# ANOVA Practice ```ANOVA Practice
MA 223
1. (a) The ANOVA table is
Source
Factor
Error
Total
df
3
16
19
SS
237.49
54.02
291.51
MS
79.16
3.38
F
23.45
p
0.000
(b) Soundly reject H0 : τ1 = τ2 = τ3 = τ4 = 0 at signiﬁcance level α = 0.05.
We can use M SE as an estimate of σ 2 . For notational convenience let &micro;k = &micro; + τk ; for each k
(each column/treatment mean) the quantity
x̄k,&middot; − &micro;k
t= √
M SE /5
should follow a t-distribution with 16 df. Thus, for example, for the ﬁrst treatment mean (k = 1)
we have x̄1,&middot; = 46.08. This means a 95 percent conﬁdence interval is given by (note M SE /5 =
0.676)
46.08 − &micro;1
−2.12 &lt; √
&lt; 2.12
0.676
which leads to 44.3 &lt; &micro;1 &lt; 47.8. Similar computations lead to 49.2 &lt; &micro;2 &lt; 52.7, 40.4 &lt; &micro;3 &lt; 43.9,
and 41.3 &lt; &micro;4 &lt; 44.8.
(c) Prototypes 3 and 4 look best. In this case we should have that
x̄3&middot; − x̄4&middot; − (&micro;3 − &micro;4 )
√
2M SE /m
t=
follows a t-distribution with 16 df. Fill in the columns averages and M SE to ﬁnd that a 95 percent
conﬁdence interval on the diﬀerence is given by
−2.12 &lt;
−0.9 − (&micro;3 − &micro;4 )
&lt; 2.12
1.163
leading to −3.37 &lt; &micro;3 − &micro;4 &lt; 1.57. Thus the two lowest kicking chains are not clearly diﬀerent.
2. (a) The null hypothesis is H0 : τ1 = τ2 = τ3 = 0, versus the alternative Ha : at least one τi ̸= 0.
(b) We’ll use an F statistic with 2 df in the numerator and 9 df in the denominator. The critical
value will be F &gt; 4.26.
(c) The ANOVA table looks like
Source
Factor
Error
Total
df
2
9
11
SS
1.1816
0.305
1.487
MS
0.5908
0.0339
F
17.43
p
&iexcl;0.001
We reject the null hypothesis.
(d) We assume that in the model xij = &micro; + τi + eij the eij are IID, normal with zero mean and
common variance.
3. The ANOVA table from Minitab is
Source
Factor
Error
Total
df
3
36
39
SS
5152
8348
13499
MS
1717
232
F
7.41
p
&iexcl;0.001
We reject H0 .
The data easily passes the equality of variances test in Minitab (p around 0.95) and the normality
test.
1
Steel wire made by four manufacturers is to be tested and compared. Ten random samples are
drawn from a batch from each manufacturer and tested for strength. The results are in the ﬁle
wire strength.mtw that I sent you.
Test the hypothesis that there is no diﬀerence in the wire strength among the manufacturers. Also,
assess the assumptions of normality and equality of variance in the data.
2
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