91Ó°ÊÓ

Given in the question that, A Harris poll asked Americans whether states should be allowed to conduct random drug tests on elected officials. Of $21355$respondents, $79\%$said yes. we have to Determine the margin of error for a $99\%$confidence interval.

The formula of margin of error :

$E=z_{\frac{\mathrm{Î\pm }}{2}}\sqrt{\frac{\hat{p}×\hat{q}}{n}}$

Here

$$\begin{gathered} \hat{p}=\frac{x}{n} \\ =0.79 \end{gathered}$$

$n=21355$

Confidence interval $99\%$

Therefore, significance level $\mathrm{Î\pm }=0.01,$$zvalue=2.576$

The margin of error for population proportion given by

$$\begin{gathered} E=2.576\sqrt{\frac{(0.79)×(1-0.79)}{21355}} \\ =0.00717 \end{gathered}$$

We have to find out that Without doing any calculation, indicate whether the margin of error is large or smaller for a $90\%$confidence interval.

When the level of confidence rises, the related critical value rises as well. The margin of error is the product of the statistic's critical value and standard error. As a result, as the level of confidence rises, so does the margin of error. Furthermore, the confidence interval grows bigger. In addition, as the level of confidence drops, the margin of error diminishes.

As a result, the margin of error for a$90\%$confidence interval will be smaller than for a$99\%$confidence interval.