Biostatistics in Practice

1. Biostatistics in Practice Session 6: Case Study Peter D. Christenson Biostatistician http://gcrc.humc.edu/Biostat

2. Case Study Hall S et al: A comparative study of Carvedilol, slow release Nifedipine, and Atenolol in the management of essential hypertension. J of Cardiovascular Pharmacology 1991;18(4)S35-38. Data is available at the class website: http://gcrc.humc.edu/Biostat Select Courses > Biostatistics in Practice 2004 > Session 6 > Download Data

3. Case Study Outline Subjects randomized to one of 3 drugs for controlling hypertension: A: Carvedilol (new) B: Nifedipine (standard) C: Atenolol (standard) Blood pressure and HR measured at baseline and 5 post-treatment periods. Primary analysis ? “The present study compares … A, B, and C for the management of … hypertension.”

4. Data Collected for Sitting dbp * 1 hour after 1st dose. We do not have data for this visit.

5. Sitting dbp from Figure 2 of the Paper Baseline A: Carvedilol B: Nifedipine C: Atenolol A 2 Weeks B C

6. Question #1 Describe dbp at baseline for the study population. Give an appropriate graphical display, and summarize dbp with just a few numbers. Is the mean appropriate? Would the median be better? Is a transformation necessary?

7. Answer #1 N = 255 Mean = 102.68 SD = 4.63 SEM = 0.29 Min = 92 Max =117 Median = 102. Log-transformation gives geometric mean = 102.58. No transformation is necessary. Mean is best. 95% of subjects between ~ 102.68 ± 2(4.63) = 93.42 to 111.94

8. Question #2 It appears that group B may have had lower dbp at baseline than group A, on the average. Is there evidence for this? Is the lower group B mean dbp lower (relative to A) than expected by chance? Write out a formal test for this question, and use software to perform the test.

9. Answer #2, Part 1 Drug Mean ± SD A 102.9 ± 4.8 B 102.2 ± 4.3 C 103.0 ± 4.8 So, the mean for B is low, as in the earlier figure, but the overall distribution is similar to that for A and C, so this is entirely due to chance, but we will formally test B vs. A on the next slide. [Would use ANOVA to include C.]

10. Answer #2, Part 2 We are formally testing, where μx represents the mean baseline dbp among those who eventually receive treatment x: H0: μA = μB vs. HA: μA≠μB Since μA – μB is estimated by 0.75 with a SE of 0.71 , tc = 0.75/0.71 = 1.05 is not larger (~ >2) than expected by random fluctuation (p=0.29), so there is not sufficient evidence that the A and B groups differed in their baseline dbp. Note that we do not expect A and B to differ at baseline due to the randomization in the study design.

11. Question #3 How much can a patient’s dbp be expected to be lowered after 2 weeks of therapy with A? We are 95% sure that this lowering will be between what two values? Repeat for drug C. Do the intervals for A and for C overlap considerably? Can this overlapping be used to compare A and C in their dbp lowering ability?

12. Answer #3 How much can a patient’s dbp be expected to be lowered after 2 weeks of therapy with A? with C? We are 95% sure that this lowering will be between what two values? Ans: Drug Estimated Δ ~95% Prediction Interval A 8.13 8.13 ± 2*9.1 = -10.1 to 26.3 C 11.5 11.5 ± 2*8.7 = - 5.9 to 28.9 The intervals for A and for C do overlap considerably. However, to compare A and C, we need to examine not these expected intervals for individuals, but rather the precision of ΔC – ΔA estimated from this study, which incorporates the Ns.

13. Question #4 Is there evidence that A and C differ in their dbp lowering ability at 2 weeks post-therapy? Formally test for this. Give a 95% confidence interval for the C-A difference in change in dbp after 2 weeks.

14. Answer #4 Is there evidence that A and C differ in their dbp lowering ability at 2 weeks post-therapy? Ans: Test H0: ΔA-ΔC = 0 vs. HA: ΔA-ΔC ≠ 0with t-test: Estimate ΔA-ΔC with 3.39, with SE of 1.36. Since tc = 3.39/1.36 = 2.50 exceeds ~2, choose HA. 95% CI for ΔA-ΔC is 3.39±2*1.36 = 0.67 to 6.11, which does not include 0, so choose HA.

15. Question #5 Is there evidence that B and A differ in their dbp lowering ability at 2 weeks post-therapy? We want to examine whether the study was large enough to detect a difference in 2 week changes in dbp between B and A. To do so, we need the SD of these changes among subjects receiving B and among subjects receiving A. Find these SDs.

16. Answer #5 Is there evidence that B and C differ in their dbp lowering ability at 2 weeks post-therapy? Ans: Test H0: ΔB-ΔA = 0 vs. HA: ΔB-ΔA ≠ 0with t-test: Estimate ΔB-ΔA with 0.96, with SE of 1.35. Since tc = 0.96/1.35 = 0.71 < ~2, choose H0 (p=0.48). SD for B is 8.29 and SD for A is 9.08.

17. Question #6 • Estimate the true minimal difference in 2 week changes in dbp between B and C that this study was able to detect. • Use the conventional risks of making incorrect conclusions that the FDA typically requires. • Set both risks of an incorrect conclusion at ≤5%.

18. Typical Statistical Power Software

19. Answer #6 • Use the conventional risks of making incorrect conclusions that the FDA typically requires. • Use α=0.05, power=0.80, NA=83, NB=82, SDA=9.08, SDB=8.29. Find Δ from a power calculation to be 3.8. • Set both risks of an incorrect conclusion at ≤5%. • Use α=0.05, power=0.95, NA=83, NB=82, SDA=9.08, SDB=8.29. Find Δ from a power calculation to be 4.9.

20. Question #7 Suppose that differences in 2 week changes in dbp between B and C of <2 mmHg is clinically irrelevant, but we would like to detect larger differences with 80% certainty. How large should such a study be?

21. Answer #7 Suppose that differences in 2 week changes in dbp between B and C of <2 mmHg is clinically irrelevant, but we would like to detect larger differences with 80% certainty. How large should such a study be? Ans: Use α=0.05, power=0.80, SDA=9.08, SDB=8.29, Δ=2. From a power calculation , NA = NB = 297.