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Math Help - chi-square and normal distribution

  1. #1
    lpd
    lpd is offline
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    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.
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  2. #2
    MHF Contributor matheagle's Avatar
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    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}
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