# Thread: SD from confidence interval

1. ## SD from confidence interval

Hi

Wanting help with dissertation - performing meta analysis as part of my medical degree - hope I am posting on correct line.

Need to work out Standard deviation from confidence interval for both placebo and treatment group think the equation is:

SD = SQRT of N x (upper limit - lower limit)/3.92.

CI = 95%

Is this correct?

Treatment group: Number = 52 - upper limit 55 - lower limit 24.3

Control group: Number 57 - upper limit 28.1 - lower limit 13.7

So are the answers: Treatment = 56
Control = 28

Any help would be great

Cheers

2. Kinda, what you have here is a one sample CI for a mean.
If n is small you have a problem.
And I do not know if this SD is the sample st dev, known as s or the population sd, known as $\sigma$.
If its s the correct CI is...

$(\bar X-t_{n-1,\alpha/2}s/\sqrt{n}, \bar X+t_{n-1,\alpha/2}s/\sqrt{n})$.
where we are assuming normality of our sample.

And if we do know $\sigma$, then the correct CI is...
$(\bar X-z_{\alpha/2}\sigma/\sqrt{n}, \bar X+z_{\alpha/2}\sigma/\sqrt{n})$.
where we are assuming normality of our sample, or approximating via the Central Limit Theorem.

Now $2z_{.025}=(2)(1.96)=3.92$, so if we take the second interval here and subract the lower limit from the upper limit
we have...

$2z_{\alpha/2}\sigma/\sqrt{n}$. Multiplying this by $\sqrt{n}/3.92$ does reveal the $\sigma$.