics, 4 th edition – For AP* STARNES, YATES, MOORE Chapter 10: Comparing Two Populations or Groups Section 10.1 Comparing Two Proportions

2. + Chapter 10 Comparing Two Populations or Groups 10.1Comparing Two Proportions 10.2Comparing Two Means

3. + Section 10.1 Comparing Two Proportions After this section, you should be able to… DETERMINE whether the conditions for performing inference are met. CONSTRUCT and INTERPRET a confidence interval to compare two proportions. PERFORM a significance test to compare two proportions. INTERPRET the results of inference procedures in a randomized experiment. Learning Objectives

4. + Comparing Two Proportions Introduction Suppose we want to compare the proportions of individuals with a certain characteristic in Population 1 and Population 2. Let ’ s call these parameters of interest p 1 and p 2. The ideal strategy is to take a separate random sample from each population and tocompare the sample proportions with that characteristic. What if we want to compare the effectiveness of Treatment 1 and Treatment 2 in a completely randomized experiment? This time,the parameters p 1 and p 2 that we want to compare are the true proportions of successful outcomes for each treatment. We usethe proportions of successes in the two treatment groups to makethe comparison. Here ’ s a table that summarizes these two situations.

5. + The Sampling Distribution of a DifferenceBetween Two Proportions In Chapter 7, we saw that the sampling distribution of a sample proportion has the following properties: Comparing Two Proportions Shape Approximately Normal if np ≥ 10 and n(1 - p) ≥ 10 To explore the sampling distribution of the difference between two proportions, let’s start with two populations having a known proportion of successes. At School 1, 70% of students did their homework last night At School 2, 50% of students did their homework last night. Suppose the counselor at School 1 takes an SRS of 100 students and records the sample proportion that did their homework. School 2’s counselor takes an SRS of 200 students and records the sample proportion that did their homework.

6. + The Sampling Distribution of a DifferenceBetween Two Proportions Using Fathom software, we generated an SRS of 100 students from School 1 and a separate SRS of 200 students from School 2. Thedifference in sample proportions was then calculated and plotted. Werepeated this process 1000 times. The results are below: Comparing Two Proportions

7. + The Sampling Distribution of a DifferenceBetween Two Proportions Comparing Two Proportions Choose an SRS of size n 1 from Population 1 with proportion of successes p 1 and an independent SRS of size n 2 from Population 2 with proportion of successes p 2. The Sampling Distribution of the Difference Between Sample Proportions

8. + Shape Approximately Normal if n 1 p 1 ≥ 10 and n 1 (1 – p 1 ) ≥ 10, and n 2 p 2 ≥ 10 and n 2 (1 – p 2 ) ≥ 10 Sampling Distribution of the Difference between 2 proportions:

9. + Example: Who Does More Homework? Suppose that there are two large high schools, each with more than 2000 students, in a certain town. At School 1, 70% of students did their homework last night. Only 50% of the students atSchool 2 did their homework last night. The counselor at School 1 takes an SRS of 100students and records the proportion that did homework. School 2 ’ s counselor takes an SRS of 200 students and records the proportion that did homework. School 1 ’ s counselor and School 2 ’ s counselor meet to discuss the results of their homework surveys. After the meeting, they both report to their principals that Comparing Two Proportions

10. + Example: Who Does More Homework? Comparing Two Proportions

11. + Confidence Intervals for p 1 – p 2 Comparing Two Proportions If the Normal condition is met, we find the critical value z* for the given confidence level from the standard Normal curve. Our confidence interval for p 1 – p 2 is:

12. + Two-Sample z Interval for p 1 – p 2 Comparing Two Proportions Two-Sample z Interval for a Difference Between Proportions

