In each of Exercises $8.123-8.128$, we provide a sample mean, sample size, sample standard deviation, and confidence level. In each exercise,

a. use the one-mean $t-$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 the formula $t_{u/2}+s/\sqrt{n}$.

$8.123\overline{x}=20,n=36,s=3$, confidence level $=95\%$

$8.123\overline{x}=20,n=36,s=3$, confidence level $=95\%$

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

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

Consider $\overline{x}=20,n=36,s=3$a $95\%$confidence level.

The needed value of $t_{\frac{\alpha }{2}}$for $95\%$confidence with $35(=36-1)$degrees of freedom is $2.030$, according to "Table IV Values of $t_{\alpha }$"

Thus, the confidence interval is,

$$\begin{gathered} \overline{x}\pm t_{\frac{\alpha }{2}}\left(\frac{s}{\sqrt{n}}\right)=20\pm 2.030\left(\frac{3}{\sqrt{36}}\right) \\ =20\pm 2.030(0.5) \\ =20\pm 1.015 \\ =(18.985,21.015) \end{gathered}$$

Therefore, the $95\%$confidence interval for $\mu$is $(18.985,21.015)$

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

$$\begin{gathered} \text{Margin of error}=\frac{21.015-18.985}{2} \\ =\frac{2.03}{2} \\ =1.015 \end{gathered}$$

Thus, the margin of error by using the half-length of the confidence interval is $1.015$.

Obtain the margin of error by using the formula.

$$\begin{gathered} Margin\text{of error}\left(E\right)=t_{\frac{\alpha }{2}}\left(\frac{s}{\sqrt{n}}\right) \\ =2.030\left(\frac{3}{\sqrt{36}}\right) \\ =2.030(0.5) \\ =1.015 \end{gathered}$$

Thus, the margin of error by using the formula is $1.015$