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.
H_{0} : π = 50 percent of all youth in the constituency lack university education.
H_{1} : 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.