A one-way analysis of variance experiment produced the following ANOVA table. (You may find it useful to reference the q table).
SUMMARY | ||||||
Groups | Count | Average | ||||
Column 1 | 3 | 0.66 | ||||
Column 2 | 3 | 1.24 | ||||
Column 3 | 3 | 2.85 | ||||
Source of Variation | SS | df | MS | F | p-value | |
Between Groups | 8.02 | 2 | 4.01 | 4.09 | 0.0758 | |
Within Groups | 5.86 | 6 | 0.98 | |||
Total | 13.88 | 8 | ||||
a. Conduct an ANOVA test at the 5% significance level to determine if some population means differ.
Do not reject H_{0}; we cannot conclude that some population means differ.
Reject H_{0}; we cannot conclude that some population means differ.
Do not reject H_{0}; we can conclude that some population means differ.
Reject H_{0}; we can conclude that some population means differ.
b. Calculate 95% confidence interval estimates of μ_{1} − μ_{2,}μ_{1} − μ_{3,} and μ_{2} − μ_{3} with Tukey’s HSD approach. (If the exact value for n_{T} – c is not found in the table, then round down. Negative values should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places. Round your answers to 2 decimal places.)
c. Given your response to part b, which means significantly differ?
a)since p value >0.05
Do not reject H_{0}; we cannot conclude that some population means differ.
b)
MSE= | 0.9800 | ||
df(error)= | 6 | ||
number of treatments = | 3 | ||
pooled standard deviation=Sp =√MSE= | 0.990 |
critical q with 0.05 level and k=3, N-k=6 df= | 4.34 | ||
Tukey's (HSD) =(q/√2)*(sp*√(1/ni+1/nj) = | 2.48 |
Lower bound | Upper bound | differ | |||
(x_{i}-x_{j})-ME | (x_{i}-x_{j})+ME | ||||
μ1-μ2 | -3.06 | 1.90 | not significant difference | ||
μ1-μ3 | -4.67 | 0.29 | not significant difference | ||
μ2-μ3 | -4.09 | 0.87 | not significant difference |
c)
none of the means differ significantly
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