A random sample of 47 manuscripts typed by Katlyn showed tha

A random sample of 47 manuscripts typed by Katlyn showed that 13 of them had errors. A random sample of 85 manuscripts typed by Dara showed that 31 of them had errors. Find a 99% confidence interval for the difference in the proportion of all manuscripts with errors typed by Katlyn compared to those typed by Dara.

Solution

p1=13/47 = 0.2765957

p2=31/85 = 0.3647059

Given a=0.01, Z(0.005) = 2.58 (from standard normal table)

So 99% confidence interval is

(p1-p2) +/- Z*sqrt(p1*(1-p1)/n1+p2*(1-p2)/n2)

--> (0.2765957-0.3647059) +/- 2.58*sqrt(0.2765957*(1-0.2765957)/47+0.3647059*(1-0.3647059)/85)

--> (-0.3037074, 0.127487)