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I have an exercise with which I'm currently struggling:

Consider a random sample $\big\{X_n\big\}_{n=1}^N$ from a normal distribution having mean $\mu$ and variance $\sigma^2$. What is the distribution of $\frac{\sum_{n=1}^N (X_n - \overline{X}_N)^2}{\sigma^2}$?

Could anyone give me some hints in the right direction on how to achieve this?

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    $\begingroup$ Do you know anything about the distribution of the sample variance? $\endgroup$
    – mark999
    Commented Nov 8, 2016 at 7:40
  • $\begingroup$ Not quite. However, as far as i've seen during my reading on wikipedia, the variance of $\overline{X}_N$ should be $\frac{s}{\sqrt{N}}$ with $s$ the sample standard deviation... $\endgroup$
    – raphael
    Commented Nov 8, 2016 at 7:50
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    $\begingroup$ Are you studying from a text book or from lecture notes? If so, there's probably something there about the distribution of (a multiple of) $S^2$. $\endgroup$
    – mark999
    Commented Nov 8, 2016 at 7:55
  • $\begingroup$ Hmm, just required work ahead of class.So there are basically no lecture notes nor a book... $\endgroup$
    – raphael
    Commented Nov 8, 2016 at 7:57
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    $\begingroup$ "Required work" sounds like you should have enough information to answer the question. Otherwise, have a look at en.wikipedia.org/wiki/… $\endgroup$
    – mark999
    Commented Nov 8, 2016 at 8:05

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mark999 has cracked it. The numerator is the standard error. If you have a look at the wiki chi squaredyou will see that formula for the chi squared distribution, which is the same as the formula you have given, if you substitute the formula for the standard error in for the numerator. I think your answer is chi square distribution

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