Biostatistics Case Studies 2010

Peter D. Christenson

Biostatistician

http://gcrc.labiomed.org/biostat

Session 1:

Study Size Tutorial

First, Use software for three papers.

Then, Discuss some logic.

Paper #1

How was N=498 determined?

What reduction in CVD events can 224 + 224 subjects detect? Nevertheless

How many subjects would be needed to detect this Δ?

Software Output for % of CVD Events

224 + 224 → detect 6.7% vs. 1.13%, i.e., 88% ↓.

Need 3115 + 3115 to detect 25% ↓ from 6.7% to 5%, i.e., a total of (3115+3115)/0.9 = 6922.

From earlier design paper (Russell 2007):

Δ = 0.85(0.05)

mm = 0.0425 mm

Software Output for Mean IMT

Each group N for 10% Dropout → 0.9N = 224

→ N = 224/0.9 = 249. Total study size = 2(249)=498

Paper #2

Williamson paper

Software Output - Percentages

Slightly larger Ns due to slightly different test to be used.

Software Output - Means

Can detect 0.4 SDs. Units? Since normal range =~ 6SD, this corresponds to ~0.4/6=7% shift in normal range.

Applies to any continuously measured outcome.

Paper #3

From Nance paper

Δ = ~8%

Δ

SD√(1/N1 + 1/N2)= 2.82

Solve for SD to get SD =~ 6.8%

Software Output for Gilchrist Paper

Some Logic

How was 498 determined?

Back to:

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

SD√(1/N1 + 1/N2)

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).

≥~1.96 if (with N = N1 = N2):

Δ ≥ 1.96SD√(2/N) or N ≥ 2SD2

Δ2

(1.96)2

t =

Need to Increase N for Power

Need to increase N to:

2SD2

Δ2

(1.96 + 0.842)2

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

Δ2

(1.96)2N = for 50% power.

for 80% power.N =

N =2SD2

Δ2(1.96 + 1.282)2 for 90% power.

from Normal Tables

Info Needed for Study Size: Comparing Means

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)

(1.96 + 0.842)22SD2

Δ2

N =

Same four quantities, but different formula, if comparing %s, hazard ratios, odds ratios, etc.

(1.96 + 0.842)2 2(0.16)2

(0.0425)2N = = 224

Each group N for 10% Dropout → 0.9N = 224

→ N = 224/0.9 = 249. Total study size = 2(249)=498

2SD2

Δ2

N = (1.96 + 0.842)2

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.

Recall: Software Output for % of CVD Events

224 + 224 → detect 6.7% vs. 1.13%, i.e., 88% ↓.

Need 3115 + 3115 to detect 25% ↓ from 6.7% to 5%, i.e., a total of (3115+3115)/0.9 = 6922.

Comparing Survival

We just saw that:

Need 3115 + 3115 to detect 25% ↓ from 6.7% to 5%, i.e., a total of (3115+3115)/0.9 = 6922.

This does not use the 10% of 6922 = 692 subjects lost to follow-up (in the analysis).

Recall that survival analysis, e.g., Kaplan-Meier curves, does use the info from these 692 subjects for as long as they were observed.

So, fewer than 6922 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, compared to 6922.

Some Study Size Software

Free Study Size Software

www.stat.uiowa.edu/~rlenth/Power

Study Size Software in GCRC Lab

ncss.com ~$500

nQuery - Used by Most Drug Companies