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# Sample Size Estimation

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1. Sample Size Estimation Chulaluk Komoltri DrPH (Bios) Faculty of Medicine Siriraj Hospital

2. Sample Size Estimation • Why? • n is large enough to provide a reliable answer to the • question • too small n  a waste of time • too many n  A waste of money & other resources • May be unethical • e.g., delayed beneficial therapy • Study Objective • - Hypothesis generating (Pilot study) •  No sample size estimation • - Estimation of parameter, Hypothesis confirmation •  Sample size estimation

3. n is usually determined by the primary objective of the study More than one primary outcomes If one of these endpoints is regarded as more important than others, then calculate n for that primary endpoint. If several outcomes are regarded as equally important, then calculate n for each outcome in turn, and select the largest n. Method of calculating n should be given in the proposal, together with the assumptions made in the calculation

4. Caution Calculation of sample size needs a number of assumptions and ‘guesstimates’, so such calculation only provides a guideto the number of subjects required.

5. Pilot study Example: (Mar 23,1999, #1515) This study for n=20 eligible burn patients will generate hypothesis about the predictive values of various patient characteristics for predicting number of days to return to work.

6. Example: (Oct 27, 1998, #1465) This is a pilot study providing preliminary descriptive statistics thatwill be used to design a larger, adequately powered study. N=24 normal healthy volunteers will be randomized to parallel groups to study the effect of 4 antidepressant drugs…

7. Estimation study, Hypothesis confirmation study Sample size determination: 2 Objectives I. Estimation of parameter(s)  Precision (95% CI) Specify  error - Estimate prevalence, sensitivity, specificity - Estimate single mean, single proportion - etc.

8. II. Test H0 Statistical power (1- ) Specify ,  error 2.1 Single group - Test of single proportion, mean - Test of Pearson’s correlation - etc. 2.2 Two groups -Test difference of - 2independentproportions, means, survival curves - 2dependent proportions, means - Test equivalence of - 2 independent proportions, means - etc. 2.3 > 2 groups

9. ,  (Efficacy trial) Truth H0 true H0 false (A=B) (AB, Difference) Decision Accept H0No error (1- ) (from p-value) (p > ) Reject H0No error (1- ) (p  ) Power = Pr (incorrect conclusion of difference ) = False positive (FP)  = Pr (incorrect conclusion of equivalence) = False negative (FN) 1 -  = Pr ( correct conclusion of difference ) = True positive (TP)

10. Sample size for estimating parameter, testing hypothesis

11. nQuery Advisor PASS

12. PS (Power and Sample Size)

13. STATCALC

14. 1. Estimation 1.1) Estimate single proportion  95% CI of  = p ± d where  = Pr. of type I error = 0.05 (2-sided) z0.025 = 1.96 p = Estimated proportion of … q = 1-p d = Margin of error in estimating  1.2) Estimate single mean  95% CI of  =x ± d n = [z/2 SD / d]2 where SD = Standard deviation of …

15. 95% CI for  = p  z/2 SE(p) = p  z/2 p(1-p)/n = p  d Thus, d = z/2 p(1-p) n d2 = z/22 p(1-p) n n = z/22 p(1-p) d2 Derivation of formulas in (1.1), (1.2) 95% CI of  = x  z/2 SE(x) = x  z/2 SD / n = x  d Thus, d = z/2 SD n n = z/2 SD 2 d

16. n = z/22 pq / d2 -----------(1) a) If no idea about p, let p = 0.5  biggest pq = 0.25 At 2-sided  = 0.05 then n = 1.962 (0.5)(0.5)  1 -----------(2) d2 d2 b) If finite pop’n (N is known), adjust n from eqn. (1) n = n ------------(3) 1+ n/N = z2pq N ------------(4) d2N + z2pq If n = 1/d2 as in eqn. (2), then n = N ------------(5) (1+d2N)

