Thompson Photo Works purchased several new, highly sophisticated processing mach

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:

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

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

Carl Knox applied for a job at Photo Works. He has been in the business for 6 years and scored 300 on the mechanical aptitude test. Carl’s prior on-the-job performance rating is 75, and he is 54 years old. Estimate Carl’s performance on the new machine. (Round your answer to 3 decimal places.)

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:

Explanation / Answer

Question a)

What is this equation called?

Multiple regression equation

Question b)

How many dependent and independent variables are there?

One dependent

Four independent

Question c)

What is the number 0.612 called?

Regression coefficient

Question d)

As age increases by one year, how much does estimated performance on the new machine increases?

As age increases by one year the estimated performance on the new machine increases by 0.005 unit.

Question e)

y^ = 10.2 + 02X1 + 0.786X2 +0.612x3 + 0.005X4

    = 10.2 + (02*6) + (0.786*300) + (0.612*75) + (0.005*54)

    = 293.370

The estimated Carl’s performance on the new machine is 293.370

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