If the two samples are drawn from different populations, we may be interested in finding out whether the difference between the proportion of successes is significant or not. In such a case we take the hypothesis that the difference between p1, i.e. the proportion of successes in other sample, is due to the fluctuations of random sampling.

The standard error of the difference between the proportions is calculated by applying the following formula:


Where p = the pooled estimation of the actual proportion in the population. The value of p is obtained as shown:

P = (n1p1 + n2p2)/(n1 + n2)    or    p = (x1 + x2)/(n1 + n2)

Where x1 and x2 stands  for the number of occurrences in the two samples of sizes n1 and n2 respectively.

If (p1 - p2)/S.E. is less than 1.96 S.E. (5% level of the significance), the difference is regarded as due to the random sampling variation, that is  as not significant.

Illustration: In a random sample of 1000 persons from town A, 400  persons are found to be consumers of Rice. In a sample of 800 from town B, 400 persons are found to be consumers of Rice. Do these data reveal a significant difference between town A and town B, so far as the proportion of Rice consumer is concerned?

Solution: let us set up the hypothesis that the two towns do not differ so far as a proportion of rice consumers is concerned,

i.e. H0 : p1 = p2 against Ha : p1 ≠ p2.

On computing the standard error of the difference of proportions:


n1 = 1000, p1 = 400/1000 = 0.4; n2 = 800; p2 = 400/800 = 0.5

p = (1000 × 0.4) + (800 × 0.5)/(1000 + 800) = 400 + 400/1800

simply p = (x1 + x2)/(n1 + n2) = (400 + 400)/(1000 + 800)/1800 = 4/9

q = 1 -(4/9) = 5/9

p1 - p2 = 0.4 - 0.5 = -0.1

difference/S.E. = 0.1/0.024 = 4.17.

As the difference is greater than 2.58 S.E. (1% level of the significance). It could not have arisen due to fluctuations of the sampling. Hence the data reveal a significant difference between town A and town B so far as the proportion of rice consumers is concerned.