1. Today’s lesson (Chapter 12) Paired experimental designs Paired t-test Confidence interval for E(W-Y)

2. Paired Design Find two experimental units that are more like each other than randomly selected units. –Two animals from the same litter –Before and after measurements on the same subject. Apply the A treatment to one unit randomly selected and the B treatment to the other.

3. Analysis of Paired Design There are n pairs of experimental units. For unit i, W i represents the A treatment observation from the i-th pair. For unit i, Y i represents the B treatment observation from the i-th pair. Calculate the difference D i =W i -Y i. Do a one-sample t-test on the differences.

4. Analysis of Paired Design That is, null hypothesis is E(W-Y)=0. Alternative hypothesis may be left, right, or two-sided. Test statistic is the mean of the differences. Estimated standard error of the mean difference is the standard deviation of the D values divided by the square root of n. Standardize the test statistic as usual.

5. Example Problem 1 A research team evaluated a medicine to determine whether it lowered a blood component. They measured B, the amount of the component in six patients before a protocol using the medicine was followed and then measured A, the amount of the component after the administration of the medicine.

6. Example Problem 1 They wished to test the null hypothesis that E(A)=E(B) against the alternative that E(A)<E(B). Their experimental results are given in the following table. Which of the following is a correct decision? Usual options: reject at 0.01 level, accept at 0.01 and reject at 0.05, accept at 0.05 and reject at 0.10, accept at 0.10.

7. Data for Problem 1

8. Solution of Problem 1 Recognize that this problem requires a paired t-test (before and after comparison). Compute the six differences A-B: –-20, -30, -40, -30, -10, -50. Compute mean difference –sum of differences is -180 –mean difference is -30

9. Solution of Problem 1 Compute standard deviation of the six differences –six deviations from mean are -20-(-30)=10, 0, -10, 0, 20, -20 –check that they sum to zero –find the squared deviations from the mean 100, 0, 100, 0, 400, 400

10. Solution of Problem 1 Compute standard deviation of the six differences (continued) –sum the six squared deviations 1000 –find the degrees of freedom (pairs-1=5) –find the variance (sum of squared deviations per degree of freedom) = 1000/5 =200 –take the square root of the variance = 14.1

11. Solution of Problem 1 Compute the estimated standard error of the mean difference; that is, the standard deviation over the square root of the number of pairs –14.1/6 0.5 =5.77 Compute the t-statistic (standard score value of the test statistic). –T=(-30-0)/5.77=-5.20

12. Solution of Problem 1 Decide on the side of the test: left sided! Determine the degrees of freedom: # pairs- 1=5. Stretch the critical values: –-2.326 to -3.365, -1.645 to -2.015, and -1.282 to -1.476. Make your decision. –-5.20 is to the left of -3.365; reject at 0.01 level.

13. Most Fundamental Design Advice Pair what you can, randomize what you cannot. That is, always used a paired design when possible. There is a generalization of a pair. It is called a block. ADVICE: Block what you can, randomize what you cannot.

14. Why this advice? Var(W-Y) equals var(W)+var(Y)- 2cov(W,Y). ASS-U-ME var(W)=var(Y)=σ 2 Then, cov(W,Y)=ρσ 2. Var(W-Y)=σ 2 +σ 2 -2ρσ 2 =2σ 2 (1-ρ). When there is a large positive correlation within the units, the variance of the difference is small.

15. Extension to Hedging Strategies It is true that Var(W+Y) equals var(W)+var(Y) +2cov(W,Y). Same assumptions. Then Var(W+Y)=σ 2 +σ 2 +2ρσ 2 =2σ 2 (1+ρ). When W and Y are negatively correlated, variance of W+Y is reduced. That is, risk is lessened.

16. Problem 2 Based on the data given in problem 1, what is a 99 percent confidence interval for the difference in expected values E(A-B)?

17. Solution to Problem 2 Find the degrees of freedom for your estimate of the standard deviation: #pairs-1, here 5. Stretch the normal factor for this level of confidence (2.576) for the correct df: 4.032. Use the estimated standard error for the unknown standard deviation of the mean. –Left endpoint is -30-4.032(5.77)=-53.3 –Right endpoint is -30+4.032(5.77)=-6.7

18. Using SPSS to get paired t-test Statistics, compare means, paired t-test.

19. Example Computer Problem Two statistical procedures exist to determine the estimated location of a gene. Which procedure comes closer to the correct location of the gene? Use the results of a simulation study to answer the question.

20. Data of Study Genetic model specified (recessive, all families affected by the same genetic pattern--no heterogeneity). Use simulation to generate the results of 100 independent studies. Apply two statistics (maximum hlod and Kong and Cox correct to the Nonparametric Linkage (NPL) statistic of Genehunter) to each study.

21. Design of the Comparison For each of the 100 replicates, calculate the difference D of the distance from the maximum hlod analysis and the distance from the maximum NPL statistic. Use paired t-test to test the null hypothesis that the expected distance using the maximum hlod is the same as the expected distance using the maximum NPL. Closer is better.

22. Results Maximum hlod is significantly closer than the maximum NPL for this genetic model.

23. Summary Paired t-test design Discussion of why paired t-test likely to be better. Block what you can, randomize what you cannot. Illustrated computations of the paired t-test on both test and confidence interval.