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 571 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.)____ to_____(d) Develop a 95% confidence interval for this population proportion. (Round your answer to four decimal places.)___ to ____

Answer 1

Answer:

a) 0.567

b) (0.5414,0.5926)

c) (0.5364,0.5976)                  

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 1007

Number of respondent that  they will be able to afford health insurance in the future, x = 571

a)  point estimate of the population proportion

$\hat{p} = \displaystyle\frac{x}{n} = \frac{571}{1007} = 0.567$

b) 90% Confidence Interval.

$\hat{p} + z_{\text{critical}}\times \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

$z_{critical}\text{ at}~\alpha_{0.10} = \pm 1.64$

Putting values, we get,

$0.567 \pm 1.64\times \sqrt{\dfrac{0.567(1-0.567)}{1007}}\\\\0.567 \pm 0.0256=(0.5414,0.5926)$

c) 95% Confidence Interval.

$\hat{p} + z_{\text{critical}}\times \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

$z_{critical}\text{ at}~\alpha_{0.05} = \pm 1.96$

Putting values, we get,

$0.567 \pm 1.96\times \sqrt{\dfrac{0.567(1-0.567)}{1007}}\\\\0.567 \pm 0.0306=(0.5364,0.5976)$