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,
(k − 1)(b − 1) A. Zhensykbaev ANOVA where b  number of blocks, the sum of squares for error
SSE = SS (Total) − SST − SSB,
the sum of squares for treatments
k SST = b(¯Ti − x),
x
¯
i=1 the sum squares for blocks SSB is
b (¯Bi − x)2 ,
x
¯ SSB = k
i=1 the total sum of squares, SS (Total),is the sum of squared diﬀerences
of each measurement from overall mean (total number of measurement is
n = b · k)
k b (xij − x)2 .
¯ SS (Total) = SSE + SST + SSB =
i=1 j =1 The reject region is
RR = {F > Fα },
where Fα is deﬁned from the Tables VIIIXI (McClave and Benson) with
(k − 1) numerator degree of freedom and df = kb − k − b +1 = (k − 1)(b − 1)
for denominator.
Example 4. Let ﬁve observations on crop yield be taken from a diﬀerent
plot of land and that observation i always referred to plot i for all treatments. It is likely that the nature of the plot (slope, drainage, etc.) would
aﬀect the crop yield. Possibly the eﬀect is such that the eﬀect of the
treatments is masked. Let us given the following data (on crop yield):
Plot 1
Plot 2
Plot 3
Plot 4
Plot 5
Mean Treatment 1 Treatment 2 Treatment 3
Mean
4
3
7
4.67
7
2
5
4.67
5
4
6
5.0
6
8
4
6.0
3
3
8
4.67
5
4
6
Total=5.0
8 A. Zhensykbaev ANOVA Calculate test statistic. The sum squares for blocks: x = 5, xi are
¯
¯
presented in last column of the table and
SST = 5 · ((5 − 5)2 + (4 − 5)2 + (6 − 5)2 ) = 10,
SSB = 3 · (3 · (4.67 − 5)2 + (5.0 − 5)2 + (6.0 − 5)2 ) = 4,
The total sum of squares is
3 5 (xij − x)2 = (4 − 5)2 + (3 − 5)2 + · · · + (8 − 5)2 = 52,
¯ SS (Total) =
i=1 j =1 SSE = SS (Total) − SST − SSB = 52 − 10 − 4 = 38.
The results of an ANOVA can be summarized in a tabular format
Variation
Treatment df
k−1=2 SS
SST=10 Block b−1=4 SSB=4 Error
Total (k − 1)(b − 1) SSE=38
=8
n − 1 = 14 SS (Total)
=52 MS
F
M ST =
M ST /M SE =
10/2=5
5/4.75 = 1.05
M SB =
M SB/M SE =
4/4 = 1
1/4.75 = 0.21
M SE =
38/8 = 4.75 Let α = 0.05, than tα = t0.05 ≈ 4.46 (ν1 = 2, ν2 = 8). Since F = 1.05 <
4.4 = F0.05 we do not reject H0 (the fertilizers are not essential, even if one
takes into account eﬀects of blocks).
Example 5. Teaching material for economics
School 1
School 2
School 3
School 4
School 5
School 6
Mean Material A Material B Material C
Mean
64
61
55
60
69
63
63
65
72
68
70
70
58
65
60
61
64
60
59
61
63
61
59
61
65
63
61
Total=63
9 A. Zhensykbaev ANOVA The results of an ANOVA we summarize in the table
Variation
Treatment df
k−1=2 SS
SST=48 Block b−1=5 SSB=222 Error
Total (k − 1)(b − 1) SSE=74
=10
n − 1 = 17 SS (Total)
=344 MS
F
M ST =
M ST /M SE =
48/2=24
24/7.4 = 3.24
M SB =
M SB/M SE =
222/5 = 44.4 44.4/7.4 = 6.0
M SE =
74/10 = 7.4 Let α = 0.05, than Fα = F0.05 ≈ 4.1 (ν1 = 2, ν2 = 10). Since F =
3.24 < 4.1 = F0.05 we do not reject H0 (no signiﬁcant diﬀerences between
the eﬀects of the teaching materials, even if to take into account the possible
diﬀerence between schools). 10...
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This note was uploaded on 02/11/2014 for the course MATH 1390 taught by Professor Christopherstocker during the Spring '13 term at Marquette.
 Spring '13
 ChristopherStocker
 Math, Variance

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