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I know that the sum of square of normal random variables follow a chi-square distribution. But when I learn how to do a goodness-fit test I don't know why the ratio of (O-E)^2/E follows a chi-square. I tried to square root of it first so that I might get something looks like a normal, but my new question arises : why (O-E)/sqrt(E) follows a normal-distribution?

I know from sampling distribution that if the sample is from the same distribution as the population then the sampling distribution of sample mean is approximately normal if n is large according to CLT, are those two concept related in some way?

Really struggle with this, appreciate for any comments and answer!!

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  • $\begingroup$ You'll need to be more specific under what conditions you've seen (O-E)^2/E. Usually that formula is used with discrete distributions where O is a count. Your first sentence discusses a continuous random variable. $\endgroup$ – JimB Aug 1 at 2:52

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