Question

A survey of 2000 Ford owner found that 450 chose their car as their favorite car they ever owned. In a similar survey involving 1200 Ford owners, 280 of them chose Ford as well. Compute a 95% confidence interval for the difference between the proportions of owners favoring Ford in the two surveys.

Answer 1

Here, , n1 = 2000 , n2 = 1200
p1cap = 0.225 , p2cap = 0.2333

Standard Error, sigma(p1cap - p2cap),
SE = sqrt(p1cap * (1-p1cap)/n1 + p2cap * (1-p2cap)/n2)
SE = sqrt(0.225 * (1-0.225)/2000 + 0.2333*(1-0.2333)/1200)
SE = 0.0154

For 0.95 CI, z-value = 1.96
Confidence Interval,
CI = (p1cap - p2cap - z*SE, p1cap - p2cap + z*SE)
CI = (0.225 - 0.2333 - 1.96*0.0154, 0.225 - 0.2333 + 1.96*0.0154)
CI = (-0.0385 , 0.0219)