Confidence Intervals- Two population proportions

[Newcomer101]

I am having difficulty deriving the pivotal quantity for the above confidence interval. (pg 24 chapter 11)

Taking two random samples:
A~bin(n1,p1) approximates to N(n1p1,n1p1q1)
B~bin(n2,p2) approximates to N(n2p2,n2p2q2)

Then A-B approximates to N(n1p1-n2p2,n1p1q1+n2p2q2)

Therefore, should it not follow the pivotal quantity should be:
(n1p1-n2p2)-(theta1-theta2)
sqrt(n1p1q1+n2p2q2)

Any help most appreciated.

 

[freddie]

I think you might be getting confused between the distribution of X and the distribution of p(hat).

You have defined the distribution of A and B, which are equivalent to X1 and X2. These are binomial random variables with expected values n1p1 and n2p2, and variances n1p1q1 and n2p2q2. (You've used p where the Core Reading uses theta.)

Using a normal approximation, if you standardise the distribution of X1-X2 (to get the number of standard deviations X1-X2 is from its mean), you'd get:

[(X1-X2)-(n1p1-n2p2)]/sq rt(n1p1q1+n2p2q2) is approx N(0,1)

We can't use this to get a confidence interval for p1-p2 directly, so we need to use the distribution of the difference between the sample proportions p1(hat) - p2(hat), (where p1(hat) =X1/n1).

The expected value of p1(hat) - p2(hat) = p1-p2 and the variance of p1(hat) - p2(hat) = p1q1/n1 + p2q2/n2.

Standardising, we get:

[( p1(hat) - p2(hat)) - (p1-p2)]/sq rt (p1q1/n1 + p2q2/n2) is approx N(0,1)

If we want to find a 95% confidence interval for p1-p2, we use:

p1(hat) - p2(hat) +/- 1.96 sq rt (p1(hat)q1(hat)/n1 + p2(hat)q2(hat)/n2

(We have to use the sample proportions to estimate the standard deviation.)