1. Chapter 10: Inferences Involving Two Populations

2. Independent and Dependent Samples The object is to compare means of two samples and draw conclusions about the differences in population means. Two basic kinds of samples: independent and dependent (paired). Which kind you have depends on the sources of two samples and how the data was collected.

3. Dependent Samples If one observation is collected for each sample from the same source, the samples are dependent. This is often called “pair data” because you get a pair of observations from one individual or experimental unit. Examples include pretest & posttest scores, weight before and after a diet, left eye and right eye acuity, etc. There is a one-to-one correspondence between an observation in one sample and an observation in the other sample.

4. Independent Samples Two samples are independent if there is no connection between an observation of one sample with a particular observation of the other. Also, there can be no connection in the sampling procedure (an individual selected for one in no way affects the selection of any individual in the other, including by exclusion)

5. Dependent Sample Examples The same test is given to all students at the beginning and end of a course to measure learning (one pair of scores per person). IQ tests are given to husband & wife pairs. A medical treatment is given to patients matched for condition, age, sex, race, weight, and other characteristics with patients in a control group.

6. Independent Sample Examples The same test is given to all students in two classes (no pairing of scores occurs). IQ tests are given to men and women without consideration of relationship between any of them. Subjects are randomly assigned to a treatment and a control group to test a new drug. No attempt is made to “match” them.

7. Difference of means for Paired Data When dependent samples are involved, the data is paired data. Paired data results from: –before and after studies, –a common source, or –from matched pairs. We will denote the random variables from the two samples by X 1 and X 2.

8. The meaning of X 1 and X 2 Two samples are being taken. For example, the pretest is one sample, and the posttest is another sample. Then X 1 represents a pretest score and X 2 represents a posttest score. However, there is an X 1 and an X 2 for each person taking the test.

9. Paired Data X1X1 X2X2 x 11 x 21 x 12 x 22 x 13 x 23 x 14 x 24 x 15 x 25 x 16 x 26 x 17 x 27 …… Here the random variable names are given by capital letters, and individual observations by small letters. The data appear side-by- side, each pair having the same second subscript. You cannot change the order of one column without destroying the relation- ship between the columns. That is what makes it “paired data.” There are the same number of observations, n, in each sample.

10. What do we want to know? In paired data studies, the parameter of interest is the mean difference between the groups. This is conceptually different from the difference between the means of the groups. In other words, the population of interest is actually the differences between X 1 and X 2. We define a new value, d=x 1 -x 2 as one observation taken from this population.

11. Why is this important? The mean difference between the groups and the difference between the means of the groups are the same number. But their sampling distributions are different! From, we calculate, the mean difference. Now, will have a normal distribution if X 1 and X 2 are normal or n>30 (approx).

12. Distribution of mean differences If we know is normally distributed, then we can use the same tests and confidence intervals that we learned for. We won’t bother with the “variance known” situation this time. We will calculate the variance from the sample and use the t distribution. In other words, treat the d’s as the sample. Find their mean and standard deviation.

13. Distribution of mean differences There is a population parameter,, that we are trying to estimate. The point estimate is, taken from a sample of n differences (d’s). The d’s have a standard deviation, which is calculated in the same way as s. The standard deviation of is This is no difference from what we had before, except for symbols!

14. Confidence Interval for Paired Differences A (1-α)100% CI for is given by:, where.

15. Example: Salt-free diets are often prescribed for people with high blood pressure. The following data was obtained from an experiment designed to estimate the reduction in diastolic blood pressure as a result of following a salt-free diet for two weeks. Assume diastolic readings to be normally distributed. Find a 99% confidence interval for the mean reduction. Question: How do you decide which way to subtract?

