Find the margin of error.
-$99 \%$ confidence interval; $\mathrm { n } = 201 ; \overline { \mathrm { x } } = 217 ; \mathrm { s } = 34$
A) $6.2$
B) $8.4$
C) $4.7$
D) $5.6$

Question

Quizplus · Accepted Answer

The margin of error is:$$E=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$$where $z_{\alpha/2}$ is the critical value from the standard normal distribution, $\sigma$ is the population standard deviation, and $n$ is the sample size.For a 99% confidence interval, $z_{\alpha/2}=2.576$.Substituting the given values into the formula, we get:$$E=2.576\frac{34}{\sqrt{201}}=6.2$$Therefore, the margin of error is 6.2.

Find the Margin of Error 99%99 \%99% Confidence Interval; n=201;x‾=217;s=34\mathrm { n } = 201 ; \overline { \mathrm { x } } = 217 ; \mathrm { s } = 34n=201;x=217;s=34

Find the Margin of Error $99 \%$ Confidence Interval; $\mathrm { n } = 201 ; \overline { \mathrm { x } } = 217 ; \mathrm { s } = 34$