Standard Normal and T-Distribution

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Is the attached statistically sound? I want to use the t-distribution's moments to infer on a standard normal distribution. I am saying if Z is standard normal then Z also follows a t-distribution with n - 1 degrees of freedom.

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Dason

I am saying if Z is standard normal then Z also follows a t-distribution with n - 1 degrees of freedom.
I didn't read the attachment but this quote is about as correct as saying "if x=10 then x also equals 11"

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I thought as much Dason. If possible may you go over the attachment just to confirm.

Archidamus

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The T-distribution converges in distribution to a Z as the degrees of freedom(N) approaches infinity. So long as your N is massive you can use the T-dist as a substitute. But I have found the moment generating function of the normal to be much easier to work with than the T; not sure what algebraic advantage it gives you.

I have to disagree with the last paragraph. The Standard Normal distribution does not depend on mu and sigma, because they are already known as 0 and 1. A major practical advantage of the T-Dist over the Z is in situations with few observations and/or the population variance is unknown. I like to think of the T-Dist as a 'safer' version of Z, as it has wider tails to account for lower observations and unknown variance.