Answer:
b) Then z(s) is in the rejection region for H₀. We reject H₀. The p-value is smaller than α/2
c)CI 95 % = ( 0.00002 ; 0.09998)
Step-by-step explanation: In both cases, the size of the samples are big enough to make use of the approximation of normality of the difference of the proportions.
Recent Sample
Sample size n₁ = 1000
Number of events of people with financial fitness more than fair
x₁ = 410
p₁ = 410/ 1000 = 0.4 then q₁ = 1 - p₁ q₁ = 1 - 0.4 q₁ = 0.6
Sample a year ago
Sample size n₂ = 1200
Number of events of people with financial fitness more than fair
x₂ = 420
p₂ = 420/1200 p₂ = 0.35 q₂ = 1 - p₂ q₂ = 1 - 0.35 q₂ = 0.65
Test Hypothesis
Null Hypothesis H₀ p₁ = p₂
Alternative Hypothesis Hₐ p₁ ≠ p₂
CI 95 % then significance level α = 5% α = 0.05 α/2 = 0.025
To calculate p-value:
SE = √ (p₁*q₁)/n₁ + (p₂*q₂)/n₂
SE = √ 0.4*0.6/1000 + 0.65*0.35/1200
SE = √ 0.00024 + 0.000189
SE = 0.021
z(s) = ( p₁ - p₂ ) / SE
z(s) = ( 0.4 - 0.35 )/0.021
z(s) = 0.05/ 0.021
z(s) = 2.38
We find p-value from z-table to be p-value = 0.00842
Comparing
p-value with α/2 = 0.025
α/2 > p-value
Then z(s) is in the rejection region for H₀. We reject H₀
CI 95 % = ( p₁ - p₂ ) ± 2.38*SE
CI 95 % = ( 0.05 ± 2.38*0.021 )
CI 95 % = ( 0.05 ± 0.04998)
CI 95 % = ( 0.00002 ; 0.09998)
CI 95 % does not contain the 0 value affirming what the hypothesis Test already demonstrate
