Chi Square Distribution Problem

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Nexttime35
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Homework Statement


Suppose that X has normal distribution. Find the distribution of n(Sample Mean - μ)22.

The Attempt at a Solution


I honestly have no idea where to begin with this problem. Any ideas?
 
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Well I know that the Sample mean has a normal distribution ~ N(μ,σ2/n), which I think is useful to solve this problem. Now, I am confused about how to use this normal distribution for the sample mean to solve the problem. Any thoughts, using this idea?
 
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LCKurtz said:
If ##\bar X \sim N(\mu, \frac {\sigma^2}{n})## what is the distribution of$$
Z=\frac{\bar X -\mu}{\frac\sigma {\sqrt n}}$$Once you answer that, you need to work out the distribution of ##Z^2##.
 
Hi LCKurtz,

Thanks for the help. I am wondering where you got this equation from? I know it's the z-score, converting the sample mean into the z-score, but how did you come up with the equation? Thanks.
 
Actually I believe I know where this comes from.

the Sum from i=1 to n of Zi2 = the Sum from i=1 to n of ((Xi-μ)/σ)2 = chi square distribution with n degrees of freedom.
 
So, would the distribution of this be N(0,n)?
 
Nexttime35 said:

Homework Statement


Suppose that X has normal distribution. Find the distribution of n(Sample Mean - μ)22.

Nexttime35 said:
Hi LCKurtz,

Thanks for the help. I am wondering where you got this equation from? I know it's the z-score, converting the sample mean into the z-score, but how did you come up with the equation? Thanks.

Now you are confusing me. It is your question asking for the distribution of$$
\frac {n(\bar X - \mu)^2}{\sigma^2} =\frac{(\bar X - \mu)^2}{(\frac \sigma {\sqrt n})^2}$$
So if ##Z = \frac{(\bar X - \mu)}{\frac \sigma {\sqrt n}}##, you know ##Z\sim n(0,1)## and your problem is to find the distribution of ##Z^2##.