I am learning elementary statistics.

I found an exercise, which asks to compute the desired sample size for some interval for standard error.

The solution, in class slides, first assumes the sample size to be computed is big enough to be approximated by a normal distribution (central limit theorem), and uses it to get an answer that matches the assumption — the sample size is big enough.

Is this is a sound answer? I don't think so; it may be a wrong assumption first, and that result deduced from it is not qualified to do any validation, is it?

If it is not the case, how can we prove that assumption first? Or put another way,if I want to compute a desirable sample size, is it that you can never allow using the central limit theorem on that answer/question?

*I have no teacher to ask because I'm self-studying.

edit I think the post is not quite clear, the C.T.L. is used upon a random variable that is the sum of all samples to be estimated.

  • $\begingroup$ It's a good question. But let me ask you this, to provoke some reflection on the issue: if you are unwilling to estimate a sample size using a Normal approximation, what method do you propose to use instead for estimating the sample size? $\endgroup$
    – whuber
    Jan 15, 2014 at 15:40

3 Answers 3


The normal approximation is not some 'on/off' thing - it improves slowly as sample size increases and gets slowly worse as it decreases.

Which is to say, when you assume it, though it may lead you to a wrong sample size, if the sample size it leads you to is large enough*, the normal approximation at any nearby sample size should also be reasonable. If the sample size it leads you to is small, the normal approximation may be poor.

* I don't know that you would yet have a basis to conclude that in general; simulation will generally be a useful way to check

That is, if you're a little bit conservative about invoking the calculated sample size as a reason to use the normal-assumption value you calculated, it will usually work quite well.

When in doubt, you can sometimes use algebra or simulation to check how things work at the sample size you calculate; it depends on the exact details of the situation.

To my mind some of the other assumptions are more likely to cause you problems than that one.

  • $\begingroup$ OK! thanks your reply. I think I somewhat mistaken the exercise as asking for a true lower bound(it actually asks only for 'at least'). and by that reason I just worried anxiously that the wrong assumption may lead me to too large size answer. By the way, I'm curious what other assumption are more likely to cause problem than this??? $\endgroup$ Jan 17, 2014 at 1:54
  • $\begingroup$ Leaving aside the normal approximation issue, the use of 'at least' is because a larger sample size will lead to a smaller expected interval size, which should also be sufficient. That is, it wants the minimum sample size that will achieve the probabilistic bound. More details about what standard error you were calculating would allow a more complete answer. $\endgroup$
    – Glen_b
    Jan 17, 2014 at 8:57
  • $\begingroup$ Oh, assumptions that tend to cause bigger problems include independence & equality of variance $\endgroup$
    – Glen_b
    Jan 17, 2014 at 9:00

In real life, although the actual distribution is unknown, there are some hints if the normal approximation ("the $t$-statistic is $t_n$-distributed as if the data were normally distributed") is reasonable at moderate sample sizes or not.

I assume you had to find the sample size $n$ for a $1-\alpha$-confidence interval for the parameter $E(X_i), i=1,\ldots,n$ that had to be $a\sqrt{E(X_i -E(X_i))^2}$ large.

If the true distribution is not much apart from the normal distribution (in particular quite symmetric) and $\alpha$ is large enough, you can proceed the way your textbook solution suggested. In this case the normal approximation is pretty fast. Just simulate some distributions and watch how far the actual coverage of $\mu$ is missed by the $t$-based CI. This is also statistician's experience.

If the true distribution is quite right-skewed, you can take the Berry-Esseen-Bound to adjust for possible liberalness due to nonormality and small sample size. Your conservative (i.e. $n$ may be too large, true coverage probability may be larger than $1-\alpha$) solution would then be the minimum integer $n$ solving $$ 2 t_{n-1,1-\frac{\alpha}{2}+\frac{C\rho}{\sigma^3 n}}\leq a\sigma$$

  • $\begingroup$ THANKS for reply!! Yes, I am talking about a sample size for given confidence level. However your answer surpass my current level, so I am not quite understand about. The Berry-Esseen-Bound, I first time known of, has me learned that there is such bound for CLT. To me it's a cool, beautiful bound. $\endgroup$ Jan 17, 2014 at 2:22

It's only a problem if you got a sample size smaller than what is required to assume for normality.

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    $\begingroup$ This isn't really an answer by CV's standards, @SaraWong123. Although in a sense it does provide an answer to the OP's question, it is still more of a comment than an answer at present. Would you mind elaborating it into more of an answer? Alternatively, we can convert it into a comment to the question. $\endgroup$ Jan 15, 2014 at 17:27
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    $\begingroup$ You have a good point here, Sara. (BTW, welcome to our site!) It seems to me though that estimating too large a sample size is a problem, too, because it can lead to spending more time and money than needed and can even cause the investigation not to be performed. $\endgroup$
    – whuber
    Jan 15, 2014 at 18:47
  • $\begingroup$ (in my knowledge normality test is to test if it is normal)normality test is for a given samples has been available, but my question is to estimate how many sample will must be. $\endgroup$ Jan 16, 2014 at 1:43
  • $\begingroup$ Sometimes when you use one of these formulas, you will get a sample size that results in a decimal, like n=40.2. But, we can't have a sample size that's a non-integer value, so we round up if we want to get a confidence interval as tight or tighter than what is asked for. But, the formula didn't violate logic in giving us something that we couldn't actually carry out.. In the same sense, if you get an 'n' which is too small, you have to round up to a point where you can safely assume the sample mean will behave 'normally'. $\endgroup$ Jan 16, 2014 at 2:29

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