Question

In Exercises 21–24, find the mean and median for each of the two samples, then compare the two sets of results.

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 Data and Story Library (DASL) website. Do the limited data listed here show evidence of stealing by Brinks employees?

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 summarized results for the mean and median are :

Measures	Contractor was Brinks	Contractor was not Brinks
Mean	$1.55 millions	$1.73millions
Median	$1.55millions	$1.65millions

Yes, the sample results are suggestive of stealing by Brinks employees.

Step-by-step solution

Given information

The amounts collected by Brinks and others are recorded in millions of dollars.

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

Compute mean for each data set

The formula for the mean of a data set is:

$\overline{x}=\frac{\sum x}{n}$, where xis the observations and nis the count of the observations.

The mean value when the collection contractor was Brinks is given as:

$$\begin{gathered} {\overline{x}}_B=\frac{1.3+1.5+1.3+...+1.6}{10} \\ =\frac{15.5}{10} \\ \approx 1.55 \end{gathered}$$

Thus, when the contractor was Brinks, the mean value was $1.55 million.

The mean value when the collection contractor was not Brinks.

$$\begin{gathered} {\overline{x}}_O=\frac{2.2+1.9+1.5+...+1.8}{10} \\ =\frac{17.3}{10} \\ =1.73 \end{gathered}$$

Thus, the mean value when the contractor was not Brinks was $1.73 million.

Compute the median for each set of measurements

The steps to compute the median measure are as follows:

• Obtain counts of measurement; n.
• When n is even, the median is the mean of the middle values.
• When n is odd, the median is the middle value.

Compute the median for the amounts when the contractor was Brinks.

The number of observations is10.

Arrange the observations in ascending order.

1.3
1.3
1.4
1.5
1.5
1.6
1.7
1.7
1.7
1.8

The middlemost observations are1.5 and 1.6.

The median is given as:

$$\begin{gathered} M_B=\frac{1.5+1.6}{2} \\ =\frac{3.1}{2} \\ =1.55 \end{gathered}$$

Thus, the median for the amounts when the contractor was Brinks is $1.55 million.

Compute the median for the amounts when the contractor was not Brinks.

The number of observations is10.

Arrange the observations in ascending order.

1.5
1.5
1.6
1.6
1.6
1.7
1.8
1.9
1.9
2.2

The middlemost observations are 1.6 and 1.7.

The median is given as:

$$\begin{gathered} M_O=\frac{1.6+1.7}{2} \\ =\frac{3.3}{2} \\ =1.65 \end{gathered}$$

Thus, the median for the amounts when the contractor was not Brinks is $1.65 million.

Summarize and derive conclusions from the results

The summarized results are:

Measures	Contractor was Brinks	Contractor was not Brinks
Mean	1.55	1.73
Median	1.55	1.65

The mean and median values when the collection contractor was Brinks remained the same. On the other hand, the results vary in the case of the other contractor.

Also, both measures have lower values and exactly the same value when the collection contractor was Brinks. The observation may be suggestive of stealing by Brinks employees.