Answer:
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%
Step-by-step 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 )
