1. Sample Size Algorithms in Clinical Trials Chien-Hua Wu Chung-Yung Christian University Nov. 6, 2007

2. FEV1 Changes 1.A new compound, ABC-123, is being developed for long-term treatment of patients with chronic asthma. 2.Asthmatic patients were enrolled in a double-blind study and randomized to receive daily oral doses of ABC-123 or a placebo for 6 weeks. 3.The primary measurement of interest is the resting FEV1 (forced expiratory volume during the first second of expiration), which is measured before and at the end of the 6-week treatment period. 4.Does administration of ABC-123 appear to have any effect on FEV1 ?

3. ABC-123 Group Patient NumberBaseline Week 6 101 103 106 108 109 110 113 116 118 120 121 124 1.35 3.22 2.78 2.45 1.84 2.81 1.90 3.00 2.25 2.86 1.56 2.66 n/a 3.55 3.15 2.30 2.37 3.20 2.65 3.96 2.97 2.28 2.67 3.76

4. Placebo Group Patient NumberBaseline Week 6 102 104 105 107 111 112 114 115 117 119 122 123 3.01 2.24 2.25 1.65 1.95 3.05 2.75 1.60 2.77 2.06 1.71 3.54 3.90 3.01 2.47 1.99 n/a 3.26 2.55 2.20 2.56 2.90 n/a 2.92

5. Two-Sample t-Test null hypothesis: H 0 : μ 1 = μ 2 alt.hypothesis: H A : μ 1 ≠μ 2 test statistic: t= =0.974 conclusion: Because 0.974 is not ＞ 2.093, you cannot reject H 0. You conclude that the samples fail to provide significant evidence of any effect of ABC-123 on FEV. This test is based on a significance level of α ＝ 0.05.

6. 2-sample t-test: equality %let mu1=0.503; ********* mean of treatment 1; %let mu2=0.284; ********* mean of treatment 2; %let sigma=0.265; ********* standard deviation; %let alpha=0.05; ********* type I error; %let power=0.9; ********* given power; data Ssize; theta=abs(&mu1-&mu2)/σ ******* effect size; n1=2; ******* initial value; m=1; ******* equal sample size; do while (power<&power); if power<&power then n1=n1+1; n2=n1*m; ******** sampe size n1:n2=1:m; lambda=theta/sqrt((1/n1)+(1/n2)); ******** noncentrality parameter; t=tinv(1-(&alpha)/2,n1+n2-2); ******** critical value under H0; power1=1-probt(t,n1+n2-2,lambda); ******** probability of power function > critical value t; power2=probt(-t,n1+n2-2,lambda); ******** probability of power function < critical value -t; power=power1+power2; ******** power ; end; Answer: n1=n2=32

7. 2-sample t-test: superiority %let mu1=0.503; ********* mean of treatment 1; %let mu2=0.284; ********* mean of treatment 2; %let sigma=0.265; ********* standard deviation; %let delta=0.05; ********* superiority or non-inferiority margin; %let alpha=0.05; ********* type I error; %let power=0.9; ********* power; data Ssize; theta=abs(abs(&mu1-&mu2)-&delta)/σ ******* effect size; n1=2; ******* initial value; m=1; ******* equal sample size; do while (power<&power); if power<&power then n1=n1+1; n2=n1*m; ******** sampe size n1:n2=1:m; lambda=theta/sqrt((1/n1)+(1/n2)); ******** noncentrality parameter; t=tinv(1-(&alpha),n1+n2-2); ******** critical value under H0; power=1-probt(t,n1+n2-2,lambda); ******** probability of power function > critical value t; end; Answer: n12=n2=43

8. 2-sample t-test: non-inferiority %let mu1=0.503; ********* mean of treatment 1; %let mu2=0.284; ********* mean of treatment 2; %let sigma=0.265; ********* standard deviation; %let delta=-0.05; ********* superiority or non-inferiority margin; %let alpha=0.05; ********* type I error; %let power=0.9; ********* power; data Ssize; theta=abs(abs(&mu1-&mu2)-&delta)/σ ******* effect size; n1=2; ******* initial value; m=1; ******* equal sample size; do while (power<&power); if power<&power then n1=n1+1; n2=n1*m; ******** sampe size n1:n2=1:m; lambda=theta/sqrt((1/n1)+(1/n2)); ******** noncentrality parameter; t=tinv(1-(&alpha),n1+n2-2); ******** critical value under H0; power=1-probt(t,n1+n2-2,lambda); ******** probability of power function > critical value t; end; Answer: n1=n2=18

