Homework5 Answers
Exercise 12.14
(a)
ANOVA table
Source of var
d. of f.
Sum of squares
Mean Squares
Value of test stat
Between
k1=2
SSB=38.5626
MSB
=SSB/(k1)
=19.2813
Within
nk=10
SSW=89.3677
MSW
=SSW/(nk)
=8.9368
F=MSB/MSW
=2.16
Total
k1+nk
=n1=12
SST=127.9303
(b)
Step 1
Since we want to test if the three means of the populations are equal, the null hypothesis will be given
by:
3
2
1
0
:
μ
μ
μ
=
=
H
:
1
H
At least one of the means is different
Step 2
Since we are testing for equality of means among three groups, use the F distribution.
Step 3
Degrees of freedom for the numerator=2
Degrees of freedom for the denominator=10
Thus, for
α
=0.01, the critical value is given by
F
c
=7.56 (this value comes from finding
F
c
in P(F>
F
c
)=0.01 where F(2,10)).
Step 4
From the ANOVA table, we have F=2.16.
Step 5
Since 2.16<7.56, (observed value of the test statistic<critical value) we do not reject the null hypothesis.
Conclude that the means of the three populations are equal.
Exercise 12.17
(a)
Since we want to test if the three means of the populations are equal, the null hypothesis will be given
by:
3
2
1
0
:
μ
μ
μ
=
=
H
:
1
H
Not all three population means are equal.
1
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Given that we want to test if more than 2 population means are equal, we will use the OneWay
ANOVA test to test this hypothesis.
(b)
The degrees of freedom of the numerator are given by k1 and the ones for the denominator are nk.
Thus, since k=3 (we have 3 samples), and
25
6
9
10
3
2
1
=
+
+
=
+
+
=
n
n
n
n
, degrees of freedom of the
numerator=31=2 and degrees of freedom of the denominator=253=22.
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 Spring '07
 Guggenberger
 Null hypothesis, critical value, american league

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