Question

The following table contains observed frequencies for a sample of 200.

	Column Variable
Row Variable	A	B	C
P	20	44	50
Q	30	26	30

Test for independence of the row and column variables using a= .05

Compute the value of the  test statistic (to 2 decimals).

Answer 1

The following cross-tabulation have been provided. The row and column total have been calculated and they are shown below:

	A	B	C	Total
P	20	44	50	114
Q	30	26	30	86
Total	50	70	80	200

The expected values are computed in terms of row and column totals. In fact, the formula is , where Ri​ corresponds to the total sum of elements in row i, Cj​ corresponds to the total sum of elements in column j, and T is the grand total. The table below shows the calculations to obtain the table with expected values:

Expected Values	A	B	C	Total
P				114
Q				86
Total	50	70	80	200

Based on the observed and expected values, the squared distances can be computed according to the following formula: (E - O)^2/E. The table with squared distances is shown below:

Squared Distances	A	B	C
P
Q

Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

H_0H0​: The two variables are independent

H_aHa​: The two variables are dependent

This corresponds to a Chi-Square test of independence.

Rejection Region

Based on the information provided, the significance level is α = 0.05 , the number of degrees of freedom is df = (2 - 1)*(3 - 1) = 2.

Test Statistics

The Chi-Squared statistic is computed as follows:

Decision about the null hypothesis

Since it is observed that = 7.863 > = 5.991, it is then concluded that the null hypothesis is rejected.

Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the two variables are dependent, at the 0.05 significance level.