Chapter 10.3, Problem 80E

To determine

To construct and interpret a $95\%$ confidence interval for the true mean difference (Bottom minus Top) in the zinc concentrations for the wells.

Answer to Problem 80E

We are $95\%$ confident that the true mean zinc concentration at the bottom of the well is between $0.0430\text{ and }0.1178$ higher than the true mean zinc concentration at the top of the well.

Explanation of Solution

T is given:

  $\begin{array}{l}n=10\\ c=0.95\end{array}$

The mean is:

  $\begin{array}{l}\overline{x}=\frac{{\displaystyle \sum _{i=1}^n{x}_i}}{n}\\ =\frac{\begin{array}{l}0.015+0.028+0.177+0.121+0.102+0.107+0.019+0.066+\\ 0.058+0.111\end{array}}{10}\\ =\frac{0.804}{10}\\ =0.0804\end{array}$

The sample variance is then as:

  $\begin{array}{l}{s}^2=\frac{{\displaystyle \sum {(x-\overline{x})}^2}}{n-1}\\ =\frac{\begin{array}{}$ {(0.015-0.0804)}^2+{(0.028-0.0804)}^2+{(0.177-0.0804)}^2+{(0.121-0.0804)}^2+{(0.102-0.0804)}^2+{(0.107-0.0804)}^2+$$ {(0.019-0.0804)}^2+{(0.066-0.0804)}^2+{(0.058-0.0804)}^2+{(0.111-0.0804)}^2$\end{array}}{10-1}\\ =0.0027\end{array}$

The sample standard deviation is then,

  $\begin{array}{l}s=\sqrt{s^2}\\ =\sqrt{0.0027}\\ =0.0523\end{array}$

Now, the degree of freedom will be:

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

Now, the value of t will be:

  $t_{\alpha /2}=2.262$

Now, the confidence interval will be:

  $\begin{array}{l}\overline{x}-t_{\alpha /2}\times \frac{s}{\sqrt{n}}\\ =0.0804-2.262\times \frac{0.0523}{\sqrt{10}}\\ =0.0430\\ \overline{x}+t_{\alpha /2}\times \frac{s}{\sqrt{n}}\\ =0.0804+2.262\times \frac{0.0523}{\sqrt{10}}\\ =0.1178\end{array}$

Thus we conclude that we are $95\%$ confident that the true mean zinc concentration at the bottom of the well is between $0.0430\text{ and }0.1178$ higher than the true mean zinc concentration at the top of the well.