# Chapter Five: Paired Samples Methods 1/38

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1 Chapter Five: Paired Samples Methods 1/38

2 5.1 Introduction 2/38 Introduction Paired data arise with some frequency in a variety of research contexts. Patients might have a particular type of laser surgery performed on one eye with a second form performed on the remaining eye. Visual acuity measures taken on the two eyes would constitute paired data. Patients might have a baseline blood pressure measure taken, receive some treatment with a subsequent blood pressure measure then being taken. The two blood pressure measures would constitute paired data. Twins might be exposed to two different types of vaccine against some disease. An assessment as to whether the twins developed the disease or not would constitute paired data.

3 Methods Related To Mean Difference 3/38 The Paired Samples (Difference) t Test The paired samples t test is conducted on difference scores obtained by subtracting one paired observation from the other. The null hypothesis tested is H 0 : µ d = µ d0 where µ d is the mean of a difference score population and µ d0 is the hypothesized value of µ d (usually but not always zero).

4 Methods Related To Mean Difference 4/38 The Paired Samples t Test (continued) One-tailed alternatives are either H A : µ d > µ d0 or H A : µ d < µ d0 The two-tailed alternative is H A : µ d µ d0

5 Methods Related To Mean Difference 5/38 The Test Statistic The paired samples t test is nothing more than a one mean t test conducted on difference (d) scores. Therefore, the test statistic takes the form t = d µ d0 s d n where d is the sample mean of the difference scores, s 0 the sample standard deviation of the difference scores and n the number of paired observations (or difference scores).

6 Methods Related To Mean Difference 6/38 Example Researchers are interested in comparing the effectiveness of two laser surgery treatments for diabetic retinopathy. Patients who manifest the disease in both eyes have one eye treated by the first surgical method with the remaining eye then being treated by the second method. After a period of time, visual acuity is measured (via LogMar scores) for each eye.

7 Methods Related To Mean Difference 7/38 Example (continued) The question of interest is Does one treatment method produce better vision (as measured by acuity) than does the other? Use the data provided in the next slide (#8) to perform a two-tailed paired samples t test with α =.05 in order to assist in determining whether one treatment was more effective than the other. Based on this test, what do you conclude?

8 Methods Related To Mean Difference 8/38 Example (continued) Table: 5.3 LogMar Scores Treatment One Treatment Two d d σ

9 Methods Related To Mean Difference 9/38 Solution Using the sums from Table 5.3, s d =.81 (.5) = =.129 = and d =.5 7 =.071. By Equation 5.1 obtained t is t = d µ d0 n s d =.071 = =

10 Methods Related To Mean Difference 10/38 Solution (continued) Because the critical values for a two-tailed t test with six degrees of freedom conducted at the.05 level are and 2.447, the null hypothesis is not rejected. We conclude that there was insufficient evidence to allow for a conclusion that the treatments differed in their effectiveness.

11 Methods Related To Mean Difference 11/38 Introduction It sometimes happens that researchers would like to establish that there is no practical difference between treatments or between preand post-treatment means rather than showing that there is a difference. In such instances the paired samples t test can be used to establish equivalence.

12 Methods Related To Mean Difference 12/38 Two-Tailed Null And Alternative Hypotheses If we let EI U represent the upper end of EI and EI L the lower end, the equivalence null hypothesis for the (paired samples) two-tailed equivalence test is The alternative is H 0E : µ d EI L or µ d EI U H AE : EI L < µ d < EI U Notice that the null hypothesis states that the population difference score mean is not in EI while the alternative states that this mean is in EI. The null hypothesis, then, is an assertion of non-equivalence while the alternative asserts equivalence.

13 Methods Related To Mean Difference 13/38 Testing The Equivalence Null Hypothesis Testing the null hypothesis for a two-tailed equivalence test requires that two one-tailed tests be conducted. The null and alternative hypotheses for the two one-tailed tests for paired samples means are as follows Test One Test Two H 01 : µ d = EI U H 02 : µ d = EI L H A1 : µ d < EI U H A2 : µ d > EI L Both of these tests must be significant in order to reject the equivalence null hypothesis.

