CHAPTER 6: Solutions to Selected Exercises

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CHAPTER 6: Solutions to Selected Exercises
6.1
a.
The data plot suggests that there is no long-run growth or decline in the time series. Thus
the no trend model yt   0   t is reasonable.
b.
yˆ t  y 
35,404  36,762  ...  37,153
 35,651.9
30
The 95% prediction interval is
1
[ y  t .30
025 s 1 
1
]
n
= [ 35651.9  2.045 (2037.35993) 1 
1
]
30
= [31416.2, 39887.6]
6.2
c.
No, the residual plot does not display any well-defined cyclical or alternating pattern.
a.
The data plot suggests that there is a long-run straight-line growth. Therefore, the linear
trend model yt   0  1t   t is reasonable.
b.
We can reject H 0 : 1  0 because p-value = .000 < .001. We have very strong evidence
1  0. Since r 2  .853 , 84.5% of the variation in the watch sales can be explained by
this linear trend model.
c./d.
ŷ 21 = 290.089474 + 8.667669 (21) = 472.1
ŷ 22 = 290.089474 + 8.667669 (22) = 480.8
ŷ 23 = 290.089474 + 8.667669 (23) = 489.4
95% prediction intervals are
e.
For y 21 :
[421.5, 522.7]
For y 22 :
[429.5, 532.1]
For y 23 :
[437.4, 541.5]
d = 1.367 is between d L ,.05  1.20 and d U ,.05  1.41
Thus the test is inconclusive. The residual plot is also inconclusive – it could have a
vague cyclical pattern.
6.3
a.
The trend appears to be linear.
b.
The seasonal variation appears to be constant. No transformation is required.
c.
All of the independent variables in the model seem to be important. This is because the
p-values associated with Time, Q2, Q3, and Q4 are all much less than   .05 or
  .01.
45
d.
1 if time period for sales quarter is 2
0 otherwise
Q2 = 
1 if time period for sales quarter is 3
0 otherwise
Q3 = 
1 if time period for sales quarter is 4
0 otherwise
Q4 = 
e.
yˆ17  17.250, yˆ18  38.750, yˆ19  51.750, yˆ 20  23.250
f.
yˆ t  8.75  .5t  21Q2  33.5Q3  4.5Q4
yˆ17  8.75  .5(17)  210  33.50  4.50  17.25
yˆ18  8.75  .5(18)  211  33.50  4.50  38.750
g.
95% P.I. for y17 : [15.395, 19.105]
We are 95% confident that sales of the TRK-50 Mountain Bike in the first quarter of year
5 will be between 15.395 and 19.105 (15 to 19 bikes)
95% P.I. for y18 : [36.895, 40.605]
95% P.I. for y19 : [49.895, 53.605]
95% P.I. for y 20 : [21.395, 25.105]
h.
n p  1 = number of parameters –1
= number of independent variables = 4
d  2.20  d U ,.05  1.93. Hence, there is no evidence of positive correlation at
  .05.
6.4
a.
Quadratic trend.
b.
Constant seasonal variation. No transformation required.
c.
1.
2.
1 if time period for energy bill is quarter 1
Q1  
0 otherwise
1 if time period for energy bill is quarter 2
Q2  
0 otherwise
1 if time period for energy bill is quarter 3
Q3  
0 otherwise
All independent variables except Q2 seem to be important because their
associated p-values are less than   .05. The p-value for testing  4 is .1713.
However, since at least one quarter is important, we will keep Q2 , Q3 , and Q4 in
the model.
46
3.
yˆ t  276.636  7.458t  .301t 2  65.771Q1  37.870Q  127.611Q3
2
yˆ 41  276.646  7.45841  .30141  65.7711  37.8700  127.6110
= 542.61
(Use more decimal places for greater accuracy.)


