For a contingency table with two rows and two columns and frequencies of a and b in the first row and frequencies of c and d in the second row, verify that the test statistic becomes

Where

and

is such that z2 = x2 [the same result as in part (a)|. (This result shows that the chi-square test involving a 2 X 2 table is equivalent to the test for the difference between two proportions, as described in Section 8 - 6.)

a+b+c dad- bc) (a+ bc + d)b + d)la + c) x2 a + b a + b c+d q=1-p