13. + Plan: We should use a two-sample z interval for p 1 – p 2 if the conditions are satisfied. Random The data come from a random sample of 800 U.S. teens and a separate random sample of 2253 U.S. adults. Normal We check the counts of “successes” and “failures” and note the Normal condition is met since they are all at least 10: Independent We clearly have two independent samples—one of teens and one of adults. Individual responses in the two samples also have to be independent. The researchers are sampling without replacement, so we check the 10% condition: there are at least 10(800) = 8000 U.S. teens and at least 10(2253) = 22,530 U.S. adults. Example: Teens and Adults on Social Networks As part of the Pew Internet and American Life Project, researchers conducted two surveys in late 2009. The first survey asked a random sample of 800 U.S. teens about their use of socialmedia and the Internet. A second survey posed similar questions to a random sample of 2253U.S. adults. In these two studies, 73% of teens and 47% of adults said that they use social-networking sites. Use these results to construct and interpret a 95% confidence interval for thedifference between the proportion of all U.S. teens and adults who use social-networking sites. Comparing Two Proportions State: Our parameters of interest are p 1 = the proportion of all U.S. teens who use social networking sites and p 2 = the proportion of all U.S. adults who use social- networking sites. We want to estimate the difference p 1 – p 2 at a 95% confidence level.

14. + Example: Teens and Adults on Social Networks Comparing Two Proportions Do: Since the conditions are satisfied, we can construct a two- sample z interval for the difference p 1 – p 2. Conclude: We are 95% confident that the interval from 0.223 to 0.297 captures the true difference in the proportion of all U.S. teens and adults who use social-networking sites. This interval suggests that more teens than adults in the United States engage in social networking by between 22.3 and 29.7 percentage points. Take Pulse and Record!

15. + An observed difference between two sample proportions can reflect an actual difference in the parameters, or it may just be due to chance variation in random sampling or random assignment. Significance tests help us decide which explanation makes more sense. The null hypothesis has the general form H 0 : p 1 - p 2 = hypothesized value We’ll restrict ourselves to situations in which the hypothesized difference is 0. Then the null hypothesis says that there is no difference between the two parameters: H 0 : p 1 - p 2 = 0 or, alternatively, H 0 : p 1 = p 2 The alternative hypothesis says what kind of difference we expect. H a : p 1 - p 2 > 0, H a : p 1 - p 2 < 0, or H a : p 1 - p 2 ≠ 0 Significance Tests for p 1 – p 2 Comparing Two Proportions If the Random, Normal, and Independent conditions are met, we can proceed with calculations.

16. + Significance Tests for p 1 – p 2 Comparing Two Proportions If H 0 : p 1 = p 2 is true, the two parameters are the same. We call their common value p. But now we need a way to estimate p, so it makes sense to combine the data from the two samples. This pooled (or combined) sample proportion is:

17. + Two-Sample z Test for The Difference Between Two Proportions If the following conditions are met, we can proceed with a two- sample z test for the difference between two proportions: Comparing Two Proportions Two-Sample z Test for the Difference Between Proportions

18. + Plan: We should perform a two-sample z test for p 1 – p 2 if the conditions are satisfied. Random The data were produced using two simple random samples—of 1010 adults in August 2009 and 1024 in September 2010. Normal We check the counts of “successes” and “failures” and note the Normal condition is met since they are all at least 10: Independent We have two independent samples—one from each year. The researchers are sampling without replacement, so we check the 10% condition: there are at least 10(1024) = 10240 adults in 2010 and at least 10(1010) = 10100 adults in 2009. Comparing Two Proportions State: Our hypotheses are H 0 : p 1 - p 2 = 0 H a : p 1 - p 2 ≠ 0 where p 1 = the true proportion of U.S. adults that approve of Obama in 2009 and p 2 = the true proportion of U.S. adults that approve of the job Obama is doing in 2010.

19. + Example: Presidential Approval Comparing Two Proportions Do: Since the conditions are satisfied, we can perform a two-sample z test for the difference p 1 – p 2. P-value Using Table A or normalcdf, the desired P-value is 2P(z ≤ -1.35) = 2(0.0885) = 0.1770. Conclude: Since our P-value, 0.1770, is greater than the chosen significance level of α = 0.10,we fail to reject H 0. There is not sufficient evidence to conclude that the proportions of U.S. adults that approve of Obama is different in 2010. Calculator will give slightly different answer: 0.1802

20. + Example: Presidential Approval Create a 90% confidence interval to confirm our findings from the significance test. Comparing Two Proportions Do: Since the conditions are satisfied (previously), we can calculate a 90% confidence interval for p 1 – p 2. Since 0 falls in the 90% confidence interval, we fail to reject the hypothesis that there is no difference in approval ratings.

