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Math Help - Chi-Square Test

  1. #1
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    [SOLVED] Chi-Square Test

    Two different teaching procedures were used in two different groups of students. Each group had 100 students of similiar ability. These are the results:

    Group A B C D F
    I 15 25 32 17 11
    II 9 18 29 28 16

    Data is independent from two respective multinomial distribution with k = 5. Test at the 5% level the hypothesis that the two teaching methods are equally effective.

    So let X be the students in group 1, and Y students in group 2.

    So the test statistic would be: Q = \Sigma \frac{(X_i - np_i)^2}{np_i} - \frac{(Y_i - np_i)^2}{np_i}.

    Since it's a multinomial distribution, I know that the expectation is np_i, but how would I find p_i?
    Last edited by statmajor; November 19th 2009 at 05:57 PM.
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  2. #2
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    Quote Originally Posted by statmajor View Post
    Two different teaching procedures were used in two different groups of students. Each group had 100 students of similiar ability. These are the results:

    Group A B C D F
    I 15 25 32 17 11
    II 9 18 29 28 16

    Data is independent from two respective multinomial distribution with k = 5. Test at the 5% level the hypothesis that the two teaching methods are equally effective.

    So let X be the students in group 1, and Y students in group 2.

    So the test statistic would be: Q = \Sigma \frac{(X_i - np_i)^2}{np_i} - \frac{(Y_i - np_i)^2}{np_i}.

    Since it's a multinomial distribution, I know that the expectation is np_i, but how would I find p_i?
    Under the null hypothesis the maximum likelihood estimate of the expected value of cell frequency n_{ij} for a contigency table is \frac{r_i \cdot c_j}{n} where r_i is the row total (for row i) and c_j is the column total (for column j) and n is the total frequency.
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  3. #3
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    Got it; thanks.
    Last edited by statmajor; November 19th 2009 at 05:51 PM.
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