Question

A study reports that freshmen at public universities work 10.2 hours a week for pay, on average, and the sn is 8.5 hours; at private universities, the average is 8.1 hours and the sn is 6.9 hours. Assume these data are based on two independent simple random samples, each of size 1,000.

Required:
Construct a 95% confidence interval for the difference of the hours worked.

Answer 1

The 95% confidence interval for the difference of the hours worked is [-0.052 , 0.0012]

Explanation:

Formula for Confidence interval based on difference =

p1 - p2 ± z × √[p1(1 - p1)/n1] + [p2(1 - p2)/n2]

p1 = proportion for the first group = x/n

= 8.5/10.2

= 0.8333333333

≈ 0.83

n1 = 1000

p2 = proportion for the second group = 6.9/8.1

= 0.8518518519

≈ 0.85

n2 = 1000

Confidence Interval = p1 - p2 ± z × √[p1(1 - p1)/n1] + [p2(1 - p2)/n2]

= 0.83 - 0.85 ± 1.96 √[0.83(1 - 0.83)/1000] + [0.85(1 - 0.85)/1000]

= -0.02 ± 1.96 × √0.83 × 0.17/1000 + 0.85 × 0.15/1000

= -0.02 ± 1.96 × √0.0001411 + 0.0001275

= -0.02 ± 1.96 × √0.0002686

= -0.02 ± 1.96 × 0.0163890207

= -0.02 ± 0.0321224806

-0.02 - 0.0321224806

= -0.0521224806

≈ -0.052

-0.02 + 0.0321224806

= 0.0121224806

≈ 0.012

Therefore, the 95% confidence interval for the difference of the hours worked is [-0.052 , 0.0012]