An experiment has a single factor with five groups and three values in each group. In determining the among-group variation, there are 4 degrees of freedom. In determining the within-group variation, there are 20 degrees of freedom. In determining the total variation, there are 24 degrees of freedom. Also, note that SSA= 96, SSW=160, SST=256, MSA=24, MSW=8, and FSTAT=3.
Complete parts (a) through (d).Click here to view page 1 of the F table.
a. Construct the ANOVA summary table and fill in all values in the table.
Source |
Degrees of Freedom |
Sum of Squares |
Mean Square(Variance) |
F |
||||
---|---|---|---|---|---|---|---|---|
Among groups |
? |
? |
? |
? |
||||
Within groups |
? | ? |
? |
|||||
Total |
? |
? |
? |
(Simplify your answers.)
b. At the 0.005 level of significance, what is the upper-tail critical value from the F distribution?
F0.005=_________
(Round to two decimal places as needed.)
c. State the decision rule for testing the null hypothesis that all five groups have equal population means.
Reject H0 if ______________
d. What is your statistical decision?
Since FSTAT is _____ (less/more) than the upper-tail critical value, ______ (do not reject/reject) H0. There is _______ (insufficient/sufficient) evidence to conclude there is a difference in the population means for the five groups.
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