91影视

The graph is given below:

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

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,

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

$\text{Term=483}$

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

$\text{Term=536.5}$

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

$\text{Term=644.25}$

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

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

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

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.

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

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

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

Mean = 641.75 and Standard Deviation = 314.77

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

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

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.

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.