Question

A certain financial services company uses surveys of adults age 18 and older to determine if personal financial fitness is changing over time. Suppose that in February 2012, a sample of 1,000 adults showed 410 indicating that their financial security was more than fair. In February 2010, a sample of 700 adults showed 245 indicating that their financial security was more than fair.

1.Conduct the hypothesis test and compute the p-value. At a 0.05 level of significance, what is your conclusion?

Find the value of the test statistic.

Find the p-value.

2.What is the 95% confidence interval estimate of the difference between the two population proportions? (Round your answers to four decimal places.)

Answer 1

P1 :-   Their financial security  in February 2012

P2 :-  Their financial security  in February 2010

To Test :-

H0 :- P1 = P2

H1 :- P1 P2

Test Statistic :-

is the pooled estimate of the proportion P
= ( x1 + x2) / ( n1 + n2)
= ( 410 + 245 ) / ( 1000 + 700 )
= 0.3853

Z = 2.5

Test Criteria :-
Reject null hypothesis if

= 2.5 > 1.96, hence we reject the null hypothesis
Conclusion :- We Reject H0

Decision based on P value
P value = 2 * P ( Z > 2.5 )
P value = 0.02
Reject null hypothesis if P value <
Since P value = 0.02 < 0.05, hence we reject the null hypothesis
Conclusion :- We Reject H0

There is sufficient evidence to support the claim that personal financial fitness is changing over time at 5% level of significance.

n1 = 1000
n2 = 700


Lower Limit =
upper Limit =
95% Confidence interval is ( 0.0133 , 0.1067 )
( 0.0133 < ( P1 - P2 ) < 0.1067 )

Since 0 does not lie in the interval, hence we can conclude to accept alternative hypothesis.