Difference of proportions - Let \[P_1\] and \[P_2\] be the sample proportions obtained in large samples sizes \[n_1\] and \[n_2\] drawn from respective populations having proportions \[P_1\] and \[P_2\]. Consider the null hypothesis that there is no difference between the population proportions, thus \[P_1\]=\[P_2\] , we see that sampling distribution of differences in proportions in approximately normal with mean and standard deviations given by
\[\mu _{P_1-P_2}=0\] and \[ \sigma _{P_1-P_2}= \sqrt{p(1-p)(\frac{1}{m}+\frac{1}{n})}\]
where \[ p=\frac{n_1p_1+n_2p_2}{n_1+n_2}\]
Assumption made in the below equations
\[ n_1=m\] \[ n_2=n\] \[ p_1=p\] \[ p_2=q\]
\[ \sigma _{P_1-P_2}=Y
Example- A sample polls of 300 voters from district A and 200 voters from district B showed that 56% and 48% , respectively , were in favour of a given candidate . At a level of significance of 0.05 ,test the hypothesis that (a) there is a difference between the districts ,(b) the candidate is preferred in district A .(c) Find the respective P values of the test.
solution- Let \[ p_1\] and \[ p_2\] denote the proportions of all voters of districts A and B, respectively , who are in favor of the candidate.
Under the hypothesis \[ H_0: p_1=p_2\], we have
\[ \mu _{P_1-P_2}=0\]
\[ P=\frac{n_1p_1+n_2p_2}{n_1+n_2}\]=\[ P=\frac{0.56*300+0.48*200}{200+300}=0.528\]
q=1-0.528
\[ \sigma _{P_1-P_2}=\sqrt{p(1-p)(\frac{1}{n_1}+\frac{1}{n_2}})\]
=\[ \sqrt{(0.5248)(0.472)(\frac{1}{300}+\frac{1}{200})}\]
=0.0456
Using standarized variable Z=\[ \frac{P_1-P_2}{\sigma _{P_1-P_2}}=\frac{0.56-0.48}{0.0456}=1.75\]
