NCSU
ST512
QUIZ 4
Sum2 2011
Q 1. We have the resuls of an experiment conducted to compare two sweet potato cultivars (Hawaii,
NG7570) for drought tolerance1 and two levels of water regimes: control (maintained by regular
watering at or close to field capacity) and Stress (water stressed plants : soil was saturated then allowed
to dry progressively to permanent wilting point). Experimental design: CRD with four repetitions for
each treatment combination. Response is SLA = Specific leaf area (cm2 g-1)
a) Identify the effect. Please fill the blank with one of the following : Main, Interaction, Simple
effect.
o __Main_____ Mean Difference between Hawaii and NG7570 on average water stress
level
o __Interaction Mean difference between Hawaii and NG7570 when water stress is high
stress compared to the mean difference between Hawaii and NG7570 when water stress
is at control level .
o __ Simple _ Differences between Water stress at high level and control level for cultivar
NG7570
b) The following graph present the means for the four treatment combinations Explain graphically
the interaction effect.
346
350
344
340
Control
340
330
320
310
300
290
value
S
L
A
280
270
260
250
240
Stress
230
220
215
210
Hawaii
NG7570
cultivar
Hawaii
treat
Control
Stress
NG7550
For cultivar Hawaii the difference between SLA under Control and Stress conditions is 129 cm2/g (344-215), while for cultivar
NG7550 this difference is 4 cm2/g (346-340), which show the interaction between Cultivar and Water Stress Treatment: some
cultivars respond better to strees than others, with NG7550 having a miminal decrease in SLA when subjected to water stress
conditions, and Hawaii showing a large decrease in SLA when subjected to the same water stress conditions.
1
Identification of drought tolerant sweet potato (Ipomoea batatas (L.) Lam) cultivars
Prabawardani Saraswati, Mark Johnston, Ross Coventry and Joseph Holtum.
Proceedings of the 4th International Crop Science Congress Brisbane, Australia, 26 Sep – 1 Oct 2004 |
1
July 28, 2011
NCSU
ST512
QUIZ 4
Sum2 2011
c) Calculate
a. the simple effect of Water Stress (A) when cultivar is Hawaii, 215-344= -129
b.
the simple effect of Water Stress (A) when cultivar is N7550. 340-346 = -6
c.
How do you interpret their difference? Difference between (-129) and (-4) is due to cultivar
NG7550 better responding to conditions of water stress than cultivar Hawaii, which reflects the interaction
between CULTIVAR and WATER STRESS: no all cultivar respond the same to Conditions of water stress.
d) Calculate
a. main effect for Water Stress (A) : -67.5
: 63.5
b. main effect for Cultivar (B)
c. interaction effect Water Stress*Cultivar (A*B) : 61.5
SLA
Water Stress
Control
Cultivar
Hawaii
NG7570
344
346
Mean
345.0
Simple Effects for Cultivar
(346-344) = 2
(340-215) = 125
Stress
215
340
277.5
Mean
279.5
343.0
311.25
(215-344) = 129
(340-346) = -6
Simple effect
for Water Stress
Main Effect Water Stress
= -67.5 = ((-129)+(-6))/2
= 343-279.5
Main Effect Cultivar = 63.5
= (2+125)/2 = 343-279.5
Interaction effect
Cultivar*WaterStress = 61.5
= 123/2 = (125 – 2)/2
= ((-6) – (-129))/2 =
e) Fill the blanks in the analysis of variance table
Source
DF
Sum of Squares
Mean Square
F Value
Pr > F
Model
3
49483.00
16494.33
10.77
0.001
Water Stress
1
18225
18225
11.90
*
Cultivar
1
16129
16129
10.53
*
Water Stress*Cultivar
1
15129
15129
9.88
*
Error
12
18375.00
1531.25
Corrected Total
15
67858
SS(Water Stress) =
SS(Cultivars) =
SS(Treatment) =
2
24 345.0311.252  277.5311.25 2  18225


24 279.5311.25 2  343.0311.25 2  16129


24 345.0311.252  277.5311.252  279.5311.252  343.0311.252   49483


July 28, 2011
NCSU
ST512
QUIZ 4
Sum2 2011
SS(Water Stress * Cultivar) = 49483 – 18225 – 16129 = 15129
Use information in table below
a. Calculate the SS for the main effect of Water Stress
Q   1 1376   1 1384  1  860  1 1360   540
Divisor  4*( 1   1  1  1 )  16
2
2
SS Water Stress  
2
2
 540   18225
Q2

divisor
16
2
b. Calculate the SS for the interaction effect of Water Stress*Cultivar
Q  1 1376   1 1384   1  860  1 1360   492
c.
Divisor  4*( 1   1  1  1 )  16
2
2
2
2
 492   15129
Q2
SS Water Stress * Culti var  

divisor
16
2
d. Test the hypothesis of no interaction effect. Use tabular value t(df, 0.05/2) = 2.18 or
F(df1, df2, 0.05) = 4.75.
H o :  11   12   21   22  0
H1 : at least one  ij  0
Fcalc 
i  1, 2
for
j  1, 2
MSE Water Stress * Culti var  15129

 9.88
MSE
1531.25
Fcalc =9.88 > 4.74 thus, reject null hypothesis at a 0.05 significance level, there is
statistical evidence that the interaction effects are not all equal to 0.
Water Stress
Cultivar
Total
Control
Hawaii
Control
NG7570
Stress
Hawaii
Stress
NG7570
1376
1384
860
1360
Water Stress
Cultivar
-1
-1
-1
1
1
-1
1
1
Water Stress*Cultivar
1
-1
-1
1
r= 4
Q
-540
508
492
r  cij2
ij
4*4=16
4*4=16
4*4=16
Sum
Q2 r  cij2
18225
16129
15129
49483
Q2. A experiment was conducted to study the influence of time of bleeding and diethylstilbestrol (an
estrogenic compound) on plasma phospholipid in lambs. Five lambs were assigned at random to each
3
July 28, 2011
NCSU
ST512
QUIZ 4
Sum2 2011
of the four treatments groups: treatment combinations are for morning and afternoon times of bleeding
with and without diethylstilbestrol treatment.
a) Identify the design of the study : Completely Randomized Design (CRD)
b) Identify the factors and their levels : Factor A= Time of Bleeding (morning, afternoon)
and Factor B = Diethylstilbestrol Treatment (With diethylstilbestrol, without diethylstilbestrol)
c) Write the linear model to be used in the analysis of the data. Indicate the assumption of
the model
i=1,2
j=1,2 k= 1,2,3,4
Yijk     i   j   ij  eijk ,
 
eij ~iidN 0, e2
4
  i 0 ,   j 0 ,   ij 0
i
i, j
j
  ij 0   ij 0
j
i
July 28, 2011