Chapter 11

1. Chapter 11 Comparing Two Populations or Treatments

2. Suppose we have a population of adult men with a mean height of 71 inches and standard deviation of 2.5 inches. We also have a population of adult women with a mean height of 65 inches and standard deviation of 2.3 inches. Assume heights are normally distributed. Suppose we take a random sample of 30 men and a random sample of 25 women from their respective populations and calculate the difference in their heights (man’s height – woman’s height). If we did this many times, what would the distribution of differences be like? On the next slide we will investigate this distribution.

3. Male Heights Female Heights 71 65 Suppose we took repeated samples of size n = 25 from the population of female heights and calculated the sample means. We would have the sampling distribution of xF Suppose we took repeated samples of size n = 30 from the population of male heights and calculated the sample means. We would have the sampling distribution of xM. 65 71 Doing this repeatedly, we will create the sampling distribution of (xM – xF) xM - xF 6 Randomly take one of the sample means for the males and one of the sample means for the females and find the difference in mean heights. sM = 2.5 sF = 2.3

4. 6 Heights Continued . . . • Describe the sampling distribution of the difference in mean heights between men and women. • What is the probability that the difference in mean heights of a random sample of 30 men and a random sample of 25 women is less than 5 inches? The sampling distribution is normally distributed with

5. Properties of the Sampling Distribution of x1 – x2 If the random samples on which x1 and x2 are based are selected independently of one another, then Mean value of x1 – x2 The sampling distribution of x1 – x2 is always centered at the value of m1 – m2, so x1 – x2is an unbiased statistic for estimating m1 – m2. 3. In n1 and n2 are both large or the population distributions are (at least approximately) normal, x1 and x2 each have (at least approximately) normal distributions. This implies that the sampling distribution of x1 – x2 is also (approximately) normal. 1. 2. and The variance of the differences is the sum of the variances.

6. The properties for the sampling distribution of x1 – x2 implies that x1 – x2can be standardized to obtain a variable with a sampling distribution that is approximately the standard normal (z)distribution. When two random samples are independently selected and n1 and n2 are both large or the population distributions are (at least approximately) normal, the distribution of We must know s1 and s2 in order to use this procedure. If s1ands2 is unknown we must use t distributions. is described (at least approximately) by the standard normal (z) distribution.

7. Two-Sample t Test for Comparing Two Populations Null Hypothesis: H0: m1 – m2 = hypothesized value Test Statistic: The appropriate df for the two-sample t test is The computed number of df should be truncated to an integer. A conservative estimate of the P-value can be found by using the t-curve with the number of degrees of freedom equal to the smaller of (n1 – 1) or (n2 – 1). The hypothesized value is often 0, but there are times when we are interested in testing for a difference that is not 0. where and

8. Two-Sample t Test for Comparing Two Populations Continued . . . Null Hypothesis: H0: m1 – m2 = hypothesized value Area under the appropriate t curve to the left of the computed t Ha: m1 – m2 < hypothesized value 2(area to right of computed t) if +t or 2(area to left of computed t) if -t Ha: m1 – m2 ≠ hypothesized value

9. H0: m1 - m2 = 0 Ha: m1 - m2 < 0 Ha: m1 - m2 > 0 Ha: m1 - m2 ≠ 0 Another Way to Write Hypothesis Statements: H0: m1 = m2 When the hypothesized value is 0, we can rewrite these hypothesis statements: Be sure to define BOTHm1 and m2! Ha: m1 < m2 Ha: m1 > m2 Ha: m1 ≠ m2

10. Two-Sample t Test for Comparing Two Populations Continued . . . Assumptions: • The two samples are independently selected random samples from the populations of interest • The sample sizes are large (generally 30 or larger) or the population distributions are (at least approximately) normal. When comparing two treatment groups, use the following assumptions: • Individuals or objects are randomly assigned to treatments (or vice versa) • The sample sizes are large (generally 30 or larger) or the treatment response distributions are approximately normal.

