14.36
43210 :H ===
:H1
At least two means differ.
14.37 H0: µ1 = µ2 = µ3 = µ4 = µ5 = µ6 = µ7
H1: At least two means differ.
14.38 H0: µ1 = µ2 = µ3
H1: At least two means differ.
14.39 H0: µ1 = µ2 = µ3
H1: At least two means differ.
14.40 H0: µ1 = µ2 = µ3 = µ4 = µ5 =µ6 = µ7 = µ8
H1: At least two means differ.
14.41 H0: µ1 = µ2 = µ3 = µ4 = µ5 = µ6
H1: At least two means differ.
14.42 H0: µ1 = µ2 = µ3
H1: At least two means differ.
14.43 a
= .05:
==
−27,025.kn,2/ tt
2.052
Treatment Means Difference
i = 1, j = 2 128.7 101.4 27.3
LSD =
+
−
ji
kn,2/ n
1
n
1
MSEt
= 2.552
+10
1
10
1
700
= 30.20
Treatment Means Difference
i = 1, j = 2 128.7 101.4 27.3
c
=
),k(q
)27,3(q 05.
3.53
g
n
MSE
),k(q =
= 3.53
10
700
= 29.53
Treatment Means Difference
i = 1, j = 2 128.7 101.4 27.3
14.44 a
= .05:
==
−20,025.kn,2/ tt
2.086
LSD =
+
−
ji
kn,2/ n
1
n
1
MSEt
= 2.086
+5
1
5
1
125
= 14.75
Treatment Means Difference
i = 1, j = 2 227 205 22
i = 1, j = 3 227 219 8
Conclusion: The following pairs of means differ.
1
and
2
,
1
and
4
,
1
and
5
,
2
and
4
,
3
and
4
,
3
and
5
, and
4
and
5
.
Treatment Means Difference
i = 1, j = 2 227 205 22
i = 1, j = 3 227 219 8
i = 1, j = 4 227 248 −21
Conclusion: The following pairs of means differ.
1
and
5
,
2
and
4
,
3
and
4
, and
4
and
5
.
c
=
),k(q
05.
q
(5, 20) = 4.23
g
n
MSE
),k(q =
= 4.23
5
125
= 21.15
Treatment Means Difference
i = 1, j = 2 227 205 22
i = 1, j = 3 227 219 8
i = 1, j = 4 227 248 −21
1
2
1
5
2
4
3
4
4
5
=
),k(q
05.
q
Treatment Means Difference
i = 1, j = 2 1.33 2.50 −1.17
14.46 a. LSD =
+
−
ji
kn,2/ n
1
n
1
MSEt
= 1.782
+5
1
5
1
06.1
= 1.16
Treatment Means Difference
i = 1, j = 2 3.94 4.78 −.84
==C/
E
LSD =
+
−
ji
kn,2/ n
1
n
1
MSEt
= 2.404
+5
1
5
1
06.1
= 1.57
Treatment Means Difference
i = 1, j = 2 3.94 4.78 −.84
14.47 LSD method: C = 4(3)/2 = 6,
E
= .05,
==C/
E
.0083
LSD =
+
−
ji
kn,2/ n
1
n
1
MSEt
(LSD must be calculated for each pair of treatments.)
Treatment Means Difference LSD
i = 1, j = 2 68.83 65.08 3.75 5.74
i = 1, j = 3 68.83 62.01 6.82 6.47
14.48 Tukey’s method:
=
),k(q
05.
q
(4, 116)
3.68
= 3.68
30
0.961
= 20.83
LSD method with the Bonferroni adjustment: C = 4(3)/2 = 6,
E
= .05,
==C/
E
.0083
Treatment Means Difference Tukey LSD
i = 1, j = 2 90.17 95.77 −5.60 20.83 21.13
i = 1, j = 3 90.17 106.8 −16.67 20.83 21.13
14.49a
==210 :H
3
=
4
:
1
H
At least two means differ.
