I'm trying to calculate SD from 95% CI to do a meta analysis.
i have a mean change of 6MWD test from pre intervention to post intervention of pulmonary rehabilitation groups which is 12.7(-27.1, 40.0)metres
how do you convert it to SD?
As you can see the CI isn't symmetrical around the mean
Confidence intervals for means can be used to calculate standard deviations. But this applies to confidence intervals for mean values calculated within an intervention group and not for estimates of differences between interventions. Most confidence intervals are 95% confidence intervals. If the sample size is large (say bigger than 100 in each group), the 95% confidence interval is 3.92 standard errors wide (3.92 = 2 × 1.96). The standard deviation for each group is obtained by dividing the length of the confidence interval by 3.92, and then multiplying by the square root of the sample size: SD = square root of N multiplied by the difference between upper and lower limit/3.92.
For 90% confidence intervals 3.92 should be replaced by 3.29, and for 99% confidence intervals, it should be replaced by 5.15. If sample size is small, them confidence interval should be calculated using a value from the t-distribution. I hope this helps. Also see what the expert, Martin Stausholm, said 3 years ago about this kind of question:
SD for 1 group = ((CI)/2)/TINV(0.05;n-1)*SQRT(n)
SD for 2 groups = ((CI))/2/TINV(0.05;n1+n2-2)*SQRT(n1+n2)
Maurice Ekpenyong Thank you very much for the answer!
one more question can i use this formula although as you can see my confidence interval isn't symmetrical around the mean 40-12.7 = 27.3, and 12.7-(-27.1) = 39.8
Yes you can. I recommend to read more in Cochrane Handbook:
https://handbook-5-1.cochrane.org/chapter_7/7_7_3_2_obtaining_standard_deviations_from_standard_errors_and.htm
or
https://training.cochrane.org/handbook/current/chapter-06#section-6-5
- Badges
- Science topic