Power Analysis

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The statistical power can be thought of as the probability of accepting an alternative hypothesis, when the alternative hypothesis is true. When interpreting statistical power, we seek experiential setups that have high statistical power.

• Low Statistical Power: Large risk of committing Type II errors, e.g. a false negative.
• High Statistical Power: Small risk of committing Type II errors.

It is common to design experiments with a statistical power of 80% or better, e.g. 0.80.

Problem 1

Part 1

Problem: An investigator is planning a clinical trial to evaluate the efficacy of a new drug designed to reduce systolic blood pressure. The plan is to enroll participants and to randomly assign them to receive either the new drug or a placebo. Systolic blood pressures will be measured in each participant after 12 weeks on the assigned treatment.
If the new drug shows a 5 unit reduction in mean systolic blood pressure, this would represent a clinically meaningful reduction. How many patients should be enrolled in the trial to ensure that the power of the test is 80% to detect this difference? A two sided test will be used with a 5% level of significance.

In order to compute the effect size, an estimate of the variability in systolic blood pressures is needed. Analysis of data from the Framingham Heart Study showed that the standard deviation of systolic blood pressure was 19.0. This value can be used to plan the trial.

Solution:
Effect size (Cohen’s d), d = \(\frac{|\mu_1 – \mu_2|}{pooled\ SE}\) = \(\frac{|5|}{19}\) = 0.26
Required sample size,
n = \(2(\frac{z_{1-\frac{\alpha}{2}} + z_{1-\beta}}{ES})\) = \(2(\frac{1.96 – 0.84}{0.26})^2\) = 232.22

Using R :

> library(pwr)
> pwr.t.test(n = NULL,
+            d = 0.26, 
+            sig.level = 0.05, 
+            power = 0.8, 
+            type = "two.sample", 
+            alternative = "two.sided")

     Two-sample t test power calculation 

              n = 233.1791
              d = 0.26
      sig.level = 0.05
          power = 0.8
    alternative = two.sided

NOTE: n is number in *each* group

Therefore, samples of size n1=232 and n2= 232 will ensure that the test of hypothesis will have 80% power to detect a 5 unit difference in mean systolic blood pressures in patients receiving the new drug as compared to patients receiving the placebo.

Part 2

Problem: Based on prior experience with similar trials, the investigator expects that 10% of all participants will be lost to follow up or will drop out of the study.

Solution:
The investigators hypothesized a 10% attrition rate (in both groups), and to ensure a total sample size of 232 they need to allow for attrition:
N (number to enroll) * (% retained) = desired sample size

Therefore, N (number to enroll) = desired sample size/(% retained)

N = 232/0.90 = 258

The investigator must enroll 258 participants to be randomly assigned to receive either the new drug or placebo.

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References:

• Statistical Power analysis in Python - Cohen’s D

• What does effect size tell you?

• Sample Sizes for Two Independent Samples, Continuous Outcome