2
yˆ 42  276.646  7.458 42  .30142  65.7710  37.8701  127.6110
= 456.49
4.
95% P.I. for y 41 : [400.94, 685.03]
95% P.I. for y 42 : [312.33, 601.45]
95% P.I. for y 43 : [237.91, 532.68]
95% P.I. for y 44 : [381.15, 682.16]
5.
np 1  6 1  5
d  .84  d L ,.05  1.23. We have evidence of positive autocorrelation at
  .05
d.
1.
1  .59408.
Yes, p-value = .0003 <   .001 (Very strong evidence)
2.
p-value = .1011 for  1 . However, if t 2 is important, we keep t in model.
p-value = .0787 for  4 . However if Q1 and Q3 are important, we keep Q2 .
The associated p-values for t 2 , Q1 , and Q3 are less than   .05.
3.
Point Forecasts
yˆ 41  605.33 yˆ 42  505.67 yˆ 43  426.94 yˆ 44  569.97
95% P.I. for y 41 : [506.84, 703.82]
95% P.I. for y42 : [391.11, 620.23]
95% P.I. for y 43 : [307.22, 546.66]
95% P.I. for y44 : [448.49, 691.46]
4.
2
yˆ 40  ˆ  ˆ1ˆ40 1  283.949  9.22040     .35340   
 70.107Q1  35.429Q2  127.525Q3  .594ˆ40 1
where for   1
ˆ40  y 40  ˆ 40  y 40  [283.949  9.220(40)  .353(40) 2 ]
and for   1
ˆ40 1  yˆ 40 1  ˆ 40 1  yˆ 40 1  [283.949  9.220(40    1)
 .353(40    1) 2 +70.107 Q1  35.429 Q2  127.525 Q3 ]
= .594 ˆ40  2
47
5.
For period 41,
y 41   41   41   41  1 40  a41 
Thus, the predicted energy bill for period 41 is yˆ 41  ˆ 41  ˆ1ˆ40 where
ˆ40  y 40  ˆ 40
Here
ˆ40  539.78  [283.94906  9.21968(40)
 .35348(40) 2  70.10688(0)
 35.42856(0)  126.52509 (0)]
 539.78 - 480.72986  59.05014
and yˆ 41  ˆ 41  ˆ1ˆ40
= [283.94906 – 9.21968 (41) + .35348 (41) 2
+ 70.10688 (1) – 35.42856 (0)
 126.52509 (0)] + .59408 (59.05014)
= 570.24894 + .59408 (59.05014)
= 605.3285
An approximate 95% prediction interval of y 41 is
6 
[ yˆ 41  t.40
025 s ]
= [605.3285  1.96 (50.2513211)]
= [506.8380, 703.8191]
For time period 42
y42   42   42   42  (1 41  a42 )
Therefore, the predicted energy bill for period 42 is
y 42  ˆ 42  ˆ1 41 where ˆ41  yˆ 41  ˆ 41
Here, ˆ41  yˆ 41  ˆ 41 = 605.3285 – 570.24894
= 35.07956
and yˆ 42  ˆ 42  ˆ1ˆ41
= [283.94906 – 9.21968 (42) +
.35348 (42) 2 + 70.10688 (0) –
35.42856 (1) – 126.52509 (0)]
+ .59408 (35.07956)
= 484.83266 + .59408 (35.07956)
48
= 505.6717
An approximate 95% prediction interval for y 42 is
 
6 
ˆ
[ yˆ 42  t .40
025 s 1  1
2
]
= [505.6717  1.96 (50.2513211)
= [505.6717  1.96 (58.4502)]
= [391.1116, 620.2318]
1 .59408 ]
2
In a similar fashion we find that a point prediction of y 43 is yˆ 43  426.9411 and
that an approximate 95% prediction interval for y 43 is
   ˆ 
[ yˆ 43  1.96 s 1  ˆ1
2
4
1
]
 [426.9411 
1.96 (50.2513211) 1  .59408  .59408 ]
2
4
 [307.2235, 546.6588]
Finally, we find that a point prediction of y44 is yˆ 44  569.9732 and that an
approximate 95% prediction interval for y 44 is
[ yˆ 44  1.96 s 1  1   1   1  ]
2
4
6
 [569.9732 
1.96(50.2513211) ( 1  (.59408) 2  (.59408) 4  (.59408) 6 ) ]
 [448.4875, 691.4589]
6.5
a.
b.
Yes, all the associated p-values are less the   .05
*
yˆ169
 5.3489
*
yˆ170
 5.2641
c.
yˆ t*  4.80732  .00352t  .05247 M 1  .14079M 2  .10710M 3  .04988MM 4  .02542M 5
 .19017M 6  .38245M 7  .41377M 8  .07142M 9  .05064M 10  .14194M 11
*
yˆ169
 4.80732  .00352(169)  .05247  5.3497 (round-off error)
*
yˆ170
 4.80732  .00352(170)  .14079  5.2649 (roundoff error)
49
d.
   5.3489
*
yˆ169  yˆ169
4
4
 818.5739
(819 rooms)
95% P.I.: [(5.2913) 4 , 5.4065 ]  [783.8799, 854.4071]
4
e.
   5.2641
*
yˆ170  yˆ170
4
4
 767.8856 768 rooms 
95% P.I. [(5.2065) 4 , (5.3217) 4 ] = [734.8243, 802.0502]
f.
d =1.262
n p -1 = 13-1=12
n = 168
Table A5 does not provide d L , .05 for these values. However, from the values in the table
it appears likely that there is positive autocorrelation.
6.6
a.
The use of a growth curve model seems appropriate since it appears that the data might
be described by the model
 
y t   0 1t  t (see Figure 6.22)
b.
Yes
c.
y t   0 1t  t
 
lny t  ln  0   tln ( 1 )  ln  t
lny t   0  1t  u t
where
 0  ln (  0 ),
d.
1  ln 1 , and u t  ln  t
1.
ˆ 0  .54334 and ˆ1  .38997
2.
A point estimate of
 1 is ˆ1  eˆ  e .38997  1.4769
1
Growth rate = 100 ( 1  ˆ1 )  100(1  1.4769)  47.69 %
3.
The point prediction of ln y 12 is ln yˆ12  4.1363
Thus, the point forecast of
y12 is yˆ12  e 4.1363  62.5709
4.
The 95% prediction interval for ln y12 is [3.8303, 4.4423]. Thus, a 95%
Prediction interval for y12 is
[e 3.8303, e 4.4423 ] = [46.0764, 84.9701].
We can be 95% confident that the number of reported cased of the disease in
Month 12 will be at least 46.0764 (or roughly 46) cases and will be at most
84.9701 (or roughly 85) cases.
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