I would very much appreciate if someone can give me an idea of how to solve this exercise.

I need to show that the t-test and the z-test are equivalent when $\nu$ (degrees of freedom) tends to infinity.

So far, I have the probability distribution for the T test given by:

$$f(t, ν) = \frac{1}{\sqrt{\nu\ \cdot \pi}}\cdot\frac{Γ[\frac{(\nu + 1)}{2}]}{Γ[(\nu/2)]}\left(1+ \frac{t^2}{\nu}\right)^{-\frac{\nu+1}{2}}$$

I took the limit and after some tricky calculations I got the Gaussian pdf.

Do not know how to proceed from now.

  • 1
    $\begingroup$ It's not 100% clear if this is what the question you refer to is asking, but (aside from the distribution you look up) what is the difference between a z test and a t-test? $\endgroup$
    – Glen_b
    Oct 17 '16 at 18:19
  • $\begingroup$ @Glen_b First, I had to prove that the T-test distribution given above, tends to the Gaussian distribution in the limit of a large number of degrees of freedom. And then , the next part is to show that the t-test and z test are equivalent as the degrees of freedom tends to infinity. Apparently the main difference between them both is that you must know the standard deviation of the population and your sample size must be above 30 in order for you to be able to use the z-score. Otherwise, we use the t-score. But still, I guess this is not helping too much in having an idea to solve it. $\endgroup$
    – GM Andrea
    Oct 17 '16 at 18:27
  • 1
    $\begingroup$ @GMAndrea I took the liberty of editing your post to improve mathJax formatting, but it appears to still be missing some brackets. As I don't want to ambiguously change anything, could you revise it? Thanks. $\endgroup$
    – Firebug
    Oct 17 '16 at 18:35
  • $\begingroup$ @GMAndrea So if you can show that the sample standard deviation converges to the population standard deviation, would that not establish that they go to the same value of the test statistic (t-stat -> z-stat) being evaluated on the same distribution (t-dist -> z-dist) - so that in the limit the two tests are the same? $\endgroup$
    – Glen_b
    Oct 17 '16 at 18:54
  • $\begingroup$ Thank you @Glen_b ! I already put the extra brackets. Could you please show me how to start? It's like I still do not get it $\endgroup$
    – GM Andrea
    Oct 17 '16 at 19:43

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