Just as the difference between two sample means is normally distributed for large samples, so is the difference between two sample proportions. That is, if Y1 and Y2 are independent binomial random variables with parameters (n1, p1) and (n2, p2), respectively, then (Y1/n1) − (Y2/n2) is approximately normally distributed for large values of n1 and n2.

Find E((Y1/n1) − (Y2/n2)) and V((Y1/n1) − (Y2/n2)).

I know that the mean of the sum of two random variables X and Y is the sum of their means and that for independent random variables X and Y, the variance of their sum or difference is the sum of their variances. I'm not sure how to exactly construct the procedure for calculating the expected mean and variance.


Actually is like you said it.....

$\hat{p_1}=\frac{Y_1}{n_1}\sim N\left(\ p_1,\frac{p_1(1-p_1)}{n_1}\right)$ and $\hat{p_2}=\frac{Y_2}{n_2}\sim N\left(\ p_2,\frac{p_2(1-p_2)}{n_2}\right)$

By independence the difference between two normal variables is again normal and in this case is:

$\hat{p_1}-\hat{p_2} \sim N \left(\ p_1-p_2,\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2}\right)$

So the mean is the difference of the means and the variance is the sums of the variances

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