Q. 3.173 - Elementary Statistics

Question

In given exercise.

a. obtain and interpret the quartiles.

b. determine and interpret the interquartile range.

c. find and interpret the five-number summary.

d. identify potential outliers, if any:

e. construct and interpret a boxplot.

An issue of Brokerage Report discussed the capital spending of telecommunications companies in the United States and Canada. The capital spending in thousands of dollars for each of $27$ telecommunications companies is shown in the following table.

Step-by-Step Solution

Answer

a) The quartiles are $660, 1800, 4749.5.$From the result we can observe that $25\%$of the capital spending of telecommunications companies in the U.S and Canada are less than $4749.5$dollars.

b) The interquartile range for the given data set is $4089.5.$ From the result we can observe that $50\%$of the capital spending of telecommunications companies in the U.S and Canada are over $4089.5$ thousands of dollars.

c) The variation between the quartiles is huge and the largest variation in the whole data set is the fourth quartile. 

d) The potential outliers are $11189 and 17341.$

e) The boxplot is 

Step 1/10

Given Information:

An issue of Brokerage Report discussed the capital spending of telecommunications companies in the United States and Canada. The capital spending in thousands of dollars for each of $27$ telecommunications companies is shown in the following table.

Step 2/10

a) We have to find the quartiles.

Arrange the data in increasing order.

$\begin{array}{llllllllllllll}21& 70& 125& 195& 389& 649& 656& 664& 682& 1006& 1300& 1403& 1433& 1800\\ 1982& 2205& 2515& 3027& 3634& 4200& 5299& 5947& 7886& 8543& 9310& 11189& 17341& \end{array}$

The number of observations is $27.$ The middle term of the given data is the median.

Therefore median is $1800.$

The second quartile is $1800.$

Step 3/10

Consider the first part of the data which is below the median.

$\begin{array}{|llllllllllllll|}\hline 21& 70& 125& 195& 389& 649& 656& 664& 682& 1006& 1300& 1403& 1433& 1800\\ \hline\end{array}$

The number of observations is $14.$

Median is

$\begin{array}{r}=\frac{(n+1)}{2}\\ =\frac{15}{2}\\ =7.5\end{array}$

Hence the first quartile is

$\begin{array}{r}{Q}_1=\frac{656+664}{2}\\ =660\end{array}$

The first quartile is $660.$

Step 4/10

Now consider the second part of the data set which is greater than the median.

$\begin{array}{|lllllllll|}\hline 1800& 1982& 2205& 2515& 3027& 3634& 4200& 5299& 5947\\ 7886& 8543& 9310& 11189& 17341& & & & \\ \hline\end{array}$

The number of observations is $14.$

So the median will be at

$\begin{array}{r}=\frac{(n+1)}{2}\\ =\frac{15}{2}\\ =7.5\end{array}$

Step 5/10

The third quartile of the data set is 

$\begin{array}{r}{Q}_3=\frac{4200+5299}{2}\\ =4749.5\end{array}$

We can see that $25\%$of the capital spending of telecommunications companies in the U.S and Canada are less than $4749.5$dollars.

Step 6/10

b) We can now find the interquartile range.

The interquartile range is the difference between the first and third quartile.

$\begin{array}{r}IQR=Q_3-Q_1\\ =4749.5-660\\ =4089.5\end{array}$

Now we can see that $50\%$of the capital spending of telecommunications companies in the U.S and Canada are over $4089.5$ thousands of dollars.

Step 7/10

c) We find the five-number summary is shown below:

The minimum value in the data set is $21.$

The lower quartile of the data, $Q_1=660.$

The median of the data set $Q_2=1800.$

The upper quartile of the data,$Q_3=4749.5$

The maximum value in the data set is $17341.$

Step 8/10

The measure of variation in the middle quarter is

$\begin{array}{r}=Q_2-Q_1\\ =1800-660\\ =1140\end{array}$

The measure of variation in the third quarter is

$\begin{array}{r}=Q_3-Q_2\\ =4749.5-1800\\ =2949.5\end{array}$

The measure of variation in the first quarter is

$\begin{array}{r}=Q_1-Min\\ =660-21\\ =639\end{array}$

The measure of variation in fourth quarter is

$\begin{array}{r}=Max-Q_3\\ =17341-4749.5\\ =12591.5\end{array}$

We can observe that the variation between the quartiles is huge and the largest variation in the whole data set is the fourth quartile.

Step 9/10

d) Find the upper and lower limits of the data set.

$\begin{array}{r}\text{ Lower limit }=Q_1-1.5\left(IQR\right)\\ =660-1.5\left(4089.5\right)\\ =-5474.25\\ \text{ Upper limit }=Q_3+1.5\left(IQR\right)\\ =4749.5+1.5\left(4089.5\right)\\ =10883.75\end{array}$

Potential outliers are the observations that lie above or below the limit. Here $11189, 17341$are the potential outliers.

Step 10/10

e) Using MINITAB we can construct the boxplot.

The asterisk represents the potential outlier of the data.