A poll in 2017 reported that 699 out of 1027 adults in a certain country believe that marijuana should be legalized. When this poll about the subject

Given Information:
Sample size $(n)=1027$
$X=699$
Sample proportion is given by the formula:
$(\hat{p})=\frac{X}{n}$
$=\frac{699}{1027}$
$\approx 0.6806232$
(a) To construct $99\mathrm{\%}$ confidence interval:
Confidence level $=0.99$
Significance level $\alpha =1-0.99=0.01$
Confidence interval for proportion is obtained using the formula:
$C.I=\hat{p}{z}_{a/2}\frac{\sqrt{\hat{p}(1-\hat{p})}}{n}$
Using a standard normal table, Z-critical value at $\frac{0.01}{2}=0.005$ significance level is 2.576
Substitute the values in the formula:
$C.I=0.6806232\pm 2.576\frac{\sqrt{0.6806232(1-0.6806232)}}{1027}$
$=0.6806232\pm 2.576\times 0.01455$
$=(0.6806232-0.037481,0.6806232+0.037481)$
$=(0.643,0.718)$
Thus, we are 99% confident that the population proportion of adults in the country in 2017 who believe marijuana should be legalized is between (0.643, 0.718)
(b) To construct 90% confidence interval:
Confidence level $=0.90$
Significance level $\alpha =1-0.90=0.10$
Confidence interval for proportion is obtained using the formula:
$C.I=\hat{p}{z}_{a/2}\frac{\sqrt{\hat{p}(1-\hat{p})}}{n}$
Using a standard normal table, Z-critical value at $\frac{0.10}{2}=0.05$ significance level is 1.645
Substitute the values in the formula:
$C.I=0.6806232\pm 1.645\frac{\sqrt{0.6806232(1-0.6806232)}}{1027}$
$=0.6806232\pm 1.645\times 0.01455$
$=(0.6806232-0.023935,0.6806232+0.023935)$
$=(0.657,0.705)$
Thus, we are 90% confident that the population proportion of adults in the country in 2017 who believe marijuana should be legalized is between (0.657, 0.705)
(c) To find the margin of error for each of the confidence Intervals:
Margin of error is given by the formula:
$M.E=z_{a/2}\frac{\sqrt{\hat{p}(1-\hat{p})}}{n}$
Margin of error of 99% confidence interval is:
$M.E=2.576\frac{\sqrt{0.6806231743(1-0.6806231743)}}{1027}$
$=2.576\times 0.01455$
$=0.037481$
$\approx 0.037$
Thus, margin of error for 99% confidence interval is 0.037
Margin of error of 90% confidence interval is:
$M.E=1.645\frac{\sqrt{0.6806231743(1-0.6806231743)}}{1027}$
$=1.645\times 0.01455$
$=0.023935\approx 0.024$
Thus, margin of error for 99% confidence interval is 0.024
(d) The margin of error of 80% confidence interval will be less than the margin of error of 90% and 99% confidence intervals because intervals get wider with an increasing confidence level.
To construct 80% confidence interval:
Confidence level $=0.80$
Significance level $\alpha =1-0.80=0.20$
Confidence interval for proportion is obtained using the formula:
$C.I=\hat{p}{z}_{a/2}\frac{\sqrt{\hat{p}(1-\hat{p})}}{n}$
Using a standard normal table, Z-critical value at $\frac{0.20}{2}=0.10$ significance level is 1.2816
Substitute the values in the formula:
$C.I=0.6806232\pm 1.2816\frac{\sqrt{0.6806232(1-0.6806232)}}{1027}$
$=0.6806232\pm 1.2816\times 0.01455$
$=(0.6806232-0.0186473,0.6806232+0.0186473)$
$=(0.662,0.699)$ Thus, we are 80% confident that the population proportion of adults in the country in 2017 who believe marijuana should be legalized is between (0.662, 0.699)
The width of 80% confidence interval is narrower when compared to the confidence intervals of 99% and 90% levels.