A two-by-two table is a two-proportion test

Topic: Chi-squared tests · more questions on this topic
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Of 70 viewers of one channel, 40 are left-wing; of 100 viewers of another, 50 are. Compute the Pearson statistic for independence of channel and allegiance, and check that it equals the square of the pooled two-proportion z statistic, 0.918.

Worked answer (click to show)

The table and its fitted counts R_iC_j/170 are

	Left	Not left	Total
Channel 1	40 (37.06)	30 (32.94)	70
Channel 2	50 (52.94)	50 (47.06)	100
Total	90	80	170

Every cell differs from its fitted count by 2.94 in absolute value, so

T=2.94^2\Bigl(\frac1{37.06}+\frac1{32.94}+\frac1{52.94}+\frac1{47.06}\Bigr)=0.843.

The pooled two-proportion statistic is z=0.918, and 0.918^2=0.843. The two statistics agree, and \chi^2_1 is the law of the square of a standard normal. The chi-squared p-value, 0.36, is the two-sided z test’s; the one-sided z test halves it.