Chapter 8.3, Problem 64E

To determine

To calculate: The population mean for the confidence interval of 95%.

Answer to Problem 64E

Thepopulation mean is from16.4872% to 26.5128%.

Explanation of Solution

Given information:

Sample size $n=8$

Confidence interval $c=95\%$

Table for sample is:

1	26
2	31
3	23
4	22
5	11
6	22
7	14
8	31

Formula used:

  $\text{Sample mean}\left(\overline{x}\right)=\frac{\text{sum}\text{of}\text{the}\text{samples}}{\text{total}\text{number}\text{of}\text{sample}}$

  $\text{Standard deviation}\left(s_x\right)=\sqrt{\frac{{\displaystyle \sum {\left(x_i-\overline{x}\right)}^2}}{n-1}}$

Degree of freedom $df=n-1$

Margin of error $E=t^{*}\frac{s_x}{\sqrt{n}}$

Calculation:

Find the mean, use the formula $\text{Sample mean}\left(\overline{x}\right)=\frac{\text{sum}\text{of}\text{the}\text{samples}}{\text{total}\text{number}\text{of}\text{sample}}$ .

  $\begin{array}{c}\overline{x}=\frac{\text{sum}\text{of}\text{the}\text{samples}}{\text{total}\text{number}\text{of}\text{sample}}\\ =\frac{26+31+23+22+11+22+14+31}{8}\\ =22.5\end{array}$

Find the standard deviation, use the formula $\text{Standard deviation}\left(s_x\right)=\sqrt{\frac{{\displaystyle \sum {\left(x_i-\overline{x}\right)}^2}}{n-1}}$ .

  $\begin{array}{c}{s}_x=\sqrt{\frac{{\displaystyle \sum {\left(x_i-\overline{x}\right)}^2}}{n-1}}\\ =\sqrt{\frac{{\left(26-22.5\right)}^2+\cdots +{\left(31-22.5\right)}^2}{8-1}}\\ =7.191\end{array}$

Now, find the degree of freedom, use the formula $df=n-1$ .

  $\begin{array}{c}df=n-1\\ =8-1\\ =7\end{array}$

Know that, the confidence level is 95%.

Convert 95% into decimal.

  $\frac{95}{100}=0.95$

Find the column.

  $\begin{array}{c}\frac{1-c}{2}=\frac{1-0.95}{2}\\ =0.025\end{array}$

From table B find critical value $t^{*}$ , use the row $df$ and column.

Thus, the critical value of $t^{*}\text{is}2.365$ .

Find the margin of error, use the formula $E=t^{*}\frac{s_x}{\sqrt{n}}$ .

  $\begin{array}{c}E=t^{*}\frac{s_x}{\sqrt{n}}\\ =2.365\times \frac{7.191}{\sqrt{8}}\\ =6.0128\end{array}$

Now, find the population mean.

  $\begin{array}{l}\overline{x}-E<\mu <\overline{x}+E\\ 22.5-6.0128<\mu <22.5+6.0128\\ 16.4872<\mu <28.5128\end{array}$

Hence, the required population mean is from 16.4872% to 26.5128%.