Question

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)

Answer 1

The correct answer is option (D) (-0.21, 0.01)

Explanation:

Solution:

Data Given;

Total number of people in country S = 100

people with blue eyes in country S = 15

Total number of people in country B = 100

People with blue eyes in country B = 25

Let the Y represent number of people with blue eyes in both countries and let Z represent total number of people in both country.

Therefore,

Y₁= 15

Y₂= 25

Z₁ = 100

Z₂ = 100

Sample proportion of country (S₁) = Y1/Z1 = 15/100 = 0.15

Sample proportion of B country (S₂) = Y2/Z2 = 25/100 = 0.25

To calculate the 95% confidence interval, we use the formula;

Confidence interval (CI) =

(S₁ -S₂) ± Zₐ₋₂ * √[S₁( 1-S₁)/z₁ + S₂(1-S₂)/z₂]

At 95% confidence interval, the z-score from standard normal table is 1.96.

Substituting into the formula, we have;

CI = (0.15-0.25) ± 1.96*√[0.15( 1-0.15/100) + 0.25(1-0.25)/100]

= -0.1 ± 1.96*√[(0.15*0.85)/100 + (0.25*0.75)/100]

= -0.1 ± 1.96*√[0.1275/100+ 0.1875/100]

= -0.1 ± 1.96*√[0.001275 + 0.001875]

= -0.1 ± 1.96* √0.00315

= -0.1 ± 1.96*0.0561

= -0.1 +0.11

= -0.1 -0.11 , -0.1 + 0.11

CI = (-0.21, 0.01)