Question

You may need to use the appropriate appendix table or technology to answer this 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.
(a)
State the hypotheses that can be used to test for a significant difference between the population proportions for the two years. (Let p1 = population proportion most recently saying financial security more than fair and p2 = population proportion from the year before saying financial security more than fair. Enter != for ≠ as needed.)
H0:  

Ha:  

(b)
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. (Use
p1 − p2.
Round your answer to two decimal places.)
  

Find the p-value. (Round your answer to four decimal places.)
p-value =   
Incorrect:
State your conclusion.
Do not reject H0. There is insufficient evidence to conclude the population proportions are not equal. The data do not suggest that there has been a change in the population proportion saying that their financial security is more than fair.
Do not reject H0. There is sufficient evidence to conclude the population proportions are not equal. The data suggest that there has been a change in the population proportion saying that their financial security is more than fair.   
Reject H0. There is insufficient evidence to conclude the population proportions are not equal. The data do not suggest that there has been a change in the population proportion saying that their financial security is more than fair.
Reject H0. There is sufficient evidence to conclude the population proportions are not equal. The data suggest that there has been a change in the population proportion saying that their financial security is more than fair.
Correct: Your answer is correct.
(c)
What is the 95% confidence interval estimate of the difference between the two population proportions? (Round your answers to four decimal places.)
  
_____
to   
_____
What is your conclusion?
The 95% confidence interval zero, so we can be 95% confident that the population proportion of adults saying that their financial security is more than fair .

Answer 1

a)
Below are the null and alternative Hypothesis,
Null Hypothesis, H0: p1 = p2
Alternate Hypothesis, Ha: p1 ≠ p2

b)

p1cap = X1/N1 = 410/1000 = 0.41
p1cap = X2/N2 = 420/1200 = 0.35
pcap = (X1 + X2)/(N1 + N2) = (410+420)/(1000+1200) = 0.3773

Test statistic
z = (p1cap - p2cap)/sqrt(pcap * (1-pcap) * (1/N1 + 1/N2))
z = (0.41-0.35)/sqrt(0.3773*(1-0.3773)*(1/1000 + 1/1200))
z = 2.89

P-value Approach
P-value = 0.0039
As P-value < 0.05, reject the null hypothesis.

Reject H0. There is sufficient evidence to conclude the population proportions are not equal. The data suggest that there has been a change in the population proportion saying that their financial security is more than fair.

c)

Here, , n1 = 1000 , n2 = 1200
p1cap = 0.41 , p2cap = 0.35

Standard Error, sigma(p1cap - p2cap),
SE = sqrt(p1cap * (1-p1cap)/n1 + p2cap * (1-p2cap)/n2)
SE = sqrt(0.41 * (1-0.41)/1000 + 0.35*(1-0.35)/1200)
SE = 0.021

For 0.95 CI, z-value = 1.96
Confidence Interval,
CI = (p1cap - p2cap - z*SE, p1cap - p2cap + z*SE)
CI = (0.41 - 0.35 - 1.96*0.021, 0.41 - 0.35 + 1.96*0.021)
CI = (0.0188 , 0.1012)

0.0188 to 0.1012

The 95% confidence intervaldoes not zero, so we can be 95% confident that the population proportion of adults saying that their financial security is more than fair