16. Solution: Population Parameter of Interest: The mean reduction (difference) in diastolic blood pressure. Determine the distribution to use: Assumptions: Both sample populations are assumed normal, σ unknown. Use t with df = 8  1 = 7. Confidence level: 1   = 0.99 Two-tailed situation,  /2 = 0.005 t (df,  /2) = t (7, 0.005) = 3.50 Sample evidence: Sample information:

17. Calculate the error bound: A 99% confidence interval for  d is

18. Hypothesis Testing: When testing a null hypothesis about the mean difference, the test statistic is where t* has a t distribution with df = n  1. Example: The corrosive effects of various chemicals on normal and specially treated pipes were tested by using a dependent sampling plan. The data collected is summarized by where d is the amount of corrosion on the treated pipe subtracted from the amount of corrosion on the normal pipe.

19. Example (continued): Does this sample provide sufficient evidence to conclude the specially treated pipes are more resistant to corrosion? Use  = 0.05 a.Solve using the classical approach. b.Solve using the p-value approach. Solution: 1.State the hypotheses (you must say something about the direction of the difference): Test for the mean difference in corrosion, normal pipe - treated pipe. The null and alternative hypothesis: H 0 :  d = 0 (did not lower corrosion) H a :  d > 0 (did lower corrosion)

20. 2.Determine the appropriate type of test: Assumptions: Assume corrosion measures are approximately normal, σ unknown. Use t-test for paired differences. 3. Define the rejection region: a. Right tailed test, Reject H 0 if t*>t (16,0.05) = 1.75. b. Reject H 0 if p<.05. 4.Calculate the value of the test statistic:

21. 5.State the conclusion: a.Decision: Reject H 0 because t*=4.896>1.75 b. Decision: Reject H 0 because p<.0001<α=.05. Conclusion: The treated pipes do not corrode as much as the normal pipes when subjected to chemicals.

22. Two Independent Samples Compare the means of two populations Parameter of interest: (  1 -  2 ) Base inferences on The parentheses indicate that we are thinking of the difference as one parameter Consider the general confidence interval formula, P±TS. We know what P is now. We need to know the distribution of to find T and S.

23. Distribution of. The sampling distribution of has a mean, The point estimate of is The standard deviation of is Since the variances are hardly ever known, we will have to estimate them.

24. Sample Standard Deviation The sample standard deviation of is The following assumptions are needed to use the above formula: –The samples are randomly selected from normally distributed populations –The samples are independent –There is no reason to believe σ 1 =σ 2 –The populations (not samples) are “large”

25. Distribution The t distribution will be used. Degrees of freedom: –If n 1 =n 2, no problem, df=n 1 -1. –Otherwise, df may be calculated by a complicated formula. Statistical computer software will do this automatically. –Alternatively, the smaller of n 1 -1 and n 2 -1 can be used as an approximation. (conservative— actual confidence level will be higher, actual p- value will be lower)

26. Confidence Interval Now we have all the information we need. P= T=t (df,α/2) S= A (1-α)100% confidence interval for (  1 -  2 ) is given by

27. Example: A recent study reported the longest average workweeks for non-supervisory employees in private industry to be chef and construction. Find a 95% confidence interval for the difference in mean length of workweek between chef and construction. Assume normality for the sampled populations. Solution: Parameter of interest:  1 -  2 where  1 is the mean hours/week for chefs and  2 is the mean hours/week for construction workers.

28. df = 11, the smaller of: n 1  1 = 18  1 = 17 and n 2  1 = 12  1 = 11.  = 0.05 t (df,  /2) = t (11, 0.025) = 2.20 A 95% Confidence interval for  1 -  2 is

29. Note: 1.Using a calculator, the confidence interval is.55 to 7.65. 2.This confidence interval is narrower than the approximate interval computed on the previous slide. This illustrates the conservative (wider) nature of the confidence interval when approximating the degrees of freedom.

30. Hypothesis Tests: To test a null hypothesis about the difference between two population means, use the test statistic where df is the smaller of df 1 or df 2 when computing t* without the aid of a computer. Note: The hypothesized difference between the two population means (  01   02 ) can be any specified value. The most common value is zero.