21. + Plan: We should perform a two-sample z test for p 1 – p 2 if the conditions are satisfied. Random The data were produced using two simple random samples—of 100 women and 200 men. Normal We check the counts of “successes” and “failures” and note the Normal condition is met since they are all at least 10: Independent We clearly have two independent samples—one from men and one from women. Individual responses in the two samples also have to be independent. It is safe to assume that one volunteers response to the drug is independent of another volunteer. Example: All drugs created Equal? Suppose the Acme Drug Company develops a new drug, designed to prevent colds.The company states that the drug is equally effective for men and women. To test thisclaim, they choose a a simple random sample of 100 women and 200 men from apopulation of 100,000 volunteers. At the end of the study, 38% of the women caught acold; and 51% of the men caught a cold. Based on these findings, can we reject thecompany's claim that the drug is equally effective for men and women? Use a 0.05level of significance. Comparing Two Proportions State: Our hypotheses are H 0 : p 1 - p 2 = 0 H a : p 1 - p 2 ≠ 0 where p 1 = the true proportion of women that caught a cold and p 2 = the true proportion of men who caught a cold.

22. + Example: All drugs created equal? Comparing Two Proportions Do: Since the conditions are satisfied, we can perform a two-sample z test for the difference p 1 – p 2. P-value Using Table A or normalcdf, the desired P-value is 2P(z ≤-2.13) = 2(0.017) = 0.034 Conclude: Since our P-value, 0.034, is less than the chosen significance level of α = 0.05,we reject H 0. There is evidence to conclude there is a difference in effectiveness of the drug between men and women.

23. + Example: Significance Test in an Experiment High levels of cholesterol in the blood are associated with higher risk of heart attacks. Willusing a drug to lower blood cholesterol reduce heart attacks? The Helsinki Heart Studyrecruited middle-aged men with high cholesterol but no history of other serious medicalproblems to investigate this question. The volunteer subjects were assigned at random to oneof two treatments: 2051 men took the drug gemfibrozil to reduce their cholesterol levels, and acontrol group of 2030 men took a placebo. During the next five years, 56 men in thegemfibrozil group and 84 men in the placebo group had heart attacks. Is the apparent benefitof gemfibrozil statistically significant? Perform an appropriate test to find out. State: Our hypotheses are H 0 : p 1 - p 2 = 0ORH 0 : p 1 = p 2 H a : p 1 - p 2 < 0H a : p 1 < p 2 where p 1 is the actual heart attack rate for middle-aged men like the ones in this study who take gemfibrozil, and p 2 is the actual heart attack rate for middle-aged men like the ones in this study who take only a placebo. What are the implications of a Type I error? What are the implications of a Type II error? We reject the null hypothesis when we should not have. This implies Gemfibrozil reduces heart attacks when it actually does not. We fail to reject the null hypothesis when we should have. This implies Gemfibrozil does not reduce heart attacks when it actually does. Considering these implications, what is an appropriate significance level?

24. + Example: Cholesterol and Heart Attacks Comparing Two Proportions Do: Since the conditions are satisfied, we can perform a two-sample z test for the difference p 1 – p 2. P-value Using Table A or normalcdf, the desired P- value is 0.0068 Conclude: Since the P-value, 0.0068, is less than 0.01, the results are statistically significant at the α = 0.01 level. We can reject H 0 and conclude that there is convincing evidence of a lower heart attack rate for middle-aged men like these who take gemfibrozil than for those who take only a placebo. Plan: We should perform a two-sample z test for p 1 – p 2 if the conditions are satisfied. Random The data come from two groups in a randomized experiment Normal The number of successes (heart attacks!) and failures in the two groups are (56, 1995), (84,1946). These are all at least 10, so the Normal condition is met. Independent Due to the random assignment, these two groups of men can be viewed as independent. Individual observations in each group should also be independent: knowing whether one subject has a heart attack gives no information about whether another subject does.

