Question

The Consumer Reports National Research Center conducted a telephone survey of 2,000 adults to learn about the major economic concerns for the future. The survey results showed that 1,740 of the respondents think the future health of Social Security is a major economic concern.

Required:
a. What is the point estimate of the population proportion of adults who think the future health of Social Security is a major economic concern.
b. At 90% confidence, what is the margin of error?
c. Develop a 90% confidence interval for the population proportion of adults who think the future health of Social Security is a major economic concern.
d. Develop a 95% confidence interval for this population proportion.

Answer 1

a. 0.87

b. The margin of error is of 0.0124.

c. The 90% confidence interval for the population proportion of adults who think the future health of Social Security is a major economic concern is (0.8576, 0.8824).

d. The 95% confidence interval for this population proportion is (0.8553, 0.8847).

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence level of
`1-\alpha`, we have the following confidence interval of proportions.

`\pi \pm z\sqrt{(\pi(1-\pi))/(n)}`

In which

z is the z-score that has a p-value of
`1 - (\alpha)/(2)`.

The margin of error is of:

`M = z\sqrt{(\pi(1-\pi))/(n)}`

The confidence interval is:

`\pi \pm M`

a. What is the point estimate of the population proportion of adults who think the future health of Social Security is a major economic concern.

The sample proportion, that is, 1740 out of n = 2000. So

`\pi = (1740)/(2000) = 0.87`

0.87 is the point estimate.

b. At 90% confidence, what is the margin of error?

90% confidence level

So
`\alpha = 0.1`, z is the value of Z that has a p-value of
`1 - (0.1)/(2) = 0.95`, so
`Z = 1.645`.

Then the margin of error is of:

`M =  z\sqrt{(\pi(1-\pi))/(n)}`

`M =  1.645\sqrt{(0.87*0.13)/(2000)}`

`M = 0.0124`

The margin of error is of 0.0124.

c. Develop a 90% confidence interval for the population proportion of adults who think the future health of Social Security is a major economic concern

`\pi - M = 0.87 - 0.0124 = 0.8576`

`\pi + M = 0.87 + 0.0124 = 0.8824`

The 90% confidence interval for the population proportion of adults who think the future health of Social Security is a major economic concern is (0.8576, 0.8824).

d.

95% confidence level

So
`\alpha = 0.05`, z is the value of Z that has a p-value of
`1 - (0.05)/(2) = 0.975`, so
`Z = 1.96`.

Then the margin of error is of:

`M =  z\sqrt{(\pi(1-\pi))/(n)}`

`M =  1.96\sqrt{(0.87*0.13)/(2000)}`

`M = 0.0147`

The confidence interval will be of:

`\pi - M = 0.87 - 0.0147 = 0.8553`

`\pi + M = 0.87 + 0.0147 = 0.8847`

The 95% confidence interval for this population proportion is (0.8553, 0.8847).