17. Conclusion: n = N ---------(5) (1+d2N) where N = Population size d = Margin of error in estimating pop’n proportion Formula # 5 is appropriate under: 1) Objective: To estimate a single proportion 2) Assume p = 0.5 to get the biggest sample size 3) Assume population is finite (known pop’n size N) 4) Assume sample is selected by simple random sampling (SRS)

19. 2)Test 2.1) Test of difference in 2 independent proportions H0 : 1 - 2 = 0 H1 : 1 - 2  0 where  = Probability of type I error  = Probability of type II error p1 = Proportion of … in gr. 1 q1 = 1-p1 p2 = Proportion of … in gr. 2 q2 = 1-p2 p = (p1+p2)/2 q = 1-p

20. 2.2) Test of difference in 2 independent means H0 : 1 - 2 = 0 H1 : 1 - 2  0 where  = Probability of type I error  = Probability of type II error  = Common standard deviation of … in group 1, 2  = Difference in mean … between 2 groups = 1 - 2

21. 2.3) Test of significance of 1 proportion H0 :  = p0 H1 :  = p1 where p0 = Proportion of … under H0 p1 = Proportion of … under H1 2.4) Test of significance of 1 mean H0 :  = 0 H1 :  = 1 where  = Standard deviation of …  = 1 - 0

22. 2.5) Test of significance of 1 correlation H0 :  = 0 H1 :  = 1 n = (Z/2 + Z) 2 [F(Z0) - F(Z1)] where F(Z0) = Fisher’s Z transformation of 0 = 0.5 ln [(1+0)/(1-0)] F(Z1) = Fisher’s Z transformation of 1 = 0.5 ln [(1+1)/(1-1)] ln = Natural logarithm + 3

23. Examples

24. Example 1: (Mar 18, 2000, #1688) This is a cross-sectional study of the prevalence of pulmonary hypertension (PHT) in patients aged 15-70 years with sickle cell disease. The primary endpoint is PHT diagnosis based on observed pulmonary pressure by droppler echocardiogram. A sample of n = 140 will provide 95% CI for true prevalence rate of PHT of 0.10  0.05. = 1.962 (0.1)(0.9) / 0.052 = 139

25. 2-sided CI = p ± d • 2) 1-sided CI = p + d • or 1-sided CI = p - d

26. 2-sided 95% CI for  = p ± d d  level of precision p = 0.10, d = 0.15  95% CI = 0.10 ± 0.15 = -0.05, 0.25 ???

27. pq d (error) n  • How big is d ? • Absolute d • 2. Relative d: d  20% of prevalence(p) p d 95% CI n 0.80 0.05*p = 0.04 0.76, 0.84 384 0.05 0.75, 0.85 246 0.10*p = 0.08 0.72, 0.88 96 0.10 0.70, 0.90 62 0.15*p = 0.12 0.68, 0.92 43 0.15 0.65, 0.95 28 0.20*p = 0.16 0.64, 0.96 24 0.20 0.60, 1.00 16

28. Example 2 จากวัตถุประสงค์ของการศึกษาเพื่อประมาณค่าสัดส่วนการเกิดภาวะตาบอด ถ้าประมาณว่า95% confidence interval (CI) ของค่าสัดส่วน การเกิดภาวะตาบอดจริงจะเท่ากับ20%  3% หรือ 17% - 23% จะต้องทำการศึกษาในผู้ป่วยต้อหินจำนวน 683คนดังรายละเอียดการคำนวณคือ n = z/22 p(1-p) / d2 เมื่อ p =ค่าประมาณของสัดส่วนการเกิดภาวะตาบอดในผู้ป่วยต้อหิน= 0.2 d =ความคลาดเคลื่อนในการประมาณค่าสัดส่วน= 0.03  =โอกาสที่จะเกิดtype I error = 0.05(2-sided) z0.025 = 1.96 ดังนั้น n = 1.962 (0.2)(0.8)/0.032 = 682.95 = 683