11. Are women still paid less than men for comparable work? A study was carried out in which salary data was collected from a random sample of men and from a random sample of women who worked as purchasing managers and who were subscribers to Purchasing magazine. Annual salaries (in thousands of dollars) appear below (the actual sample sizes were much larger). Use a = .05 to determine if there is convincing evidence that the mean annual salary for male purchasing managers is greater than the mean annual salary for female purchasing managers. H0: m1 – m2 = 0 Ha: m1 – m2 > 0 If we had defined m1 as the mean salary for female purchasing managers and m2 as the mean salary for male purchasing managers, then the correct alternative hypothesis would be the difference in the means is less than 0. Where m1 = mean annual salary for male purchasing managers and m2 = mean annual salary for female purchasing managers State the hypotheses:

12. Men Women 80 60 Salary War Continued . . . H0: m1 – m2 = 0 Ha: m1 – m2 > 0 Assumptions: • Given two independently selected random samples of male and female purchasing managers. Where m1 = mean annual salary for male purchasing managers and m2 = mean annual salary for female purchasing managers Even though these are samples from subscribers of Purchasing magazine, the authors of the study believed it was reasonable to view the samples as representative of the populations of interest. Verify the assumptions 2) Since the sample sizes are small, we must determine if it is plausible that the sampling distributions for each of the two populations are approximately normal. Since the boxplots are reasonably symmetrical with no outliers, it is plausible that the sampling distributions are approximately normal.

13. Salary War Continued . . . H0: m1 – m2 = 0 Ha: m1 – m2 > 0 Test Statistic: P-value =.004a= .05 Since the P-value < a, we reject H0. There is convincing evidence that the mean salary for male purchasing managers is higher than the mean salary for female purchasing managers. Where m1 = mean annual salary for male purchasing managers and m2 = mean annual salary for female purchasing managers What potential type error could we have made with this conclusion? Truncate (round down) this value. Type I Now find the area to the right of t = 3.11 in the t-curve with df = 15. Compute the test statistic and P-value To find the P-value, first find the appropriate df.

14. The Two-Sample t Confidence Interval for the Difference Between Two Population or Treatment Means The general formula for a confidence interval for m1 – m2 when • The two samples are independently selected random samples from the populations of interest • The sample sizes are large (generally 30 or larger) or the population distributions are (at least approximately) normal. is The t critical value is based on df should be truncated to an integer. For a comparison of two treatments, use the following assumptions: 1) Individuals or objects are randomly assigned to treatments (or vice versa) 2) The sample sizes are large (generally 30 or larger) or the treatment response distributions are approximately normal. where and

15. x4 = 3924 s4 = 829.67 x8 = 4069.27 s8 = 952.90 In a study on food intake after sleep deprivation, men were randomly assigned to one of two treatment groups. The experimental group were required to sleep only 4 hours on each of two nights, while the control group were required to sleep 8 hours on each of two nights. The amount of food intake (Kcal) on the day following the two nights of sleep was measured. Compute a 95% confidence interval for the true difference in the mean food intake for the two sleeping conditions. Find the mean and standard deviation for each treatment.

16. 4-hour 8-hour 4000 x4 = 3924 s4 = 829.67 x8 = 4069.27 s8 = 952.90 Food Intake Study Continued . . . • Assumptions: • Men were randomly assigned to two treatment groups Verify the assumptions. 2) The assumption of normal response distributions is plausible because both boxplots are approximately symmetrical with no outliers.

17. x4 = 3924 s4 = 829.67 x8 = 4069.27 s8 = 952.90 Food Intake Study Continued . . . Based upon this interval, is there a significant difference in the mean food intake for the two sleeping conditions? No, since 0 is in the confidence interval, there is not convincing evidence that the mean food intake for the two sleep conditions are different. Calculate the interval. We are 95% confident that the true difference in the mean food intake for the two sleeping conditions is between -814.1 Kcal and 523.6 Kcal. Interpret the interval in context.

18. Pooled t Test • Used when the variances of the two populations are equal (s1 = s2) • Combines information from both samples to create a “pooled” estimate of the common variance which is used in place of the two sample standard deviations • Is not widely used due to its sensitivity to any departure from the equal variance assumption P-values computed using the pooled t procedure can be far from the actual P-value if the population variances are not equal. When the population variances are equal, the pooled t procedure is better at detecting deviations from H0 than the two-sample t test.

19. Suppose that an investigator wants to determine if regular aerobic exercise improves blood pressure. A random sample of people who jog regularly and a second random sample of people who do not exercise regularly are selected independently of one another. Can we conclude that the difference in mean blood pressure is attributed to jogging? What about other factors like weight? One way to avoid these difficulties would be to pair subjects by weight then assign one of the pair to jogging and the other to no exercise.

