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Math Help - Chi-Square Test

  1. #1
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    Chi-Square Test

    Let the result of a random experiment be classified as one of the mutually exclusive and exhaustive ways A1, A2, A3 and also one of the mutually exclusive and exhaustive ways B1, B2, B3, B4. 200 independent trials of the experiment result in the following data:

    B1 B2 B3 B4
    A1 10 21 15 6
    A2 11 27 21 13
    A3 6 19 27 24

    Test at 0.05 significance level the hypothesis of independence. Null hypothesis H_0 : P(A_i \cap B_j) = P(A_i)P(B_j) against the alternate hypothesis (dependence exists)

    Would the test statistic be:

    Q(X) = \Sigma \Sigma \frac{(X_{ij} - \frac{C_j R_i}{n})^2}{\frac{C_j R_i}{n}}
    where Cj = the sum of column j, and Ri = sum of row i

    so for X_{11} = \frac{(10 - \frac{(27)(52)}{200})^2}{\frac{(27)(52)}{200}}

    and so forth?
    Last edited by statmajor; December 2nd 2009 at 08:04 AM.
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  2. #2
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    Quote Originally Posted by statmajor View Post
    Let the result of a random experiment be classified as one of the mutually exclusive and exhaustive ways A1, A2, A3 and also one of the mutually exclusive and exhaustive ways B1, B2, B3, B4. 200 independent trials of the experiment result in the following data:

    B1 B2 B3 B4
    A1 10 21 15 6
    A2 11 27 21 13
    A3 6 19 27 24

    Test at 0.05 significance level the hypothesis of independence. Null hypothesis H_0 : P(A_i \cap B_j) = P(A_i)P(B_j) against the alternate hypothesis (dependence exists)

    Would the test statistic be:

    Q(X) = \Sigma \Sigma \frac{(X_{ij} - \frac{C_j R_i}{n})^2}{\frac{C_j R_i}{n}}
    where Cj = the sum of column j, and Ri = sum of row i

    so for X_{11} = \frac{(10 - \frac{(27)(52)}{200})^2}{\frac{(27)(52)}{200}}

    and so forth?

    \begin{array}{cccccc}\ &B1&B2&B3&B4&total\\<br />
A1&10&21&15&6&52\\<br />
A2&11&27&21&13&72\\<br />
A3&6&17&27&24&76\\<br />
\hline<br />
Total&27&67&63&43&200\end{array}

    \chi^2 = \frac{200}{52}(\frac{10}{27}+\frac{21}{67}+\frac{1  5}{63}+\frac{6}{43})+\frac{200}{72}(\frac{11}{27}+  \frac{27}{67}+\frac{21}{63}+\frac{13}{43})+\frac{2  00}{76}(\frac{6}{27}+\frac{17}{67}+\frac{27}{63}+\  frac{24}{43})



    \chi^2 = 11.9

    Degree of dependency, \nu = (h-1)(k-1)=(3-1)(2-1)=2

    P(\chi^2>11.9)=0.997


    Since the P-value (0.997) is larger than the significance level (0.95), we cannot accept the null hypothesis. Thus, we conclude that there is a relationship between A and B.
    Last edited by novice; December 2nd 2009 at 11:37 AM. Reason: change < to > on last line
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