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I am conducting a meta analysis and one of studies doesn't have the standard deviation so i've been stuggling to work out the standard error. CI's are of 95%

I was wondering if I could calculate the standard error given the mean and sample sizes without the standard deviation or the data set.

I've looked online for guides and formulas but I'm not too sure. Any help would be much appreciated, thanks.

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  • $\begingroup$ Work backwards from what.$ Might be possible if you have a 95% CI. Why not tells us the whole story behind your question? $\endgroup$ – BruceET Jul 12 at 22:28
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    $\begingroup$ Yes I have a 95% CI $\endgroup$ – Fusiozii Jul 12 at 22:35
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    $\begingroup$ Do you have a formula that could do this? $\endgroup$ – Fusiozii Jul 12 at 22:36
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A 95% CI for $\mu,$ given a random sample of size $n$ from a normal population, is of the form $$\bar X \pm t_cS/\sqrt{n},$$ where $\bar X$ and $S$ are the sample mean and sample standard deviation, respectively, and where $t_c$ cuts probability $0.025$ from the upper tail of Student's t distribution with $n-1$ degrees of freedom.

The 'margin of error' $t_cS/\sqrt{n}$ is half the width of the CI. So if you have that, and you know $n.$ you can find $t_c$ from a printed table of t distributions. Then you can solve for the (estimated) standard error $S/\sqrt{n}$ and for $S.$

Example:

n = 20;  x = rnorm(20, 50, 3)
t.test(x)$conf.int
[1] 48.66417 51.47860
attr(,"conf.level")
[1] 0.95
t.c = qt(.975, 19); t.c
[1] 2.093024  # look in printed table (line 19) to verify
  • The 95% CI is $(48.66417, 51.47860).$
  • Its half width (margin of error) is $t_cS/\sqrt{n}=(51.47860 - 48.66417)/2 = 1.407215.$
  • Then the (estimated) standard error is $S/\sqrt{20}= 1.407215/2.093024 = 0.6723358.$

As a check, this is in substantial agreement (within rounding error) of the standard error calculated directly from the data : both are $0.6723.$

sd(x)/sqrt(20)
[1] 0.6723355
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