Question

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?

Answer 1

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