264
Chapter 13: The Analysis of Variance
13.1
The summary statistics are:
1
y
= 1.875,
2
1
s
= .6964286,
2
y
= 2.625,
2
2
s
= .8392857, and
n
1
=
n
2
= 8.
The desired test is:
H
0
:
μ
1
=
μ
2
vs.
H
a
:
μ
1
≠
μ
2
, where
μ
1
,
μ
2
represent the
mean reaction times for Stimulus 1 and 2 respectively.
a.
SST = 4(1.875 – 2.625)
2
= 2.25, SSE = 7(.696428) + 7(.8392857) = 10.75.
Thus,
MST = 2.25/1 = 2.25 and MSE = 10.75/14 = .7679.
The test statistic
F
= 2.25/.7679
= 2.93 with 1 numerator and 14 denominator degrees of freedom.
Since
F
.05
= 4.60,
we fail to reject
H
0
: the stimuli are not significantly different.
b.
Using the Applet,
p
–value =
P
(
F
> 2.93) = .109.
c.
Note that
2
p
s
= MSE = .7679.
So, the two–sample
t
–test statistic is |
t
| =
⎟
⎠
⎞
⎜
⎝
⎛
−
8
2
7679
.
625
.
2
875
.
1
=
1.712 with 14 degrees of freedom.
Since
t
.025
= 2.145, we fail to reject
H
0
.
The two
tests are equivalent, and since
F
=
T
2
, note that 2.93
≈
(1.712)
2
(roundoff error).
d.
We assumed that the two random samples were selected independently from normal
populations with equal variances.
13.2
Refer to Ex. 10.77.
The summary statistics are:
1
y
= 446,
2
1
s
= 42,
2
y
= 534,
2
2
s
= 45,
and
n
1
=
n
2
= 15.
a.
SST = 7.5(446 – 534)
2
= 58,080, SSE = 14(42) + 14(45) = 1218.
So, MST = 58,080
and MSE = 1218/28 = 1894.5.
The test statistic
F
= 58,080/1894.5 = 30.64 with 1
numerator and 28 denominator degrees of freedom.
Clearly,
p
–value < .005.
b.
Using the Applet,
p
–value =
P
(
F
> 30.64) = .00001.
c.
In Ex. 10.77,
t
= –5.54.
Observe that (–5.54)
2
≈
30.64 (roundoff error).
d.
We assumed that the two random samples were selected independently from normal
populations with equal variances.
13.3
See Section 13.3 of the text.
13.4
For the four groups of students, the sample variances are:
2
1
s
= 66.6667,
2
2
s
= 50.6192,
2
3
s
= 91.7667,
2
4
s
= 33.5833 with
n
1
= 6,
n
2
= 7,
n
3
= 6,
n
4
= 4.
Then, SSE = 5(66.6667)
+ 6(50.6192) + 5(91.7667) + 3(33.5833) = 1196.6321, which is identical to the prior
result.
13.5
Since
W
has a chi–square distribution with
r
degrees of freedom, the mgf is given by
2
/
)
2
1
(
)
(
)
(
r
tW
W
t
e
E
t
m
−
−
=
=
.
Now,
W
=
U
+
V
, where
U
and
V
are independent random variables and
V
is chi–square
with
s
degrees of freedom.
So,
2
/
2
/
)
(
)
2
1
(
)
2
1
)(
(
)
(
)
(
)
(
)
(
)
(
r
s
tU
tV
tU
V
U
t
tW
W
t
t
e
E
e
E
e
E
e
E
e
E
t
m
−
−
+
−
=
−
=
=
=
=
.
Therefore,
2
/
)
(
2
/
2
/
)
2
1
(
)
2
1
(
)
2
1
(
)
(
)
(
s
r
s
r
tU
U
t
t
t
e
E
t
m
−
−
−
−
−
=
−
−
=
=
.
Since this is the mgf for a chi–
square random variable with
r
–
s
degrees of freedom, where
r
>
s
, by the Uniqueness
Property for mgfs
U
has this distribution.