25. + Section 10.1 Comparing Two Proportions In this section, we learned that… Choose an SRS of size n 1 from Population 1 with proportion of successes p 1 and an independent SRS of size n 2 from Population 2 with proportion of successes p 2. Confidence intervals and tests to compare the proportions p 1 and p 2 of successes for two populations or treatments are based on the difference between the sample proportions. When the Random, Normal, and Independent conditions are met, we can use two- sample z procedures to estimate and test claims about p 1 - p 2. Summary

26. + Section 10.1 Comparing Two Proportions In this section, we learned that… The conditions for two-sample z procedures are: An approximate level C confidence interval for p 1 - p 2 is where z* is the standard Normal critical value. This is called a two-sample z interval for p 1 - p 2. Summary

27. + Section 10.1 Comparing Two Proportions In this section, we learned that… Significance tests of H 0 : p 1 - p 2 = 0 use the pooled (combined) sample proportion The two-sample z test for p 1 - p 2 uses the test statistic with P-values calculated from the standard Normal distribution. Inference about the difference p 1 - p 2 in the effectiveness of two treatments in a completely randomized experiment is based on the randomization distribution of the difference of sample proportions. When the Random, Normal, and Independent conditions are met, our usual inference procedures based on the sampling distribution will be approximately correct. Summary

28. + Chapter 10 Comparing Two Populations or Groups 10.1Comparing Two Proportions 10.2Comparing Two Means

29. + Section 10.2 Comparing Two Means After this section, you should be able to… DESCRIBE the characteristics of the sampling distribution of the difference between two sample means CALCULATE probabilities using the sampling distribution of the difference between two sample means DETERMINE whether the conditions for performing inference are met USE two-sample t procedures to compare two means based on summary statistics or raw data INTERPRET computer output for two-sample t procedures PERFORM a significance test to compare two means INTERPRET the results of inference procedures Learning Objectives

30. + Comparing Two Means Introduction In the previous section, we developed methods for comparing two proportions. What if we want to compare the mean of some quantitativevariable for the individuals in Population 1 and Population 2? Our parameters of interest are the population means µ 1 and µ 2. Once again, the best approach is to take separate random samples from each population and to compare the sample means. Suppose we want to compare the average effectiveness of two treatments in a completely randomized experiment. In this case, the parameters µ 1 and µ 2 are the true mean responses for Treatment 1 and Treatment 2, respectively. We use the mean response in the two groups to make the comparison. Here’s a table that summarizes these two situations:

31. + The Sampling Distribution of a Difference BetweenTwo Means In Chapter 7, we saw that the sampling distribution of a sample mean has the following properties: Comparing Two Means To explore the sampling distribution of the difference between two means, let’s start with two Normally distributed populations having known means and standard deviations. Based on information from the U.S. National Health and Nutrition Examination Survey (NHANES), the heights (in inches) of ten-year-old girls follow a Normal distribution N(56.4, 2.7). The heights (in inches) of ten-year-old boys follow a Normal distribution N(55.7, 3.8). Suppose we take independent SRSs of 12 girls and 8 boys of this age and measure their heights. Shape Approximately Normal if the population distribution is Normal or n ≥ 30 (by the central limit theorem).

