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I am currently taking an introductory statistics course on Udacity and I'm having trouble understanding the concept of the t-statistic. My current understanding of the t-Statistic is as follows:

If the population standard deviation is unknown, we can't calculate the Z-score. So, instead, we calculate a very similar statistic (i.e the t-score) using the sample Standard deviation in place of the population standard deviation. Is Bessels correction used when the sample standard deviation is used to estimate the population standard deviation?

Also, I've seen two forms of the equation for calculating the t-statistic.

$$ t=\frac{\overline{x}- \mu _{0}}{S/ \sqrt{n} } $$

and

$$ t=\frac{\overline{x}- \mu _{0}}{S/ \sqrt{n-1} } $$ mu = population mean x = sample mean n = sample size S = sample Standard deviation Which on of these is correct?

mahela007
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1 Answers1

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You want $$ t=\frac{\bar{x}-\mu_0}{S/\sqrt{n}} $$ since if $x_1,\ldots,x_n$ are i.i.d normally distributed with mean $\mu_0$ and variance $\sigma^2$, then $t$ follows a $t$-distribution with $n-1$ degrees of freedom which enables you calculate $p$-values and confidence intervals. Note, however, that the factor $n-1$ appears within the sample standard deviation $S$ as $$ S=\sqrt{\frac{1}{n-1}\sum_{i=1}^n(x_i-\bar{x})^2}. $$ This is to ensure that ${\rm E}[S^2]=\sigma^2$, i.e. that the estimator of the variance $S^2$ is unbiased.

Stefan Hansen
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  • Thanks for the reply. Does this mean that the second formula for t that I have given is incorrect?

    What does i.i.d mean?

     – mahela007 Jul 24 '13 at 04:37
  • @mahela007: Yes, the second isn't used for statistical tests. I.i.d. means independent and identically distributed. – Stefan Hansen Jul 24 '13 at 08:56
  • hm.. I came across this web page about the t-score. It appears that they do no use Bessel's correction when approximating the population SD. (Scroll down to the section titled "Solution A") http://stattrek.com/probability-distributions/t-distribution.aspx

    Am I correct in assuming that they haven't used Bessel's correction? If so, is it just a mistake or is there some reason for doing so?

     – mahela007 Jul 26 '13 at 12:43
  • What makes you think that? They nowhere state hos the estimated standard deviation is calculated, do they? – Stefan Hansen Jul 26 '13 at 13:16
  • they say the standard deviation of the sample is 50 days for a sample of 15 bulbs.. 50 / 15 is used in the equation to calculate the t-score. (instead of 50 / (15-1) ) – mahela007 Jul 26 '13 at 14:30
  • The standard deviation is $S=50$ and hence $S$ is already calculated. So you can't really know if they have used $n$ and $n- 1$ in the formula for $S$. Bessel's correction is the use of $n-1$ instead of $n$ in the formula for $S$ (second equation in my answer) and has nothing to do with the $\sqrt{n}$ when calculating $t$. Furthermore, you have a typo. They use $50/\sqrt{15}$ and not $50/15$ in the answer. – Stefan Hansen Jul 26 '13 at 15:36
  • Ok.. I understand now. :D Thanks for your help – mahela007 Jul 27 '13 at 15:43