# Why not use the T-distribution to estimate the mean when the sample is large?

Basic statistics courses often suggest using a normal distribution to estimate the mean of a population parameter when the sample size n is large (typically over 30 or 50). Student's T-distribution is used for smaller sample sizes to account for the uncertainty in the standard deviation of the sample. When sample size is large, the sample standard deviation gives good information on the population standard deviation, allowing for a normal-distribution estimate. I get that.

But why use an estimate when you can get your confidence interval exactly? Regardless of sample size, what's the point of using the normal distribution if it's just an estimate of something you can get exactly with the T-distribution?

• @Glen_b Yes, that would be interval estimators. Regarding these intervals: "You must use the t-distribution table when working problems when the population standard deviation (σ) is not known and the sample size is small (n<30)" (from web.pdx.edu/~stipakb/download/PA551/NormalVersusTdistribution.doc). Why don't people use the T-distribution all the time when the population standard deviation is not known (even when n>30)? Commented Aug 17, 2014 at 3:39

Just to clarify on relation to the title, we aren't using the t-distribution to estimate the mean (in the sense of a point estimate at least), but to construct an interval for it.

But why use an estimate when you can get your confidence interval exactly?

It's a good question (as long as we don't get too insistent on 'exactly', since the assumptions for it to be exactly t-distributed won't actually hold).

"You must use the t-distribution table when working problems when the population standard deviation (σ) is not known and the sample size is small (n<30)"

Why don't people use the T-distribution all the time when the population standard deviation is not known (even when n>30)?

I regard the advice as - at best - potentially misleading. In some situations, the t-distribution should still be used when degrees of freedom are a good deal larger than that.

Where the normal is a reasonable approximation depends on a variety of things (and so depends on the situation). However, since (with computers) it's not at all difficult to just use the $$t$$, even if the d.f. are very large, you'd have to wonder why the need to worry about doing something different at n=30.

If the sample sizes are really large, it won't make a noticeable difference to a confidence interval, but I don't think n=30 is always sufficiently close to 'really large'.

There is one circumstance in which it might make sense to use the normal rather than the $$t$$ - that's when your data clearly don't satisfy the conditions to get a t-distribution, but you can still argue for approximate normality of the mean (if $$n$$ is quite large). However, in those circumstances, often the t is a good approximation in practice, and may be somewhat 'safer'. [In a situation like that, I might be inclined to investigate via simulation.]

• I have read somewhere in this document that $n=30$ is good when $\alpha=5\%$. But I'm not sure it is sufficient. Commented Aug 17, 2014 at 10:34
• @StéphaneLaurent For most purposes it should be fine at 5%, but such judgements are very much up to the individual. There are situations - I encountered one only today - where that level of error might be enough to matter. Commented Aug 17, 2014 at 11:28
• @StéphaneLaurent You might get some decent insight from Johnson, V. E. (2013). Revised standards for statistical evidence. Proceedings of the National Academy of Sciences, 110(48):19313–19317. This article fits in the the post-Why most published research findings are false critique of research (a la How Science Goes Wrong) Commented Aug 17, 2014 at 18:27
• @StéphaneLaurent Your article answers my question. For the record, a rough translation of it's conclusion: "The use of the normal distribution as an approximation to Student's t-distribution is exclusively the product of 20th century technological limitations. These limitations have disappeared with modern statistical software, and there is no longer any reason to use these non-conservative approximations". Commented Aug 17, 2014 at 22:04
• @TheThunderChimp Caveat: if population variance is known (e.g. estimating population proportion — mean of a dichotomous variable), then the standard normal (z), and not the t distribution is appropriate. Commented Aug 18, 2014 at 20:12

It's a historical anachronism. There are many of them in statistics.

If you didn't have a computer, it was hard to use the t-distribution, and much easier to use a normal distribution. Once the sample size gets large, they two distributions become similar (how large is 'large' is another question).

• That seems a pretty shallow answer for a deeper question. Commented Aug 17, 2014 at 14:39
• Not sure what you mean. You don't think that's the reason? (The most upvoted answer makes the same point - although more eloquently and elaborately.) Commented Aug 17, 2014 at 18:20
• I downvoted because your answer reads to me like: Because history. Brief recapitulation of your question. Commented Aug 17, 2014 at 18:29
• Thanks for letting me know - it's nicer than an anonymous downvote that I didn't know the reason for. Commented Aug 20, 2014 at 20:20
• Historically, one "used" these distributions by looking up values in tables. The only way in which it would have been any easier to use a Normal distribution would have been that one didn't have to pick the column corresponding to the degrees of freedom. That's scarcely a concern. What did limit usage was that at some point it makes little sense to expand the tables to large degrees of freedom: the books would become too large.
– whuber
Commented Jun 18, 2016 at 15:44

Because in either case (using the normal distribution or the t-distribution), cumulative distribution values are derived numerically (there is no closed form for the integral of $e^{-x^2}$ , or the integral of the t-density). The cumulative distribution function of t distribution with n-degrees of freedom tends to the CDF of a standard normal as $n \rightarrow \infty$. If n is large, the numerical error in approximating the integral is less than the error made by replacing the t-density by the normal density.
In other words, the "exact" t-value is not "exact", and within the approximation error, the value is the same as the CDF value for the standard normal.

• At which sizes do the numerical errors in estimating t outweigh the gains from using it?
– jona
Commented Aug 17, 2014 at 7:24
• surely you can calculate the t-values to arbitrary precision, and so they can be as precise as the quantities you're comparing them to. Commented Aug 17, 2014 at 10:32
• "In other words, the "exact" t-value is not "exact", and within the approximation error, the value is the same as the CDF value for the standard normal." I'm not sure this is a reliable rule of thumb. Commented Aug 17, 2014 at 11:49
• This answer misses the point. As an example, the values of the cumulative Normal distribution and cumulative Student t distribution at $-2$ become indistinguishable in the 16th significant figure (that is, approximately to double precision) only when the sample size exceeds $5.9325 \times 10^{16}$. This indicates that numerical error is not an issue in any practical problem.
– whuber
Commented Jun 18, 2016 at 15:40
• Whuber, you are right. I used "numerical error" improperly. I meant all the errors handling numbers: numerical approximation of the integrals, numerical errors for working with finite precision, and numerical errors due to truncation. If one could work with infinite precision, there would be no justification for replacing the t-distribution with the normal Commented Jun 20, 2016 at 19:14