Question

In Exercises 21–24, find the coefficient of variation for each of the two samples; then compare the variation. (The same data were used in Section 3-1.) 21.

Parking Meter Theft Listed below are amounts (in millions of dollars) collected from parking meters by Brinks and others in New York City during similar time periods. A larger data set was used to convict five Brinks employees of grand larceny. The data were provided by the attorney for New York City, and they are listed on the DASL Website. Do the two samples appear to have different amounts of variation?

Collection Contractor Was Brinks 1.3 1.5 1.3 1.5 1.4 1.7 1.8 1.7 1.7 1.6

Collection Contractor Was Not Brinks 2.2 1.9 1.5 1.6 1.5 1.7 1.9 1.6 1.6 1.8

Short answer

The coefficient of variation for data collected byBrinksis equal to 11.5%.

The coefficient of variation for data collected by other contractors is equal to 12.8%.

The amount of variation in the data collected by Brinks and by other contractors is approximately the same.

Step-by-step solution

Given information

The data is collected from parking meters by Brinks and by other contractors and is listed in two samples containing 10 observations each.

Coefficient of variation

The sample coefficient of variation is the quotient of the sample standard deviation to the sample mean. The formula of coefficient of variation is as follows:

$C.V.=\frac{s}{\overline{x}}\times 100$

, where

$s$is the sample standard deviation;

$\overline{x}$is the sample mean.

Computation of the first sample

The first sample is defined as collections made when the contractor was Bricks.

The mean of sample 1 is given as follows:

$$\begin{gathered} {\overline{x}}_1=\frac{\left(1.3+1.5+1.3+1.5+1.4+1.7+1.8+1.7+1.7+1.6\right)}{10} \\ =\frac{15.5}{10} \\ =1.55 \end{gathered}$$

The standard deviation of sample 1 is given as follows:

$$\begin{gathered} s_1=\sqrt{\frac{{\sum }_{i=1}^n{\left(x_i-{\overline{x}}_1\right)}^2}{n_1-1}} \\ =\sqrt{\frac{{\left(1.3-1.55\right)}^2+{\left(1.5-1.55\right)}^2+....+{\left(1.6-1.55\right)}^2}{10-1}} \\ =0.18 \end{gathered}$$

Thus, the mean for sample 1 is equal to 1.55 million dollars, and the standard deviation of sample 1 is equal to 0.18 million dollars.

The coefficient of variation for sample 1 is computed as follows:

$$\begin{gathered} C{V}_1=\frac{s_1}{{\overline{x}}_1}\times 100 \\ =\frac{0.18}{1.55}\times 100 \\ =11.5\% \end{gathered}$$

Therefore, the coefficient of variation for sample 1 is equal to 11.5%.

Computation of the second sample

The second sample is defined as collections made when the contractor was not Bricks.

The mean of sample 2 is given as follows:

$$\begin{gathered} {\overline{x}}_2=\frac{\left(2.2+1.9+1.5+1.6+1.5+1.7+1.9+1.6+1.6+1.8\right)}{10} \\ =\frac{17.3}{10} \\ =1.73 \end{gathered}$$

The standard deviation of sample 2 is given as follows:

$$\begin{gathered} s_2=\sqrt{\frac{{\sum }_{i=1}^n{\left(x_i-{\overline{x}}_2\right)}^2}{n_2-1}} \\ =\sqrt{\frac{{\left(2.2-1.73\right)}^2+{\left(1.9-1.73\right)}^2+....+{\left(1.8-1.73\right)}^2}{10-1}} \\ =0.22 \end{gathered}$$

Thus, the mean for sample 2 is equal to 1.73 million dollars, and the standard deviation of sample 2 is equal to 0.22 million dollars.

The coefficient of variation for sample 2 is computed as follows:

$$\begin{gathered} C{V}_2=\frac{s_2}{{\overline{x}}_2}\times 100 \\ =\frac{0.22}{1.73}\times 100 \\ =12.8\% \end{gathered}$$

Therefore, the coefficient of variation for sample 2 is equal to 12.8%.

Comparing the measures for two samples

For the given two samples, the coefficients of variation for the two samples are comparable in magnitude.

Therefore, the two samples appear to have the sameamount of variation.