A two-factor factorial experiment is conducted to compare fleece weights of Meri
ID: 3312627 • Letter: A
Question
A two-factor factorial experiment is conducted to compare fleece weights of Merino, Suffolk, and Dorset ewes fed one of two diets. Two ewes of each breed are randomly assigned to each diet. The fleece weights (in pounds) are as follows:
Merino Suffolk Dorset
Diet 1 14 9 8
15 10 8
Diet 2 13 8 11
12 9 12
The partially completed ANOVA table is as follows:
Source df SS MS F
Total 66.2500
Diet 0.0833 0.0833 0.19992
Breed 46.5000 23.2500 55.79955
Diet x Breed 17.1667
Error 2.5000 0.41667
The correct values for the diet x breed mean squares and F value, respectively, are:
A 2 x 4 factorial experiment is conducted to compare yields of 4 varieties of soybeans that are planted in rows either 15 inches or 30 inches apart. Two plots of ground are randomly assigned to each combination of soybean variety and row spacing. The yields of soybeans (in bushels per acre) are as follows:
The partially completed ANOVA table is as follows:
Should we reject or not reject the null hypothesis for the interaction between variety and row spacing? Use a significance level of = 0.05.
8.58335 and 20.59988Explanation / Answer
1) here diet *breed df =(type of diet-1)*(type of breed-1) =(2-1)*(3-1) =2
tehrefore MS =SS/df =17.1667/2=8.58335
and test statistic F =MS/MSE =8.58335/0.41667=20.59988
therefiore correct option =8.58335 and 20.59988
2)
degree of freedom for interaction =(type of variety-1)*(type of row spacing-1)=(4-1)*(2-1)=3
tehrefore MS =SS/df =31.5/3=10.5
and test statistic F =MS/MSerror =10.5/2.75=3.818
for critical value at 0.05 level and (3,8) degree of freedom =4.066
correct option is :
Because the calculated F value (3.818) is not greater than the F value from the table (4.07), we do not reject the null hypothesis.
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