biostatistics case studies 2014
DESCRIPTION
Biostatistics Case Studies 2014. Session 1: Sample Size & Power for Inequality and Equivalence Studies I. Youngju Pak, PhD. Biostatistician [email protected]. Class Schedule. Announcements. All class materials will be uploaded in the following website - PowerPoint PPT PresentationTRANSCRIPT
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Biostatistics Case Studies 2014
Youngju Pak, PhD.
Biostatistician
Session 1:
Sample Size & Power for Inequality and Equivalence Studies I
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Class Schedule
Date Topic Related Paper
Session 1, Sept 16 Sample Size & Power for Inequality and Equivalence Studies I
Howard Paper, Gilchrist Paper, Williamson Paper
Session 2, Sept 23 Sample Size & Power for Inequality and Equivalence Studies II
Diestelhorst Paper
Session 3, Sept 30 Research Study Designs To Be Determined
Session 4, Oct 7 Regression Models and Multivariate Analyses
TBD
Session 5, Oct 14 Survival Analysis Fundamentals
TBD
Session 6, Oct 21 Free Topics & Discussion TBD
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Announcements
• All class materials will be uploaded in the following website
• http://research.labiomed.org/Biostat/Education/CaseStudies_Fall2014/CaseStudies2014Outline.htm
• Try to read posted articles before each as best as you can and pay more attention to statistical components when you read them
• Send me an e-mail ([email protected]) so I can communicate with you if necessary.
• Send me a copy of article that you want to discuss if you have one. This might be used for the last session
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Inequality study:• Two or more treatments are assumed equal (H0)and
the study is designed to find overwhelming evidence of a difference (Superiority and/or Inferiority).
• Most common comparative study type.
• It is rare to assess only one of superiority or inferiority (“one-sided” statistical tests), unless there is biological impossibility of one of them.
• Hypotheses:Ha: | mean(treatment ) - mean (control ) | ≠ 0H0: | mean(treatment ) - mean (control ) | = 0
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Insignificnat p-values for Inequality tests
• Insignificant p-values (> 0.05) usually mean that you don’t find a statistically sufficient evidence to support Ha and this doesn’t necessary mean H0 is true.
• H0 might or might not be true => Your study is still “INCONCLUSIVE”.
• Insignificant p-values do NOT prove your null !
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Equivalence Study:Two treatments are assumed to differ (H0) and the study is designed to find overwhelming evidence that they are equal.
• Usually, the quantity of interest is a measure of biological activity or potency(the amount of drug required to produce
an effect) and “treatments” are drugs or lots or batches of drugs.
• AKA, bioequivalence.
• Sometimes used to compare clinical outcomes for two active treatments if neither treatment can be considered standard or accepted. This usually requires LARGE numbers of subjects.
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Hypotheses for equivalence tests• Ha : mean (trt 1) – mean (trt 2) = 0
• H0: mean(trt 1) - mean (trt 2 ) ≠ 0
• With a finite sample size, it is very hard to find two group means are exactly the same.
• So we put a tolerability level for the equivalence, AKA, the equivalence margin, usually denoted as Δ
• Practical hypotheses would be • Ha : Δ 1< mean(trt 1) – mean (trt2) < Δ2
• H0 : mean(trt 1) – mean (trt2) ≤ Δ 1
or mean(trt 1) – mean (trt2) ≥ Δ2
Non-inferiority
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Today, we are going to learn how to determine sample size for Inequality tests using software
for three papers.
Then, Discuss some logic.
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Paper #1
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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?
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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.
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From earlier design paper (Russell 2007):
Δ = 0.85(0.05)
mm = 0.0425 mm
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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
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Paper #2
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Williamson paper
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Software Output - Percentages
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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.
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Paper #3
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From Nance paper
Δ = ~8%
Δ
SD√(1/N1 + 1/N2)= 2.82
Solve for SD to get SD =~ 6.8%
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Software Output for Gilchrist Paper
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Some Logic
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How was 498 determined?
Back to:
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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
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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 =
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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
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Info Needed for Study Size: Comparing Means
1. Effect
2. Subject variability
3. Type I error (1.96 for α=0.05; 2.58 for α=0.01)
4. Power (0.842 for 80% power; 1.645 for 95% power)
(1.96 + 0.842)22SD2
Δ2N =
Same four quantities, but different formula, if comparing %s, hazard ratios, odds ratios, etc.
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(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
Δ2N = (1.96 + 0.842)2
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Change Effect Size to be Detected
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SD Estimate Could be Wrong
Should examine SD as study progresses.
May need to increase N if SD was underestimated.
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Some Study Size Software
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Free Study Size Software
www.stat.uiowa.edu/~rlenth/Power
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Study Size Software in GCRC Lab
ncss.com ~$500
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nQuery - Used by Most Drug Companies