Chapter 12
Analysis of Variance
122
OneWay ANOVA
1. a. Oneway analysis of variance is appropriate for these data because they represent three or
more populations categorized by a single characteristic that distinguishes the populations
from each other.
The distinguishing characteristic in this case is epoch.
b. Oneway analysis of variance tests the equality of two or more population means by
analyzing sample variances.
It finds a difference in the population means if the variance
between the sample means is larger than can be expected considering the variance within the
samples.
2. One test is better than three tests because a single test allows the conclusion to be made at the
stated level of significance.
Suppose, for example, all tests are conducted at the
α
= 0.05 level
of significance: for a single test, P(no type I error) = 1–
α
= 0.95; for three independent tests,
P(no type I error) = (1–
α
)
3
= (0.95)
3
= 0.857 – and so P(type I error) = 0.143 instead of 0.05.
Simply put, multiple tests increase the likelihood of chance differences between the samples
causing rejection of H
o
when there are actually no differences between the populations.
3. We should reject the hypothesis that the three epochs have the same mean skull breadth.
There
is sufficient evidence to conclude that at least one of the means is different from the others.
4. No.
Rejection of the hypothesis that all means are the same implies only that at least one of
the means is not the same as the others. The hypothesis that all the means are the same is
rejected when the variation between the means is larger than it is expected to be were all the
means equal, but there is nothing in the mathematics of the test statistic that can identify which
individual contributions to the variation between means are significant.
NOTE: When testing the hypothesis that three or more groups have the same mean, the test statistic
is F, where F is the ratio is the ratio of the variance between the groups to the variance within the
groups as defined in the text.
This manual generally uses the generic notation F =
22
Bp
s/
s .
As in
previous chapters, the superscripts and subscripts (the numerator df and denominator df) may be
used to identify which F distribution to look up in the tables.
5. H
o
:
μ
1
=
μ
2
=
μ
3
H
1
: at least one
μ
i
is different
α
= 0.05 and df
num
= 2, df
den
= 33
C.V.
F = F
α
= F
0.05
= 3.3158
calculations:
F =
s
= 669.0011/70.6481
= 9.4695 [TI83/84+]
Pvalue = P(
2
33
F >9.4695) = 0.0006 [TI83/84+]
conclusion:
Reject Ho; there is sufficient evidence to reject the claim that
μ
1
=
μ
2
=
μ
3
.
There is
sufficient evidence to reject the claim that the three books have the same mean Flesch
Reading Ease score.
F
1
3.3158
2
33
0.05
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CHAPTER 12
Analysis of Variance
6.
H
o
:
μ
1
=
μ
2
=
μ
3
H
1
: at least one
μ
i
is different
α
= 0.05 and df
num
= 2, df
den
= 33
C.V.
F = F
α
= F
0.05
= 3.3158
calculations:
F =
22
Bp
s/
s
= 133.3/34.1
= 3.91
[Minitab]
Pvalue = P(
2
33
F >3.91) = 0.030
[Minitab]
conclusion:
Reject Ho; there is sufficient evidence to reject the claim that
μ
1
=
μ
2
=
μ
3
.
There is
sufficient evidence to reject the claim that the three books have the same mean numbers
of words per sentence.
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 Spring '11
 Dr.Kalluri
 Variance, sufficient evidence

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