Answered

Get expert insights and community-driven knowledge on IDNLearn.com. Get step-by-step guidance for all your technical questions from our knowledgeable community members.

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?

Sagot :

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

Thank you for using this platform to share and learn. Keep asking and answering. We appreciate every contribution you make. Find the answers you need at IDNLearn.com. Thanks for stopping by, and come back soon for more valuable insights.