Question

This week I decided to conduct a random poll of people who enjoy chocolate over vanilla. I believe that the proportion of people who enjoy chocolate over vanilla in Switzerland is greater, than those who live in Hershey, PA. In a random, independent poll, 344 out of 520 people in Switzerland prefer chocolate over vanilla. In Hershey, PA, 215 people out of 400 stated that they preferred chocolate over vanilla. Calculate the two samples with a 95% interval.

Answer 1

Here, , n1 = 520 , n2 = 400
p1cap = 0.6615 , p2cap = 0.5375

Standard Error, sigma(p1cap - p2cap),
SE = sqrt(p1cap * (1-p1cap)/n1 + p2cap * (1-p2cap)/n2)
SE = sqrt(0.6615 * (1-0.6615)/520 + 0.5375*(1-0.5375)/400)
SE = 0.0324

For 0.95 CI, z-value = 1.96
Confidence Interval,
CI = (p1cap - p2cap - z*SE, p1cap - p2cap + z*SE)
CI = (0.6615 - 0.5375 - 1.96*0.0324, 0.6615 - 0.5375 + 1.96*0.0324)
CI = (0.0605 , 0.1875)