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Needing some help on the below: I believe that the proportion of people who live in...

Needing some help on the below:

I believe that the proportion of people who live in Maine and drive a four wheel drive vehicle is higher than the proportion of people who live in Florida and drive a four wheel drive vehicle. In two independent polls 256 out of 415 people in Maine drive four wheel drive vehicles and 134 out of 271 people in Florida drive four wheel drive vehicles. Calculate the two samples with a 95% confidence interval.

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Answer #1

Here, , n1 = 415 , n2 = 271
p1cap = 0.6169 , p2cap = 0.4945


Standard Error, sigma(p1cap - p2cap),
SE = sqrt(p1cap * (1-p1cap)/n1 + p2cap * (1-p2cap)/n2)
SE = sqrt(0.6169 * (1-0.6169)/415 + 0.4945*(1-0.4945)/271)
SE = 0.0386

For 0.95 CI, z-value = 1.96
Confidence Interval,
CI = (p1cap - p2cap - z*SE, p1cap - p2cap + z*SE)
CI = (0.6169 - 0.4945 - 1.96*0.0386, 0.6169 - 0.4945 + 1.96*0.0386)
CI = (0.0467 , 0.1981)

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