Assignment 22 Answers

MSF / Hours
26 / 2
34.2 / 4.17
29 / 4.42
34.3 / 4.75
85.9 / 4.83
143.2 / 6.67
85.5 / 7
140.6 / 7.08
140.6 / 7.17
40.4 / 7.17
101 / 10
239.7 / 12
179.3 / 12.5
126.5 / 13.67
140.8 / 15.08

An important element in the scheduling procedure of the Union Camp Corporation’s corrugated container plant in Trenton, New Jersey, involved forecasting the amount of processing time each job would require. One piece of information available was the number of thousands of square feet (MSF) in the job. Data for 15 randomly selected jobs processed on a particular printing press are to the left.

(The data are not available in a spreadsheet. Either key them in yourself, or copy and paste from the word document.)

We are interested in forecasting how long it will take to processes two pending jobs. Job A consists of 157.3 MSF and Job B consists of 64.7 MSF.

1. Regress Hours on MSF, report the resulting regression equation, and use that equation to create a point forecast of the number of hours it will take to process Job A. Do the same for Job B. [30 points]

Using Data Analysis/Regression gave the regression line described below.

Coefficients
Intercept / 3.312316042
MSF / 0.044489502

Upon plugging in the MSF’s of the two jobs, we get 10.31 and 6.19 hours as the point forecasts.

Job A / Job B
1 / 1
157.3 / 64.7
Point Forecast / 10.31 / 6.191


2. Create a new variable called MINUTES by multiplying Hours by 60. Regress MINUTES on MSF, report the resulting regression equation, and use that equation to create a point forecast of the number of minutes it will take to process Job A. Do the same for Job B. Comment on how the two point forecasts compare to those given in answer to question 1. [20 points]

The new regression equation “adjusts” automatically to the change from Hours to Minutes. The intercept and coefficient are 60x what they were before, and so are the point forecasts. The probabilities are identical if we express both sigma and X=8 in minutes. SO….regression does not care if the data and hours or in miutes…It can and will adjust.

Coefficients / Standard Error / Upper 95.0% / Job A / Job B
Intercept / 198.7389625 / 84.12638 / 380.483 / 1 / 1
MSF / 2.669370112 / 0.701281 / 4.184395 / 157.3 / 64.7
Point Forecast / 618.6309 / 371.4472
sigma / 166.4158 / 166.4158
X / 480 / 480
Normdist / 0.202411 / 0.742896
tstat / -0.83304 / 0.652299
dof / 13 / 13
1-t.dist.rt / 0.209932 / 0.737213

3. Assume the point forecasts in answer to 1) above are means of normal distributions. Furthermore, assume the standard deviations (σ) of those normal distributions are both known to equal 2.77 hours. Will Job A take less than 8 hours to process? Will Job B? (As always, I’m looking for more than Yes/No answers.) [20 points]

The probability of X being less than 8 is calculated using normdist(8,mean,2.77,true) to be 0.202 and 0.743 respectively.

Job A / Job B
1 / 1
157.3 / 64.7
Point Forecast / 10.31 / 6.191
sigma / 2.77 / 2.77
X / 8 / 8
Normdist / 0.202 / 0.743

4. Answer question 3 but now assuming that the 2.77 is an imperfect estimate of σ. As in other situations, the quality of the estimate of σ is measured by “degrees of freedom”. Please use 13 degrees of freedom when answering. [10 points]

The probability of X being less than 8 is now calculated using the t-distribution with 13 degrees of freedom (to account for the fact that 2.77 is an estimate of sigma). To do this requires the extra step of calculating the t-stat…which is (8-10.31)/2.77 = -.83 for job A and 0.653 for job B. The resulting left tail probabilities are 0.21 and 0.737…VERY close to those from question 3 above.

Job A / Job B
1 / 1
157.3 / 64.7
Point Forecast / 10.31 / 6.191
sigma / 2.77 / 2.77
X / 8 / 8
Normdist / 0.202 / 0.743
tstat / -0.83 / 0.653
dof / 13 / 13
1-t.dist.rt / 0.21 / 0.737

5. One might expect that fuel economy (as measured by miles per gallon) would decrease with the size of a car’s engine (as measured by displacement in liters). Use regression and the exam2 data (linked to the assignment for class 19) to comment (Briefly…yes or no will suffice) on this assumption. Also, provide a point forecast of MPG for a car with a 2.9 liter engine. [20 points]

Regressing MPG on Displacement confirms that there is a negative relationship in the data….the best fit line has a coefficient of displacement of -2.97. The point forecast of MPG goes down 2.97 MPG for each additional liter of engine size. A car with an engine with 2.9 liters has a point forecast of 26.83 MPG

Coefficients / Standard Error
Intercept / 35.44919576 / 0.850190215
Displacement / -2.971357736 / 0.24640952
Point forecast = 35.44 - 2.97*2.9 / 26.83225832