1. Design and Analysis of Clinical Study 4. Sample Size Determination Dr. Tuan V. Nguyen Garvan Institute of Medical Research Sydney, Australia

2. Practical Difference vs Statistical Significance Outcome Group A Group B Improved 9 18 No improved21 12 Total 30 30 % improved30%60% Chi-square: 5.4; P < 0.05 “Statistically significant” Outcome Group A Group B Improved 6 12 No improved14 8 Total 20 20 % improved30%60% Chi-square: 3.3; P > 0.05 “Statistically insignificant”

3. The Classical Hypothesis Testing Define a null hypothesis (H0) and a null hypothesis (H1) Collect data (D) Estimate p-value = P(D | H0) If p-value > , accept H0; if p-value < , reject H0

4. P-value là gì ? “Alendronate treatment was associated with a 5% increase in BMD compared to placebo (p<0.05)” 1.It has been proved that alendronate is better than placebo? 2.If the treatment has no effect, there is less than a 5% chance of obtaining such result 3.The observed effect is so large that there is less than 5% chance that the treatment is no better than placebo 4.I don’t know

6. P value is NOT the likelihood that findings are due to chance the probability that the null hypothesis is true given the data P-value is 0.05, so there is 95% chance that a real difference exists With low p-value (p < 0.001) the finding must be true The lower p-value, the stronger the evidence for an effect

7. P-value Grew out of quality control during WWII Question: the true frequency of bad bullets is 1%, what is the chance of finding 4 or more bad bullets if we test 100 bullets? Answer: With some maths (binomial theorem), p=2% So, p-value is the probability of getting a result as extreme (or more extreme) than the observed value given an hypothesis

8. Process of Reasoning The current process of hypothesis testing is a “proof by contradiction” If the null hypothesis is true, then the observations are unlikely. The observations occurred ______________________________________ Therefore, the null hypothesis is unlikely If Tuan has hypertension, then he is unlikely to have pheochromocytoma. Tuan has pheochromocytoma ______________________________________ Therefore, Tuan is unlikely to have hypertension

9. What do we want to know? Clinical P(+ve | Diseased): probability of a +ve test given that the patient has the disease P(Diseased | +ve): probability of that the patient has the disease given that he has a +ve test Research P(Significant test | No association): probability that the test is significant given that there is no association P(Association | Significant test): probability that there is an association given that the test statistic is significant

10. Diagnostic and statistical reasoning DiagnosisResearch Absence of diseaseThere is no real difference Presence of diseaseThere is a difference Positive test resultStatistical significance Negative test resultStatistical non-significance Sensitivity (true positive rate) Power (1-  ) False positive rateP-value Prior probability of disease (prevalence) Prior probability of research hypothesis Positive predictive valueBayesian probability

11. Case study Một phụ nữ 45 tuổi Không có tiền sử ung thư vú trong gia đình Đi xét nghiệm truy tầm ung thư bằng mammography Kết quả dương tính Hỏi: Xác suất người phụ nữ này bị ung thư là bao nhiêu? Các yếu tố liên quan cần biết để có câu trả lời: - Tần suất ung thư trong quần thể - Độ nhạy của test chẩn đoán - Độ đặc hiệu của test chẩn đoán

12. 10000 Phụ nữ Ung thư 100 Không 9900 +ve 95 -ve 5 +ve 990 -ve 8910 Tần suất ung thư trong quần thể: 1% Độ nhạy của xét nghiệm: 95%Độ đặc hiệu xét nghiệm: 90% Tổng số bệnh nhân +ve: 95 + 990 =1085 Xác suất bị K nếu có kết quà dương tính: 95/1085= 0.087 Kết quả +ve= +ve mammography -ve= -ve mammography Giải đáp

13. 10000 nghiên cứu Vit C Hiệu quả 5000 Không 5000 +ve 4000 -ve 1000 +ve 250 -ve 4750 Vit C = giả dược, 50% Power: 80%Alpha: 5% Tổng số kết quả nghiên cứu +ve: 4000+250 =4250 Xác suất Vit C có hiệu quả vói điều kiện +ve kết quả: 4000/4250= 0.94 βα1-β1-α Sai lầm loại IISai lầm loại I +ve: p <0.05 -ve: p>0.05 Suy luận trong nghiên cứu khoa học

14. What Are Required for Sample Size Estimation? Parameter (or outcome) of major interest –Blood pressure Magnitude of difference in the parameter –10 mmHg is an important difference / effect Variability of the parameter –Standard deviation of blood pressure Bound of errors (type I and type II error rates) –Type I error = 5% –Type II error = 20% (or power = 80%)

15. The Normal Distribution 0.95 0.025 0 -1.961.96 0.95 0.05 0 1.64 Prob.Z1Z2 0.800.841.28 0.901.281.64 0.951.641.96 0.992.332.81 Z1Z1 Z2Z2

16. The Normal Deviates AlphaZ  c 0.201.28 0.101.64 0.051.96 0.012.81 PowerZ  0.800.84 0.901.28 0.951.64 0.992.33

17. Study Design and Outcome Single population Two populations Continuous measurement Categorical outcome Correlation

18. Sample Size for Estimating a Population Proportion How close to the true proportion Confidence around the sample proportion. Type I error. N = (Z  ) 2 p(1-p) / d 2 –p: proportion to be estimated. –d: the accuracy of estimate (how close to the true proportion). –Z  /2 : A Normal deviate reflects the type I error. Example: The prevalence of obesity is thought to be around 20%. We want to estimate the preference p in a community within 1% with type I error of 5%. Solution N = (1.96) 2 (0.2)(0.8) / 0.01 2 = 6146 individuals.

