A researcher computes a 2 x 3 chi-square test for independence. What is the critical value for this test at a.05 level of significance?
Please let me know in case you've doubts. Here' the answe below.
Lets solve this in 2 ways.
First by using the Chi-Square cumulative distribution table.
1.
The cumulative area is 0.95 ( 1-.05 = .95)
The Chi-Square can be got from the ChiSquare table. Open it look for df = (k1-1)*(k2-1), i.e. (2-1)*(3-1) = 1*3 = 3
cumualtive probability = .95.
So, look for df = 3 ( df1 = 1, df2 = 2), with probability = 0.95
It is Chi-Square is 7.8147
The critical value of 7.8147
2.
The cumulative area is 0.95 ( 1-.05 = .95)
The Chi-Square = CHISQ.INV(0.95, 3)
X^2 = 7.8147
The critical value of 7.8147
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