2. Do Exercise 4.7.6, but change the value 10 in the upper left corner to 14 and

Question

2. Do Exercise 4.7.6, but change the value 10 in the upper left corner to 14 and the value 24 in the lower right corner to 20. Explicitly show the calculations involved in the test.

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

Test, at the 0.05 significance level, the hypothesis of independence of the A attribute and the B attribute, namely, H0 : P(Ai Bj) = P(Ai)P(Bj), i = 1,2,3 and j = 1, 2, 3, 4, against the alternative of dependence.

Explanation / Answer

Given table data is as below

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calculation formula for E table matrix

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expected frequecies calculated by applying E - table matrix formulae

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calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above

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set up null vs alternative as

null, Ho: attribute A and B are independent

alternative, H1: attribute A and B are dependent

level of significance, = 0.05

from standard normal table, chi square value at right tailed, ^2 /2 =12.592

since our test is right tailed,reject Ho when ^2 o > 12.592

we use test statistic ^2 o = (Oi-Ei)^2/Ei

from the table , ^2 o = 12.941

critical value

the value of |^2 | at los 0.05 with d.f (r-1)(c-1)= ( 3 -1 ) * ( 4 - 1 ) = 2 * 3 = 6 is 12.592

we got | ^2| =12.941 & | ^2 | =12.592

make decision

hence value of | ^2 o | > | ^2 | and here we reject Ho

^2 p_value =0.044

ANSWERS

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null, Ho: attribute A and B are independent

alternative, H1: attribute A and B are dependent

test statistic: 12.941

critical value: 12.592

p-value:0.044

decision: reject Ho

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