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 6 0.71 Column 2 6 1.43 Column 3 6 2.15 Source of Variation SS df MS F p-value Between Groups 10.85 2 5.43 20.88 0.0000 Within Groups 3.86 15 0.26 Total 14.71 17
 SUMMARY Groups Count Average Column 1 6 0.71 Column 2 6 1.43 Column 3 6 2.15 ANOVA Source of Variation SS df MS F p-value Between Groups 10.85 2 5.43 20.88 0.0000 Within Groups 3.86 15 0.26 Total 14.71 17

a. Conduct an ANOVA test at the 1% significance level to determine if some population means differ.

• Reject H0; we can 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.

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

b. Calculate 99% 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.)

 Population Mean Differences Confidence Interval μ1 − μ2 [ , ] μ1 − μ3 [ , ] μ2 − μ3 [ , ]

c. Given your response to part b, which means significantly differ?

 Population Mean Differences Can we conclude that the population means differ? μ1 − μ2 μ1 − μ3 μ2 − μ3

a)since p value <0.01

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

b)

 MSE= 0.26 df(error)= 15 number of treatments = 3 pooled standard deviation=Sp =√MSE= 0.51
 critical q with 0.01 level and k=3, N-k=15 df= 4.84 Tukey's (HSD) =(q/√2)*(sp*√(1/ni+1/nj)         = 1.01
 Lower bound Upper bound (xi-xj ) ME (xi-xj)-ME (xi-xj)+ME μ1-μ2 -0.72 1.01 -1.73 0.29 μ1-μ3 -1.44 1.01 -2.45 -0.43 μ2-μ3 -0.72 1.01 -1.73 0.29

c)

 μ1-μ2 no μ1-μ3 YEs μ2-μ3 No

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