Question

Construct the indicated confidence interval for the difference between population proportions p1 - p2. Assume that the samples are independent and that they have been randomly selected. A marketing survey involves product recognition in New York and California. Of 558 New Yorkers surveyed, 193 knew the product while 196 out of 614 Californians knew the product. Construct a 99% confidence interval for the difference between the two population proportions.

Answer 1

-0.0423≤p≤0.0977

Explanation:

The margin error formula is expressed as;

ME = p±√p(1-p)/n

n is the sample size

For the first population proportion p1:

p1 = 193/558 = 0.3469

n1 = 558

p2 = 196/614 = 0.3192

n2 = 614

Taking the difference in the confidence of the proportion

p = p1-p2

p = 0.3469-0.3192

p= 0.0277

Get the margin of error

Margin of error = z*√p(1-p)/n

Margin of error = 2.5758*√0.35(1-0.35)/558 + 2.5758*√0.32(1-0.32)/614

Margin of error = 2.5758*√0.35(0.65)/558 + 2.5758*√0.32(0.68)/614

Margin of error = 2.5758*(0.02019)+2.5758(0.0188)

Margin of error = 0.07

Get the confidence interval

CI = p±ME

CI = (p-ME, p+ME)

CI = (0.0277-0.07, 0.277+0.07)

CI = (-0.0423, 0.0977)

Hence a 99% confidence interval for the difference between the two population proportions is -0.0423≤p≤0.0977