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Math Help - test statistics/critical regions

  1. #1
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    test statistics/critical regions

    Please help me! I am stuck on this chapter and this question seemed to have it all, so its probably the best one to ask. Thank you!


    Let X and Y denote the weights in grams of male and female gallinules. Assume that X is N(\mu_{x},\sigma^{2}_{x}) and Y is N(\mu_{y},\sigma^{2}_{y})


    a) Given n=16 observations of X and m=13 observations of Y, define a test statistic and critical region for testing the null hypothesis H_{0}: \mu_{x}=\mu_{y} against the one-sided alternative hypothesis H_{1}: \mu_{x} > \mu_{y}. Let \alpha=0.01 and assume variances are equal.

    b) Given that \bar{x}=415.16, s^{2}_{x}=1356.75, \bar{y}=347.40 and s^{2}_{y}=692.21, calculate the value of the test statistic and make conclusion

    c) Test whether the assumption of equal variances is valid. Let \alpha=0.05.

    d) Despite the fact that \sigma^{2}_{x}=\sigma^{2}_{y} is accepted in part (c), let us say we suspect the equality is not valid. Thus use the test proposed by Welch.


    Thank you for even looking! I am ready to tear my hair out trying to understand this entire chapter. I hope someone out there can help me!
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  2. #2
    Super Member Random Variable's Avatar
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    a) The test statistic is  T = \frac {\bar{X} - \bar{Y}}{S_{p} \sqrt{1/n + 1/m}}

    where  S_{p} = \sqrt{\frac{(n-1)S^{2}_{X} + (m-1)S^{2}_{Y}}{n+m-2}}

    T follows a t distribution with n+m-2 degrees of freedom

    For this problem the critical region is  t \ge t_{(0.01, 27)}


    c) The alternative hypothesis is  H_{1}: \sigma^{2}_{X} \neq \sigma^{2}_{Y}

    The test statistic is  F = \frac {S^{2}_{X}}{S^{2}_{Y}}

    F follows an F distribution with  r_{1} = n-1 and  r_{2}=m-1 degrees of freedom respectively

    The critical region is  f \ge F_{(0.025, 15,12)} or  \frac {1}{f} \ge F_{(0.025,12,15)}


    I have to review how to do part (d).
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  3. #3
    MHF Contributor matheagle's Avatar
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    a-c look fine, d is just Satherwaite's approximation.

    You use as your test stat

     T = \frac {\bar{X} - \bar{Y}}{\sqrt{S_X^2/n + S_Y^2/m}}

    The NASTY degrees of freedom (which is not an integer) can be found under...
    Unequal sample sizes, unequal variance at http://en.wikipedia.org/wiki/Student's_t-test
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  4. #4
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    I calculated the test statistic using the formula you provided and got 5.767 and using the degrees of freedom formula, I got r=26.628. I did some searching through the link you sent and it seems I have to use these two values for yet another test, but I can't seem to find an equation. Do you have any idea what equation that might be? Thank you very much.
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  5. #5
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    I have a couple more questions. Can the critical region in part a be found on the t test table? If so it looks like 2.473 if I'm not mistaken.

    Also, how would I look up the f critical regions?

    Thank you very VERY much
    Last edited by madgab; June 27th 2009 at 04:36 PM. Reason: another question
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  6. #6
    Super Member Random Variable's Avatar
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    If so it looks like 2.473 if I'm not mistaken.
    Yes, that is correct.

    Also, how would I look up the f critical regions?
    F-Distribution Tables

     F_{(0.025, 15,12)} = 3.1772
     F_{(0.025,12,15)} = 2.9633
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  7. #7
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    Thank you. I'm sorry to bother you again. I calculated the value of the test statistic as 5.5071 but I don't know what conclusion this gives. I got the critical region of 2.473. so what does 2.473<5.5071 mean? or do I have even more calculations for this problem. i keep confusing myself with all the different values.
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  8. #8
    Super Member Random Variable's Avatar
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    Quote Originally Posted by madgab View Post
    Thank you. I'm sorry to bother you again. I calculated the value of the test statistic as 5.5071 but I don't know what conclusion this gives. I got the critical region of 2.473. so what does 2.473<5.5071 mean? or do I have even more calculations for this problem. i keep confusing myself with all the different values.
    It makes more sense to call the critical region the "rejecting region."

    The value you calculated for t (5.5071) falls in that region (t > 2.473). That means that there is enough evidence at the  \alpha = 0.01 significance level to reject the null hypothesis in favor of the alternative hypothesis.

    If your calculated value had not fallen in the rejecting region, then there would not have been enough evidence at that  \alpha level to reject the null hypothesis.
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  9. #9
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    Thank you! That makes sense. And part c ended up showing that the assumption of equal variances IS valid. I think I have all the conclusions required. Thanks again
    Last edited by madgab; June 28th 2009 at 11:13 AM. Reason: fixed errors
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