# Biostatistics Case Studies 2008 by gjy28315

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```									Biostatistics Case Studies 2008

Session 4:
Study Size Tutorial

Peter D. Christenson
Biostatistician
http://gcrc.labiomed.org/biostat
How was 498
determined?
How IMT Change Comparison Will be Made

Strength of Treatment
Effect:
Signal:Noise Ratio t=
Observed Δ
SD√(1/N1 + 1/N2)

Δ = Aggressive - Standard Mean Diff in IMT changes
SD = Std Dev of within group IMT changes
N1 = N2 = Group size
| t | > ~1.96   ↔   p<0.05
Could Solve for N
Observed Δ
t=                         ≥~1.96      if (with N = N1 = N2):
SD√(1/N1 + 1/N2)

2SD2
Δ ≥ 1.96SD√(2/N)       or N ≥          (1.96)2
Δ2
This is not quite right.
The Δ is the actual observed difference.
This sample Δ will vary from the real Δ in “everyone”.
Need to increase N in case the sample happens to
have a Δ that is lower than the real Δ (50% possibility).
Need to Increase N for Power
Power is the probability that p<0.05 if Δ is the real
effect, incorporating the possibility that the Δ in our
sample could be smaller.
2SD2
N=             (1.96)2    for 50% power.
Δ2
Need to increase N to:
2SD2
N=             (1.96 + 0.842)2     for 80% power.
Δ2

2SD2
N=             (1.96 + 1.282)2      for 90% power.
Δ2
from Normal Tables
Info Needed for Study Size: Comparing Means

2SD2
N=           (1.96 + 0.842)2
Δ2

1. Effect
2. Subject variability
3. p-value (1.96 for p=0.05; 2.58 for p=0.01)
4. Power (0.842 for 80% power; 1.645 for 95% power)

Same four quantities, but different formula, if
comparing %s, hazard ratios, odds ratios, etc.
Need SD from
earlier paper
From earlier design paper (Russell 2007):

Δ=
0.85(0.05)
mm =
0.0425 mm
2SD2
N=           (1.96 + 0.842)2
Δ2

2(0.16)2
N=                (1.96 + 0.842)2   = 224
(0.0425)2

Each group N for 10% Dropout → 0.9N = 224
→ N = 224/0.9 = 249. Total study size = 2(249)=498
Software Input
Software Output

Each group N for 10% Dropout → 0.9N = 224
→ N = 224/0.9 = 249. Total study size = 2(249)=498
Change Effect Size to be Detected
SD Estimate Could be Wrong

Should examine SD as study progresses.
May need to increase N if SD was underestimated.
Comparing Percentages

...

What reduction in CVD events can 224 + 224 subjects detect?
Software Input

For %s, the
SD is
determined
from the %s.
Thus, do not
need to
supply SD.
Software Output

224 + 224 → detect 6.7% vs. 1.14%, i.e., 88% ↓.
Need 3107 + 3107 to detect 25% ↓ from 6.7% to 5%,
i.e., a total of (3107+3107)/0.9 = 6904.
Comparing Survival

We just saw that:
Need 3107 + 3107 to detect 25% ↓ from 6.7% to 5%,
i.e., a total of (3107+3107)/0.9 = 6904.
This does not use the 10% of 6904 = 690 subjects
lost to follow-up (in the analysis).
Recall that survival analysis, e.g., Kaplan-Meier
curves, does use the info from these 690 subjects for
as long as they were observed.
So, fewer than 6909 subjects are needed using
survival analysis - we calculate that N now.
Comparing Survival

Really 0.
This
software
requires
>0.
Comparing Survival

Total of 6304, includes 10% loss.
SD of BMD % change =~ 6% (all body sites)
Software Input
Software Output

Probably higher than 80% since SDs are not reported
directly and I estimated SD=0.06 from total body BMD
only.
Software Input - Percentages
Software Output - Percentages

Total N = 218 for 80% power or 290 for 90% power.
Software Input - Means

Will
give #
of SD
units
Software Output - Means

Can detect 0.4 SDs. Would prefer to express in the units of
the outcome.
Applies to any continuously measured outcome.
Some Study Size Software
Free Study Size Software
www.stat.uiowa.edu/~rlenth/Power
Free Study Size Software: Example
Pilot data: SD=8.19 in 36 subjects.
We propose N=40 subjects/group in order to provide
80% power to detect (p<0.05) an effect Δ of 5.2:
Study Size Software in GCRC Lab
ncss.com   ~\$500
nQuery - Used by Most Drug Companies

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