29. Dizzy pts. 1. Dix-Hallpike test 2. Side-lying test BPPV No BPPV Example 3 Title: Diagnosis of Benign Paraxysmal Positional Vertigo (BPPV) by Side-lying test as an alternative to the Dix-Hallpike test Investigator: Dr. Saowaros Asawavichianginda Design: Diagnostic study Subjects: Dizzy patients, aged 18-80 yrs, onset < 2 wks

30. Sample size: Based on 95% CI of true sensitivity (Se) = 0.9 ± 0.1 where p = expected sensitivity = 0.9 q = 1-p = 0.1 d = allowable error = 0.1  = 0.05 (2-sided), Z0.025 = 1.96 So, n = 34.56 = No. of patients with BPPV from Dix-Hallpike test Since prevalence of BPPV among dizzy patients = 40% Thus, no. of dizzy patients = 34.56 = 86.4 = 87 0.4 Dix-Hallpike test (Gold std) + (BPPV) - (No BPPV) Side-lying test + Se - 1 – Se 35 52 87

31. Example 4 การศึกษานี้มีวัตถุประสงค์เพื่อประมาณค่าเฉลี่ยของsubcarinal angle ในคนไทยปกติ และจากการศึกษาของ ... ในคนปกติจำนวน100 รายอายุ ... ปี พบว่าค่าเฉลี่ยของsubcarinal angleเท่ากับ60.8(SD=11.8) ถ้ากำหนดให้95% confidence interval (CI)ของค่าเฉลี่ยของ subcarinal angleในประชากรไทย () มีค่าเท่ากับ61  2 (SD=13) จะต้องทำการศึกษาในคนไทยปกติจำนวน163คนดังรายละเอียดการคำนวณดังนี้ n = [z/2 SD / d]2 เมื่อ SD = Standard deviation ของ subcarinal angle = 13 d = Margin of error ในการประมาณค่าเฉลี่ย= 2 = Probability of type I error (2-sided) = 0.05 z0.025 = 1.96 ดังนั้น n = [1.96*13/2]2 = 162.31 = 163

32. Example 5 Title: Efficacy of polyethylene plastic wrap for the prevention of hypothermia during the immediate postnatal period in low birth weight premature infants Investigator: Dr. Santi Punnahitananda Design: RCT, 2-parallel arms Subjects: Infants with  34 gestational wks, birth weight  1800 gms Outcome: Infant’s body temperature taken on nursery admission Infants,  34 gestational wks, BW  1800 gms Randomization Plastic wrap No Plastic wrap Body temp.  Hypothermia Body temp.  Hypothermia

33. Sample size estimation: Based on Test of 2 independent proportions Our unit  hypothermia in low birth weight, premature infants = 55% (p1 = 0.55) Assume that plastic wrap would reduce hypothermia to 20% (p2 = 0.2)

34. Example 6 Title: Relationship between microalbuminuria (MAU) and diabetic retinopathy (DR) in type 2 diabetes Investigator: Dr. Attasit Srisubat Design: Cross-sectional study Subjects: Type 2 diabetic pts., DM duration  5 yrs Diabetic retinopathy DR No DR Microalbuminurea With p1 1-p1 n1 Without p2 1-p2 n2

35. where p1 = Proportion of DR in type 2 DM w/ microalbuminuria = 0.43 q1 = 1 - p1 p2 = Proportion of DR in type 2 DM w/o microalbuminuria = 0.28 q2 = 1 - p2 p = (p1 + rp2) / (r+1) q = 1 - p r = n2/n1 = 2

36. Example 7 Title: Can knee immobilization after total knee replacement (TKA) save blood from wound drainage Investigator: Dr. Vajara Wilairatana Design: Randomized controlled trial Subjects: Pts. with hip disease that require TKA Pts. with hip disease that require TKA Randomization Knee elevation 40° A-P splint and Knee elevation 40° Blood loss Blood loss

37. where  = Difference in mean postoperative blood loss between 2 groups   = SD of postoperative blood loss Kim YH et al. Knee splint in 69 knees, mean wound drainage = 436 ml, SD = 210 ml Ishii et al. 30 non-splint knees, mean blood loss = 600 ml, SD = 293