9. 2-sample t-test: equivalence %let mu1=0.503; ********* mean of treatment 1; %let mu2=0.284; ********* mean of treatment 2; %let sigma=0.265; ********* standard deviation; %let delta=0.05; ********* equivalence margin; %let alpha=0.05; ********* type I error; %let power=0.9; ********* given power; data Ssize; theta1=abs(&delta-abs(&mu1-&mu2))/σ ******* effect size 1; theta2=abs(&delta+abs(&mu1-&mu2))/σ ******* effect size 2; n1=2; ******* initial value; m=1; ******* equal sample size; do while (power<&power); if power<&power then n1=n1+1; n2=n1*m; ******** sampe size n1:n2=1:m; lambda1=theta1/sqrt((1/n1)+(1/n2));******** noncentrality parameter 1; lambda2=theta2/sqrt((1/n1)+(1/n2));******** noncentrality parameter 2; t=tinv(1-(&alpha),n1+n2-2); ******** critical value under H0; power1=probt(t,n1+n2-2,lambda1); ******** probability of power function < critical value t; power2=probt(t,n1+n2-2,lambda2); ******** probability of power function < critical value -t; power=1-power1-power2; ******** power ; end; Answer: n1=n2=44

10. 2-sample t-test: summary Sample size increases as the mean difference decreases when the standard deviation is fixed. Sample size increases as the standard deviation increases when the mean difference is fixed. Sample size increases as the power increase. Smaller equivalence margin, larger sample size. Smaller non-inferiority/superiority margin, smaller sample size. Sample size of equivalence > superiority > non- inferiority

11. Cautions: SAS codes Non-inferiority/superiority: margin < mean difference Equivalence: margin > mean difference

12. HAM-A Scores in GAD 1.A new serotonin-uptake inhibiting agent, SN-X95, is being studied in subjects with general anxiety disorder (GAD). 2.Fifty-two subjects diagnosed with GAD of moderate or greater severity consistent with the “Diagnostic and Statistical Manual, 3rd Edition” (DSMIIIR) were enrolled and randomly assigned to one of three treatment groups: 25 mg SN-X95, 100 mg SN-X95, or placebo. 3.After 10 weeks of once-daily oral dosing in a double-blind fashion, a test based on the Hamilton Rating Scale for Anxiety (HAM-A) was administered. 4.This test consists of 14 anxiety-related items (e.g., ‘anxious mood’, ‘tension’, ‘insomnia’, ‘fears’, etc.), each rated by the subject as ‘not present’, ‘mild’, ‘moderate’, ‘severe’, or ‘very severe’. 5.HAM-A test scores were found by summing the coded values of all 14 items using the numeric coding scheme of 0 for ‘not present’, 1 for ‘mild’, 2 for ‘moderate’, 3 for ‘severe’, and 4 for ‘very severe’. 6.Are there any differences in mean HAM-A test scores among the three groups?

13. Lo-Dose(25mg) Hi-Dose(100mg) Patient NumberHAM-A 101 104 106 110 112 116 120 121 124 125 130 136 137 141 143 148 152 21 18 19 n/a 28 22 30 27 28 19 23 22 20 19 26 35 n/a Patient NumberHAM-A 103 105 109 111 113 119 123 127 128 131 135 138 140 142 146 150 151 16 21 31 25 23 25 18 20 18 16 24 22 21 16 33 21 17

14. Placebo Patient NumberHAM-A 102 107 108 114 115 117 118 122 126 129 132 133 134 139 144 145 147 149 22 26 29 19 n/a 33 37 25 28 26 n/a 31 27 30 25 22 36 32

15. One-Way ANOVA null hypothesis: H 0 : μ 1 = μ 2 =μ 3 alt.hypothesis: H A : not H 0 test statistic: F=173.6/24.9=6.97 decision rule: Reject H 0 if F>F(0.05,2,45) =3.2 conclusion: Because 6.97 ＞ 3.2, you reject H 0 and conclude that there is a significant difference in mean HAM-A scores among the 3 does groups.

