we are conducting a meta-analysis on community-based intervention trials. Some of the studies don't have the standard deviation, but only the 95% CI and mean. Sample sizes are large. I have tried to calculate by taking the square root of the sample size and multiplying it by the 95% CI then dividing by 3.92 but the result does not seem right. Is there any key tip that I missed in this regard?
Thank you all
Masoumeh Jabbari
Divide the length of the confidence interval by 3.92 and multiply by the square root of the sample size to get the standard deviation for each group: 3.92 should be substituted with 3.29 for 90 percent confidence intervals and 5.15 for 99 percent confidence intervals.
Take the following steps:
1. Determine the mean or average value for each data collection.
2. Subtract the mean from each value to calculate the deviation of each item of data.
3. Take the sum of the deviations and square them.
4. Add all of the squared deviations together.
5. Take this number and divide it by one less than the number of elements in the data collection.
In conclusion:
1. First, calculate the sum of the squares of the differences between each data point and the sample mean.
2. Next, divide the total by the sample size minus one to get the variance.
3. Finally, to get the SD, take the square root of the variance.
@Jabbari Same query has been discussed on the below thread on RG.
https://www.researchgate.net/post/Is-it-possible-to-calculate-SD-from-95CI-and-mean
In a meta-analysis, one needs the (sample) mean and the standard deviation of the (sample) mean (i.e. standard error (SE), also known as standard uncertainty) from each study (or group). SE=(up-limit - low-limit )/(2*1.96) under the assumption of normality and 95% level.
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