Solved – Calculate standard deviation from sample size, mean, and confidence interval

confidence intervaldescriptive statisticsmathematical-statisticsstandard deviationstandard error

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).

Best Answer

  • The standard deviation for percentage/proportion is:
    \begin{align} \sigma &= \sqrt{p(1-p)} \\[5pt] &= \sqrt{0.642(1-0.642)} \\[5pt] &= 0.4792 \end{align} 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 ${\rm prob} = (1-\alpha)/2$.

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

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