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 1,100 adults showed 385 indicating that their financial security was more than fair.
Required:
a. State the hypothesis that can be used to test for a significant difference between the population proportions for the two years?
b. What is the sample proportion indicating that their financial security was more that fair in 2012?In 2010?
c. Conduct the hypothesis test and compute the p-value.At a .05 level of significance what is your conclusion?
d. What is the 95% confidence interval estimate of the difference between the two population proportion?

Answer 1

Answer:

The calculated z = 2.85 falls in the critical region z > 1.96 so we accept the null hypothesis that there is no difference between the population proportions for the two years.

Step-by-step explanation:

Let p1 be the sample of 1000 adults chosen in 2012 and

p2 be the sample of 1100 adults chosen in 2010

Part a:

The hypothesis that can be used to test for a significant difference between the population proportions for the two years is:

H0: p1-p2= 0           against the claim      Ha: p1-p2≠0

Part b:

The sample proportion indicating that their financial security was more that fair in 2012 is

p1^= 410/1000= 0.41

In 2010 is:

p2^= 385/1100= 0.35

Part c:

p^c= 410+385/1000+1100= 0.3785

q^c= 1-p^c= 1-0.3785= 0.6124

The critical region is z > ±1.96

The test statistic is

z= p1^- p2^ / sqrt ( p^cq^c( 1/n1+ 1/n2)

z= 0.41-0.35 /sqrt( 0.3785*0.6124(1/1000 +1/1100)

z=  0.06 / 0.021036

z= 2.85

Conclusion

The calculated z = 2.85 falls in the critical region z > 1.96 so we accept the null hypothesis that there is no difference between the population proportions for the two years.

P- value :  0.00466

The result is significant( accept the null hypothesis) for  value less than 0.05

Part d:

                                                     2012                         2010

Sample                                       1000                           1100

Proportions                               0.41                             0.35

Standard Error               √0.41( 0.59)/1000        √0.35( 0.65)/1100

                                               = 0.0155                0.01438

Standard Error for the difference = √0.0155² + 0.01438² =0.0211

p1-p2 ± z* standard error for the difference

0.41-0.35 ± 1.96 *( 0.0211)

0.06 ± 0.041356

-0.018644,   0.101356

The 95 % confidence interval estimate is (-0.018644,   0.10135