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Agronomy 112

Problem Set 1

Due Wednesday, 31 July 2013 (in class)

NAME: RENERIO P. GENTALLAN JR. Signature: ______________________________

Student No.: 2011-39346

The effect on the growth of seedlings of two sources of light (A and B), each at two intensities (low and

high) were compared. The experimenter chose 16 identical seedlings. Four randomly selected seedlings

were grown under each of four treatments: AL (source A at low intensity), AH (source A at high

intensity), BL (source B at low intensity) and BH (source B at high intensity). All management practices

were applied uniformly across the treatments. The heights of the 16 plants were measured after four

weeks. Let AL, AH, BL and BH be the true mean seedling height for the four treatments.

Treatment Four-Week Height

AL 30.55 32.64 32.37 32.04

AH 31.23 31.09 30.62 30.42

BL 34.41 34.88 34.07 33.87

BH 35.61 35 33.65 32.91

LET:

FACTOR A= SOURCE OF LIGHT (A, B)

FACTOR B= LIGHT INTENSITY (HIGH, LOW)

1. State the statistical linear model completely and define each term in the context of the experiment.TWO-FACTOR FACTORIAL IN RANDOM COMPLETE BLOCK DESIGN(RCBD)

STATISTICAL LINEAR MODEL: Yijk=+i+ k+J+ij+ijkYijk=observation from plot given kth level of light intensity B and ith type of source of light

=overall mean of seedling height

k=effect of kth level of block

i =effect of source of lightj=effect of jth level of light intensity

jk= effect of interaction of the ith type of source of light and the kth level of light intensityijk=random error associated with the whole plot given the ith type of source of light and kth level of light intensity

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2. Give the treatment and design structure of the experiment.LET:

FACTOR A= SOURCE OF LIGHT (A, B)

FACTOR B= LIGHT INTENSITY (HIGH(H), LOW(L))

TWO-FACTOR FACTORIAL IN RANDOM COMPLETE BLOCK DESIGN(RCBD)

LAYOUT:

REPLICATE 1 (block1) REPLICATE 2 (block2) REPLICATE 3(block3) REPLICATE 4(block4)

Replicates/blocks=4

2x2 factorial treatment combination of 2 light sources and 2 light intensities

Treatments: AH= Light source A, High IntensityAL= Light source A, Low Intensity

BH= Light source B, High Intensity

BL= Light source B, Low Intensity

AL BH

BH AL

BL BH

AL AH

AH AL

BL BH

BH BL

AH AL

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3. Complete the analysis of variance table manually. Show all computations in the space provided.

Treatment Four-Week Height

Treatment

Total(T)

AL 30.55 32.64 32.37 32.04127.6

AH 31.23 31.09 30.62 30.42123.36

BL 34.41 34.88 34.07 33.87137.23

BH 35.61 35 33.65 32.91137.17

Rep

total(R) 131.8 133.61 130.71 129.24

Grand

total(G)

525.36

Source of

Variation

Df Sum of Squares Mean Squares Fc Ftab

5%

Replication 3 2.54285 0.84762 1.3493 3.86

Treatment

Source of light(A)

Light Intensity(B)

AXB

3

(1)

(1)

(1)

36.587251

34.3396

1.155625

1.092025

12.19575

34.3396

1.155625

1.092025

19.41411*

54.66435*

1.8396

1.73837

3.86

5.12

5.12

5.12

ERROR 9 5.6537 0.62819

TOTAL 15 44.7838*=significant at 5% level

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Computations:

CF= G2/rab= (525.36)2/(4)(2)(2)=17250.19560

Total SS= x2-CF= ((30.55)2+.+(32.91)

2)- 17250.19560=44.7838

Replication SS= (R2/ab)-CF= ((131.8)

2++(129.24)

2)/(2)(2))= (69010.9538/4)- 17250.19560=2.54285

Treatment SS=(T2/r)-CF=((127.6)

2++(137.17)

2)/(4))= (69147.1314/4)- 17250.19560=36.58725

Error SS=Total SS- Replication SS- Treatment SS =44.7838-2.54285-36.58725=5.6537

Repl. MS= Repl. SS/(r-1)= 2.54285/3=0.84762

Treatment MS=Treatment SS/(ab-1)= 36.58725/3=12.19575

Error MS=Error SS/((r-1)(ab-1))= 5.6537/9=0.62819

F(Treatment)= Treatment MS/ Error MS= 12.19575/0.62819=1.3493

F(Replication)= Repl. MS / Error MS= 0.84762/ 0.62819=19.41411

A SS=(A2/rb)-CF=((250.96)

2+(274.4)2)/(4*2))- 17250.19560=34.3396

B SS=(B2/ra)-CF=((264.83)

2+(260.53)

2)/(4*2))- 17250.19560=1.155625

AXB SS= Treatment SSA SS - B SS=36.58725-34.3396-1.155625=1.092025

AXB MS= AXB SS/(a-1)(b-1) =1.092025/1=1.092025

F(AXB)= AXB MS / Error MS= 1.092025/ 0.62819=1.73837

A MSS= A SS/(a-1)= 34.3396/1= 34.3396

B MS= B SS/(b-1)= 1.155625/1=1.155625

F(A)= A MS / Error MS= 34.3396/0.62819=54.66435

F(B)= B MS / Error MS=1.155625/0.62819=1.8396

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a) Do the responses of seedlings to light intensity vary with the source of light?State the relevant null and alternative hypotheses (in words and symbols). Give the appropriate test

statistic, the decision rule and conclusion.

Ho: AL = AH = BL = BH ; ________________________________________________________

Ha: _________________________________________________________________________

Test Statistic: Fc=__AXB MS__

ERROR MS

Decision Rule: Reject Ho if Fc > Ftab=F(1,9)=5.12

Decision: Since, Fc=1.73837

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b) Based on the results of the test in a) , would it be appropriate to test the main effects of light sourceand light intensity? Justify.

Yes, because we have sufficient evidence to say that the interaction effect between the two factors, light

source and intensity, is not significant. If interaction effect is not significant, the effect per factor should

be examined.

Answer only if the answer to be is Yes.

c) Do the two sources of light differ in their effects on seedling height?Ho:

Ha: AB;___________________________________________________________________

Test Statistic: Fc=___A MS___

ERROR MS

Decision Rule: Reject Ho if Fc > Ftab=F(1,9)=5.12

Decision: Since, Fc=54.66435>Ftab=5.12, reject Ho

Conclusion: At 5% level of significance, we have sufficient evidence to say__________________________

_____________________________________________________________________________

Do the two light intensities differ with respect to their effects on seedling height?

Ho: H = L;___________________________________________________________________

Ha: HL ; ___________________________________________________________________

Test Statistic: Fc=___B MS___

ERROR MS

Decision Rule: Reject Ho if Fc > Ftab=F(1,9)=5.12

Decision: Since, Fc=1.8396

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d) Give a measure of reliability of the experiment. Do you think the results of the experiment are reliable?cv(coefficient of variation)=

=

=2.414%

Rule of thumb: Lower cv, nearer to the true value and more reliable

Based on the coefficient of variation, since we have a very low coefficient of variation of 2.414%, we can

therefore infer that the experiment is reliable.

e) Give a measure of the adequacy of fit of the statistical linear model. Interpret.Measure of adequacy of fit=R-square(coefficient of determination)

R-square= Treatment SS/Total SS=36.587251/44.7838=0.816975

Rule of thumb: Higher R-square value, statistical linear model more adequately fit

This only means that 81.6975% of the total variation is contributed by the treatment.

Based on the high R-square(>80%) of 81.6975%, we can infer that the statistical linear model is

adequately fit.