Using the Student's T test, T-Critical is calculated via:

$t = \frac{\bar{X} - \mu_{0}}{s / \sqrt{n}}$

Looking at Wikipedia article on the unbiased Estimation of the standard deviation, there section Result for the normal Distribution which mentions a correction factor $c_4(n)$ for the sample's measured standard deviation, s, based on the sample size. Questions:

(1) Is this correction factor included in the Student's T table data since it is by degrees of freedom?

(2) If (1) is no why not?


1 Answer 1


1) No it isn't.

2) because the calculation of the distribution of the test statistic relies on using the square root of the ordinary Bessel-corrected variance to get the estimate of standard deviation.

If it were included it would only scale each t-statistic - and hence its distribution - by a factor (a different one at each d.f.); that would then scale the critical values by the same factor.

So, you could, if you like, construct a new set of "t"-tables with $s*=s/c_4$ used in the formula for a new statistic, $t*=\frac{\overline{X}-\mu_0}{s*/\sqrt{n}}=c_4(n)t_{n-1}$, then multiply all the tabulated values for $t_\nu$ by the corresponding $c_4(\nu+1)$ to get tables for the new statistic. But we could as readily base our tests on ML estimates of $\sigma$, which would be simpler in several ways, but also wouldn't change anything substantive about testing.

Making the estimate of population standard deviation unbiased would only make the calculation more complicated, and wouldn't save anything anywhere else (the same $\bar{x}$, $\overline{x^2}$ and $n$ would still ultimately lead to the same rejection or non-rejection). [To what end? Why not instead choose MLE or minimum MSE or any number of other ways of getting estimators of $\sigma$?]

There's nothing especially valuable about having an unbiased estimate of $s$ for this purpose (unbiasedness is a nice thing to have, other things being equal, but other things are rarely equal).

Given that people are used to using Bessel-corrected variances and hence the corresponding standard deviation, and the resulting null distributions are reasonably straightforward, there's little - if anything at all - to gain by using some other definition.

  • 1
    $\begingroup$ Thanks for the thoughtful answer. Small sample statistics are so imprecise that it didn't seem that the correction would magically fix the small sample problem. $\endgroup$
    – MaxW
    May 26, 2015 at 16:51
  • 1
    $\begingroup$ Sorry, I'm not sure what you're getting at there. The $c_4$ term simply leads to an estimator where $E(\hat\sigma)=\sigma$ under normal sampling. Which small sample problem are we talking about (presumably, since the question is about the t-test there's some problem with the $t$ I missed?), and how was it 'fixed'? $\endgroup$
    – Glen_b
    May 26, 2015 at 18:33
  • $\begingroup$ Sorry to be obtuse. I was referring to situation in which one has taken 3 measurements and calculated mean, std. dev. and 95% confidence interval based on that data. With such a small sample the confidence interval is huge. Making the standard deviation correction isn't going to magically change the accuracy and precision of a 3-shot sample significantly. $\endgroup$
    – MaxW
    May 26, 2015 at 23:21
  • $\begingroup$ Ah. Thanks, I see now. You're correct that nothing magic happens; with small samples the confidence interval will be wide, and scaling a statistic doesn't impact any of that at all; indeed we can show that formally via the pivotal quantities used to construct the usual confidence intervals. $\endgroup$
    – Glen_b
    May 27, 2015 at 0:03

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