20. Summary of the Paired t test for Comparing Two Population or Treatment Means Null Hypothesis: H0: md = hypothesized value Test Statistic: Where n is the number of sample differences and xdand sd are the mean and standard deviation of the sample differences. This test is based on df = n – 1. Alternative Hypothesis: P-value: Ha: md> hypothesized value Area to the right of calculated t Ha: md< hypothesized value Area to the left of calculated t Ha: md≠ hypothesized value 2(area to the right of t) if +t or2(area to the left of t) if -t Where md is the mean of the differences in the paired observations The hypothesized value is usually 0 – meaning that there is no difference.

21. Summary of the Paired t test for Comparing Two Population or Treatment Means Continued . . . Assumptions: • The samples are paired. • The n sample differences can be viewed as a random sample from a population of differences. • The number of sample differences is large (generally at least 30) or the population distribution of differences is (at least approximately) normal.

22. An engineering association wants to see if there is a difference in the mean annual salary for electrical engineers and chemical engineers. A random sample of electrical engineers is surveyed about their annual income. Another random sample of chemical engineers is surveyed about their annual income. Is this an example of paired samples? No, there is no pairing of individuals, you have two independent samples

23. A pharmaceutical company wants to test its new weight-loss drug. Before giving the drug to volunteers, company researchers weigh each person. After a month of using the drug, each person’s weight is measured again. Is this an example of paired samples? Yes, you have two observations on each individual, resulting in paired data.

24. Can playing chess improve your memory? In a study, students who had not previously played chess participated in a program in which they took chess lessons and played chess daily for 9 months. Each student took a memory test before starting the chess program and again at the end of the 9-month period. If we had subtracted Post-test minus Pre-test, then the alternative hypothesis would be the mean difference is greater than 0. H0: md= 0 Ha: md < 0 Where mdis the mean memory score difference between students with no chess training and students who have completed chess training First, find the differences pre-test minus post-test. State the hypotheses.

25. Playing Chess Continued . . . H0: md= 0 Ha: md < 0 Assumptions: 1) Although the sample of students is not a random sample, the Where md is the mean memory score difference between students with no chess training and students who have completed chess training Verify assumptions investigator believed that it was reasonable to view the 12 sample differences as representative of all such differences. 2) A boxplot of the differences is approximately symmetrical with no outliers so the assumption of normality is plausible.

26. Playing Chess Continued . . . H0: md= 0 Ha: md < 0 Test Statistic: Where md is the mean memory score difference between students with no chess training and students who have completed chess training State the conclusion in context. Compute the test statistic and P-value. P-value ≈ 0 df = 11a= .05 Since the P-value < a, we reject H0. There is convincing evidence to suggest that the mean memory score after chess training is higher than the mean memory score before training.

27. Paired t Confidence Interval for md When • The samples are paired. • The n sample differences can be viewed as a random sample from a population of differences. • The number of sample differences is large (generally at least 30) or the population distribution of differences is (at least approximately) normal. the paired t interval for md is Where df = n - 1

28. Playing Chess Revisited . . . Compute a 90% confidence interval for the mean difference in memory scores before chess training and the memory scores after chess training. We are 90% confident that the true mean difference in memory scores before chess training and the memory scores after chess training is between -201.5 and -87.69.

29. Large-Sample Inferences Concerning the Difference Between Two Population or Treatment Proportions

30. Let’s investigate the sampling distribution of pfreeze - ptape Some people seem to think that duct tape can fix anything . . . even remove warts! Investigators at Madigan Army Medical Center tested using duct tape to remove warts versus the more traditional freezing treatment. Suppose that the duct tape treatment will successfully remove 50% of warts and that the traditional freezing treatment will successfully remove 60% of warts.

31. Suppose we repeatedly treated 100 warts using the duct tape method and calculated the proportion of warts that are successfully removed. We would have the sampling distribution of ptape. Suppose we repeatedly treated 100 warts using the traditional freezing treatment and calculated the proportion of warts that are successfully removed. We would have the sampling distribution of pfreeze .6 .5 Doing this repeatedly, we will create the sampling distribution of (pfreeze – ptape) pfreeze - ptape .1 pfreeze = the true proportion of warts that are successfully removed by freezing pfreeze = .6 ptape = the true proportion of warts that are successfully removed by using duct tape ptape = .5 Randomly take one of the sample proportions for the freezing treatment and one of the sample proportions for the duct tape treatment and find the difference.

