Question

The given data represent the total compensation for 10 randomly selected CEOs and their​ company's stock performance in 2009. Analysis of this data reveals a correlation coefficient of zequalsnegative 0.2426. What would be the predicted stock return for a company whose CEO made​ $15 million?

Answer 1

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Answer with Explanation:

For simplicity, we denote the compensations with variable x and the stock return with variable y. Let us first find the mean and standard deviation for both compensation (x) and return (y).

Mean of Compensation (x) will be,

$X\;=\;\frac{26.43\;+\;12.03\;+\;19.74\;+\;13.54\;+\;11.97\;+\;11.41\;+\;25.94\;+\;14.46\;+\;17.13\;+\;14.71}{10}$

$X\; = \;16.737$

Mean of Stock Return (y) will be,

$Y\;=\;\frac{5.43\;+\;30.89\;+\;31.89\;+\;80.06\;-\;8.22\;+\;2.89\;+\;4.39\;+\;10.95\;+\;4.18\;+\;11.94}{10}$

$Y\;=\;17.44$

Standard Deviation for Compensation (x) is given by,

$\sigma _{x}\;=\;\sqrt{\frac{\sum (x_{i}-X)^{2}}{N}}$

$\sigma _{x}\;=\;\sqrt{\frac{(26.43-16.73)^{2}+(12.03-16.73)^{2}+....+(14.71-16.73)^{2}}{10}}$

$\sigma _{x}\;=\;5.30$

Standard Deviation for Compensation (x) is given by,

$\sigma _{y}\;=\;\sqrt{\frac{\sum (y_{i}-Y)^{2}}{N}}$

$\sigma _{y}\;=\;\sqrt{\frac{(5.43-17.44)^{2}+(30.89-17.44)^{2}+....+(11.94-17.44)^{2}}{10}}$

$\sigma_{y}\;=\;23.97$

To find the predicted stock return, we have to use the equation for of line of regression,

$y_{required}\;-\;Y\;=\;z\;*\;\frac{\sigma _{x}}{\sigma _{y}}\;*\;(x_{required}-X)\;.........\;(1)$

where,

$z\; =\;Correlation\;coefficient\;=\;-0.2426\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(given)$

$x_{required}\;=\;Compensation\;of\;\ 15\;million\;=\;15$

$y_{required}\;=\;Stock\;return\;at\;x_{required}$

Equation (1) will become,

$y_{required}\;-\;17.44\;=\;-0.2426\;*\;\frac{5.30}{23.97}\;*\;(15\;-\;16.73)$

$y_{required}\;=\;-0.054*\;(-1.73)\;+\;17.44\;$

$y_{required}\;=\;17.533$.