The sampling distribution of the number of successes follows a binomial probability distribution. Hence its standard error is given 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. probability of failure

Illustration: a coin was tossed 400 times and the head turned up 216 times. Test the hypothesis that the coin is unbiased.

Solution: let us take the hypothesis that the coin is unbiased. On the basis of 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 and 200 tails.

Observed number of heads = 216

Difference between observed number of heads and 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) = 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). Hence our hypothesis is true and we conclude that coin is unbiased.

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

Solution: let us 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

Since the difference is more than 1.96 at 5 per cent level of significance, the hypothesis rejected. Hence we cannot say that the dice is fair at 5 percent level of significance.

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