32. Properties of the Sampling Distribution of p1 – p2 If two random samples are selected independently of one another, the following properties hold: 1. This says that the sampling distribution of p1 – p2 is centered at p1 – p2 so p1 – p2 is an unbiased statistic for estimating p1 – p2. Since the value for p1 and p2 are unknown, we will combine p1 and p2 to estimate the common value of p1 and p2 Use: 3. If both n1 and n2 are large (that is, if n1p1> 10, n1(1 – p1) > 10, n2p2> 10, and n2(1 – p2) > 10), then p1 and p2 each have a sampling distribution that is approximately normal, and their difference p1 – p2 also has a sampling distribution that is approximately normal. When performing a hypothesis test, we will use the null hypothesis that p1 and p2 are equal. We will not know the common value for p1 and p2. 2.

33. Use: Summary of Large-Sample z Test for p1 – p2 = 0 Null Hypothesis: H0: p1 – p2 = 0 Test Statistic: Alternative Hypothesis: P-value: Ha: p1 – p2 > 0 area to the right of calculated z Ha: p1 – p2 < 0 area to the left of calculated z Ha: p1 – p2 ≠ 0 2(area to the right of z) if +z or 2(area to the left of z) if -z

34. Another Way to Write Hypothesis statements: H0: p1 - p2 = 0 Ha: p1 - p2 > 0 Ha: p1 - p2 < 0 Ha: p1 - p2 ≠ 0 Be sure to define both p1 & p2! H0: p1 = p2 Ha: p1 > p2 Ha: p1 < p2 Ha: p1 ≠ p2

35. Since p1 and p2 are unknown we must use p1 and p2 to verify that the samples are large enough. 2) Both sample sizes are large n1p1> 10, n1(1 – p1) > 10, n2p2> 10, n2(1 – p2) > 10 Summary of Large-Sample z Test for p1 – p2 = 0 Continued . . . Assumption: • The samples are independently chosen random samples or treatments were assigned at random to individuals or objects

36. Investigators at Madigan Army Medical Center tested using duct tape to remove warts. Patients with warts were randomly assigned to either the duct tape treatment or to the more traditional freezing treatment. Those in the duct tape group wore duct tape over the wart for 6 days, then removed the tape, soaked the area in water, and used an emery board to scrape the area. This process was repeated for a maximum of 2 months or until the wart was gone. The data follows: Do these data suggest that freezing is less successful than duct tape in removing warts?

37. 2) The sample sizes are large enough because: n1p1 = 100(.6) = 60 > 10 n1(1 – p1) = 100(.4) = 40 > 10 n2p2 = 100(.85) = 85 > 10 n2(1 – p2) = 100(.15) = 15 > 10 Duct Tape Continued . . . H0: p1 – p2 = 0 Ha: p1 – p2 < 0 Assumptions: 1) Subjects were randomly assigned to the two treatments. Where p1 is the true proportion of warts that would be successfully removed by freezing and p2 is the true proportion of warts that would be successfully removed by duct tape

38. Duct Tape Continued . . . H0: p1 – p2 = 0 Ha: p1 – p2 < 0 P-value ≈ 0 a = .01 Since the P-value < a, we reject H0. There is convincing evidence to suggest the proportion of warts successfully removed is lower for freezing than for the duct tape treatment.

39. 2) Both sample sizes are large n1p1> 10, n1(1 – p1) > 10, n2p2> 10, n2(1 – p2) > 10 A Large-Sample Confidence Interval for p1 – p2 When • The samples are independently chosen random samples or treatments were assigned at random to individuals or objects a large-sample confidence interval for p1 – p2 is

40. The article “Freedom of What?” (Associated Press, February 1, 2005) described a study in which high school students and high school teachers were asked whether they agreed with the following statement: “Students should be allowed to report controversial issues in their student newspapers without the approval of school authorities.” It was reported that 58% of students surveyed and 39% of teachers surveyed agreed with the statement. The two samples – 10,000 high school students and 8000 high school teachers – were selected from schools across the country. Compute a 90% confidence interval for the difference in proportion of students who agreed with the statement and the proportion of teachers who agreed with the statement.

41. p1 = .58 p2 = .39 2) Both sample sizes are large enough n1p1 = 10000(.58) > 10, n1(1 – p1) = 10000(.42) > 10, n2p2 = 8000(.39) > 10, n2(1 – p2) = 8000(.61) > 10 Newspaper Problem Continued . . . Based on this confidence interval, does there appear to be a significant difference in proportion of students who agreed with the statement and the proportion of teachers who agreed with the statement? Explain. 1) Assume that it is reasonable to regard these two samples as being independently selected and representative of the populations of interest. We are 90% confident that the difference in proportion of students who agreed with the statement and the proportion of teachers who agreed with the statement is between .178 and .202.