# Solution Manual for Forecasting and Predictive Analytics with Forecast X, 7th Edition

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Solutions to Chapter 2 Exercises
1. The mean volume of sales for a sample of 100 sales representatives is $25,350 per month.
The sample standard deviation is $7,490. The vice president for sales would like to know
whether this result is significantly different from $24,000 at a 95 percent confidence
level. Set up the appropriate null and alternative hypotheses, and perform the appropriate
statistical test.
H0: ยต = 24,000 vs.
H1: ยต โ 24,000
?? = (25,350 โ 24,000) รท (7,490 รท โ100)
?? = 25,350 โ 24,000 รท 7,490 รท 10
?? = 1,350 รท 749 = 1.802
At n = 100 and for a 95% confidence level and a two tailed t-test the table value
of t = 1.96. Since 1.802 is less than 1.96 we would fail to reject the null
hypothesis (H0:) and conclude that there is not sufficient statistical evidence to say
that the two values are significantly different.
2. Larry Bomser has been asked to evaluate sizes of tire inventories for retail outlets of a
major tire manufacturer. From a sample of 120 stores, he has found a mean of 310 tires.
The industry average is 325. If the standard deviation for the sample was 72, would you
say that the inventory level maintained by this manufacturer is significantly different
from the industry norm? Explain why. (Use a 95 percent confidence level.)
H0: ยต = 325 vs.
H1: ยต โ 325
?? = (310 โ 325) รท (72 รท โ120)
?? = 310 โ 325 รท 72 รท 10.95
?? = โ25 รท 6.58 = โ3.80
At n = 120 and for a 95% confidence level and a two tailed t-test the table value
of t = 1.96. Since -3.80 is less than -1.96 we would reject the null hypothesis (H0:)
and conclude that there is sufficient statistical evidence to say that the sample
mean is significantly different from 325. This is because -3.80 is quite far into the
left tail of the t-distribution.
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without the prior written consent of McGraw-Hill Education.
3. Twenty graduate students in business were asked how many credit hours they were
taking in the current quarter. Their responses are shown as follows (c2p3):
Student
Number
1
2
3
4
5
6
7
8
9
10
Credit
Hours
2
7
9
9
8
11
6
8
12
11
Student
Number
11
12
13
14
15
16
17
18
19
20
Credit
Hours
6
5
9
13
10
6
9
6
9
10
1. Determine the mean, median, and mode for this sample of data. Write a sentence
explaining what each means.
Mean =
8.3
Median =
9
Mode =
9
The mean is the arithmetic average which is equal to the sum of the
observations (166) divided by the number of observations (20). That is,
(166/20) = 8.3.
The median is the middle value that splits the data into two equal parts
when the data are arrayed from low to high (or high to low). In this case
sorting from low to high would show us that observations 10 through 14
are all equal to 9. When there are an even number of observations there is
no middle value. In such a case one uses the average of the two most
middle values, which in this case are both 9 (observations 10 and 11).
Thus, the median is 9.
The mode is the most frequently occurring value. In this sample there are
five observations equal to 9. The next most would be a value of 6 which
occurs four times. Thus the mode for this sample is 9.
2. It has been suggested that graduate students in business take fewer credits per
quarter than the typical graduate student at this university. The mean for all
graduate students is 9.1 credit hours per quarter, and the data are normally
distributed. Set up the appropriate null and alternative hypotheses, and determine
whether the null hypothesis can be rejected at a 95 percent confidence level.
Copyright ยฉ 2019 McGraw-Hill Education. All rights reserved. No reproductionor distribution
without the prior written consent of McGraw-Hill Education.
To find a solution you first need to calculate the standard deviation of the
20 sample values. It is 2.64.
H0: ยต โฅ 9.1 vs.
H1: ยต 9.1
?? = (7.25 โ 7.01) รท (2.51 รท โ400)
?? = 7.25 โ 7.01 รท 2.51 รท 20
?? = 0.24 รท 0.1255 = 1.91
The t-value for a one tailed test at a 95% confidence level and df=399 (i.e. 400-1)
is 1.645. Since 1.91 is greater than 1.645 you would reject the null hypothesis and
conclude that the rating for Ms. Westonโs bank is greater than the norm of 7.01.
2. Draw a diagram to illustrate your result.
Copyright ยฉ 2019 McGraw-Hill Education. All rights reserved. No reproductionor distribution
without the prior written consent of McGraw-Hill Education.
1.645
1.91
3. How would your result be affected if the sample size had been 100 rather than
400, with everything else being the same?
H0: ยต โค 9.1 vs.
H1: ยต > 9.1
?? = (7.25 โ 7.01) รท (2.51 รท โ100)
?? = 7.25 โ 7.01 รท 2.51 รท 10
?? = 0.24 รท 0.251 = 0.956
In this case you would fail to reject the null hypothesis so would not have
evidence at a 95% confidence level that Ms. Westonโs bank had ratings
above the norm of 7.01.
