Test Of Hypothesis On Proportions

It follows a similar method to the one for means except that the standard error utilized in this case:

Sp = √(pq/n) 

Z score is computed as, Z = (P - Π)/Sp    

Whereas P = Proportion found in the sample.

            Π - The hypothetical proportion.

Illustration

A member of parliament as MP claims that in his constituency only 50 percent of the total youth population lacks university education. A local media company wanted to ascertain that claim hence they conducted a survey taking a sample of 400 youths, of these 54 percent lacked university education.

Required:

At 5 percent level of significance confirms if the Member of Parliament's claim is wrong.

Solution

Note:   it is a two tailed tests because we wish to test the hypothesis that the hypothesis is different (≠) and not against a specific alternative hypothesis for illustration < less than or > more than.

      H₀ :  π = 50 percent of all youth in the constituency lack university education.

      H₁ :  π ≠ 50 percent of all youth in the constituency lack university education.

Sp = √(pq/n) = √((0.5 * 0.5)/400) = 0.025

      Z = ¦{(.54 - 0.50)/0.025}¦ = 1.6

At 5percent level of significance for a two-tailored test the critical value is 1.96 because calculated Z value < tabulated value (1.96).

That is 1.6 < 1.96 we accept the null hypothesis.

Hence the Member of Parliament's claim is accurate.