Assuming the population standard deviation σ = 3, how large should a sample be to estimate the population mean µ with a margin of error not exceeding 0.5?

My answer:

SE<= 0.5=3/√n ---> n=36

But probably the correct answer is:

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So, can anyone explain more about the second solution?

  • 2
    $\begingroup$ Hint: consider where the "$1.96$" came in and notice that your answer doesn't use this factor at all. Why not? $\endgroup$
    – whuber
    Oct 31, 2018 at 18:53
  • 2
    $\begingroup$ The difference come from the explanation of "a margin of error". You explain the a margin of error as SE (standard error). The second explanation of a margin of error is the half of the width of the 95% CI (confidence interval) which is 1.96*SE. $\endgroup$
    – user158565
    Oct 31, 2018 at 18:55

1 Answer 1


A 95% confidence interval for normal population $\mu,$ where the SD is $\sigma,$ based on a sample of size $n$ is of the form $\bar X \pm 1.96\sigma/\sqrt{n}.$ As in @statistician's Comment, the 'standard error' of $\bar X$ is $SD(\bar X) = \sigma/\sqrt{n}.$ And the 'margin of error' of the confidence interval is $\Delta = 1.96\sigma/\sqrt{n}.$ You know $\sigma = 3$ and you want to find $n$ so that $\Delta = 0.5$ (or perhaps a bit smaller).

So solve $\Delta = 1.96\sigma/\sqrt{n}$ for $n$ in terms of $\Delta$ and $\sigma.$ That gives the equation $n = (1.96\sigma/\Delta)^2,$ to which you refer in your question. So $n \approx 11.76^2 = 138.4.$ Rounding up to the nearest integer, you get $n = 139$ as the sample size that gives $\Delta$ just a bit smaller than the required 0.5.

Note: Your solution $n = 36$ gives SE$ =3/\sqrt{36} = 0.5,$ equivalent to $\Delta = 1.96(0.5),$ but not to $\Delta = 0.5.$ Roughly speaking, to cut $\Delta$ in half requires a four-fold increase in $n,$ and dividing by 1.96 is almost cutting $\Delta$ in half, which explains why $4(36) = 144$ is not far from the correct answer $n = 139.$

  • $\begingroup$ (+1), I just want to add that this is an estimation only. It is entirely possible to collect 139 samples (or even more) and still not it your intended margin of error. $\endgroup$ Oct 31, 2018 at 21:31
  • $\begingroup$ Because $\sigma$ is known, the width of the CI is determined by $\alpha$ and $n,$ but of course not 100% of the CIs will cover $\mu.$ $\endgroup$
    – BruceET
    Oct 31, 2018 at 23:35
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    $\begingroup$ Fair point, but in practice that assumption doesn't always hold, yet people will use this approach to estimate the number of samples required (e.g. in simulation). Just wanted the caveat here for practitioners. Thanks for clarifying. $\endgroup$ Nov 1, 2018 at 3:10

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