Question

A turfgrass scientist is looking for an optimal approach to control a certain plant disease in Kentucky blue grass. He compares three different management strategies designed so that they would reduce spread of the disease. He has set up a field study with a total of 20 experimental plots and 4 treatments (3 disease prevention treatments and a control treatment. Each treatment has been assigned to 5 randomly selected experimental plots. Plant biomass is then measured from each pot at the end of the experiment. ANOVA table and the treatment means are shown below.

ANOVA:

	DF	Sum of Squares	Mean Square	F Value	Pr > F
Treatment	3	5413.1
Error		536.0
Total

                

Treatment	Mean values of the plant biomass	Letters for part a)	Letters for part b)
Fertilizer rate 1	6.6
Fertilizer rate 2	21.2
Fertilizer rate 3	31.4
Control (no fertilizer)	2.2

a) Do all pairwise comparisons between the treatment means using LSD, (a=0.05). Present the results using letters assigned to treatment means (Put the letters in the column Letters for part a) in the above table)

b) Do all pairwise comparisons between the treatment means using Tukey’s HSD (a=0.05). Present the results using letters assigned to treatment means. (Put the letters in the column Letters for part b) in the above table)

c) Comment on differences in conclusions obtained using the two methods. Which method would you use for this analysis? For full credit, provide an explanation of your choice.

Answer 1

1)

LSD Method

Where:
t = critical value from the t-distribution table
MSW = mean square within

t_critical = 2.12 when at df = 16 alpha = 0.05

MSW = 33.5

Hence, LSD = 11.90

From Data,

Mean _{Fertilizer rate 1} = 6.6

Mean _{Fertilizer rate 2} = 21.2

Mean _{Fertilizer rate 3} = 31.4

Mean _control = 2.2

When mod(Mean _{Group 1} - Mean _{Group 2} ) > LSD, then means are not same.

Pairs (Mean _{Fertilizer rate 1} & Mean _{Fertilizer rate 2})

mod(Mean _{Fertilizer rate 1} - Mean _{Fertilizer rate 2} ) = mod(6.6-21.2) = 14.6 > LSD. Hence they are not same

Pairs (Mean _{Fertilizer rate 1} & Mean _{Fertilizer rate 3})

mod(Mean _{Fertilizer rate 1} - Mean _{Fertilizer rate 3} ) = mod(6.6-31.4) = 24.8 > LSD. Hence they are not same

Pairs (Mean _{Fertilizer rate 1} & Mean _Control)

mod(Mean _{Fertilizer rate 1} - Mean _Control ) = mod(6.6-2.2) = 4.4 < LSD. Hence they are same

Pairs (Mean _{Fertilizer rate 2} & Mean _{Fertilizer rate 3})

mod(Mean _{Fertilizer rate 2} - Mean _{Fertilizer rate 3} ) = mod(21.2-31.4) = 10.2 < LSD. Hence they are same

Pairs (Mean _{Fertilizer rate 2} & Mean _Control)

mod(Mean _{Fertilizer rate 2} - Mean _Control ) = mod(21.2-2.2) = 19 > LSD. Hence they are not same

Pairs (Mean _{Fertilizer rate 3} & Mean _Control)

mod(Mean _{Fertilizer rate 3} - Mean _Control ) = mod(31.4-2.2) = 29.2 > LSD. Hence they are not same

b)

HSD Method

HSD = q *Sqrt (MS _Within /n)          

q is calculated from q table using (No of Groups vs degrees of freedom)

where q is 4.05,

Hence, HSD = 10.48

When mod(Mean _{Group 1} - Mean _{Group 2} ) > HSD, then means are not same.

Pairs (Mean _{Fertilizer rate 1} & Mean _{Fertilizer rate 2})

mod(Mean _{Fertilizer rate 1} - Mean _{Fertilizer rate 2} ) = mod(6.6-21.2) = 14.6 > HSD. Hence they are not same

Pairs (Mean _{Fertilizer rate 1} & Mean _{Fertilizer rate 3})

mod(Mean _{Fertilizer rate 1} - Mean _{Fertilizer rate 3} ) = mod(6.6-31.4) = 24.8 > HSD. Hence they are not same

Pairs (Mean _{Fertilizer rate 1} & Mean _Control)

mod(Mean _{Fertilizer rate 1} - Mean _Control ) = mod(6.6-2.2) = 4.4 < HSD. Hence they are same

Pairs (Mean _{Fertilizer rate 2} & Mean _{Fertilizer rate 3})

mod(Mean _{Fertilizer rate 2} - Mean _{Fertilizer rate 3} ) = mod(21.2-31.4) = 10.2 < HSD. Hence they are same

Pairs (Mean _{Fertilizer rate 2} & Mean _Control)

mod(Mean _{Fertilizer rate 2} - Mean _Control ) = mod(21.2-2.2) = 19 > HSD. Hence they are not same

Pairs (Mean _{Fertilizer rate 3} & Mean _Control)

mod(Mean _{Fertilizer rate 3} - Mean _Control ) = mod(31.4-2.2) = 29.2 > HSD. Hence they are not same

c)

LSD method is preferable to HSD as the value of HSD is close to difference in means between pairs (Mean _{Fertilizer rate 2} & Mean _{Fertilizer rate 3}). Hence for this pair, the test is inconclusive.