Question

Suppose you have a balanced one-way completely randomized experiment, with a=3a=3 and n=4n=4. Furthermore, assume that MSE=32.028MSE=32.028. Calculate the difference between two sample treatment means that would result in a p-value of 0.05, using the following methods: (a) no multiple comparisons correction, (b) Tukey, (c) Bonferroni, and (d) Scheffe. Use R or another software package to calculate the appropriate multipliers. Hint: Do ungraded problem 1 first. Use confidence intervals to help answer this question.

Answer 1

here number of treatment=a=3 and

number of replication=n=4

so total df=a*n-1=3*4-1=12-1=11

treatment df=a-1=3-1=2 and error df=(a-1)(r-1)=(3-1)(4-1)=8

(a) here we can use

least significant difference=LSD=sqrt(2*MSE/n)*t(alpha,error df)=sqrt(2*32.028/4)*t(0.05,8)=sqrt(2*32.028/4)*2.31=9.24

(b) Tukey HSD=sqrt(MSE/n)*q(alpha,a,error df)=sqrt(32.028/4)*q(0.05,3,8)=sqrt(32.028/4)*4.53=12.82

(c)bonferroni=sqrt(2*MSE/n)*t(alpha/k,error df)=sqrt(2*32.028/4)*t(0.05/3,8)=sqrt(2*32.028/4)*3.016=12.07

since here we have three treatment and total number of pair for comparison=k=³C₂=3

so effective alpha=0.05/3

(d)sheffe=sqrt((a-1)F(alpha,a-1,error df)*MSE*(1/n1+1/n2))=sqrt((3-1)*F(0.05,2,8*32.028*(1/4+1/4))=

=sqrt((3-1)*4.07*32.028*(1/4+1/4))=11.42