14 Methods Related To Mean Difference 14/38 Example A manufacturer of blood pressure monitoring devices wishes to demonstrate that a new, less expensive, model produces equivalent results to the older, more expensive, device. To this end, blood pressures of 10 subjects are assessed by means of both devices. Difference scores (d) are calculated for the paired readings with the following results. d = 16 d 2 = 94 Use these results to perform a two-tailed equivalence test with the equivalence interval of 4 to 4. Begin by stating the null and alternative equivalence hypotheses.

15 Methods Related To Mean Difference 15/38 Solution The null and alternative equivalence hypotheses are as follows. H 0E : µ d 4 or µ d 4 We calculate d 2 (P d) 2 and s d = n 1 H AE : 4 < µ d < 4 n = d d = n 94 (16) = = = = 2.280

16 Methods Related To Mean Difference 16/38 Solution (continued) Values of obtained t for Test One and Two are then t 1 = = and t 2 = ( 4.0) = Critical t values based on 16 1 = 15 degrees of freedom for the two one-tailed tests are and It follows that both results are significant leading to rejection of the null hypothesis that the two devices are not equivalent insofar as mean blood pressures are concerned in favor of the alternative that maintains equivalence.

17 Methods Related To Mean Difference 17/38 One-Tailed Null And Alternative Hypotheses The null and alternative hypotheses for the one-tailed equivalence test are one of the following: H 0E : µ d EI U H AE : µ d < EI U OR H 0E : µ d EI L H AE : µ d > EI L The first one-tailed equivalence hypothesis given above is tested by Test One with the second being carried out by means of Test Two.

18 Methods Related To Mean Difference 18/38 Example Eighteen subjects who routinely take a standard dose of a cholesterol lowering drug are switched to a lower does that is believed to produce less severe side effects. The question of interest is, Will the reduced dose produce cholesterol levels that are not worse than those obtained with the higher dose? Cholesterol is measured while patients are still on the higher dose and again after they have been on the lower dose for a specified period of time. The researchers decide that if average cholesterol rises by less than six points the reduced dose can be declared to produce results that are no worse than those produced by the higher dose.

19 Methods Related To Mean Difference 19/38 Example (continued) Difference scores are generated by subtracting cholesterol levels obtained from the higher dosage from those obtained from the lower dosage. The following results are obtained. d = 40, d 2 = 730 Use a one-tailed equivalence test with α =.05 to make this determination.

20 Methods Related To Mean Difference 20/38 Solution Because the researchers wish to determine if the rise in cholesterol is less than six points, a one-tailed equivalence test of the form H 0E : µ d 6 H AE : µ d < 6

21 Methods Related To Mean Difference 21/38 Solution (continued) From the previously provided sums we calculate d 2 (P d) 2 s d = n 1 n = 730 (40) = = and Then, d = d n = = t 1 = d µ d0 n s d = =

22 Methods Related To Mean Difference 22/38 Solution (continued) Critical t for a one-tailed test conducted at α =.05 with 17 degrees of freedom is The null hypothesis is rejected leading to the conclusion that the reduced dose is not worse than the higher dose insofar as control of cholesterol is concerned.

23 Methods Related To Mean Difference 23/38 Introduction The paired samples t test attempts to determine whether there is a difference between the means of two paired variables. A related and usually more informative question is, How large is the difference between the means of the paired variables? This difference can be estimated with a confidence interval.

24 Methods Related To Mean Difference 24/38 Construction Of The Interval Construction of L and U are as follows L = d t s d n U = d + t s d n where s d is the sample standard deviation of the difference scores and t is the appropriate value from the t table with n 1 degrees of freedom.

25 Methods Related To Mean Difference 25/38 Example Use the data in slide number 8 to form a two-sided 95% confidence interval. What is the meaning of this interval?

