Question

11. + -/4.16 points ASWSBE14 8.E.035.MI. My Notes + Ask Your Teacher You may need to use the appropriate appendix table or technology to answer this 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 585 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. (Round your answer to four decimal places.) (d) Develop a 95% confidence interval for this population proportion. (Round your answer to four decimal places.) to

Answer 1

Solution :

Given that,

n = 1007

x = 585

a) Point estimate = sample proportion = $\hat p$ = x / n = 585 / 1007 = 0.5809

1 - $\hat p$ = 1 - 0.5809 = 0.4191

b) At 90% confidence level

$\alpha$ = 1 - 90%

$\alpha$ =1 - 0.90 =0.10

$\alpha$/2 = 0.05

Z$\alpha$/2 = Z_0.05 = 1.645

Margin of error = E = Z_{$\alpha$ / 2} * $\sqrt$ (($\hat p$ * (1 - $\hat p$ )) / n)

= 1.645 ($\sqrt$((0.5809 * 0.4191) / 1007)

= 0.0256

c) A 90% confidence interval for population proportion p is ,

$\hat p$ ± E

= 0.5809 ± 0.0256

= ( 0.5553, 0.6065 )

d) At 95% confidence level

$\alpha$ = 1 - 95%

$\alpha$ =1 - 0.95 =0.05

$\alpha$/2 = 0.025

Z$\alpha$/2 = Z_{0.025 = 1.960}

Margin of error = E = Z_{$\alpha$ / 2} * $\sqrt$ (($\hat p$ * (1 - $\hat p$ )) / n)

= 1.96 ($\sqrt$((0.5809 * 0.4191) /1007 )

= 0.0305

A 95% confidence interval for population proportion p is ,

$\hat p$ ± E  

= 0.5809 ± 0.0305

= ( 0.5504, 0.6114)