In each exercise $8.63-8.68$, we provide a sample mean, sample size, population standard deviation, and confidence level. In each case, perform the following tasks:

a. Use the one-mean $z-$interval procedure to find a confidence interval for the mean of the population from which the sample was drawn.

b. Obtain the margin of error by taking half the length of the confidence interval.

c. Obtain the margin of error by using Formula $8.1$on page $325$

$\tilde{x}=20,n=36,\sigma =3$, confidence level $=95\%$

$\tilde{x}=20,n=36,\sigma =3$, confidence level $=95\%$

The formula used: the confidence interval $\overline{x}\pm z_{\frac{a}{2}}\left(\frac{\sigma }{\sqrt{n}}\right)$and$\text{Margin of error}\left(E\right)=z_{\frac{a}{2}}\left(\frac{\sigma }{\sqrt{n}}\right)$

Compute the $95\%$confidence interval for $\mu$.

Consider $\overline{x}=20,n=36,\sigma =3$, and confidence level is $95\%$.

The required value of $z_{\frac{a}{2}}$ with a$99\%$ confidence level is $1.96$ based on "Table II Areas under the standard normal curve."

Thus, the confidence interval is,

$$\begin{gathered} \overline{x}\pm z_{\frac{a}{2}}\left(\frac{\sigma }{\sqrt{n}}\right)=20\pm 1.96\left(\frac{3}{\sqrt{36}}\right) \\ =20\pm 1.96(0.5) \\ =20\pm 0.98 \\ =(19.02,20.98) \end{gathered}$$

Therefore, the $95\%$confidence interval for $\mu$is $(19.02,20.98)$

Using the half-length of the confidence interval, calculate the margin of error.

$$\begin{gathered} \text{Margin of error}=\frac{20.98-19.02}{2} \\ =\frac{1.96}{2} \\ =0.98 \end{gathered}$$

Thus, the margin of error by using the half-length of the confidence interval is $0.98$

Using a formula, calculate the margin of error.

Thus, the margin of error by using the formula is $0.98$