Chapter 8.3, Problem 63E

To determine

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

Answer to Problem 63E

Thepopulation mean is from17.0218% to 19.9382%.

Explanation of Solution

Given information:

Sample size $n=20$

Confidence interval $c=95\%$

Table for sample

1	15.8
2	13.6
3	15.6
4	19.1
5	22.4
6	15.6
7	22.5
8	17.2
9	19.4
10	22.6
11	19.4
12	18.0
13	14.6
14	18.7
15	21.0
16	14.8
17	22.6
18	21.5
19	14.3
20	20.9

Formula used:

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

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

Degree of freedom $df=n-1$

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

Calculation:

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

  $\begin{array}{c}\overline{x}=\frac{\text{sum}\text{of}\text{the}\text{samples}}{\text{total}\text{number}\text{of}\text{sample}}\\ =\frac{15.8+13.6+\cdots +14.3+20.9}{20}\\ =18.48\end{array}$

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

  $\begin{array}{c}{s}_x=\sqrt{\frac{{\displaystyle \sum {\left(x_i-\overline{x}\right)}^2}}{n-1}}\\ =\sqrt{\frac{{\left(15.8-18.48\right)}^2+\cdots +{\left(20.9-18.48\right)}^2}{20-1}}\\ =3.1158\end{array}$

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

  $\begin{array}{l}df=n-1\\ =20-1\\ =19\end{array}$

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, critical value $t^{*}\text{is}2.093$ .

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

  $\begin{array}{c}E=t^{*}\frac{s_x}{\sqrt{n}}\\ =2.093\times \frac{3.1158}{\sqrt{20}}\\ =1.4582\end{array}$

Now, find the population mean.

  $\begin{array}{l}\overline{x}-E<\mu <\overline{x}+E\\ 18.48-1.4582<\mu <18.48+1.4582\\ 17.0218<\mu <19.9382\end{array}$

Hence, the required population mean is from 17.0218% to 19.9382%.