Chapter 11
Inferential Methods in Regression and Correlation
Section 11.1
1.
(a) The slope of the estimated regression line (
β
= .095) is the expected change
of in the response variable
y for each one-unit increase
in the x variable.
This, of course, is just the usual interpretation of the
slope of a straight line.
Since x is measured in inches, a one-unit increase in x corresponds to a one-
inch increase in pressure drop.
Therefore, the expected change in flow rate is .095 m
3
/min.
(b) When the pressure drop, x, changes from 10 inches to 15 inches, then a 5 unit increase in x has
occurred.
Therefore, using the definition of the slope from (a), we expect about a 5(.095) = .475
m
3
/min. increase in flow rate (it is an
increase
since the sign of
β
= .095 is
positive
).
(c) For x = 10,
μ
y.10
= -.12 + .095(10) = .830.
For x = 15,
μ
y.15
= -.12 + .095(15) = 1.305.
(d)
When x = 10, the flow rate y is normally distributed with a mean value of
μ
y.10
= .830 and a
standard
deviation of
σ
y.10
=
σ
= .025.
Therefore, we standardize and use the z table to find:
P(y > .835) = P(z
>
025
.
830
.
835
.
−
) = P(z > .20) = 1 - P(z
≤
.20) = 1- .5793 = .4207 (using Table I).
2.
(a) The slope of the estimated regression line (
β
= -.01) is the expected change in reaction time for a one
degree Fahrenheit increase in the temperature of the chamber.
So, with a one degree Fahrenheit increase in temperature, the true average reaction time will decrease
by .01 hours.
With a 10 degree increase in temperature, the true average reaction time will decrease by (10)(.01) = 1
hour.
(b)
When x = 200,
3
)
200
(
01
.
5
200
=
−
=
•
y
μ
When x = 250,
5
.
2
)
250
(
01
.
5
250
=
−
=
•
y
(c)
=
=
<
<
)
250
when
6
.
2
4
.
2
(
x
y
P
8164
.
0918
.
9082
.
)
33
.
1
(
)
33
.
1
(
)
33
.
1
33
.
1
(
075
.
5
.
2
6
.
2
075
.
5
.
2
4
.
2
=
−
=
−
<
−
<
=
<
<
−
=
⎟
⎠
⎞
⎜
⎝
⎛
−
<
<
−
z
P
z
P
z
P
z
P
Next, the probability that all five observed reaction times are between 2.4 and 2.6 is (.8164)
5
= .3627
3.
(a) The simple linear regression model states that y =
α
+
β
x.
Here, y = ln(V) and x = 1/T, so the model
becomes ln(V) =
α
+
β
(1/T) + e.
Exponentiating both sides gives:
exp(ln(V)) = exp(
α
+
β
(1/T)+e),
or, V =
ε
=
ε
=
γ
T
e
e
/
βα
T
e
e
/
1
)
(
0
(
γ
1
)
1/T
ε
, where
ε
is the antilog of the error term e,
γ
0
= e
α
, and
γ
1
=