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 | 91 | |
119 | 115 | 81 | |
113 | 125 | 85 | |
106 | 105 | 102 | |
130 | 108 | 88 | |
115 | 109 | 117 | |
129 | 96 | 109 | |
112 | 115 | 121 | |
104 | 99 | ||
85 | 107 | ||
x_{j} |
120 | 107 | 100 |
s_{j}^{2} |
110.29 | 119.56 | 186.22 |
State the null and alternative hypotheses.
H_{0}: At least two of the population means are
equal.
H_{a}: At least two of the population means are
different.
H_{0}: μ_{A} ≠ μ_{B} ≠ μ_{C}
H_{a}: μ_{A} = μ_{B} = μ_{C}
H_{0}: μ_{A} =
μ_{B} = μ_{C}
H_{a}: μ_{A} ≠
μ_{B} ≠ μ_{C}
H_{0}: Not all the population means are equal.
H_{a}: μ_{A} = μ_{B} = μ_{C}
H_{0}: μ_{A} =
μ_{B} = μ_{C}
H_{a}: Not all the population means are equal.
Find the value of the test statistic. (Round your answer to two decimal places.)
test statistic =
Find the p-value. (Round your answer to four decimal places.)
p-value =
State your conclusion.
Do not reject H_{0}. There is sufficient evidence to conclude that the means of the three treatments are not equal.
Do not reject H_{0}. There is not sufficient evidence to conclude that the means of the three treatments are not equal.
Reject H_{0}. There is sufficient evidence to conclude that the means of the three treatments are not equal.
Reject H_{0}. There is not sufficient evidence to conclude that the means of the three treatments are not equal.
H_{0}: μ_{A} =
μ_{B} = μ_{C}
H_{a}: Not all the population means are equal.
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Anova: Single Factor | |||||
SUMMARY | |||||
Groups | Count | Sum | Average | Variance | |
A | 8 | 960 | 120 | 110.2857 | |
B | 10 | 1070 | 107 | 119.5556 | |
C | 10 | 1000 | 100 | 186.2222 | |
ANOVA | |||||
Source of Variation | SS | df | MS | F | P-value |
Between Groups | 1800.714 | 2 | 900.3571 | 6.39 | 0.0057 |
Within Groups | 3524 | 25 | 140.96 | ||
Total | 5324.714 | 27 |
In Excel = DATA> DATA ANALYSIS >ANOVA(SINGLE FACTOR) > SELECT THE DATA >OK
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test statistic = 6.39
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p-value = 0.0057
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Reject H_{0}. There is sufficient evidence to conclude that the means of the three treatments are not equal
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