Question

In: Statistics and Probability

Thompson Photo Works purchased several new, highly sophisticated processing machines. The production department needed some guidance...

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 115 8 340 130 31
Sue Trythall 116 8 325 119 21

The equation is:

yˆy⁢^⁢  = 10.1 + 0.2x1 + 0.386x2 + 0.912x3 + 0.005x4

  1. What is this equation called?
  • Multiple regression equation

  • Multiple standard error of estimate

  • Coefficient of determination

  1. How many dependent and independent variables are there?
  1. What is the number 0.386 called?
  • Regression coefficient

  • Coefficient of determination

  • Homoscedasticity

  • Multicollinearity

  1. As age increases by one year, how much does estimated performance on the new machine increase? (Round your answer to 3 decimal places.)
  1. Carl Knox applied for a job at Photo Works. He has been in the business for 11 years and scored 350 on the mechanical aptitude test. Carl’s prior on-the-job performance rating is 85, and he is 69 years old. Estimate Carl’s performance on the new machine. (Round your answer to 3 decimal places.)

Solutions

Expert Solution

Solution:

Given:

x1 = Length of time an employee was in the industry

x2 = Mechanical aptitude test score

x3 = Prior on-the-job rating

x4 = Age

y = Performance on the new machine

The equation is:

y⁢^⁢  = 10.1 + 0.2x1 + 0.386x2 + 0.912x3 + 0.005x4

Part a) What is this equation called?

If regression equation has one independent variable,then it is called simple linear regression equation and if it has more than one independent variables, then it is called Multiple linear regression equation.

Since we have more than 1 independent variables (x1, x2, x3 , x4), this is Multiple regression equation.

Part b) How many dependent and independent variables are there?

Dependent variable is a variable which we want to estimate or predict and independent variables are those which are used to predict to the dependent variable.

Thus we have one dependent variable = y = Performance on the new machine

and 4 independent variables:

x1 = Length of time an employee was in the industry

x2 = Mechanical aptitude test score

x3 = Prior on-the-job rating

x4 = Age

Part c) What is the number 0.386 called?

the number 0.386 is the coefficient of x2 variable, thus this is called as Regression coefficient.

Thus correct answer is: Regression coefficient

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

Variable Age is given by x4 in regression equation. Its regression coefficient is 0.005

thus if age increases by one year, then estimated performance on the new machine increase by 0.005.

Part e) Carl Knox applied for a job at Photo Works. He has been in the business for 11 years and scored 350 on the mechanical aptitude test. Carl’s prior on-the-job performance rating is 85, and he is 69 years old. Estimate Carl’s performance on the new machine.

We have to find Estimated value of Carl’s performance on the new machine.

x1 = Length of time an employee was in the industry = 11 years

x2 = Mechanical aptitude test score = 350

x3 = Prior on-the-job rating = 85

x4 = Age = 69 years

thus

y⁢^⁢  = 10.1 + 0.2x1 + 0.386x2 + 0.912x3 + 0.005x4

y⁢^⁢  = 10.1 + 0.2* 11 + 0.386 * 350 + 0.912 * 85 + 0.005 * 69

y⁢^⁢  = 10.1 + 2.2 + 135.1 + 77.52+ 0.345

y⁢^⁢  = 225.265


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