26 Methods Related To Mean Difference 26/38 Solution and s d =.81 (.5) = d =.5 7 =.071. =.359

27 Methods Related To Mean Difference 27/38 Solution (continued) Noting from Appendix B that the appropriate t value (with six degrees of freedom) for a two-sided 95% confidence interval is 2.447, Equations 5.2 and 5.3 yield L = d t s d n = =.403 and U = d + t s d n = =.261.

28 Methods Related To Mean Difference 28/38 Solution (continued) A statistical interpretation of this interval would maintain that we can be 95% confident that µ d is between.403 and.261. From the researcher s point of view, it can be stated with 95% confidence that the average difference in visual acuity between patients receiving laser treatment one and those receiving laser treatment two was between.403 and.261 LogMar units. Based on this interval can you rule out a difference of zero?

29 Methods Related To Mean Difference 29/38 Assumptions Assumptions underlying the paired samples t test, equivalence tests based on the paired samples t test, and the confidence interval for mean difference are the same as those for the one mean t test and consequently, the one mean Z test. These are normality independence You should note that these assumptions apply to the difference scores rather than to the two component distributions.

30 Methods Related To Proportions 30/38 McNemar s Test McNemar s test is used to assess outcomes for paired dichotomous data. The null hypothesis assessed is H 0 : π =.5 which is an assertion that the proportion of pairs in which one treatment is favored is equal to the proportion of pairs in which the other treatment is favored. Alternatives take one of the following forms. H A : π < 0 H A : π > 0 H A : π 0

31 1 An equivalent test utilizes the chi-square distribution. Methods Related To Proportions 31/38 McNemar s Test (continued) There are both approximate and exact forms of the test. The approximate form is based on the normal curve 1 and uses the test statistic Z = ˆp.5.5 n where ˆp is the proportion of pairs in which the designated treatment has the advantage and n is the number of pairs utilized in the analysis. The exact test is based on the binomial distribution.

32 Methods Related To Proportions 32/38 Approximate Method: Example Two forms of sunscreen are to be compared with respect to level of protection afforded against sun-induced skin damage. One form of sunscreen is applied to a randomly selected arm of each of 15 volunteer subjects with the second form then being applied to the remaining arm. After a specified period of sun exposure, the skin of each arm is examined and graded as exhibiting a satisfactory level of protection S or as not exhibiting a satisfactory level of protection S.

33 Methods Related To Proportions 33/38 Approximate Method: Example (continued) It was found that the number of pairs where both arms were graded S was 2, both were S was 3, product one produced an S while product two produced an S was 9 and the number of pairs where product one produced an S while product two produced an S was 1. Use the approximate method to perform a two-tailed McNemar s test at α =.05 on the sunscreen data. What is your conclusion concerning the relative effectiveness of the two sunscreen products?

34 Methods Related To Proportions 34/38 Solution Ignoring noninformative outcomes, we note that the proportion of outcomes favoring product one is 9 10 =.90 and n = 10. Equation 5.4 then gives Z = = Appendix A shows that the critical values for a two-tailed test conducted at α =.05 are and 1.96 thereby leading to rejection of the null hypothesis.

35 Methods Related To Proportions 35/38 Solution (continued) Viewed from a statistical point of view the null and alternative hypotheses are H 0 : π =.5 and H A : π.5 When viewed from the context of this study, the null hypothesis asserts that there is no difference in the effectiveness of the two products while the alternative maintains that such a difference does exist. In this case, we rejected the claim of no difference in favor of a statement that the two products do differ in effectiveness. Further, we can conclude that product one is superior to product two in terms of protection provided.

36 Methods Related To Proportions 36/38 Exact Method The exact method employs the binomial distribution to test the null hypothesis H 0 : π =.5

37 Methods Related To Proportions 37/38 Example Use the example outlined in slide 32 with the result reported in slide 33 to perform an exact McNemar s test with alternative H A : π >.5

38 Methods Related To Proportions 38/38 Solution Noting that of the 10 pairs of arms, the number in which treatment one afforded a satisfactory result while treatment two did not was 9. By Equation 4.5 we calculate P (9) = and P (10) = The p-value for a test with alternative H A : π >.5 is then = Because this value is less than α =.05, the null hypothesis is rejected.

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