32. + The Sampling Distribution of a DifferenceBetween Two Means Using Fathom software, we generated an SRS of 12 girls and a separate SRS of 8 boys and calculated the sample mean heights. The differencein sample means was then calculated and plotted. We repeated thisprocess 1000 times. The results are below: Comparing Two Means

33. + The Sampling Distribution of a DifferenceBetween Two Means Comparing Two Means Choose an SRS of size n 1 from Population 1 with mean µ 1 and standard deviation σ 1 and an independent SRS of size n 2 from Population 2 with mean µ 2 and standard deviation σ 2. The Sampling Distribution of the Difference Between Sample Means

34. + Example: Who ’ s Taller at Ten, Boys or Girls? Based on information from the U.S. National Health and Nutrition Examination Survey(NHANES), the heights (in inches) of ten-year-old girls follow a Normal distribution N (56.4, 2.7). The heights (in inches) of ten-year-old boys follow a Normal distribution N (55.7, 3.8). A researcher takes independent SRSs of 12 girls and 8 boys of this age and measures theirheights. After analyzing the data, the researcher reports that the sample mean height of theboys is larger than the sample mean height of the girls. Comparing Two Means

35. + Example: Who ’ s Taller at Ten, Boys or Girls? Comparing Two Means (c) Does the result in part (b) give us reason to doubt the researchers’ stated results?

36. + The Two-Sample t Statistic Comparing Two Means If the Normal condition is met, we standardize the observed difference to obtain a t statistic that tells us how far the observed difference is from its mean in standard deviation units: The two-sample t statistic has approximately a t distribution. We can use technology to determine degrees of freedom OR we can use a conservative approach, using the smaller of n 1 – 1 and n 2 – 1 for the degrees of freedom.

37. + Confidence Intervals for µ 1 – µ 2 Comparing Two Means Two-Sample t Interval for a Difference Between Means

38. + Big Trees, Small Trees, Short Trees, Tall Trees The Wade Tract Preserve in Georgia is an old-growth forest of longleaf pines that has survived in a relatively undisturbed state for hundreds of years. One question of interest to foresters whostudy the area is “ How do the sizes of longleaf pine trees in the northern and southern halves of the forest compare? ” To find out, researchers took random samples of 30 trees from each half and measured the diameter at breast height (DBH) in centimeters. Comparative boxplotsof the data and summary statistics from Minitab are shown below. Construct and interpret a90% confidence interval for the difference in the mean DBH for longleaf pines in the northernand southern halves of the Wade Tract Preserve. Comparing Two Means State: Our parameters of interest are µ 1 = the true mean DBH of all trees in the southern half of the forest and µ 2 = the true mean DBH of all trees in the northern half of the forest. We want to estimate the difference µ 1 - µ 2 at a 90% confidence level.

39. + Big Trees, Small Trees, Short Trees, Tall Trees Comparing Two Means Do: Since the conditions are satisfied, we can construct a two-sample t interval for the difference µ 1 – µ 2. We’ll use the conservative df = 30-1 = 29. Plan: We should use a two-sample t interval for µ 1 – µ 2 if the conditions are satisfied. Random The data come from a random samples of 30 trees each from the northern and southern halves of the forest. Normal The boxplots give us reason to believe that the population distributions of DBH measurements may not be Normal. However, since both sample sizes are at least 30, we are safe using t procedures. Independent Researchers took independent samples from the northern and southern halves of the forest. Because sampling without replacement was used, there have to be at least 10(30) = 300 trees in each half of the forest. This is pretty safe to assume. Conclude: We are 90% confident that the interval from 3.83 to 17.83 centimeters captures the difference in the actual mean DBH of the southern trees and the actual mean DBH of the northern trees. This interval suggests that the mean diameter of the southern trees is between 3.83 and 17.83 cm larger than the mean diameter of the northern trees.

40. + Comparing Two Means Example: Plastic Grocery Bags Do plastic bags from Target or plastic bags from Walmart hold more weight? A group of AP Statistic students decided to investigate by filling a random sample of 5 bags from each store with common grocery items until the bags ripped. Then they weighed the contents of items in each bag to determine its capacity. Here are their results, in grams: Target:12,57213,99911,21515,44710,896 Walmart: 955210,896 6983 8767 9972 Problem: (a)Construct and interpret a 99% confidence interval for the difference in mean capacity of plastic grocery bags from Target and Walmart. State: We want to estimate at the 99% confidence level where = the mean capacity of plastic bags from Target (in grams) and = the mean capacity of plastic bags from Walmart (in grams). Plan: If the conditions are met, we will calculate a two-sample t interval Random: The students selected a random sample of bags from each store. Normal: Since the sample sizes are small, we must graph the data to see if it is reasonable to assume that the population distributions are approximately Normal. Independent: The samples were selected independently and it is reasonable to assume that there are more than 10(5) = 50 plastic grocery bags at each store.

