Question

Thompson Photo Works purchased several new, highly sophisticated processing machines. The production department needed some guidance with respect to qualifications needed by an operator. Is age a factor? Is the length of service as an operator (in years) important? In order to explore further the factors needed to estimate performance on the new processing machines, four variables were listed:

x₁ = Length of time an employee was in the industry

x₂ = Mechanical aptitude test score

x₃ = Prior on-the-job rating

x₄ = Age

Performance on the new machine is designated y.

Thirty employees were selected at random. Data were collected for each, and their performances on the new machines were recorded. A few results are:

	Performance on New Machine,		Length of Time in Industry,		Mechanical Aptitude Score,		Prior On-the-Job Performance,		Age,
Name	y		x1		x2		x3		x4
Mike Miraglia	113		6		350		127		45
Sue Trythall	118		5		329		129		21

The equation is:

yˆy⁢^⁢  = 10.9 + 0.4x₁ + 0.286x₂ + 0.712x₃ + 0.003x₄

As age increases by one year, how much does estimated performance on the new machine increase? (Round your answer to 3 decimal places.)

The estimated performance increases by ________

Carl Knox applied for a job at Photo Works. He has been in the business for 3 years and scored 360 on the mechanical aptitude test. Carl’s prior on-the-job performance rating is 75, and he is 36 years old. Estimate Carl’s performance on the new machine. (Round your answer to 3 decimal places.)

Estimated performance:

Answer 1

(A) we have the regression equation

y⁢  = 10.9 + 0.4x₁ + 0.286x₂ + 0.712x₃ + 0.003x₄

we have to find the estimated value of y when x4 or age increases by one year. This means we have define the slope of x4 or age

Slope constant for age or x4 is +0.003, this means that for every one year increase in age, there will be an increase of 0.003 in the performance {Positive slope means increase in y value}

So, answer is 0.003

(B)

we have the regression equation

y⁢  = 10.9 + 0.4x₁ + 0.286x₂ + 0.712x₃ + 0.003x₄

we have to find the value of y when x1 = 3, x2 = 360, x3 = 75 and x4 = 36

setting these values in the regression equation, we get

y⁢  = 10.9 + 0.4*(3) + 0.286*(360) + 0.712*(75) + 0.003*(36)

this gives us

y = 10.9 + 1.2 + 102.96 + 53.4 + 0.108

or y = 168.568

this is the required estimated performance