Consider the partial ANOVA table shown below. Let a = .01 Source of Variation DF

ID: 3352298 • Letter: C

Question

Consider the partial ANOVA table shown below. Let a = .01

Source of Variation

Between Treatments

180

Within Treatments (Error)

Total

380

If all the samples have five observations each:

there are 10 possible pairs of sample means.

the only appropriate comparison test is the Tukey-Kramer.

all of the absolute differences will likely exceed their corresponding critical values.

there is no need for a comparison test – the null hypothesis is not rejected.

Source of Variation

Between Treatments

180

Within Treatments (Error)

Total

380

for (3,16) degree of freedom in numerator and denominator at 0.01 level criical value =5.29

as test statistic is not higher than critical value we can not reject null hypothesis

there is no need for a comparison test – the null hypothesis is not rejected.

Source of Variation DF SS MS F Between Treatments 3 180 60 4.8 Within Treatments (Error) 16 200 12.5 Total 19 380

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