Copyright ยฉ 2019 McGraw-Hill Education. All rights reserved. No reproductionor distribution
without the prior written consent of McGraw-Hill Education.
7. In a sample of 25 classes, the following numbers of students were observed (c2p7):
Class
1
2
3
4
5
6
7
8
9
10
11
12
13
Number of
students
40
50
42
20
29
39
49
46
52
45
51
64
43
Class
14
15
16
17
18
19
20
21
22
23
24
25
Number of
students
37
35
44
10
40
36
20
20
29
58
51
54
1. Calculate the mean, median, standard deviation, variance, and range for this
sample.
Mean =
40.16
Median =
42
Std, Dev. = 13.14
Variance =
172.72
2. What is the standard error of the mean based on this information?
The standard error of the mean = Std Dev / Square Root of n
The standard error of the mean = 13.14 / 5 = 2.628
3. What would be the best point estimate for the population class size?
The best point estimate would be the sample mean = 40.16.
4. What is the 95 percent confidence interval for class size? What is the 90 percent
confidence interval? Does the difference between these two make sense?
The 95% confidence interval would be the sample mean ยฑ 1.96 (the standard error
of the mean)
40.16 ยฑ 1.96(2.628) = 40.16 ยฑ 5.15 = 35.01 to 45.31
The 90% confidence interval would be the sample mean ยฑ 1.645 (the standard
error of the mean)
Copyright ยฉ 2019 McGraw-Hill Education. All rights reserved. No reproductionor distribution
without the prior written consent of McGraw-Hill Education.
40.16 ยฑ 1.645(2.628) = 40.16 ยฑ 4.32 = 35.84 to 44.48
It makes sense that if you are less confident the confidence band would be more
narrow.
8. CoastCo Insurance, Inc., is interested in forecasting annual larceny thefts in the United
States using the following data (c2p8):
Year
1972
1973
1974
1975
1976
1977
1978
1979
1980
1981
1982
1983
Larceny Thefts
4151
4348
5263
5978
6271
5906
5983
6578
7137
7194
7143
6713
Year
1984
1985
1986
1987
1988
1989
1990
1991
1992
1993
1994
Larceny Thefts
6592
6926
7257
7500
7706
7872
7946
8142
7915
7821
7876
1. Prepare a time-series plot of these data. On the basis of this graph, do you think
there is a trend in the data? Explain.
Larceny Thefts
1972
1973
1974
1975
1976
1977
1978
1979
1980
1981
1982
1983
1984
1985
1986
1987
1988
1989
1990
1991
1992
1993
1994
9,000
8,000
7,000
6,000
5,000
4,000
3,000
2,000
1,000
0
Larceny Thefts
Linear (Larceny Thefts)
Copyright ยฉ 2019 McGraw-Hill Education. All rights reserved. No reproductionor distribution
without the prior written consent of McGraw-Hill Education.
As seen in the graph a dotted trend line has been added. The trend line shows a
positive trend in the data.
2. Look at the autocorrelation structure of larceny thefts for lags of 1, 2, 3, 4, and 5.
Do the autocorrelation coefficients fall quickly toward zero? Demonstrate that the
critical value for rk is 0.417. Explain what these results tell you about a trend in
the data.
The ACF graph supports the idea that there is not a strong trend in the data. You
see that after the first few bars the values fall quickly toward zero.
rk = 2/ (square root of n) = 2/ square root 23 = 2/4.796 = 0.417.
3. On the basis of what is found in parts a andb, suggest a forecasting method from
Table 2.1 that you think might be appropriate for this series.
Since the data are annual there can be no seasonality. A Holtโs exponential
smoothing model may be best although as one can see in the part (a) graph a
linear trend is an alternative. Holtโs MAPE = 3.90%, while the linear trend MAPE
= 5.78%.
9. Use exploratory data analysis to determine whether there is a trend and/or seasonality in
mobile home shipments (MHS). The data by quarter are shown in the following table
(c2p9):
Period
Mar-81
Jun-81
Sep-81
Dec-81
Mar-82
Jun-82
MHS
54.9
70.1
65.8
50.2
53.3
67.9
Period
Dec-84
Mar-85
Jun-85
Sep-85
Dec-85
Mar-86
MHS
66.2
62.3
79.3
76.5
65.5
58.1
Period
Sep-88
Dec-88
Mar-89
Jun-89
Sep-89
Dec-89
MHS
59.2
51.6
48.1
55.1
50.3
44.5
Period
Jun-92
Sep-92
Dec-92
Mar-93
Jun-93
Sep-93
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without the prior written consent of McGraw-Hill Education.
