A certain financial services company uses surveys of adults age 18 and older to determine if personal financial fitness is changing over time. A recent sample of 1,000 adults showed 410 indicating that their financial security was more than fair. Suppose that just a year before, a sample of 1,200 adults showed 420 indicating that their financial security was more than fair.

Required:
a. State the hypotheses that can be used to test for a significant difference between the population proportions for the two years.
b. Conduct the hypothesis test and compute the p-value. At a 0.05 level of significance, what is your conclusion?
c. What is the 95% confidence interval estimate of the difference between the two population proportions?
d. What is your conclusion?

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