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8.Calculate samplesize for clinical trials (continuous outcome)
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8.Calculate samplesize for clinical trials (continuous outcome)

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Part 8 of 9; 8.Calculate samplesize for clinical trials (continuous outcome)

Part 8 of 9; 8.Calculate samplesize for clinical trials (continuous outcome)

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8.Calculate samplesize for clinical trials (continuous outcome) 8.Calculate samplesize for clinical trials (continuous outcome) Presentation Transcript

  • © Dr Azmi Mohd Tamil, 2012 Calculate Your Own Sample Size – Part 8 (outcome continuous data) Jones SR, Carley S & Harrison M. An introduction to power and sample size estimation. Emergency Medical Journal 2003;20;453-458. 2003
  • 2 © Dr Azmi Mohd Tamil, 2012 Continuous data (two independent groups) We need to specify the following; Standard deviation of the variable (s.d) Clinically relevant difference (δ) The significant level (α) – 0.05 The power (1 - β ) – 80%
  • 3 © Dr Azmi Mohd Tamil, 2012 Continuous data (two independent groups) The standardised difference is calculated as; δ , s.d
  • 4 © Dr Azmi Mohd Tamil, 2012 Example If difference between means = 10 mmHg & pop. standard deviation = 20 mm Hg Then standardised difference; 10 mm Hg/20 mm Hg = 0.5
  • 5 © Dr Azmi Mohd Tamil, 2012 Continuous data (two independent groups) We draw a straight line from the value for the standardized difference to the value of 0.80 on the scale for power. Read off the value for N on the line corresponding to α = 0.05, which gives a total sample size of eg. 128, so we required 64 samples for each group.
  • 6 © Dr Azmi Mohd Tamil, 2012
  • 7 © Dr Azmi Mohd Tamil, 2012 Or refer to a table Sdiff = 0.5, sample size = 64. So 2 groups = 128.
  • 8 © Dr Azmi Mohd Tamil, 2012 Alternative to table http://www.palmx.org/samplesize/Calc_Samplesize.xls
  • 9 © Dr Azmi Mohd Tamil, 2012 Or you can use PS2 We still end up with the same answer.
  • 10 © Dr Azmi Mohd Tamil, 2012 PS2 We are planning a study of a continuous response variable from independent control and experimental subjects with 1 control(s) per experimental subject. In a previous study the response within each subject group was normally distributed with standard deviation 20. If the true difference in the experimental and control means is 10, we will need to study 64 experimental subjects and 64 control subjects to be able to reject the null hypothesis that the population means of the experimental and control groups are equal with probability (power) 0.8. The Type I error probability associated with this test of this null hypothesis is 0.05.
  • 11 © Dr Azmi Mohd Tamil, 2012 Manual Calculation (2 groups) s = standard deviation, d = the difference to be detected, and C = constant (refer to table below); if α=0.05 & 1-β=0.8, then C = 7.85. (Snedecor and Cochran 1989)
  • 12 © Dr Azmi Mohd Tamil, 2012 Manual Calculation d = 10 mmHg s = 20 mm Hg n = 1 + 2 x 7.85 (20/10)2 = 63.8 = 64 This is similar to the table and PS2!
  • 13 © Dr Azmi Mohd Tamil, 2012 Manual Calculation (pre post) s = standard deviation, d = the difference to be detected, and C = constant (refer to table below); if α=0.05 & 1-β=0.8, then C = 7.85. (Snedecor and Cochran 1989)
  • 14 © Dr Azmi Mohd Tamil, 2012 Manual Calculation d = 10 mmHg s = 20 mm Hg n = 1 + 7.85 (20/10)2 = 32.4 = 33 This is similar to PS2!
  • 15 © Dr Azmi Mohd Tamil, 2012 What if we have 2 s.d.? Which one to choose? Answer: Combine the two values!
  • 16 © Dr Azmi Mohd Tamil, 2012 Example 1
  • 17 © Dr Azmi Mohd Tamil, 2012 Compare LDL between normal & overweight A cross-sectional study of 56 first-year students, of both genders. Assume 26 were males, so 19 males were normal weight, 7 were overweight. LDL (normal) : 82.3 + 15.7 (n1=19) LDL (overweight): 88.3 + 7.6 (n2=7)
