MGT 575
Quantitative Forecasting
Problem Set Key
Problem 13, pp. 228-229
From the information presented, we can infer that Josh believes that sales depend on population. Therefore, sales would be our dependent (Y) variable and population would be our independent (X) variable. Graphing the data in Excel shows a fairly linear trend, so linear regression would be appropriate.
Sales vs Population
1600
1400
1200
1000
800
600
400
200
0
0 5 10
Population (millions)
15 20
Using POM for Windows forecasting module yields the following:

a) Sales in thousands(Y) = 190.94 + 69.81Population in millions (X) b) Regression equation interpretation: Y intercept Slope r = .94 This says that the regression line is a good fit to the data. r
2
= .89 This says that 89% of the variation in sales can be explained by variation in population.
MAD = 100.38 This is the mean of the absolute values of the forecast errors. We would like this to be close to 0.
MSE = 12,216.6 This is the mean of the squared forecast errors. We would like this to be close to 0.
Bias = 0 This is not a surprise. Bias should = 0 when using regression.
Standard Error = 122.19 This value may be used to construct confidence intervals around our forecast.
Overall, the MAD and MSE are higher than desired, but the r and r
2
indicate this is a pretty good forecasting model. c) Territory 1 Sales = 190.94 + 69.81(7.43) = 709.6 thousand
Territory 1 Sales = 190.94 + 69.81(3.87) = 461.1 thousand

Problem 15, pp. 229-230
There are several approaches to this problem depending upon your assumptions.
1) You could do separate forecasts for each time period (7 forecasts each with n = 4)
2) You could sum the calls for each day and do one forecast for the entire day.
I will assume 2) for simplicity
Graphing the data in Excel shows no pronounced trend (this is iffy since n is so small), so exponential smoothing would be appropriate.
Total Monday Calls
540
520
500
480
460
440
420
1 2 3 4 time
Using POM for Windows forecasting module yields the following for α = .3:

Do the same for α = .7 b) Based on MAD, α = .3 is better, but upon further analysis, it is not optimum.
MAD for α = .3 is 39.26
MAD for α = .7 is 40.33
Going the next step and allowing POM for Windows to select the optimum value of α, yields α = 1 and a MAD of 37.
Probably the best way to forecast using these data is to recognize that when plotted as a single time series, there appears to be a seasonal peak around noon. Enter the entire data set (n = 28) into POM for Windows and use the Decomposition of a Time Series approach to forecasting. Using the additive approach yields the following printout:
Use the Periods 29-35 as your forecasts by time period for next Monday.
Note that MAD is 5.19 – much smaller than for the exponential smoothing approaches.

The following graph shows that this is probably the best approach.