Final Exam Practice Problem Solutions
The following table shows the weekly sales for the number of gallons of deck sealer at the local hardware store
during the spring months and the smoothing values for a 4-point moving average and simple exponential method
with α of .2.
Week
1
2
3
4
5
6
7
8
GallonCansSold MA(4)
SES of .2
106
106
110
106.8
108
107.04
97
105.25
105.03
210
131.25
126.03
136
137.75
128.02
128
142.75
128.02
134
152
129.21
1. Calculate a simple moving average with a length of 4 for the sales of cans of deck sealer.
The column labeled MA(4) lists the calculated moving averages for each week.
The moving average for week 4, the first value in the series, is
4
~
S ales =
∑ sales
1
4
=
106 + 110 + 108 + 97
= 105.25
4
The moving average for week 5 is
5
~
S ales =
∑ sales
2
4
=
110 + 108 + 97 + 210
= 131.25
4
And so on . . .
2. Using the 4-point moving average, forecast sales of cans of deck sealer for week 9.
With smoothing methods, a forecast can only be made for the next period. The forecast for week
9 is simply the last calculated moving average which includes the last value in the time series.
The forecast for week 9 is 152 cans.
3. Calculate the simple exponential smoothing factor using α of 0.2. The first value is given in the table.
The column labeled SES with a of .2 lists the calculated simple exponential smoothing values
using an initial starting value of 106. To obtain the value for week 2:
~
S ales = .2(110) + .8(106) = 106.8
~
The value for week 3 is S ales = .2(108) + .8(106.8) = 107.04 or 107 cans
And so on . . .
4. Using the exponential smoothing factor with α of 0.2, forecast sales of cans of deck sealer for week 9.
Again, the forecast is the last calculated value which includes the last value in the time series. Using
exponential smoothing, the forecast for week 9 is 129.21 or 129 cans.
Final Exam Practice Problem Solutions
Page 1
5. Using the exponential smoothing factor with α of 0.2, what is the forecast error for week 6?
Forecast error is et = yt − yˆ t
The actual sales for week 6 was 136 cans. The forecast for week 6 was the SES smoothing factor
calculated using week five, 126.03 cans
The error in the forecast for week 6 is 136 – 126.03 = 9.97 cans
The following table shows the weekly sales for the number of gallons of deck sealer at the local hardware store
during the spring months. Use this table to answer the following questions.
Week
Gallon Cans Sold
Lag1
Lag2
1
106
*
*
2
110
106
*
3
108
110
106
4
97
108
110
5
210
97
108
6
136
210
97
7
128
136
210
8
134
128
136
An autoregressive model to predict the gallons sold with two lagged variables is:
Gallon Cans Sold = 166 - 0.13 Lag1 - 0.10 Lag2
6. Use the autoregressive model to make a forecast for the gallons of paint sold in week 9.
The lag1 value for week 9 is 134. The lag2 value for week 9 is 128. Substitute these values in the
regression equation to obtain the forecast for week 9.
Predicted Gallons Sold = 166 – 0.13(134) – 0.10(128) = 135.78 or 136 gallons
7. If the actual gallons sold in week 9 were 133, what is the forecast error for week? What is the absolute
percentage error for week 9?
Forecast error is et = yt − yˆ t
The actual gallons sold were 133 and 135.78 gallons were forecast so the forecast error is 133 –
135.78 = - 2.78
The absolute percent error is
2.09% =
133 − 135.78
133
×100
8. Using the autoregressive model, what was the forecast for the gallons of paint sold in week 4?
The lag1 value for week 4 is 108 and the lag2 value is 110. The forecast for week 4 would have
been
Predicted Gallons Sold = 166 – 0.13(108) – 0.10(110 = 140.96 or 141 gallons
9. What is the forecast error for week 4? What is the absolute percentage error for week 4?
For week 4, the actual gallons sold were 97 gallons and the forecast was for 140.96 gallons.
The forecast error was 97 – 140.96 = - 43.96
The absolute percent error is
Final Exam Practice Problem Solutions
45.32% =
97 − 140.96
97
×100
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A regression model was fit to a time series that exhibits both trend and seasonality. The model, based on 60
observations is
yˆ t = 55.5 + 12.8Time − 32.5Q1 − 42.5Q2 − 20Q3
10. Is there a positive or negative trend in the series? Explain.
The slope coefficient for Time is positive, so there is a positive trend in the series.
11. Which quarter has the highest values on average? Explain.
Quarter 4 has the highest values on average. All of the slope coefficients for Quarters 1 through
3 are negative. Because Quarter 4 is the indicator variable left out of the regression, it is the
baseline for comparison. Compared to Quarter 4, all the other quarters had lower sales.
12. Use the model to forecast for the next time period, the first quarter of the next year. Use Time=61.
The predicted value for Quarter 1 is
Q1 of next year = 55.5 + 12.8(61) – 32.5(1) = 803.8
13. Use the model to forecast for the fourth quarter of the next year. Use Time=64.
The predicted value for Quarter 4 is
Q4 of next year = 55.5 + 12.8(64) = 874.7
The following table gives the monthly price of apples and gas for the year 2006. Also, a Minitab printout is
provided of a second-order autoregressive model for the gas prices.
14. What are the lag2 values for the gas prices?
Month
Jan
Feb
Mar
Apr
May
June
July
Aug
Sep
Oct
Nov
Dec
Gas prices
2.359
2.354
2.444
2.801
2.993
2.963
3.046
3.033
2.637
2.319
2.287
2.380
Lag2 Gas Prices
*
*
2.359
2.354
2.444
2.801
2.993
2.963
3.046
3.033
2.637
2.319
15. Using the values from the table, what is the forecasted gas price for January 2007?
January 2007 is the next value in the series. The lag1 value is December, 2.380 and the lag2
value is November, 2.287.
Final Exam Practice Problem Solutions
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The forecast for January 2007 using the regression equation is
Predicted January 2007gas price = 1.282 + 1.314(2.380) - 0.788(2.287) = $2.607
The following is a time series plot and regression model for the monthly revenue ($Billion) for Walmart from
November 2003 to January 2007. The regression model uses indicator variables for months and a Time variable
that counts from 1 for the first data value in the series.
16. Interpret the slope coefficient of Time.
The Time variable is the trend component of the time series. Walmart revenues have been
increasing $0.145 billion per month, on average.
17. Interpret the slope coefficient for Dec.
January is the month left off of the indicator variables so it is the baseline for comparison.
Revenues in December are 11.56 billion more, on average, than revenues in January after taking
into account the general growth trend.
18. What revenue would you predict for Wal-Mart in February 2007?
Predict Revenue for Feb 2007 = 12.032 + 0.145(40) + 1.461(1) = $19.293 billion
19. What does it mean that the slope coefficient for Oct is negative?
In comparison with the baseline month of January, October revenues are .22 billion less, on
average, than January revenues after taking in to account the general growth trend.
Final Exam Practice Problem Solutions
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