  • 18 © Dr Azmi Mohd Tamil, 2012 Need to combine both s.d. 15.7 & 7.6 Normal; 82.3 + 15.7 (n=19) Overweight; 88.3 + 7.6 (n=7) Diff = 88.3 – 82.3 = 6 Sd=SQRT(((18*15.7^2)+(6*7.6^2))/24)=14.11763 Assume 19 normal for every 7 OW. m= 2.714286
  • 19 © Dr Azmi Mohd Tamil, 2012 The combined s.d.
  • 20 © Dr Azmi Mohd Tamil, 2012 We are planning a study of comparing LDL between normal males and overweight males with 2.714286 normal per overweight subject. In a previous study the response within each subject group was normally distributed with standard deviation 14.11763 (combine s.d. 15.7 & 7.6). If the true difference of LDL means between overweight subjects and normal subjects is 6, we will need to study 60 overweight subjects and 163 normal subjects to be able to reject the null hypothesis that the population means of the overweight and normal groups are equal with probability (power) 0.8. Minimum required sample size of 223. The Type I error probability associated with this test of this null hypothesis is 0.05.
  • 21 © Dr Azmi Mohd Tamil, 2012 Example 2
  • 22 © Dr Azmi Mohd Tamil, 2012 VFI Control; 1.93 + 0.81 (n=40) Case-3; 5.5 + 3.6 (n=53) Diff = 5.5 – 1.93 =3.57 Sd = SQRT(((52*3.6^2)+(39*0.81^2))/91) Sd = 2.772526
  • 23 © Dr Azmi Mohd Tamil, 2012
  • 24 © Dr Azmi Mohd Tamil, 2012 Example 3
  • 25 © Dr Azmi Mohd Tamil, 2012 Adiponectin (Obese vs non-obese) Obese; 6.4+ 0.6 (n=53) Normal; 10.2 + 0.8 (n=30) Diff = 10.2 – 6.4 =3.8 Sd =SQRT(((52*0.6^2)+(29*0.8^2))/81) Sd = 0.678415001
  • 26 © Dr Azmi Mohd Tamil, 2012
  • 27 © Dr Azmi Mohd Tamil, 2012 What If There Is No Prior Information? Instead of saying "Sample sizes are not provided because there is no prior information on which to base them“, do this instead; Find previously published information Conduct small pre-study If a very preliminary pilot study, sample size calculations not usually necessary
  • 28 © Dr Azmi Mohd Tamil, 2012 Conclusion You can calculate your own sample size. Tools are available and most of them are free. Decide what is your study design and choose the appropriate method to calculate the sample size. If despite following ALL these notes fastidiously, your proposal is still rejected by the committee due to sample size, kindly SEE THEM, not us.
  • 29 © Dr Azmi Mohd Tamil, 2012 References (incl. for StatCalc) Fleiss JL. Statistical methods for rates and proportions. New York: John Wiley and Sons, 1981. Gehan EA. Clinical Trials in Cancer Research. Environmental Health Perspectives Vol. 32, pp. 3148, 1979. Jones SR, Carley S & Harrison M. An introduction to power and sample size estimation. Emergency Medical Journal 2003;20;453-458. 2003 Kish L. Survey sampling. John Wiley & Sons, N.Y., 1965. Krejcie, R.V. & Morgan, D.W. (1970). Determining sample size for research activities. Educational & Psychological Measurement, 30, 607-610. Snedecor GW, Cochran WG. 1989. Statistical Methods. 8th Ed. Ames: Iowa State Press.
  • 30 © Dr Azmi Mohd Tamil, 2012 References (PS2) Dupont WD, Plummer WD, Jr: Power and Sample Size Calculations: A Review and Computer Program. Controlled Clinical Trials 11:116-128, 1990 Dupont WD, Plummer WD, Jr: Power and Sample Size Calculations for Studies Involving Linear Regression. Controlled Clinical Trials 19:589-601, 1998 Schoenfeld DA, Richter JR: Nomograms for calculating the number of patients needed for a clinical trial with survival as an endpoint. Biometrics 38:163-170, 1982 Pearson ES, Hartley HO: Biometrika Tables for Statisticians Vol. I 3rd Ed. Cambridge: Cambridge University Press, 1970 Schlesselman JJ: Case-Control Studies: Design, Conduct, Analysis. New York: Oxford University Press, 1982 Casagrande JT, Pike MC, Smith PG: An improved approximate formula for calculating sample sizes for comparing two binomial distributions. Biometrics 34:483-486, 1978 Dupont WD: Power calculations for matched case-control studies. Biometrics 44:1157-1168, 1988 Fleiss JL. Statistical methods for rates and proportions. New York: John Wiley and Sons, 1981.
  • 31 © Dr Azmi Mohd Tamil, 2012 THANK YOU