1
$\begingroup$

I am conducting a meta-analysis with a skewed distribution. To address this issue, I transformed the "marker" data into a log-scale, "ln marker".

I obtained the (geometric) mean and standard deviation of ln marker from one article.

My goal is to find the 95% CI of ln marker. To do this, I transformed the geometric mean and geometric standard deviation back into log-scale, so they became the mean and standard deviation of ln marker again. I then used the formula "mean +/- 1.96 SD/sqrt(N)" to calculate the 95% CI of ln marker, assuming the normal distribution by log transformation.

However, I am unsure if this is the right way to get the 95% CI of ln marker with geometric mean and SD.

$\endgroup$

1 Answer 1

1
$\begingroup$

Yes, that approach seems reasonable. Two notes.

  • Your equation for the CI assumes a large sample size. You really should use the critical value from the t distribution (which accounts for sample size) and not 1.96 (which is correct only for large samples).
  • Your CI is in the log scale. Back transform both confidence limits to get a a confidence interval in the scale of the original data
$\endgroup$

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.