Chapter 8.3, Problem 70E

(a)

To determine

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

Answer to Problem 70E

Thepopulation mean is from0.0430% to 0.1178%.

Explanation of Solution

Given information:

Sample size $n=10$

Confidence interval $c=95\%$

Table for sample is:

Well	Bottom	Top
1	0.430	0.415
2	0.226	0.238
3	0.567	0.390
4	0.531	0.410
5	0.707	0.605
6	0.716	0.609
7	0.651	0.632
8	0.589	0.523
9	0.469	0.411
10	0.723	0.612

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

  $\begin{array}{c}\overline{x}=\frac{\text{sum}\text{of}\text{the}\text{samples}}{\text{total}\text{number}\text{of}\text{sample}}\\ =\frac{0.015+0.028+\cdots +0.058+0.111}{10}\\ =0.0804\end{array}$

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

  $\begin{array}{c}{s}_x=\sqrt{\frac{{\displaystyle \sum {\left(x_i-\overline{x}\right)}^2}}{n-1}}\\ =\sqrt{\frac{{\left(0.015-0.0804\right)}^2+\cdots +{\left(0.111-0.0804\right)}^2}{10-1}}\\ =\sqrt{0.0025}\\ =0.0523\end{array}$

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

  $\begin{array}{c}df=n-1\\ =10-1\\ =9\end{array}$

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 $t^{*}\text{is}2.262$ .

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

  $\begin{array}{c}E=t^{*}\frac{s_x}{\sqrt{n}}\\ =2.262\times \frac{0.0523}{\sqrt{10}}\\ =0.0374\end{array}$

Now, find the population mean.

  $\begin{array}{c}\overline{x}-E<\mu <\overline{x}+E\\ 0.0804-0.0374<\mu <0.0804-0.0374\\ 0.0430<\mu <0.1178\end{array}$

Hence, the required population mean is from 0.0430% to 0.1178%.

(b)

To determine

Explain whether or not the confidence interval gives convincing evidence of a difference in zinc concentrations at the top and bottom of wells in the region.

Answer to Problem 70E

The confidence interval is sufficient to give convincing evidence of a difference in zinc concentrations at the top and bottom of wells in the region.

Explanation of Solution

From 70(a)confidence interval contain the population mean from0.0430% to 0.1178%. Since the confidence interval lies above 0.

Hence, the confidence interval is sufficient to give convincing evidence of a difference in zinc concentrations at the top and bottom of wells in the region.