Question

In a recent poll of 500 13-year-olds, many indicated to enjoy their relationships with their parents. Suppose that 200 of the 13-year olds were boys and 300 of them were girls. We wish to estimate the difference in proportions of 13-year old boys and girls who say that their parents are very involved in their lives. In the sample, 93 boys and 172 girls said that their parents are very involved in their lives. What is a 96% confidence interval for the difference in proportions (proportion of boys minus proportion of girls)?

(a) Calculate a 95% confidence interval for the difference in proportions (proportion of boys minus proportion of girls)?
(b) Interpret your interval calculated above.

Answer 1

CI 96 % = ( - 0.1128 ; 0.0728 )

CI 95% = ( - 0.1087 ; 0.6878 )

The two intervals contain 0 value therefore we can support that there is not statistics difference between the two groups with confidence level of 96 % and 95%

Explanation:

Boys sample:

sample size n₁ = 200

x₁ number of boys saying their parents are very involved in their lives

= 93

p₁ = x₁/n₁ = 93/200 = 0.465 then q₁ = 1 - p₁ q₁ = 1 - 0.465 q₁ = 0.535

Girls sample

sample size n₂ = 300

x₂ number of girls saying their parents are very involved in their lives

= 172

p₂ = x₂/n₂ = 172/ 300 = 0.573 then q₂ = 1 -0.573 q₂ = 0.427

CI 96 % α = 4 % α = 0.04 α/2 = 0.02

p₁ - p₂ = 0.465 - 0.573 = - 0.168

CI 96 % = ( p₁ - p₂ ) ± z(c) * SE

z(c) for α = 0.02 z(c) = - 2.05

SE = √ (p₁*q₁)/n₁ + (p₂*q₂)/n₂

SE = √ ( 0.465*0.535)/200 + (0.573*0.427)/300

SE = √ 0.00124 + 0.000815

SE = √ 0.00205

SE = 0.0453

CI 96 % = ( -0.02 ± ( 2.05 * 0.0453 ) )

CI 96 % = ( -0.02 ± ( 0.0928 ))

CI 96 % = ( - 0.1128 ; 0.0728 )

a) CI 95% α = 5 % α = 0.05 α/2 = 0.025

SE = 0.0453

z(c) for 0.025 is from z-table z(c) = 1.96

CI 95% = ( - 0.02 ± 1.96 * 0.0453)

CI 95% = ( - 0.02 ± 0.08878 )

CI 95% = ( - 0.1087 ; 0.6878 )