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: ![\hat{p}=\dfrac{579}{1007}=0.575](https://tex.z-dn.net/?f=%5Chat%7Bp%7D%3D%5Cdfrac%7B579%7D%7B1007%7D%3D0.575)
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 :
, where z* =criticalz-value.
For 90% confidence , z*= 1.645 ( By z-table)
Then , ![E=(1.645)\sqrt{\dfrac{0.575(1-0.575)}{1007}}\approx0.0256](https://tex.z-dn.net/?f=E%3D%281.645%29%5Csqrt%7B%5Cdfrac%7B0.575%281-0.575%29%7D%7B1007%7D%7D%5Capprox0.0256)
∴ 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)](https://tex.z-dn.net/?f=%28%5Chat%7Bp%7D-E%2C%5C%20%5Chat%7Bp%7D%2BE%29%3D%280.575-0.0256%2C%5C%200.575%2B0.0256%29)
![=(0.5494,\ 0.6006)](https://tex.z-dn.net/?f=%3D%280.5494%2C%5C%200.6006%29)
∴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{\dfrac{0.575(1-0.575)}{1007}}\approx0.0305](https://tex.z-dn.net/?f=E%3D%281.96%29%5Csqrt%7B%5Cdfrac%7B0.575%281-0.575%29%7D%7B1007%7D%7D%5Capprox0.0305)
95% confidence interval for p = ![(0.575- 0.0305,\ 0.575+0.0305)](https://tex.z-dn.net/?f=%280.575-%200.0305%2C%5C%200.575%2B0.0305%29)
![=(0.5445,\ 0.6055)](https://tex.z-dn.net/?f=%3D%280.5445%2C%5C%200.6055%29)
∴a 95% confidence interval for this population proportion=(0.5445, 0.6055)