41. + Conservative df = 5 – 1 = 4, the critical value for 99% confidence is t* = 4.604. The confidence interval is: With technology and df =7.5, CI = (–101, 7285). Notice how much narrower this interval is. Do: Conclude: We are 99% confident that the interval from –1380 to 8564 grams captures the true difference in the mean capacity for plastic grocery bags from Target and from Walmart. Since the interval includes 0, it is plausible that there is no difference in the two means. Thus, we do not have convincing evidence that there is a difference in mean capacity. However, if we increased the sample size we would likely find a convincing difference since it seems pretty clear that Target bags have a bigger capacity. (b) Does your interval provide convincing evidence that there is a difference in the mean capacity among the two stores?

42. + An observed difference between two sample means can reflect an actual difference in the parameters, or it may just be due to chance variation in random sampling or random assignment. Significance tests help us decide which explanation makes more sense. The null hypothesis has the general form H 0 : µ 1 - µ 2 = hypothesized value We’re often interested in situations in which the hypothesized difference is 0. Then the null hypothesis says that there is no difference between the two parameters: H 0 : µ 1 - µ 2 = 0 or, alternatively, H 0 : µ 1 = µ 2 The alternative hypothesis says what kind of difference we expect. H a : µ 1 - µ 2 > 0, H a : µ 1 - µ 2 < 0, or H a : µ 1 - µ 2 ≠ 0 Significance Tests for µ 1 – µ 2 Comparing Two Means If the Random, Normal, and Independent conditions are met, we can proceed with calculations.

43. + Significance Tests for µ 1 – µ 2 Comparing Two Means To find the P-value, use the t distribution with degrees of freedom given by technology or by the conservative approach (df = smaller of n 1 - 1 and n 2 - 1).

44. + Two-Sample t Test for The Difference Between Two Means If the following conditions are met, we can proceed with a two-sample t test for the difference between two means: Comparing Two Means Two-Sample t Test for the Difference Between Two Means

45. + Example: Calcium and Blood Pressure Does increasing the amount of calcium in our diet reduce blood pressure? Examination of a large sample of people revealed a relationship between calcium intake and blood pressure. Therelationship was strongest for black men. Such observational studies do not establishcausation. Researchers therefore designed a randomized comparative experiment. Thesubjects were 21 healthy black men who volunteered to take part in the experiment. Theywere randomly assigned to two groups: 10 of the men received a calcium supplement for 12weeks, while the control group of 11 men received a placebo pill that looked identical. Theexperiment was double-blind. The response variable is the decrease in systolic (top number)blood pressure for a subject after 12 weeks, in millimeters of mercury. An increase appears asa negative response Here are the data: Comparing Two Means State: We want to perform a test of H 0 : µ 1 - µ 2 = 0 H a : µ 1 - µ 2 > 0 where µ 1 = the true mean decrease in systolic blood pressure for healthy black men like the ones in this study who take a calcium supplement, and µ 2 = the true mean decrease in systolic blood pressure for healthy black men like the ones in this study who take a placebo. We will use α = 0.05.

46. + Example: Calcium and Blood Pressure Comparing Two Means Plan: If conditions are met, we will carry out a two-sample t test for µ 1 - µ 2. Random The 21 subjects were randomly assigned to the two treatments. Normal With such small sample sizes, we need to examine the data to see if it’s reasonable to believe that the actual distributions of differences in blood pressure when taking calcium or placebo are Normal. Hand sketches of calculator boxplots and Normal probability plots for these data are below: The boxplots show no clear evidence of skewness and no outliers. The Normal probability plot of the placebo group’s responses looks very linear, while the Normal probability plot of the calcium group’s responses shows some slight curvature. With no outliers or clear skewness, the t procedures should be pretty accurate. Independent Due to the random assignment, these two groups of men can be viewed as independent. Individual observations in each group should also be independent: knowing one subject’s change in blood pressure gives no information about another subject’s response.