MHS
52.8
57
57.6
56.4
64.3
67.1
Sep-82
Dec-82
Mar-83
Jun-83
Sep-83
Dec-83
Mar-84
Jun-84
Sep-84
63.1
55.3
63.3
81.5
81.7
69.2
67.8
82.7
79
Jun-86
Sep-86
Dec-86
Mar-87
Jun-87
Sep-87
Dec-87
Mar-88
Jun-88
66.8
63.4
56.1
51.9
62.8
64.7
53.5
47
60.5
Mar-90
Jun-90
Sep-90
Dec-90
Mar-91
Jun-91
Sep-91
Dec-91
Mar-92
43.3
51.7
50.5
42.6
35.4
47.4
47.2
40.9
43
Dec-93
Mar-94
Jun-94
Sep-94
Dec-94
Mar-95
Jun-95
Sep-95
Dec-95
66.4
69.1
78.7
78.7
77.5
79.2
86.8
87.6
86.4
On the basis of your analysis, do you think there is a significant trend in MHS? Is there
seasonality?
The ACF values do not fall to zero quickly suggesting some trend. The relatively
larger values at 1, 4, and 8 suggest seasonality.
What forecasting methods might be appropriate for MHS according to the guidelines in
Table 2.1?
Based on Table 2.1 causal multiple regression, Wintersโ and time series
decomposition would be good methods to consider.
10. Home sales are often considered an important determinant of the future health of the
economy. Thus, there is widespread interest in being able to forecast home sales (HS).
Quarterly data for HS are shown in the following table in thousands of units (c2p10):
Date
Mar-89
Home sales
(000) Per
Quarter
161
Date
Mar-95
Home sales
(000) Per
Quarter
154
Date
Mar-01
Home sales
(000) Per
Quarter
251
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without the prior written consent of McGraw-Hill Education.
Jun-89
Sep-89
Dec-89
Mar-90
Jun-90
Sep-90
Dec-90
Mar-91
Jun-91
Sep-91
Dec-91
Mar-92
Jun-92
Sep-92
Dec-92
Mar-93
Jun-93
Sep-93
Dec-93
Mar-94
Jun-94
Sep-94
Dec-94
179
172
138
153
152
130
100
121
144
126
116
159
158
159
132
154
183
169
160
178
185
165
142
Jun-95
Sep-95
Dec-95
Mar-96
Jun-96
Sep-96
Dec-96
Mar-97
Jun-97
Sep-97
Dec-97
Mar-98
Jun-98
Sep-98
Dec-98
Mar-99
Jun-99
Sep-99
Dec-99
Mar-00
Jun-00
Sep-00
Dec-00
185
181
145
192
204
201
161
211
212
208
174
220
247
218
200
227
248
221
185
233
226
219
199
Jun-01
Sep-01
Dec-01
Mar-02
Jun-02
Sep-02
Dec-02
Mar-03
Jun-03
Sep-03
Dec-03
Mar-04
Jun-04
Sep-04
Dec-04
Mar-05
Jun-05
Sep-05
Dec-05
Mar-06
Jun-06
Sep-06
Dec-06
Mar-07
Jun-07
243
216
199
240
258
254
220
256
299
294
239
314
329
292
268
328
351
326
278
285
300
251
216
214
240
1. Prepare a time-series plot of THS. Describe what you see in this plot in terms of
trend and seasonality.
The graph shows a clear positive trend along with a regular seasonal pattern.
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without the prior written consent of McGraw-Hill Education.
2. Calculate and plot the first 12 autocorrelation coefficients for HS. What does this
autocorrelation structure suggest about the trend?
ACF of Home sales (000) Per Quarter
Obs
ACF
1.0000
1
.8632
.2278
-.2278
2
.7620
.2278
.8000
-.2278
3
.8098
.2278
4
.8618
.2278
5
.7154
.2278
6
.5929
.2278
-.2278
7
.6169
.2278
.2000
-.2278
8
.6475
.2278
-.2278
.0000
9
.5093
.2278
-.2278
10
.3876
.2278
-.2000
-.2278
11
.4087
.2278
12
.4455
.2278
-.2278
.6000
-.2278
ACF
-.2278
.4000
Upper Limit
Lower Limit
1
2
3
4
5
6
7
8
9
10 11 12
-.2278
-.4000
-.2278
The ACF diagram shows the existing of a trend. The relatively high bars at 1, 4, 8
and 12 are suggestive of seasonality.
11. Exercise 12 of Chapter 1 includes data on the Japanese exchange rate (EXRJ) by month.
On the basis of a time-series plot of these data and the autocorrelation structure of EXRJ,
would you say the data are stationary? Explain your answer. (c2p11)
Period
1
2
3
4
5
6
7
8
9
10
11
12
EXRJ
127.36
127.74
130.55
132.04
137.86
143.98
140.42
141.49
145.07
142.21
143.53
143.69
Period
13
14
15
16
17
18
19
20
21
22
23
24
EXRJ
144.98
145.69
153.31
158.46
154.04
153.7
149.04
147.46
138.44
129.59
129.22
133.89
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without the prior written consent of McGraw-Hill Education.
Both graphs suggest that there is not a significant trend in these exchange rate
data.
Copyright ยฉ 2019 McGraw-Hill Education. All rights reserved. No reproductionor distribution
without the prior written consent of McGraw-Hill Education.

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