91Ó°ÊÓ

$\stackrel{¯}{x}=30,n=25,s=4$, confidence level $=90\%$

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

Calculate the $90\%$confidence interval for $\stackrel{¯}{x}=30,n=25,s=4$and the requisite degrees of freedom are$1.711$from "Table IV Values of $t_{\mathrm{Î\pm }}$". Thus, the $90\%$confidence interval is,

$$\begin{gathered} \stackrel{¯}{x}Â\pm t_{\frac{a}{2}}\left(\frac{s}{\sqrt{n}}\right)=30Â\pm 1.711\left(\frac{4}{\sqrt{25}}\right) \\ =30Â\pm 1.3688 \\ =(30-1.3688,30+1.3688) \\ =(28.6312,31.3688) \end{gathered}$$

Therefore, the $90\%$confidence interval $\mathrm{μ}$is $(28.6312,31.3688)$

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

$$\begin{gathered} \text{Margin of error}=\frac{\text{Upper limit}-\text{Lower limit}}{2} \\ =\frac{31.3688-28.6312}{2} \\ =\frac{2.7376}{2} \\ =1.3688 \end{gathered}$$

Thus, the margin of error by using the half-length of the confidence interval is $1.3688$

Using the formula, calculate the margin of error.

$$\begin{gathered} \text{Margin of error}\left(E\right)=t_{\frac{a}{2}}\left(\frac{s}{\sqrt{n}}\right) \\ =1.711\left(\frac{4}{\sqrt{25}}\right) \\ =1.3688 \end{gathered}$$

Thus, the margin of error by using the formula is $1.3688$