47. + Example: Calcium and Blood Pressure Comparing Two Means Do: Since the conditions are satisfied, we can perform a two-sample t test for the difference µ 1 – µ 2. P-value Using the conservative df = 10 – 1 = 9, we can use Table B to show that the P-value is between 0.05 and 0.10. Conclude: Because the P-value is greater than α = 0.05, we fail to reject H 0. The experiment provides some evidence that calcium reduces blood pressure, but the evidence is not convincing enough to conclude that calcium reduces blood pressure more than a placebo.

48. + Example: Calcium and Blood Pressure Comparing Two Means We can estimate the difference in the true mean decrease in blood pressure for the calcium and placebo treatments using a two-sample t interval for µ 1 - µ 2. To get results that are consistent with the one-tailed test at α = 0.05 from the example, we’ll use a 90% confidence level. The conditions for constructing a confidence interval are the same as the ones that we checked in the example before performing the two-sample t test. With df = 9, the critical value for a 90% confidence interval is t* = 1.833. The interval is: We are 90% confident that the interval from -0.754 to 11.300 captures the difference in true mean blood pressure reduction on calcium over a placebo. Because the 90% confidence interval includes 0 as a plausible value for the difference, we cannot reject H 0 : µ 1 - µ 2 = 0 against the two-sided alternative at the α = 0.10 significance level or against the one-sided alternative at the α = 0.05 significance level.

49. + Paper Airplanes...which flies farther? Comparing Two Means 1.Design an experiment to determine which of two airplane models flies farthest. Follow principals of experimental design (Control, Random, Replication) 2.State the hypotheses you are interested in testing 3.Carry out your plan and collect the necessary data 4.Compare the flight distances for the two models (SOCS) 5.Perform a significance test (at the 0.05 level) to determine if the difference in means is significant 6.Calculate a 95% confidence interval for the average flight distance for YOUR PLANE 7.Construct a 95% confidence interval for the DIFFERENCE in means.

50. + Example: The stronger picker-upper? In commercials for Bounty paper towels, the manufacturer claims that they are the “quicker picker-upper.” But are they also the stronger picker upper? Two AP Statistics students, Wesley and Maverick, decided to find out. They selected a random sample of 30 Bounty paper towels and a random sample of 30 generic paper towels and measured their strength when wet. To do this, they uniformly soaked each paper towel with 4 ounces of water, held two opposite edges of the paper towel, and counted how many quarters each paper towel could hold until ripping, alternating brands. Here are their results: Bounty:106, 111, 106, 120, 103, 112, 115, 125, 116, 120, 126, 125, 116, 117, 114 118, 126, 120, 115, 116, 121, 113, 111, 128, 124, 125, 127, 123, 115, 114 Generic: 77, 103, 89, 79, 88, 86, 100, 90, 81, 84, 84, 96, 87, 79, 90 86, 88, 81, 91, 94, 90, 89, 85, 83, 89, 84, 90, 100, 94, 87 Problem: (a)Display these distributions using parallel boxplots and briefly compare these distributions. Based only on the boxplots, discuss whether or not you think the mean for Bounty is significantly higher than the mean for generic. The five-number summary for the Bounty paper towels is (103, 114, 116.5, 124, 128) and the five-number summary for the generic paper towels is (77, 84, 88, 90, 103). Both distributions are roughly symmetric, but the generic brand has two high outliers. The center of the Bounty distribution is much higher than the center of the generic distribution. Although the range of each distribution is roughly the same, the interquartile range of the Bounty distribution is larger. Since the centers are so far apart and there is almost no overlap in the two distributions, the Bounty mean is almost certain to be significantly higher than the generic mean. If the means were really the same, it would be virtually impossible to get so little overlap.

