Question

# A one-way analysis of variance experiment produced the following ANOVA table. (You may find it useful...

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 H0; we cannot conclude that some population means differ.

• Reject H0; we cannot conclude that some population means differ.

• Do not reject H0; we can conclude that some population means differ.

• Reject H0; we can conclude that some population means differ.

b. Calculate 95% confidence interval estimates of μ1μ2,μ1μ3, and μ2μ3 with Tukey’s HSD approach. (If the exact value for nTc 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 H0; we cannot conclude that some population means differ.

b)

 MSE= 0.98 df(error)= 6 number of treatments = 3 pooled standard deviation=Sp =√MSE= 0.99
 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 (xi-xj)-ME (xi-xj)+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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