Question

In the spring of 2017, the Consumer Reports National Research Center conducted a survey of 1,007 adults to learn about their major healthcare concerns. The survey results showed that 579 of the respondents lack confidence they will be able to afford health insurance in the future.

a. What is the point estimate of the population proportion of adults who lack confidence they will be able to afford health insurance in the future. (Round your answer to two decimal places.)
b. At 90% confidence, what is the margin of error? (Round your answer to four decimal places.)
c. Develop a 90% confidence interval for the population proportion of adults who lack confidence they will be able to afford health insurance in the future.
d. Develop a 95% confidence interval for this population proportion. (Round your answer to four decimal places.)

Answer 1

Answer: a. 0.58

b. 0.0256

c. (0.5494, 0.6006)

d. (0.5445, 0.6055)

Explanation:

Let p be the proportion of the respondents lack confidence they will be able to afford health insurance in the future.

As per given , Sample size : n= 1007

Sample proportion of respondents lack confidence they will be able to afford health insurance in the future:
`\hat{p}=(579)/(1007)=0.575`

a. The sample proportion gives the best estimate to the population proportion.

∴ The point estimate of the population proportion of adults who lack confidence they will be able to afford health insurance in the future =0.575 ≈0.58

b. Margin of error :
`E=z^*\sqrt{(p(1-p))/(n)}` , where z* =criticalz-value.

For 90% confidence , z*= 1.645 ( By z-table)

Then ,
`E=(1.645)\sqrt{(0.575(1-0.575))/(1007)}\approx0.0256`

∴ the margin of error at 90% confidence is 0.0256 .

c. 90% confidence interval for p =
`(\hat{p}-E,\ \hat{p}+E)=(0.575-0.0256,\ 0.575+0.0256)`

`=(0.5494,\ 0.6006)`

∴a 90% confidence interval for the population proportion of adults who lack confidence they will be able to afford health insurance in the future. (0.5494, 0.6006) .

d. For 95% confidence , z*= 1.96 ( By z-table)

Margin of error :
`E=(1.96)\sqrt{(0.575(1-0.575))/(1007)}\approx0.0305`

95% confidence interval for p =
`(0.575- 0.0305,\ 0.575+0.0305)`

`=(0.5445,\ 0.6055)`

∴a 95% confidence interval for this population proportion=(0.5445, 0.6055)