Answer: a. 0.58
b. 0.0256
c. (0.5494, 0.6006)
d. (0.5445, 0.6055)
Step-by-step 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: [tex]\hat{p}=\dfrac{579}{1007}=0.575[/tex]
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 : [tex]E=z^*\sqrt{\dfrac{p(1-p)}{n}}[/tex] , where z* =criticalz-value.
For 90% confidence , z*= 1.645 ( By z-table)
Then , [tex]E=(1.645)\sqrt{\dfrac{0.575(1-0.575)}{1007}}\approx0.0256[/tex]
∴ the margin of error at 90% confidence is 0.0256 .
c. 90% confidence interval for p = [tex](\hat{p}-E,\ \hat{p}+E)=(0.575-0.0256,\ 0.575+0.0256)[/tex]
[tex]=(0.5494,\ 0.6006)[/tex]
∴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 : [tex]E=(1.96)\sqrt{\dfrac{0.575(1-0.575)}{1007}}\approx0.0305[/tex]
95% confidence interval for p = [tex](0.575- 0.0305,\ 0.575+0.0305)[/tex]
[tex]=(0.5445,\ 0.6055)[/tex]
∴a 95% confidence interval for this population proportion=(0.5445, 0.6055)