16. SAS codes for ANOVA %let power=0.8; *** power value; %let alpha=0.05; *** alpha value; %let delta=1.0246950766; *** based on means and standard deviation; data Ssize; k=4; *** # of groups; n=1; *** initial value of sample size; do while (power<&power); if power<&power then n=n+1; lambda=&delta*n; *** noncentrality parameter; f1=finv(1-&alpha,k-1,k*(n-1),0); *** critial value; power=1-probf(f1,k-1,k*(n-1),lambda); *** power; end; Answer: 12 subjects per group.

17. ADR Frequency with Antibiotic Treatment 1. A study was conducted to monitor the incidence of gastro-intestinal (GI) adverse drug reactions of a new antibiotic used in lower respiratory tract infection (LRTI) during 7 days of treatment. 2.One group consisted of 66 LRTI patients randomized to receive the new treatment and a reference group of 52 LRTI patients randomized to receive erythromycin. 3.Is there evidence of a difference in GI side effect rates between the two groups?

18. Response Frequencies Number of Number of Responders Non-RespondersTotal Test Drug Control 22(33.3%) 44 28(53.8%) 24 66 52 Combined50(42.4%) 68118

19. The Chi-Square Test Null hypothesis: Ho:p1=p2 Alt. hypothesis: Ha:p1≠p2 Test statistic: X^2=( ﹣ 5.966)^2/7.102=5.012 Decision rule: reject Hо if X^2 ＞ X1^2(0.05)=3.841 Conclusion: Because 5.012 ＞ 3.841, you reject Ho and conclude there is a significant difference in the incidence of GI adverse effects between treatment groups at a 0.05 level of significance.

20. 2 proportions: equality %let p1=0.333; %let p2=0.538; %let alpha=0.05; ********* type I error; %let power=0.9; ******** power; data Ssize; n1=2; ******* initial value; m=1; ******* equal sample size; do while (power<&power); if power<&power then n1=n1+1; n2=n1*m; ******** sampe size n1:n2=1:m; norm=probit(1-(&alpha)/2); ******** critical value under H0; power=1-CDF('NORMAL',norm, (abs(&p1-&p2)/sqrt(&p1*(1-&p1)/n1+&p2*(1-&p2)/n2)),1)+ CDF('NORMAL',-norm, (abs(&p1-&p2)/sqrt(&p1*(1-&p1)/n1+&p2*(1-&p2)/n2)),1); end; Answer: n1=n2=118

21. 2 proportions: superiority or non- inferiority %let p1=0.333; %let p2=0.538; %let margin=0.10; ********* superiority/noninferiority margin < mean difference; %let alpha=0.05; ********* type I error; %let power=0.9; ********* power; data Ssize; n1=2; ******** initial value; m=1; ******** equal sample size; do while (power<&power); if power<&power then n1=n1+1; n2=n1*m; ******** sampe size n1:n2=1:m; norm=probit(1-(&alpha)); ******** critical value under H0; power=1-CDF('NORMAL',norm, (abs(abs(&p1-&p2)-&margin)/sqrt(&p1*(1-&p1)/n1+&p2*(1-&p2)/n2)),1); end; Answer: n1=n2=366

22. 2 proportions: equivalence %let p1=0.333; %let p2=0.538; %let margin=0.4; ********* equivalence margin > difference of 2 proportions; %let alpha=0.05; ********* type I error; %let power=0.9; ********* power; data Ssize; n1=2; ******** initial value; m=1; ******** equal sample size; do while (power<&power); if power<&power then n1=n1+1; n2=n1*m; ******** sampe size n1:n2=1:m; norm=probit(1-(&alpha)); ******** critical value under H0; power=1-CDF('NORMAL',norm, (abs(abs(&p1-&p2)-&margin)/sqrt(&p1*(1-&p1)/n1+&p2*(1-&p2)/n2)),1) -CDF('NORMAL',norm, (abs(abs(&p1-&p2)+&margin)/sqrt(&p1*(1-&p1)/n1+&p2*(1-&p2)/n2)),1); end; Answer: n1=n2=107

