Calculate standard deviation from sample size, mean, and confidence interval?

I wonder if I can back calculate standard deviation from mean, sample size, and confidence interval.

For example: mean age = 40.2; sample size = 427; and 95% confidence interval = (38.9-41.5)

And if so, can it be apply to percentage measure, for example: percent being male = 64.2%; sample size = 427; and 95% confidence interval = (59.4-68.7).

• If you are assuming a normal distribution then the formula for the endpoints of the confidence interval is strictly a function of the sample standard deviation. The other variables mean and sample size are given. I don't know what you mean by "percentage measure". So I can't help you with that. – Michael R. Chernick Jan 2 '18 at 1:11
• By Percentage measure I simply meant the 64.2% of sample being male. – KubiK888 Jan 2 '18 at 4:00

• The std deviation for percentage/proportion is : $\sigma = \sqrt{p*(1-p)}$ $$\sigma = \sqrt{0.642 * (1-0.642)} = 0.4792$$ Thus when given a percentage, you can directly find the std deviation.

• For back tracking, we know, CI = $p \pm z * \frac{\sigma}{\sqrt{N}}$

For 95%, $z = 1.96$, N = 427, $p=0.642$

$\sigma = ?$

Thus use the above formula and back substitute.

• If your sample size is less than 30 (N<30), you have to use a t-value instead of Z-value. T-value calculator. The T-value has degrees of freedom $df = N-1$ and $prob = (1-\alpha)/2$.

Thus the formula is : CI = $p \pm t_{(N-1)} * \frac{\sigma}{\sqrt{N}}$

• This method uses the central limit theorem and so is only accurate in the limit of large $N$. – Moss Murderer Jan 2 '18 at 1:51
• You are correct, I gave the formula since the question had large sample size > 30. So the CLT is already into effect. For smaller sample size we can use the T-distribution instead of the Z distribution with appropriate degree of freedom. – user3808268 Jan 2 '18 at 2:17

A bit late to the party, but I noticed that the second part of the question was not fully addressed - "can it be apply to percentage measure"?

Following the OPs comment, I am assuming that by "percentage measure" we are referring to some binary outcome (Male/Female, Right handed/Left handed etc.).

In that case the variables are described by a discrete probability distribution, whereas the age is a continuous variable and is described by a continuous probability distribution. A common choice for the distribution of binary variables is the binomial distribution. Confidence intervals for the binomial can be constructed in different ways (wiki). The original study should have described how they derived those confidence intervals.

Note that you can still use the formula provided by user3808268 to get the "standard deviation", but it would be difficult to meaningfully interpret it.