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 1,100 adults showed 385 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:

H_a:

(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 =

State your conclusion.

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.

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.   

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.

(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 (contains, is completely above, or is completely below zero), so we can be 95% confident that the population proportion of adults saying that their financial security is more than fair (has increased, has decreased, or may have stayed the same.)

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
p2cap = X2/N2 = 385/1100 = 0.35
pcap = (X1 + X2)/(N1 + N2) = (410+385)/(1000+1100) = 0.3786

Test statistic
z = (p1cap - p2cap)/sqrt(pcap * (1-pcap) * (1/N1 + 1/N2))
z = (0.41-0.35)/sqrt(0.3786*(1-0.3786)*(1/1000 + 1/1100))
z = 2.83

P-value Approach
P-value = 0.0047
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 = 1100
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)/1100)
SE = 0.0212

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*0.0212, 0.41 - 0.35 + 1.96*0.0212)
CI = (0.0184 , 0.1016)

The 95% confidence interval ( is completely above zero), so we can be 95% confident that the population proportion of adults saying that their financial security is more than fair ( may have stayed the same.)