Question

The table shows the percentage P of an employee's annual salary S, in thousands of dollars, spent on the process of recruiting and training a replacement.

S = annual salary (thousands of dollars)	P = percentage of salary for recruiting
20	15
30	16
40	19
60	45
70	50

Find the equation of the regression line. (Round regression line parameters to two decimal places.)

P(S) =

What cost does the regression line give for replacement of an employee who makes an annual salary of $50,000? (Round your answer to the nearest dollar.)

Answer 1

Here S (salary) is the independent variable and P (percentage) is the dependent variable.

We know that, trend line = P = a + b*S

Where,

a = {(∑P)*(∑S2) – (∑S)*(∑S*P)} / {(n)*(∑S2) – (∑S)*(∑S)}

b = {(n)*(∑S*P) – (∑S)*(∑P)} / {(n)*(∑S2) – (∑S)*(∑S)}

S (IN 1000s)	P	S*P	S^2
20	15	300	400
30	16	480	900
40	19	760	1600
60	45	2700	3600
70	50	3500	4900
220	145	7740	11400

Here,

∑P = 145

∑S = 220

∑S*P = 7740

∑S2 = 11400

n = 5

Putting everything to find a and b

a = {(145*11400) – (220*7740)}/{(5*11400) – (220*220)}

= (1653000 – 1702800)/(57000 – 48400)

= -49800/8600

= -5.79

a = -5.79

b = {(5*7740) – (220*145)}/{(5*11400) – (220*220)}

= (38700 – 31900)/(57000 – 48400)

= 6800/8600

= 0.79

b = 0.79

Hence, P = 0.79*S – 5.79

For S = 50 (in 1000s)

P = 0.79*50 – 5.79 = 39.5 – 5.79 = 33.71 = 34

Hence For Salary 50000, P = 34

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