31. Example: A recent study compared a new drug to ease post- operative pain with the leading brand. Independent random samples were obtained and the number of hours of pain relief for each patient were recorded. The summary statistics are given in the table below. Is there any evidence to suggest the new drug provides longer relief from post-operative pain? Use  = 0.05 a.Solve using the p-value approach. b.Solve using the classical approach.

32. Solution: 1.The Hypotheses: H 0 :  1   2 = 0 (new drug relieves pain no longer) H a :  1   2 > 0 (new drug works longer to relieve pain) 2.The appropriate test: Assumptions: Both populations are assumed to be approximately normal. The samples were random and independently selected. Use t*, df = 9 3.Rejection Region: Reject if t*>t(9, 0.05) = 1.83 or p<.05. 4.Calculations:

33. 4.(cont’d) The p-value: 5.The Conclusion: Decision: Reject H 0. Conclusion: There is evidence to suggest that the new drug provides longer relief from post-operative pain.

34. If independent samples of sizes n 1 and n 2 are drawn randomly from large populations with p 1 = P 1 (success) and p 2 = P 2 (success), respectively, then the sampling distribution of has these properties: 1.a mean 2.a standard error 3.an approximately normal distribution if n 1 and n 2 are sufficiently large. Difference of Two Proportions

35. Note: To ensure normality: 1.The sample sizes are both larger than 20. 2.The products n 1 p 1, n 1 q 1, n 2 p 2, n 2 q 2 are all larger than 5. Since p 1 and p 2 are unknown, these products are estimated by 3.The samples consist of less than 10% of respective populations. Confidence Intervals: 1.A confidence interval for p 1  p 2 is based on the unbiased sample statistic. 2.The confidence limits are found using the following formula:

36. Example: A consumer group compared the reliability of two similar microcomputers from two different manufacturers. The proportion requiring service within the first year after purchase was determined for samples from each of two manufacturers. Find a 98% confidence interval for p 1  p 2, the difference in proportions needing service.

37. Solution: 1.Population Parameter of Interest: p 1 -p 2 where p 1 is the proportion of computers needing service for manufacturer 1 and p 2 is the proportion of computers needing service for manufacturer 2. 2.Check the Assumptions: Sample sizes larger than 20. Products all larger than 5. should have an approximate normal distribution. Use Z distribution.

38. 3.The Sample Evidence: Sample information: Point estimate: 4.The Confidence Interval: a.Confidence coefficients: z(  /2) = z(0.01) = 2.33 b.Error Bound:

39. c.Confidence limits: Hypothesis Tests for the difference of two proportions: Look at what we’ve done before, e.g.: All of our hypothesis tests have made use of this form:

40. Parts of the Test If the null hypothesis is there is no difference between proportions, and we can assume normality, a. the test statistic is z* b. the parameter estimate is c. the hypothesized value is 0 d. the standard error is … (more to come)

41. Let’s consider how we can construct a standard error term. Now the standard deviation of p 1 '  p 2 ' is actually However, if the null hypothesis is true, p 1 = p 2, so we can say

42. But we don’t know p and q! How can we estimate these from the sample? Under the null hypothesis, the proportions of the two samples are the same. So simply take all of the data and pool it together to estimate the common proportion. The test statistic becomes

43. Example: The proportions of defective parts from two different suppliers were compared. The following data were collected. Is there any evidence to suggest the proportion of defectives is different for the two suppliers? Use  = 0.01.

44. 1.The null and alternative hypotheses: H 0 : p 1  p 2 = 0 (proportion of defectives the same) H a : p 1  p 2  0 (proportion of defectives different) 2.The type of test: Difference of proportions, with Samples are larger than 20. Products are larger than 5. Sampling distribution should be approximately normal. Use z* for difference of proportions. 3.Rejection region: Reject H 0 if z* > z(.005) = 2.575 or z* < -z(.005) = -2.575 4.Calculations:

45. 4.Calculations cont’d: 5.Conclusion: Do not reject H 0 and conclude that there is no evidence to suggest the proportion of defectives is different for the two suppliers.