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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,
[tex]X\;=\;\frac{26.43\;+\;12.03\;+\;19.74\;+\;13.54\;+\;11.97\;+\;11.41\;+\;25.94\;+\;14.46\;+\;17.13\;+\;14.71}{10}[/tex]
[tex]X\; = \;16.737[/tex]
Mean of Stock Return (y) will be,
[tex]Y\;=\;\frac{5.43\;+\;30.89\;+\;31.89\;+\;80.06\;-\;8.22\;+\;2.89\;+\;4.39\;+\;10.95\;+\;4.18\;+\;11.94}{10}[/tex]
[tex]Y\;=\;17.44[/tex]
Standard Deviation for Compensation (x) is given by,
[tex]\sigma _{x}\;=\;\sqrt{\frac{\sum (x_{i}-X)^{2}}{N}}[/tex]
[tex]\sigma _{x}\;=\;\sqrt{\frac{(26.43-16.73)^{2}+(12.03-16.73)^{2}+....+(14.71-16.73)^{2}}{10}}[/tex]
[tex]\sigma _{x}\;=\;5.30[/tex]
Standard Deviation for Compensation (x) is given by,
[tex]\sigma _{y}\;=\;\sqrt{\frac{\sum (y_{i}-Y)^{2}}{N}}[/tex]
[tex]\sigma _{y}\;=\;\sqrt{\frac{(5.43-17.44)^{2}+(30.89-17.44)^{2}+....+(11.94-17.44)^{2}}{10}}[/tex]
[tex]\sigma_{y}\;=\;23.97[/tex]
To find the predicted stock return, we have to use the equation for of line of regression,
[tex]y_{required}\;-\;Y\;=\;z\;*\;\frac{\sigma _{x}}{\sigma _{y}}\;*\;(x_{required}-X)\;.........\;(1)[/tex]
where,
[tex]z\; =\;Correlation\;coefficient\;=\;-0.2426\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(given)[/tex]
[tex]x_{required}\;=\;Compensation\;of\;\$15\;million\;=\;15[/tex]
[tex]y_{required}\;=\;Stock\;return\;at\;x_{required}[/tex]
Equation (1) will become,
[tex]y_{required}\;-\;17.44\;=\;-0.2426\;*\;\frac{5.30}{23.97}\;*\;(15\;-\;16.73)[/tex]
[tex]y_{required}\;=\;-0.054*\;(-1.73)\;+\;17.44\;[/tex]
[tex]y_{required}\;=\;17.533[/tex].
