## (Solution document) Suppose we have independent random samples of size n 1 = 604 and n 2 = 490. The proportions of success in the two samples are p 1 = .43 and p 2 = .

Suppose we have independent random samples of size n1 = 604 and n2 = 490. The proportions of success in the two samples are p1= .43 and p2 = .51. Find the 95% confidence interval for the difference in the two population proportions.

So far I have:

From table 6.1 we see that 95% confidence corresponds to z = 1.96

The sample sizes are greater than 30, so we can use:

P1 - P2 ± z ?p1(1 - p1) / n1 + p2(1 - p2) / n2

.43 - .51 ± 1.96 ?.43(1 - .43) / 604 + .51(1 - .51) / 490

-0.08 ± 1.96(.03026)

How do I find the interval?

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