Chapter 9: Comparison of Paired Samples

Transcription

1 STAT503 Lecture Notes: Chapter 9 1 Chapter 9: Comparison of Paired Samples November 2, Introduction Observations occur in pairs such as: as identical twins, two observations on the same individual (two days, two sides, before/after) a two-treatment block design where each block is of size 2 (subjects matched in pairs for similar age, sex, profession, disease status, other extraneous variables) 9.2 The Paired Sample t-test and Confidence Interval This is a combination of two things you already know. Two sample and one sample t-tests (also C.I. s) You must recognize that the study is a paired design. Do the hypothesis steps like a 2-independent sample t-test (steps 1 4 and step 10). Calculate the differences D = Y 1 Y 2. Do the t calculations for the differences like a 1-sample t-test (steps 5 9). Directionality works as before. It is the observed differences that we wish to analyze. n is the number of pairs in the sample. Example: Have 12 fruit flies from 6 different lines (2 of each) grown in vials at two different temperatures. After a specified time for development, count the number of bristles on each fly. We think that bristle number has approximately a normal distribution, but flies from the same line should have similar bristle numbers.

2 STAT503 Lecture Notes: Chapter 9 2 So we will look at the difference in bristle number between two flies of the same line. Block = line Treatment = temperature Have 12 observations but they occur in 6 pairs so n = 6. Cold Warm Difference Line y 1 y 2 d = y 1 y mean s s SE = n Does temperature affect the number of bristles on fruit flies? Let µ 1 and µ 2 be the mean bristle number for flies grown at the colder and warmer temperatures, respectively H 0 : µ 1 = µ 2 ; the mean bristle number is the same for both temperatures H A : µ 1 µ 2 ; the mean bristle number is different at the two temperatures Use a non-directional paired sample t-test. t s = d/se d has a t-distribution with 6 1 = 5 df under H 0. [df = #pairs 1]. Test at level α = 0.05; Critical value is t(5).025 = 2.571; Will reject H 0 if it s > or t s < 2.571; otherwise will not reject H 0. Do a one-sample t-test using the difference column, i.e. d = ȳ 1 ȳ 2 = = 1.0. s d = SE d = t s = 1.0/0.577 = < t s < so do not reject H 0. [Conclude like a two-sample test.] This study does not provide evidence at the.05 significance level that flies have different mean bristle numbers at the two temperatures.

3 STAT503 Lecture Notes: Chapter 9 3 Directional hypotheses are done analogously. Confidence Interval: 95% CI for mean difference µ 1 µ 2 : ȳ 1 ȳ 2 ± t(5) 0.25 SE d = 1 ± 2.571(0.577) = ( 2.48, 0.48). Remember: Even though there are 12 observations, the df is 5, and not 10. How this would be different if we used two-independent-samples setup? How to know whether to use paired or unpaired t-test to analyze the data? Easy criterion: does each data point in one sample naturally correspond to one specific data point in the other sample? How to know when to use a paired design in your experiment? Harder to decide: If we expect extraneous variables to increase variation and we can match on these variables, or may be dictated by the problem (e.g. repeated measurements). Assumptions: Paired-sample t-test is based on the idea that the differences are approximately normal. 9.3 The Paired Design Read. 9.4 The Sign Test What if the data are not normally distributed? We can look at the sign of the difference between each pair of observations If the two treatments are the same, the sign is equally likely to be positive as negative. That is, if we take a pair at random (from the population of pairs), whether the first or second of the pair is larger is like flipping a coin. What distribution does the number of + signs have? Let p =probability that, for any pair from the population, the first will be larger.

4 STAT503 Lecture Notes: Chapter 9 4 Details H 0 : p = 0.5; the two treatments are the same in their effect (be specific) H A : p 0.5; the two treatments are different in their effect (be specific) for each pair of observations, note whether y 1 y 2 is positive (+) or negative (-), that is, + if y 1 is bigger, if y 2 is bigger count the number of + signs (= N + ) and signs (= N ) (don t count zeros) test statistic B s = larger of N + and N (true only for non-directional) use critical value B from Table 7 where n =#pairs with non-zero difference table gives column headings for both non-directional and directional tests reject H 0 if B s B can also calculate a P -value by using binomial formula For directional sign test H A is either p < 0.5; for any pair, the second is more likely to be larger (B s = N ) or H A is p > 0.5; for any pair, the first is more likely to be larger (B s = N + ) Example: skin grafts (Example 9.12 in the textbook) Skin grafts applied to both sides of body in 11 recipients. One graft has good HLA match with recipient; other does not. This is paired because we have two observations from each person. Observe time to rejection of skin graft (not normally distributed so we don t use t-test.) Does a good HLA match increase graft survival time? Let p =prob. that good HLA match survives longer in any individual H 0 : p = 0.5; HLA match and non-match are equally likely to survive longer H A : p > 0.5; HLA match is more likely to survive longer [this is directional] Use a sign test. B s = N + has a binomial distribution (n = 11, p = 0.5) under H 0. [ If this were non-directional, B s would be max(n +, N ) ] Test at level α = 0.05; critical value is 9, see Table 7. [use α = 0.05 column labeled One tail since directional] Will reject H 0 if B s 9; if B s < 9 will not reject. Data: survival times (days)

5 STAT503 Lecture Notes: Chapter 9 5 good poor sign n = 11, N = 2, N + = 9, B s = 9. Since B s 9 we reject H 0. This study provides evidence (P = ) that a good HLA match survives longer than a poor match. How to calculate P-value? Remember that P-value is simply the area in the tail B s. P-value is Pr(Y 9) for a Binomial(n = 11, p = 0.5) i.e. P value = Pr(Y = 9) + Pr(Y = 10) + Pr(Y = 11) = = P-value = < α = 0.05 (again would conclude reject H0) Two more examples are given on our handout. If H A is p > 0.5, then B s = N +, and the P-value is Pr(Y B s ) If H A is p < 0.5, then B s = N, and the P-value is Pr(Y B s ) If H A is p 0.5, then B s = max(n +, N ), and the P-value is the area in both tails, i.e. 2 Pr(Y B s ), where Y is Binomial(n, 0.5). Can use the sign test when data are not normally distributed when data consist of sign but not magnitude, e.g. smaller/larger, etc preference, better/worse, If a pair is exactly equal (so difference is 0) we ignore it in calculating B s and reduce n by Further Considerations Before & After vs. Control Sometimes observe subjects before and after some treatment, and measure the effect, i.e. the change in each individual subject. gives simple paired data (see, e.g., example on the handout) Sometimes match a case to a control on extraneous variables and compare the measurements of the two subjects

6 STAT503 Lecture Notes: Chapter 9 6 gives simple paired data Sometimes have a treatment group and a control group that are not matched by extraneous variables gives simple independent samples Sometimes expect subjects to change naturally over the duration of the study. want to distinguish change due to treatment from change due to time observe treatment and control both before and after treatment controls give a baseline of how much change to expect naturally over time cases tell how much change occurred with the treatment Data: Controls Cases Obs Before After Diff Before After Diff n mean s Apparently four groups of observations, but......we want to compare the change in the Cases with the changes in the Controls. Looking at just the differences bef ore af ter, we can do a regular two-independentsamples t-test on the two datasets of differences (for Cases and for Controls).