ANOVA Table
Source Degrees of Freedom Sum of Squares Mean Squares F
Treatments
31k =−
SST = 662.7
9.220
3
7.662
1k
SST ==
−
56.3
11.62
9.220
MSE
MST ==
==C/
E
Tukey’s method:
=
),k(q
05.
q
(4, 136)
3.79
= 3.79
10
11.62
= 9.45
Treatment Means Difference
i = 1, j = 2 61.6 57.3 4.3
i = 1, j = 3 61.6 61.8 −.2
14.50 a LSD method: C = 5(4)/2 = 10,
E
= .05,
==C/
E
.005
860.2tt 120,00 25.kn,2/ ==
−
(from Excel)
=
),k(q
05.
q
Treatment Means Difference
i = 1, j = 2 164.6 185.6 −21.0
i = 1, j = 3 164.6 154.8 9.8
14.51a
==210 :H
3
:H1
At least two means differ.
ANOVA Table
Source Degrees of Freedom Sum of Squares Mean Squares F
Treatments
21k =−
SST = 1,178
0.589
2
178,1
1k
SST ==
−
70.3
0.159
0.589
MSE
MST ==
b
=
),k(q
05.
q
(3,87)
3.40
= 3.40
83.7
30
0.159 =
Treatment Means Difference
i = 1, j = 2 53.17 49.37 3.80
14.52 Tukey’s method:
=
),k(q
05.
q
(3,57)
3.40
= 3.40
50.40
20
838,2 =
LSD method: C = 3(2)/2 = 3,
E
= .05,
==C/
E
.0167
466.2tt 57,0 08 3.kn,2/ ==
−
(from Excel)
14.53
14.54
14.55
Independents.
14.56
14.57
14.59 Use the Bonferroni method with α = .05/C with C = 1.
14.60
14.61
14.62
14.63
14.64
14.65
14.66
14.67
14.68
14.69 ANOVA Table
Source Degrees of Freedom Sum of Squares Mean Squares F .
Treatments 2 100 50.00 24.04
Blocks 6 50 8.33 4.00
14.70 ANOVA Table
Source Degrees of Freedom Sum of Squares Mean Squares F
Treatments 4 1,500 375.0 16.50
14.71 ANOVA Table
Source Degrees of Freedom Sum of Squares Mean Squares F
Treatments 3 275 91.67 7.99
14.72 Rejection region:
1bkn,1k,
FF +−−−
14,2,05.
F=
= 3.74
A ANOVA Table
Source Degrees of Freedom Sum of Squares Mean Squares F
Treatments 2 1,500 750.0 7.00
b. ANOVA Table
Source Degrees of Freedom Sum of Squares Mean Squares F
Treatments 2 1,500 750.0 10.50
c. ANOVA Table
Source Degrees of Freedom Sum of Squares Mean Squares F
Treatments 2 1,500 750 21.00
14.73 a. k = 3, b = 5, Grand mean = 10.4
SS(Total) =
22222
b
1i
2
ij
k
1j
)4.1012()4.109()4.1012()4.1010()4.107()xx( −+−+−+−+−=−
==
SSE = SS(Total) – SST – SSB = 99.6− 15.6 – 48.3 = 35.7
b SS(Total) =
22222
b
1i
2
ij
k
1j
)4.1012()4.109()4.1012()4.1010()4.107()xx( −+−+−+−+−=−
==
22222 )4.1010()4.1013()4.1016()4.108()4.1012(−+−+−+−+−+
14.74 a k = 4, b = 3, Grand mean = 5.6
SS(Total) =
222222
b
1i
2
ij
k
1j
)6.56()6.55()6.55()6.57()6.58()6.56()xx( −+−+−+−+−+−=−
==
222222 )6.56()6.56()6.54()6.55()6.55()6.54( −+−+−+−+−+−+
= 14.9
b SS(Total) =
222222
b
1i
2
ij
k
1j
)6.56()6.55()6.55()6.57()6.58()6.56()xx( −+−+−+−+−+−=−
==
222222 )6.56()6.56()6.54()6.55()6.55()6.54( −+−+−+−+−+−+
= 14.9
14.75
==210 :H
3
:H1
At least two means differ.
SSB =
91.5])38.23.3()38.21.2()38.27.2()38.24.1[(3)x]B[x(k 2222
b
1i
2
i=−+−+−+−=−
=
SSE = SS(Total) – SST – SSB = 7.30 − .87 – 5.91 = .52
ANOVA Table
Source Degrees of Freedom Sum of Squares Mean Squares F
Treatments 2 .87 .44 5.06