23. 2 proportions: summary Sample size increases as the difference of 2 proportions decreases. Sample size increases as the power increase. Smaller equivalence margin, larger sample size. Smaller non-inferiority/superiority margin, smaller sample size. Sample size of equivalence > superiority > non- inferiority

24. Cautions: SAS codes Non-inferiority/superiority: margin < mean difference Equivalence: margin > mean difference

25. Bilirubin Abnormalities Following Drug Treatment 1.Eighty-six patients were treated with an experimental drug for 3 months. 2.Pre-and post-study clinical laboratory results showed abnormally high total bilirubin values (above the upper limit of the normal range ). 3.Is there evidence of a change in the pre- to post-treatment rates of abnormalities.

26. Abnormality Frequency Summary Post-Treatment Pre-treatment Y=Total bilirubin above upper limit lf normal range. N YTotal N Y 60 14 6 6 74 12 Total 66 20 86

27. McNemar’s Test Null hypothesis: Ho:p1=p2 Alt. hypothesis: Ha:p1≠p2 Test statistic: X^2=64/20=3.20 Decision rule: reject Ho if X^2 ＞ 3.841 Conclusion: Because 3.20 is not ＞ 3.841, you do not reject Ho, concluding that, at a significance level of 0.05, there is insufficient evidence that a shift in abnormality rates occurs with treatment.

28. SAS codes for McNemar %let p01=0.189; %let p10=0.5; %let alpha=0.05; %let power=0.9; data Ssize; m=10; **** initial value of the number of discordant responses; p0=0.5; **** under H0: p=0.5; phi=&p10/&p01; p1=phi/(1+phi); do while(power<&power); m=m+1; r=m-1; alpha=probbnml(p0,m,m)-probbnml(p0,m,r-1); do while(alpha<&alpha/2); ********** obtain critial value r; r=r-1; alpha=probbnml(p0,m,m)-probbnml(p0,m,r-1); end; power=probbnml(p1,m,m)-probbnml(p1,m,r); end; pd=&p10+&p01; n=m/pd; Answer: n=72

29. Genital Wart Cure Rate 1.A company markets a therapeutic product for genital warts with a known cure rate of 40% in the general population. 2.In a study of 25 patients with genital warts treated with product, patients were also given high doses of vitamin C. 3.Is this consistent with the cure rate in the general population ?

30. Patient Number Cured ? 1YES 2NO 3YES 4NO 5YES 6 7NO Patient Number Cured ? 8YES 9NO 10NO 11YES 12NO 13YES 14NO

31. Patient Number Cured ? 15YES 16NO 17NO 18YES 19YES 20NO 21YES Patient Number Cured ? 22YES 23NO 24YES 25YES

32. The Binomial Test null hypothesis: alt. hypothesis : test statistic : X=14(the number cured out of the n=25 patients treated) decision: reject if X 4 or X 15 conclusion : Do not reject the null hypothesis. Because 14 does not lie in the rejection region, you conclude that there is insufficient evidence from this study to indicate that concomitant Vitamin C treatment has an effect on the product’s cure rate.

33. SAS codes for Binomial %let margin=0.0; %let p0=0.4; %let p1=0.56; ****p1 > p0 + margin; %let alpha=0.05; %let power=0.8; data Ssize; n=10; ********** initial value of sample size; do while(power<&power); n=n+1; r=n-1; alpha=probbnml(&p0,n,n)-probbnml(&p0,n,r-1); do while(alpha<&alpha/2); ********** obtain critial value r; r=r-1; alpha=probbnml(&p0,n,n)-probbnml(&p0,n,r-1); end; power=probbnml(&p1,n,n)-probbnml(&p1,n,r); end; Answer: n=79

34. Power and Sample Size Calculation http://biostat.mc.vanderbilt.edu/twiki/bin/view /Main/PowerSampleSize