51. + State: We want to perform a test of H 0 : µ B - µ G = 0 versus H a : µ B - µ G > 0 at the 5% level of significance where µ B = the mean number of quarters a wet Bounty paper towel can hold and µ G = the mean number of quarters a wet generic paper towel can hold. Plan: If the conditions are met, we will conduct a two-sample t test for µ B - µ G. Random: The students used a random sample of paper towels from each brand. Normal: Even though there were two outliers in the generic distribution, both distributions were reasonably symmetric and the sample sizes are both at least 30, so it is safe to use t procedures. Independent: The samples were selected independently and it is reasonable to assume there are more than 10(30) = 300 paper towels of each brand. Do: P-value: Using either the conservative df = 30 – 1 = 29 or from technology (df = 57.8), the P-value is approximately 0. Conclude: Since the P-value is less than 0.05, we reject. There is very convincing evidence that wet Bounty paper towels can hold more weight, on average, than wet generic paper towels. Since the P-value is approximately 0, it is almost impossible to get a difference in means of at least 29.5 quarters by random chance, assuming that the two brands of paper towels can hold the same amount of weight when wet. (b) Use a significance test to determine if there is convincing evidence that wet Bounty paper towels can hold more weight, on average, than wet generic paper towels.

52. + Using Two-Sample t Procedures Wisely Comparing Two Means The two-sample t procedures are more robust against non-Normality than the one-sample t methods. When the sizes of the two samples are equal and the two populations being compared have distributions with similar shapes, probability values from the t table are quite accurate for a broad range of distributions when the sample sizes are as small as n 1 = n 2 = 5. Sample size less than 15: Use two-sample t procedures if the data in both samples/groups appear close to Normal (roughly symmetric, single peak, no outliers). If the data are clearly skewed or if outliers are present, do not use t. Sample size at least 15: Two-sample t procedures can be used except in the presence of outliers or strong skewness. Large samples: The two-sample t procedures can be used even for clearly skewed distributions when both samples/groups are large, roughly n ≥ 30. Using the Two-Sample t Procedures: The Normal Condition

53. + Using Two-Sample t Procedures Wisely Comparing Two Means Here are several cautions and considerations to make when using two- sample t procedures. In planning a two-sample study, choose equal sample sizes if you can. Do not use “pooled” two-sample t procedures! We are safe using two-sample t procedures for comparing two means in a randomized experiment. Do not use two-sample t procedures on paired data! Beware of making inferences in the absence of randomization. The results may not be generalized to the larger population of interest.

54. + Section 10.2 Comparing Two Means In this section, we learned that… Choose an SRS of size n 1 from Population 1 and an independent SRS of size n 2 from Population 2. The sampling distribution of the difference of sample means has: Confidence intervals and tests for the difference between the means of two populations or the mean responses to two treatments µ 1 – µ 2 are based on the difference between the sample means. If we somehow know the population standard deviations σ 1 and σ 2, we can use a z statistic and the standard Normal distribution to perform probability calculations. Summary

55. + Section 10.2 Comparing Two Means Since we almost never know the population standard deviations in practice, we use the two-sample t statistic where t has approximately a t distribution with degrees of freedom found by technology or by the conservative approach of using the smaller of n 1 – 1 and n 2 – 1. The conditions for two-sample t procedures are: Summary

56. + Section 10.2 Comparing Two Means The level C two-sample t interval for µ 1 – µ 2 is where t* is the critical value for confidence level C for the t distribution with degrees of freedom from either technology or the conservative approach. To test H 0 : µ 1 - µ 2 = hypothesized value, use a two-sample t test for µ 1 - µ 2. The test statistic is P-values are calculated using the t distribution with degrees of freedom from either technology or the conservative approach. Summary

57. + Section 10.2 Comparing Two Means The two-sample t procedures are quite robust against departures from Normality, especially when both sample/group sizes are large. Inference about the difference µ 1 - µ 2 in the effectiveness of two treatments in a completely randomized experiment is based on the randomization distribution of the difference between sample means. When the Random, Normal, and Independent conditions are met, our usual inference procedures based on the sampling distribution of the difference between sample means will be approximately correct. Don’t use two-sample t procedures to compare means for paired data. Summary