322/332 Stat Experimental Design and Sampling
Midterm
Monday, October 27, 2008, 3:30-4:30
An experiment was conducted to investigate the eectiveness of a new fungicide for grape
plants. Part of a vineyard was divided into three blocks, and four treatments were applied
at random to four plots within each block. The four treatments diered in the concentration
of fungicide applied (2, or 4 units) and the time of application (May or June). Specially, the
four treatments are:
Treatment
1
2
3
4
Concentration
2
4
2
4
Time
May
May
June
June
At harvest time, the following yields were recorded:
Block Treatment and Yield (Kg) Total
1
2
1
4
3
629 632 660
698
2619
2
3
3
2
1
729 648 683
4
634
4
2
1
565 654 680
3
679
2694
2578
7891
In addition, the treatment totals and averages are
Treatment
1
2
3
4
Total (Kg)
1995 1931 2106 1859
Average (Kg) 665.0 643.7 702.0 619.7
1. (2 marks) This experiment uses blocking. Give two reasons why blocking is useful here?
Answer: Blocking can prevent confounding due to those explanatory variables held xed
within the block. Blocking also improves the precision of treatment comparisons.
2. (2 marks) Is randomization used in this experiment? How does the random assignment to
plots help with potential confounding by unknown explanatory factors in the experiment?
Answer: Yes, randomization is used in this experiment. Randomly assigning the treatments
to plots reduces the risk of confounding by unknown explanatory variables and balances the
eect of unknown factors across the treatments.
3. (2 marks) A model for this experiment is
Yabj = + ab + j + Rabj , Rabj G(0, )
where a = 1, 2 indexes Concentration, b = 1, 2 indexes Time and j = 1, 2, 3 indexes the
blocks.
What types of constraints do we use to produce the least squares estimates of the treatment
eects ab = yab+ y+++ and the blocking eect j = y++j y+++ ?
Answer: Two constraints,
2
a
2
b ab
= 0 and
3
j
j = 0.
4. (2 marks) The owner of vineyard is puzzled by the assumption that there is a single
standard deviation in the model. Provide a suitable explanation of this assumption?
Answer: This assumption is assuming that variation in yields at the harvest time for grape
plants is the same no matter what of types treatments they received.
5. (1 marks) How could you examine the validity of this assumption?
Answer: plot the estimated residuals rabj = yabj ab j versus the treatments.
6. (2 marks) To test the hypothesis that there are no dierences among the treatments, we
use the Analysis of Variance (ANOVA). Provide a brief explanation (words, not algebra) of
the discrepancy measure used for this test.
Answer: We have two ways to estimate 2 . The rst is the least squares estimate that does
not depend on any hypothesis about the treatments. The second estimate is based on the
assumption that there are no dierences among the treatments. The discrepancy measure is
the ratio of the second estimate to the rst.
7. (4 marks) A partial ANOVA table is given below. Please ll in the blanks. Is there any
evidence of dierences among the treatments (Use 10% as the signicance level)?
Treatment
Block
Residual
Total
SS
DF MS
10977.6 ?
?
1730.2
?
?
?
?
?
19051
11
Ratio
?
?
Answer: The complete ANOVA table is listed below:
SS
DF
MS
Ratio
Treatment 10977.6 3 3659.2 3.4612
Block
1730.2
2
865
0.8182
Residual
6343.2
6 1057.2
Total
19051
11
From F table, the p-value P r (F3,6 3.4612) < 0.1, so there is weak evidence of dierences
among the treatments according to 10% signicance level.
8. (1 marks) What is the numerical value of the estimate ?
Answer:
1057.2 = 32.51.
9. (2 marks) Explain what it means to say that there is interaction between Concentration
and Time.
Answer: If there is interaction between Concentration and Time, this means the eect on
the response variable (yields) of changing Concentration depends on the levels of Time.
10. (2 marks) A further analysis shows that there is weak evidence of interaction. Prepare
a plot to demonstrate this nding.
Answer: see the plots below, either one is ok.
11. (3 marks) Find the distribution of the estimator corresponding to the contrast =
11 +12
21 +22 ?
2
2
Answer: G(, 1/3).
12. Find a 95% condence interval for the change of average yield due to the change of
Time?
Answer: The estimate of the change is = 12 +22 11 +21 = 6.5, and then the 95%
2
2
condence interval is (6.5 2.447 32.51/ 3, 6.5 + 2.447 32.51/ 3) = (39.43, 52.43).