A random sample of 100 people from Country S had 15 people with blue eyes. A separate random sample of 100 people from Country B had 25 people with blue eyes.
Assuming all conditions are met, which of the following is a 95 percent confidence interval to estimate the difference in population proportions of people with blue eyes (Country S - Country B)?

A) (-0.01, 0.21)
B) (-0.15, -0.05)
C) (-0.19, -0.01)
D (-0.21, 0.01)
E) (-0.24, 0.01)

Given Information:

Sample size of country S = ns = 100

Sample size of country B = nb = 100

Number of people with blue eyes in country S = 15

Number of people with blue eyes in country B = 25

Confidence level = 95%

Required Information:

Difference in population proportion = ?

A (-0.01, 0.21)

Step-by-step explanation:

The difference in mean is

μ = ps - pb

μ = 15/100 - 25/100

μ = 0.15 - 0.25

μ = 0.10

The difference in standard error is

ο = z*√(ps(1 - ps)/ns + pb(1 - pb)/nb)

The z-score corresponding to 95% confidence interval is 1.96

ο = 1.96*√(0.15(1 - 0.15)/100 + 0.25(1 - 0.25)/100)

ο = 1.96*√(0.15(0.85)/100 + 0.25(0.75)/100)

ο = 0.11

Therefore, the difference in population proportion with blue eyes is

μ ± ο

μ + ο, μ - ο

0.10 + 0.11, 0.10 - 0.11

0.21, -0.01

(-0.01, 0.21)

Therefore, the correct option is A.

p = 6s

step-by-step explanation:

where is abdul's work? don't see that anywhere posted : /

step-by-step explanation: