Question

Two random samples are taken, one from among UVA students and the other from among UNC students. Both groups are asked if academics are their top priority. A summary of the sample sizes and proportions of each group answering yes'' are given below:

UVA (Pop. 1): n1=100,p̂1=0.716

UNC (Pop. 2):n2=87,p̂2=0.602

Find a 92.2% confidence interval for the difference p1−p2p1−p2 of the population proportions.

Answer 1

Solution:

Confidence interval for difference between two population proportions:

Confidence interval = (P1 – P2) ± Z*sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)]

Where, P1 and P2 are sample proportions for first and second groups respectively

We are given

P1 = 0.716

P2 = 0.602

N1 = 100

N2 = 87

(P1 – P2) = 0.716 – 0.602 = 0.114

Confidence level = 92.2%

Critical Z value = 1.7624

(by using z-table)

Confidence interval = (P1 – P2) ± Z*sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)]

Confidence interval = 0.114 ± 1.7624*sqrt[(0.716*(1 – 0.716)/100) + (0.602*(1 – 0.602)/87)]

Confidence interval = 0.114 ± 1.7624* 0.0692

Confidence interval = 0.114 ± 0.1219

Lower limit = 0.114 - 0.1219 = -0.0079

Upper limit = 0.114 + 0.1219 = 0.2359

Confidence interval = (-0.0079, 0.2359)