Question

Please help how to solve this??

An agency surveyed people worldwide, asking them, "Do you live in a household with more than one car?" Of the 1243 respondents surveyed in India, 492 said they live in a household with more than one car. In Canada, of the 861 respondents, 575 said they lived in a household with more than one car. Find the margin of error for a 99% confidence interval for the difference in proportions for households with more than one car between India and Canada PI - PC (assume 20.005 = 2. 58). Note: 1- Round any intermediate numbers to 4 decimal places. 2- Round your final answer to 3 decimal places. Enter your final answer with 3 decimal places.

Answer 1

Here, , n1 = 1243 , n2 = 861
p1cap = 0.3958 , p2cap = 0.6678

Standard Error, sigma(p1cap - p2cap),
SE = sqrt(p1cap * (1-p1cap)/n1 + p2cap * (1-p2cap)/n2)
SE = sqrt(0.3958 * (1-0.3958)/1243 + 0.6678*(1-0.6678)/861)
SE = 0.0212

For 0.99 CI, z-value = 2.58

MArgin of error = z *SE
= 2.58 * 0.0212
= 0.055