Question

An agronomist is conducting a field experiment to identify the best management practice for minimizing spread of a certain plant disease in corn. He compares four different management strategies designed so that they would reduce the spread of the disease. He has set up a field study with a total of 25 experimental plots planted with corn and with 5 treatments (4 disease prevention treatments and a control treatment). Each treatment has been assigned to 5 randomly selected plots. Plant biomass was then measured from each plot at the end of the experiment. The ANOVA table and the treatment means are shown below.

ANOVA:

	DF	Sum of Squares	Mean Square	F-value
Treatment	4	345	86	2.4
Error	20	714	36

    

a)(10 points) Conduct all pairwise comparisons between the treatment means using LSD, (=0.05). Present the results using letters assigned to treatment means (Use letters in the column Letters for part a) in the below table)

             

Show LSD value:

b)(10 points) Conduct all pairwise comparisons between the treatment means using Tukey’s HSD (=0.05). Present the results using letters assigned to treatment means. (Use the column Letters for part b) in the below table)

Show HSD value:

                                                NAME:____________________________

c)(10 points) Did you expect to see differences in conclusions obtained using the two methods (LSD and Tukey’s)? Which method would you use for this analysis? For full credit, provide an explanationof your choice.

             

Treatment	Mean values of the plant biomass	Letters for part a)	Letters for part b)
Management 1	6.6
Management 2	14.6
Management 3	31.4
Management 4	24.1
Control (no management)	2.2

Step by step procedure by hand, possibly shortest way to go about this problem would be ideal. thank you in advance

Answer 1

Required table:

Treatment	Mean values of the plant biomass	Letters for part a)	Letters for part b)
Management 1	6.6	1	A
Management 2	14.6	2	B
Management 3	31.4	3	C
Management 4	24.1	4	D
Control (no management)	2.2	5	E

b)

Here we have 5 groups and total number of observations are 25. So degree of freedom is

df=25-5=20

Critical value of t for and df = 20 is 2.086.

The Fisher's LSD Value is

Following table shows the required intervals:

groups (i-j)	xbari	xbarj	ni	nj	LSD	xbari-xbarj	absolute diff	Lower limit	Upper limit	Significant(Yes/No)
1-2	6.6	14.6	5	5	7.92	-8	8	-15.92	-0.08	Yes
1-3	6.6	31.4	5	5	7.92	-24.8	24.8	-32.72	-16.88	Yes
1-4	6.6	24.1	5	5	7.92	-17.5	17.5	-25.42	-9.58	Yes
1-5	6.6	2.2	5	5	7.92	4.4	4.4	-3.52	12.32	No
2-3	14.6	31.4	5	5	7.92	-16.8	16.8	-24.72	-8.88	Yes
2-4	14.6	24.1	5	5	7.92	-9.5	9.5	-17.42	-1.58	Yes
2-5	14.6	2.2	5	5	7.92	12.4	12.4	4.48	20.32	Yes
3-4	31.4	24.1	5	5	7.92	7.3	7.3	-0.62	15.22	No
3-5	31.4	2.2	5	5	7.92	29.2	29.2	21.28	37.12	Yes
4-5	24.1	2.2	5	5	7.92	21.9	21.9	13.98	29.82	Yes

c)

Critical value for , df=20 and k=5 is

So Tukey's HSD will be

Following table shows the required intervals:

groups (i-j)	xbari	xbarj	ni	nj	HSD	xbari-xbarj	Lower limit	Upper limit	Significant(Yes/No)
A-B	6.6	14.6	5	5	11.35	-8	-19.35	3.35	No
A-C	6.6	31.4	5	5	11.35	-24.8	-36.15	-13.45	Yes
A-D	6.6	24.1	5	5	11.35	-17.5	-28.85	-6.15	Yes
A-E	6.6	2.2	5	5	11.35	4.4	-6.95	15.75	No
B-C	14.6	31.4	5	5	11.35	-16.8	-28.15	-5.45	Yes
B-D	14.6	24.1	5	5	11.35	-9.5	-20.85	1.85	No
B-E	14.6	2.2	5	5	11.35	12.4	1.05	23.75	Yes
C-D	31.4	24.1	5	5	11.35	7.3	-4.05	18.65	No
C-E	31.4	2.2	5	5	11.35	29.2	17.85	40.55	Yes
D-E	24.1	2.2	5	5	11.35	21.9	10.55	33.25	Yes