Solutions for odd number questions in 12.1-12.2
12.7
A deterministic model does not allow for random error or variation, whereas a probabilistic model
does. An example where a deterministic model would be appropriate is:
Let y =cost of a 2 × 4 piece of lumber and x =length(in feet)
The model would be y = β1 x. There should be no variation in price for the same length of wood.
An example where a probabilistic model would be appropirate is:
Let y = sales per month of a commodity and x =amount of money spent advertising
The model would be y = β0 + β1 x + ². The sales per month will probably vary even if the amount
of money spent on advertising remains the same.
12.11
ˆ 1 = −.78.
From Exercise 12.10, β̂0 = 7.10 and beta
The fitted line is ŷ = 7.10 − .78x. To obtain predicted values of y, we substitute values of x into
the prediction equation(fitted line).
a.
x
y
ŷ = 7.10 − .78x
(y − ŷ)
(y − ŷ)2
7
2
1.64
.36
.1296
4
4
3.98
.02
.0004
6
2
2.42
-.42
.1764
2
5
5.54
-.54
.2916
1
7
6.32
.68
.4624
1
6
6.32
-.32
.1024
3
5
4.76
.24
SE =
P
(y − ŷ) = .02
c.
1
.0576
SSE =
P
(y − ŷ)2 = 1.2204
x
y
ŷ = 14 − 2.5x
(y − ŷ)
(y − ŷ)2
7
2
-3.5
5.5
30.25
4
4
4
0
0
6
2
-1
3
9
2
5
9
-4
16
1
7
11.5
-4.5
20.25
1
6
11.5
-5.5
30.25
3
5
6.5
-1.5
SE =
P
(y − ŷ) = −7
2.25
SSE =
P
(y − ŷ)2 = 108.00
The line in part(b) has SSE = 108.00, which is obviously greater than the SSE(= 1.2204) for
the least squares line.
12.13
b. Looking at the scattergram, x and y appear to have a negative linear relationship.
c. The least squares estimates of β1 , β̂1 = −.9939; the least squares estimates of β0 , β̂0 = 8.543.
d. The least squares line is ŷ = 8.543 − .9939x. After plotting the line on the scattergram, we
can see it fits the data well.
ˆ 0 has no meaning other than the
e. β̂0 = 8.543, since x = 0 is not in the observed range, beta
y-intercept.
β̂1 = −.9939. The estimated change in the mean value of y for each unit change in x is -.0039.
These interpretations are valid only for values of x in the range from 2 to 8.
2