91Ó°ÊÓ

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

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

Compute the $95\%$confidence interval for $\mathrm{μ}$.

Consider $\stackrel{¯}{x}=20,n=36,\mathrm{σ}=3$, 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,

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

Therefore, the $95\%$confidence interval for $\mathrm{μ}$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 $0.98$