This document discusses calculating sample sizes for studies involving continuous outcome data from two independent groups. It provides the formula for calculating the standardized difference between groups given the clinically relevant difference and population standard deviation. It then shows how to use power, alpha level, and standardized difference to determine sample size using tables or software. Examples are provided to demonstrate calculating the combined standard deviation when two standard deviations are provided, and what to do if prior information is not available to determine sample size.

1. © 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

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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%

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Continuous data (two
independent groups)
The standardised difference is
calculated as;
δ ,
s.d

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Example
If difference between means = 10 mmHg
& pop. standard deviation = 20 mm Hg
Then standardised difference;
10 mm Hg/20 mm Hg = 0.5

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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.

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Or refer to a table
Sdiff = 0.5,
sample size =
64.
So 2 groups =
128.

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Alternative to table
http://www.palmx.org/samplesize/Calc_Samplesize.xls

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Or you can use PS2
We still
end up
with the
same
answer.

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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.

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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)

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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!

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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)

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Manual Calculation
d = 10 mmHg
s = 20 mm Hg
n = 2 + 7.85 (20/10)2
= 33.4 = 33
This is similar to PS2!

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What if we have 2 s.d.?
Which one to choose?
Answer: Combine the two values!

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Example 1

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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)

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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

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The combined s.d.

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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.

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Example 2

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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

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Example 3

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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

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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

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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.

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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.

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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.

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THANK
YOU