The sampling distribution of the number of successes follows a binomial probability distribution. Hence its standard error is shown  by the formula:

                                            S. E. of no. of successes = √npq,

                                            where n = size of a sample

                                                       p = probability of successes in each trial

                                                       q = (1 - p), i.e. the probability of failure

Illustration 1: A coin was tossed 400 times &  the head turned up 216 times. Now test the hypothesis that the coin is unbiased.

Solution: Let's take the hypothesis that the coin is unbiased. According to  this hypothesis the probability of getting head or tail would be equal, i.e. ½  hence in 400 throws of a coin we should expect 200 heads & 200 tails.

Observed number of heads = 216

Difference between the observed number of heads and the  expected number of heads = 216 - 200 = 16

S. E. of no. of heads = √npq

N = 400, p = q = ½

∴ S. E. = √(400× 1/2×1/2)

∴ S. E. = 10

Difference/S.E. = 16/10 = 1.6

The difference between observed and expected number of heads is less than 1.96 SE (5% level of significance). Therefore  our hypothesis is true and we conclude that the coin is unbiased.

Illustration 2: In 324 throws of a six-faced dice, the odd points appeared 180 times. Would you say that the dice is fair at 5 percent level of significance?

Solution: Let's take the hypothesis that the dice we would expect 162 odd points in 324 throws

S.E. = √npq = √(324× 1/2×1/2) = 9

Diff. /S.E. = (180 - 162)/9 = 2

As the difference is more than 1.96 at 5 percent level of significance, the hypothesis rejected. Therefore  we cannot say that the dice is fair at 5 percent level of significance.