Question

Trace metals found in wells affect the taste of drinking water, and high concentrations can pose a health risk. Researchers measured the concentration of zinc (in milligrams/liter) near the top and the bottom of $10$randomly selected wells in a large region. The data are provided in the following table.

a. Make a dot-plot of the difference (Bottom – Top) in the zinc concentrations for these $10$wells.

b. Describe what the graph reveals about whether the amount of zinc at the top and bottom of the wells is the same, on average.

c. Calculate the mean difference and the standard deviation of the differences. Interpret the mean difference.

Short answer

Part a. The dot plot is as follows:

Part b. 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.

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

Step-by-step solution

Part a. Step 1. Explanation

The dot plot is as follows:

Part b. Step 1. Explanation

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.

Part c. Step 1. Explanation

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:

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

The sample variance is then as:

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

The sample standard deviation is then,

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

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