Chapter 10.3, Problem 76E

(a)

To determine

To make a dot plot of the difference (Bottom minus Top) in the zinc concentrations for these ten wells.

Explanation of Solution

The dot plot is as follows:

(b)

To determine

To describe what the graph reveals about whether the amount of zinc at the top and bottom of the wells is the same on average.

Explanation of Solution

From the above graph in part (a), we have,

We then note that all ten dots in the dot plot lie to the right to zero which implies the most of the difference (Bottom minus Top) are positive and thus the mean zinc concentration at the bottom appears to be greater than the mean zinc concentration at the top.

(c)

To determine

To calculate the mean difference and the standard deviation of the differences and interpret the mean difference.

Answer to Problem 76E

The mean is $0.0804$ and the standard deviation is $0.0523$ .

Explanation of Solution

It is given that:

  $0.015,0.028,0.177,0.121,0.102,0.107,0.019,0.066,0.058,0.111$

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,

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

The difference are on average $0.0804$ milligrams which varies by $0.0523$ milligrams on average.