8
$\begingroup$

I'm conducting a power analysis to derive the required sample size for a study - basically compared exposed / non-exposed with 30-day mortality as outcome. I'll check for crude mortality rates with chi-square, but also use logistic regression with probable confounders.

When I run a power analysis - power 0.8, significance level 0.05, effect size 0.15 and estimated 10 confounders I get that I'd need only n=117 which seem quite small. comparing with chi-square - it suggest that I'd need 350.

I'm using R and pwr:

pwr.f2.test(u=10, v=NULL, f2=0.15, sig.level=0.05, power=0.8)
pwr.chisq.test(w=0.15, N=NULL, df=1 , sig.level=0.05, power=0.8 )

Is this predictable or am I misusing this?

$\endgroup$

1 Answer 1

10
$\begingroup$

The two tests (logistic regression and chi-square) are equivalent and a power analysis should give the same answer.

You are assuming that a value of 0.15 for f2 and w are the same effect size, they're not. A small value of w is 0.1, a small value of f2 is 0.02.

cohen.ES(test=c("chisq"), size=c("small"))
cohen.ES(test=c("f2"),    size=c("small"))

Edit: Elaborated on the similarity of the two approaches.

IF you give the same data to logistic regression and a chi-square test (strictly: without Yates' correction), you get the same result. Here's an example

> set.seed(1234)
> x <- rbinom(100, 1, 0.2) 
> y <- rbinom(100, 1, 0.2) 
> chisq.test(table(x, y), correct=FALSE)

    Pearson's Chi-squared test #'

data:  table(x, y)
X-squared = 0.155, df = 1, p-value = **0.694**

Warning message:
In chisq.test(table(x, y), correct = FALSE) :
  Chi-squared approximation may be incorrect
> summary(glm(y ~ x, family="binomial"))

Call:
glm(formula = y ~ x, family = "binomial")

Deviance Residuals: 
   Min      1Q  Median      3Q     Max  
-0.753  -0.753  -0.753  -0.668   1.794  

Coefficients:
            Estimate Std. Error z value Pr(>|z|)    
(Intercept)   -1.114      0.251   -4.43  9.4e-06 ***
x             -0.272      0.693   -0.39     **0.69**    
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

(Dispersion parameter for binomial family taken to be 1)

    Null deviance: 110.22  on 99  degrees of freedom
Residual deviance: 110.06  on 98  degrees of freedom
AIC: 114.1

Number of Fisher Scoring iterations: 4

The p-values are the same, so the power should be the same. I can't remember the formulas for the two different versions of the effect size. Effect size measures are a little weird because in the old days you wanted to minimize the number of tables that you put into books (so we have, for example, $f^2$ instead of $R^2$, when there's a direct relationship between them, and $R^2$ is what everyone understands).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.