Question

Best-paid CEOs.Refer to Glassdoor Economic Research firm’s 2015 ranking of the 40 best-paid CEOs in Table 2.1 (p. 65). Recall that data were collected on a CEO’s current salary, age, and the ratio of salary to a typical worker’s pay at the firm.

a.Create a scatterplot to relate a CEO’s ratio of salary to worker pay to the CEO’s age. Comment on the strength of the association between these two variables.

b.Conduct an outlier analysis of the ratio variable. Identify the highly suspect outlier in the data.

c.Remove the highly suspect outlier from the data and recreate the scatterplot of part a. What do you observe?

Short answer

a. The graph is given below:

No strong association

b.Highly suspect outlier = 1951

c. No major change

Step-by-step solution

Creating a Scatterplot

The graph is given below:

There is no visible strong association between the CEO’s age and the ratio of the CEO’s salary to that of a typical worker. The data is clustered to the left of the graph except for a few scattered points.

Conducting an outlier analysis and identifying the highly suspect outliers

We will use the box plot method to conduct an outlier analysis on the ratio variable.

We will first calculate the quartiles, IQR, Inner, and Outer fences.

Arranging the data in ascending order,

$\begin{array}{c}\text{Q1=}\frac{\text{N+1}}{\text{4}}\\ =\frac{40+1}{4}\\ =\frac{41}{4}\\ {\text{=10.25}}^{\text{th}}\end{array}$

$\text{Term=483}$

$\begin{array}{c}\text{Q2=}\frac{\text{N+1}}{\text{2}}\\ =\frac{40+1}{2}\\ =\frac{41}{2}\\ {\text{=20.25}}^{\text{th}}\end{array}$

$\text{Term=536.5}$

$\begin{array}{c}\text{Q3=}\frac{\text{3}\left(\text{N+1}\right)}{\text{4}}\\ \text{=}\frac{3\left(40+1\right)}{4}\\ =\frac{123}{4}\\ {\text{=30.75}}^{\text{th}}\end{array}$

$\text{Term=644.25}$

$\begin{array}{c}{\text{IQR=Q}}_{\text{U}}{\text{–Q}}_{\text{L}}\\ \text{=644.25–483}\\ \text{=161.25}\end{array}$

$\begin{array}{c}\text{LowerInnerFence}={\text{Q}}_{\text{L}}-\text{1}.\text{5}\left(\text{IQR}\right)\\ =\text{483}–\text{1}.\text{5}\left(\text{161}.\text{25}\right)\\ =\text{483}–\text{241}.\text{875}\\ =\text{241}.\text{125}\end{array}$

$\begin{array}{c}\text{UpperInnerFence}={\text{Q}}_{\text{U}}+\text{1}.\text{5}\left(\text{IQR}\right)\\ =\text{644}.\text{25}+\text{1}.\text{5}\left(\text{161}.\text{25}\right)\\ =\text{644}.\text{25}+\text{241}.\text{875}\\ =\text{886}.\text{125}\end{array}$

Now we will plot these,

The graph shows thatthe outliers are 1951, 1522, 1192, 1133, and 939.

To find which of these are highly suspect outliers (with a z-score > 3), we will check the z-score of these outliers.

First, we will calculate the mean and standard deviation of the data.

$\begin{array}{c}\text{Mean=}\frac{\text{Sumofallobservations}}{\text{No.ofobservations}}\\ \text{=}\frac{25670}{40}\\ =641.75\end{array}$

$\begin{array}{c}\text{Variance=}\frac{{\displaystyle \sum {\left(\text{χ}-\right)}^2}}{\text{n}-\text{1}}\\ \text{=}\frac{\text{3864170}}{\text{39}}\\ \text{=99081.28}\end{array}$

$\begin{array}{c}\text{StandardDeviation=}\sqrt{\text{Variance}}\\ =\sqrt{99081.28}\\ =314.77\end{array}$

Mean = 641.75 and Standard Deviation = 314.77

$\begin{array}{c}\text{z-scoreof1951=}\frac{\text{1951}-\text{641.75}}{\text{314.77}}\\ \text{=}\frac{\text{1309.25}}{\text{314.77}}\\ \text{=4.159}\end{array}$

$\begin{array}{c}\text{z-scoreof1522=}\frac{\text{1522}-\text{641.75}}{\text{314.77}}\\ \text{=}\frac{\text{880.25}}{\text{314.77}}\\ \text{=2.79}\end{array}$

Based on the values of the z-score, 1522 is not a highly suspect outlier. Therefore, none of the values lower than 1522 are outliers.

So, the only highly suspect outlier is 1951.

Creating a scatterplot without the highly suspect outlier.

1951 is a highly suspect outlier. Therefore, we will remove that from the data and create the scatterplot.

There is not much difference in the scatterplot after removing the highly suspect outlier.