# chi-square and normal distribution

• Oct 24th 2010, 06:20 PM
lpd
chi-square and normal distribution
Hi. I have this problem.

Making use of the fact that the chi-square distribution can be approximated by
a normal distribution when $v$, the number of degrees of freedom, is large,
show that for large samples from a normal population,

$s^2 \ge \sigma_0^2[1+z_{\alpha}\sqrt{\frac{2}{n-1}}]$

is an approximate critical region of size $\alpha$ for testing $H_0:\sigma^2=\sigma_0^2$ against $H_1:\sigma^2> \sigma^2$

Thanks.
• Oct 24th 2010, 10:32 PM
matheagle
The statement that the chi-square distribution can be approximated by a normal doesn't make sense.
The chi-square is strictly positive, while the normal isn't.
I assume you mean that you can apply the central limit theorem since $S^2$ is an average.
By that I would use

${S^2-\sigma^2_0\over \sqrt{V(S^2)}}\approx N(0,1)$

YUP, this works....

$V(S^2)}={2\sigma^4\over n-1}$

So ${S^2-\sigma^2_0\over {\sqrt{2}\sigma^2_0\over \sqrt{n-1}}}>z_{\alpha}$