Suppose that a hypothesis test on a 64 table is carried out

Suppose that a hypothesis test on a 6×4 table is carried out at an =0.043 level of significance. If the p-value for the test is p = 0.019, then what is the value of the test statistic X^2?

Solution

Here, for an a x b table, the number of degrees of freedom is

df = (a-1)(b-1) = (6-1)*(4-1) = 15

Then, as P = 0.019, then by table/technology,

chi^2 = 28.43489288 [ANSWER]