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 | ||
xj |
120 | 107 | 100 |
sj2 |
110.29 | 119.56 | 186.22 |
State the null and alternative hypotheses.
H0: At least two of the population means are
equal.
Ha: At least two of the population means are
different.
H0: μA ≠ μB ≠ μC
Ha: μA = μB = μC
H0: μA =
μB = μC
Ha: μA ≠
μB ≠ μC
H0: Not all the population means are equal.
Ha: μA = μB = μC
H0: μA =
μB = μC
Ha: 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 H0. There is sufficient evidence to conclude that the means of the three treatments are not equal.
Do not reject H0. There is not sufficient evidence to conclude that the means of the three treatments are not equal.
Reject H0. There is sufficient evidence to conclude that the means of the three treatments are not equal.
Reject H0. There is not sufficient evidence to conclude that the means of the three treatments are not equal.
H0: μA =
μB = μC
Ha: 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 H0. There is sufficient evidence to conclude that the means of the three treatments are not equal
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