Question
In: Statistics and Probability
The following data are from a completely randomized design. In the following calculations, use α =...
The following data are from a completely randomized design. In the following calculations, use
α = 0.05.
| Treatment 1 |
Treatment 2 |
Treatment 3 |
|
|---|---|---|---|
| 64 | 82 | 69 | |
| 48 | 73 | 53 | |
| 55 | 89 | 61 | |
| 37 | 68 | 45 | |
|
xj |
51 | 78 | 57 |
|
sj2 |
130.00 | 87.33 | 106.67 |
Find the value of the test statistic. (Round your answer to two decimal places.)___
Find the p-value. (Round your answer to three decimal places.)
p-value = ____
+Use Fisher's LSD procedure to determine which means are different.
Find the value of LSD. (Round your answer to two decimal places.)
LSD = ____
Find the pairwise absolute difference between sample means for
each pair of treatments
x1 − x2=
x1 − x3=
x2 − x3=
To test whether the mean time needed to mix a batch of material is the same for machines produced by three manufacturers, a chemical company obtained the following data on the time (in minutes) needed to mix the material.
| Manufacturer | ||
|---|---|---|
| 1 | 2 | 3 |
| 21 | 27 | 20 |
| 26 | 26 | 18 |
| 23 | 30 | 23 |
| 18 | 33 | 23 |
Find the value of the test statistic. (Round your answer to two decimal places.)___
Find the p-value. (Round your answer to three decimal places.)
p-value = ____
At the α = 0.05 level of significance, use Fisher's LSD procedure to test for the equality of the means for manufacturers 1 and 3.
Find the value of LSD. (Round your answer to two decimal places.)
LSD = ____
Find the pairwise absolute difference between sample means for manufacturers 1 and 3.
x1 − x3 =___
Solutions
Expert Solution
1)
Applying ANOVA:
value of the test statistic =7.44
p value =0.012
| Fisher's (LSD) for group i and j =(tN-k)*(sp*√(1/ni+1/nj) = | 16.62 | ||||
| Difference | Absolute Value | Conclusion |
| x2-x1 | 27.00 | significant difference |
| x3-x1 | 6.00 | not significant difference |
| x3-x2 | 21.00 | significant difference |
2)
| test statistic = | 8.34 |
| p value is less than 0.009 | |
| Fisher's (LSD) for group i and j =(tN-k)*(sp*√(1/ni+1/nj) = | 4.83 | ||||
x1 − x3 = 1 : not significant difference
orchestra answered 1 year agoRelated Solutions
The following data are from a completely randomized design. In the following calculations, use α =...
The following data are from a completely randomized design. In
the following calculations, use α = 0.05 (level of
significance)
Treatment 1
Treatment 2
Treatment 3
63
82
69
47
72
54
54
88
61
40
66
48
sample mean
51
77
58
sample variance
96.67
97.34
81.99
a. Use analysis of variance to test for a significant difference
among the means of the three treatments.
b. Use Fisher’s LSD procedure to determine which means are
different.
The following data are from a completely randomized design. In the following calculations, use α =...
The following data are from a completely randomized design. In
the following calculations, use
α = 0.05.
Treatment
1
Treatment
2
Treatment
3
64
81
68
48
71
55
53
88
60
43
64
49
xj
52
76
58
sj2
80.67
112.67
64.67
(a)Use analysis of variance to test for a significant difference
among the means of the three treatments.
State the null and alternative hypotheses.
H0: At least two of the population means are
equal.
Ha: At least two...
The following data are from a completely randomized design. In the following calculations, use α =...
The following data are from a completely randomized design. In
the following calculations, use
α = 0.05.
Treatment
1
Treatment
2
Treatment
3
62
83
68
46
71
55
53
88
62
39
70
47
xj
50
78
58
sj2
96.67
79.33
82.00
(a)
Use analysis of variance to test for a significant difference
among the means of the three treatments.
State the null and alternative hypotheses.
H0: At least two of the population means are
equal.
Ha: At least...
The following data are from a completely randomized design. In the following calculations, use α =...
The following data are from a completely randomized design. In
the following calculations, use
α = 0.05.
Treatment
1
Treatment
2
Treatment
3
63
82
69
46
72
54
53
87
62
46
63
47
xj
52
76
58
sj2
64.67
114.00
91.33
(a)
Use analysis of variance to test for a significant difference
among the means of the three treatments.
State the null and alternative hypotheses.
H0: μ1 =
μ2 = μ3
Ha: Not all the population means are...
The following data are from a completely randomized design. In the following calculations, use α =...
The following data are from a completely randomized design. In
the following calculations, use
α = 0.05.
Treatment
1
Treatment
2
Treatment
3
62
82
69
46
73
53
53
89
61
39
60
45
xj
50
76
57
sj2
96.67
156.67
106.67
(a)
Use analysis of variance to test for a significant difference
among the means of the three treatments.
State the null and alternative hypotheses.
H0: μ1 ≠
μ2 ≠ μ3
Ha: μ1 =
μ2 = μ3
H0:...
The following data are from a completely randomized design. In the following calculations, use α =...
The following data are from a completely randomized design. In
the following calculations, use
α = 0.05.
Treatment
1
Treatment
2
Treatment
3
64
81
68
48
73
53
53
89
60
43
69
55
xj
52
78
59
sj2
80.67
78.67
44.67
(a) Use analysis of variance to test for a significant
difference among the means of the three treatments.
State the null and alternative hypotheses.
H0: μ1 ≠
μ2 ≠ μ3
Ha: μ1 =
μ2 = μ3
H0:...
The following data are from a completely randomized design.
You may need to use the appropriate technology to answer this
question.
The following data are from a completely randomized design.
Treatment
A
B
C
162
142
126
142
157
123
166
123
138
144
142
140
148
137
150
168
157
127
Sample
mean
155
143
134
Sample
variance
135.6
166.0
108.4
(a)
Compute the sum of squares between treatments.
(b)
Compute the mean square between treatments.
(c)
Compute the sum of squares due to error.
(d)
Compute the...
Develop the analysis of variance computations for the following completely randomized design. At α = 0.05,...
Develop the analysis of variance computations for the following
completely randomized design. At α = 0.05, is there a
significant difference between the treatment means?
Treatment
A
B
C
135
106
93
120
114
82
114
124
84
106
104
102
132
107
90
115
108
117
128
96
110
102
115
120
104
99
82
93
xj
119
106
99
sj2
143.71
128.67
173.56
State the null and alternative hypotheses.
H0: Not all the population means are
equal.
Ha:...
Develop the analysis of variance computations for the following completely randomized design. At α = 0.05,...
Develop the analysis of variance computations for the following
completely randomized design. At α = 0.05, is there a
significant difference between the treatment means?
Treatment
A
B
C
136
108
92
119
115
81
112
124
85
107
103
102
131
107
88
113
108
118
129
96
111
113
115
120
103
97
81
106
xj
120
106
100
sj2
112.86
137.56
187.56
State the null and alternative hypotheses.
H0: μA =
μB = μC
Ha: μA ≠...
The following data are from a completely randomized design. Treatment A B C 163 141 127...
The following data are from a completely randomized design.
Treatment
A
B
C
163
141
127
142
156
122
166
125
137
145
142
141
149
137
151
165
145
120
Sample
mean
155
141
133
Sample
variance
118.0
102.8
146.0
(a)
Compute the sum of squares between treatments.
(b)
Compute the mean square between treatments.
(c)
Compute the sum of squares due to error.
(d)
Compute the mean square due to error. (Round your answer to two
decimal places.)...
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