19. Effect of Accuracy Example: The prevalence of disease in the general population is around 30%. We want to estimate the prevalence p in a community within 2% with 95% confidence interval. N = (1.96) 2 (0.3)(0.7) / 0.02 2 = 2017 subjects.

20. Sample Size for Difference between Two Means Hypotheses: H o : m 1 = m 2 vs. H a : m 1 = m 2 + d Let n 1 and n 2 be the sample sizes for group 1 and 2, respectively; N = n 1 + n 2 ; r = n 1 / n 2 ; s: standard deviation of the variable of interest. Then, the total sample size is given by: If we let Z = d/  be the “effect size”  then: Where Z  and Z 1-  are Normal deviates If n 1 = n 2, power = 0.80, alpha = 0.05, then (Z  + Z 1-  ) 2 = (1.96 + 1.28) 2 = 10.5, then the equation is reduced to:

21. Sample Size for Two Means vs.“Effect Size” For a power of 80%, significance level of 5%

22. Sample Size for Difference between 2 Proportions Hypotheses: H o :  1 =  2 vs. H a :  1 =  2 + d. Let p 1 and p 2 be the sample proportions (e.g. estimates of  1 and  2 ) for group 1 and group 2. Then, the sample size to test the hypothesis is: Where: n = sample size for each group ; p = (p 1 + p 2 ) / 2 ; Z  and Z 1-  are Normal deviates A better (more conservative) suggestion for sample size is:

23. Sample Size for Difference Between 2 Prevalence For most diseases, the prevalence in the general population is small (e.g. 1 per 1000 subjects). Therefore, a difference formulation is required. Let p 1 and p 2 be the prevalence for population 1 and population 2. Then, the sample size to test the hypothesis is: Where: n = sample size for each group; Z  and Z 1-  are Normal deviates.

24. Sample Size for Two Proportions: Example Example: The preference for product A is expected to be 70%, and for product B 60%. A study is planned to show the difference at the significance level of 1% and power of 90%. The sample size can be calculated as follows: –p 1 = 0.6; p 2 = 0.7; p = (0.6 + 0.7)/2 = 0.65; Z  = 2.81; Z  = 1.28. –The sample size required for each group should be: Adjusted / conservative sample size is:

25. Sample Size for Two Proportions vs. Effect Size Difference from p 1 by: 0.10.20.30.40.50.60.70.8 P 1 0.1424131674128191410 0.262517382473020149 0.375919889503019138 0.482520689472817.. 0.5825198824122... 0.67591736731.... 0.762513145..... 0.842473...... Note: these values are “unadjusted” sample sizes

26. Sample size for Estimating an Odds Ratio In case-control study the data are usually summarized by an odds ratio (OR), rather then difference between two proportions. If p 1 and p 2 are the proportions of cases and controls, respectively, exposed to a risk factor, then: If we know the proportion of exposure in the general population (p), the total sample size N for estimating an OR is: Where r = n 1 / n 2 is the ratio of sample sizes for group 1 and group2; p is the prevalence of exposure in the controls; and OR is the hypothetical odds ratio. If n1 = n 2 (so that r = 1) then the fomula is reduced to:

27. Sample Size for an Odds Ratio: Example Example: The prevalence of vertebral fracture in a population is 25%. It is interested to estimate the effect of smoking on the fracture, with an odds ratio of 2, at the significance level of 5% (one-sided test) and power of 80%. The total sample size for the study can be estimated by:

28. Some Comments The formulae presented are theoretical. They are all based on the assumption of Normal distribution. The estimator [of sample size] has its own variability. The calculated sample size is only an approximation. Non-response must be allowed for in the calculation.

29. Computer Programs Software program for sample size and power evaluation –PS (Power and Sample size), from Vanderbilt Medical Center. This can be obtained from me by sending email to (t.nguyen@garvan.org.au). Free.t.nguyen@garvan.org.au On-line calculator: –http://ebook.stat.ucla.edu/calculators/powercalc/http://ebook.stat.ucla.edu/calculators/powercalc/ References: –Florey CD. Sample size for beginners. BMJ 1993 May 1;306(6886):1181-4 –Day SJ, Graham DF. Sample size and power for comparing two or more treatment groups in clinical trials. BMJ 1989 Sep 9;299(6700):663-5. –Miller DK, Homan SM. Graphical aid for determining power of clinical trials involving two groups. BMJ 1988 Sep 10;297(6649):672-6 –Campbell MJ, Julious SA, Altman DG. Estimating sample sizes for binary, ordered categorical, and continuous outcomes in two group comparisons. BMJ 1995 Oct 28;311(7013):1145-8. –Sahai H, Khurshid A. Formulae and tables for the determination of sample sizes and power in clinical trials for testing differences in proportions for the two-sample design: a review. Stat Med 1996 Jan 15;15(1):1-21. –Kieser M, Hauschke D. Approximate sample sizes for testing hypotheses about the ratio and difference of two means. J Biopharm Stat 1999 Nov;9(4):641-50.