8 pts
Name:
Please show your work!
KEY
1) A plant science student wishes to determine factors that influence the prevalence of
Medusahead (an invasive annnual grass) on rangelands in eastern Oregon. Soil test results from 20 sites showed a positive correlation between the number of Medusahead present and naturally occuring levels of soil nitrogen. Can he conclude that high nitrogen fertility causes an increase in the prevalence of Medusahead? Explain your answer.
No. This is an observational study, not a designed experiment. It is not possible to determine cause and effect because the treatments (N levels) have not been randomly assigned to the experimental units, and other factors that could be related to nitrogen fertility and/or medusahead incidence have not been controlled.
15 pts
2) For each of the three examples below, circle the scenario (type of experiment) that would favor a larger optimum plot size. In the space on the right, provide a brief explanation for your choice.
Which will favor a larger plot size?
Reason(s) for your choice i.) ii.) irrigation treatments
OR planting density treatments fertilizer study on
Douglas fir trees OR a wheat crop
Need larger scale equipment for irrigation.
Border effects of irrigation treatment may be greater because moisture can move in the soil and there may be some drift of water in the wind.
In general, larger plants require larger plots to achieve the same level of precision. Standard errors are reduced when more plants are sampled, and large plants take up more space. iii.) testing site with uniform soil OR variable soil properties
Up to a point, plot to plot variation decreases with larger plots. On variable soils, larger plots may be needed to reduce plot to plot variation and control experimental error.
1

8 pts
12 pts
1) An agricultural consultant was hired by a chemical company to determine if a new herbicide that they have developed is safe for use on cabbage and if it effectively controls important weed species. The consultant conducted a trial and reported that the new herbicide caused no damage to the cabbage crop and that it effectively controlled germination of redroot pigweed. However, in subsequent trials conducted by the company, cabbage plants showed damage symptoms from the herbicide, and germination of redroot pigweed was not significantly reduced by the herbicide.
In evaluating damage symptoms on cabbage, the consultant appears to have made a
(circle the best option): a) Type 1 error b) Type 2 error
In evaluating the effect of the herbicide on redroot pigweed germination, the consultant appears to have made a (circle the best option): a) Type 1 error b) Type 2 error
2) You have just been hired by a seed company to evaluate 8 oat cultivars for yield potential in the Willamette Valley. Last year, the company conducted a similar experiment using a standard plot size, and obtained a CV for yield of 10%. They would like to be able to detect differences of 25% of the mean, 80% of the time using an alpha level of 0.05. You plan to use the standard plot size and a Randomized Complete Block
Design. Will 4 replicates (blocks) be sufficient to attain the precision and power desired for this experiment?
Hints: There are several ways to solve this problem – look for the easiest way!
You won’t be able to calculate power (1
) directly, but you should be able to answer the question. t = r = dfe = (r-1)(t-1) t(0.05, 21 df) = t(0.40, 21 df) =
CV =
 d =
X = d = sqrt(431.8861) =20.78
8
4
21
2.080
0.859
10
25
1 d
2 
2(t
1
 t ) CV
2
2

2(2.080

0.859) 10 2 r 4 d
2 
431.8861
d

20.78
r
 r

2(2.080

0.859) 10
2
25
2
2(2.145

0.868) 10 2
25
2

2.76

2.905
(20.78<25) Should be able to detect differences that are within 20.78 units of the mean.
Yes, they can expect to achieve the desired power with the conditions specified.
Alternatively, you could solve for r using d=25 and the same t values as above with 21 df, to see if the result for r is less than or equal to 4. The first calculation gives an estimate of 3 reps. Substituting t values for 14 df, an estimate of 3 reps is obtained again. You would only need 3 reps to meet the objectives.
2

15 pts
3) You wish to compare the effects of five soil amendments on growth of strawberries. You are concerned about possible shading from a row of trees on the west side of the field.
You intend to include three replicates of each treatment in 12 m 2 plots. Show how you would arrange your plots on the field diagram below, taking into consideration what you know about plot shape and orientation. Explain why you chose this arrangement.
Block 1 Block 2 Block 3
B
C
A
E
D
A
D
E
B
C
C
D
E
A
B
15 m
2 m
25 m
6 m

Use an RBD to control variation due to shading (light gradient). Blocks should be roughly square in shape and oriented across the gradient, to maximize the differences among blocks and minimize the variation within blocks, to allow uniform comparisons among treatments within blocks.

Long narrow plots should be arranged parallel to the gradient. In this way each plot is exposed to the same variation in light within blocks.

The treatments should be randomly assigned to plots within each block. Each treatment appears once in each block.

Use border rows around the plots to minimize border effects.
3

4) Four fertilizer formulations were evaluated for their effects on the yield of corn using a
Completely Randomized Design. a) Fill in the boxes to complete the ANOVA.
14 pts
As CRD
Source
Total
Fertilizer
Error
DF
19
3
16
SS
6568
3048
3520
MS
1016
220
F
4.62
8 pts
6 pts
8 pts
6 pts b) Based on the results, are there significant differences among the fertilizers at the

= 0.05 probability level? What is your proof?
(use the tables at the back of this exam)
F critical at

= 0.05 with 3 and 16 df is 3.24
F observed (4.62) is greater than 3.24, so reject the null hypothesis and conclude that there are differences among the fertilizers. c) Assuming that each treatment (fertilizer formulation) was replicated an equal number of times, what is the standard error of a treatment mean?
N = 20, t = 4, so r=5 s
Y

MSE

220 r 5

6.633
d) The mean corn yield of plots that received fertilizer Z is 125 bu/acre. What are the lower and upper limits of a 95% confidence interval around the mean of Z? critical t for alpha=0.05 with 16 df is 2.120 half-width of CI = se*tcrit = 6.633*2.12 = 14.06 lower limit = 125 – 14.06 = 110.94 bu/acre upper limit = 125 + 14.06 = 139.06 bu/acre
5) An experiment is conducted as a Randomized Complete Block Design (RBD). After the experiment is completed it is determined that the relative efficiency compared to a CRD is 1.60. Circle the answer below that provides the best interpretation of this result. a. A CRD would have been 60% more efficient than the RBD b. A CRD would have been 160% more efficient than the RBD c. The RBD was 60% more efficient than the CRD d. The RBD was 160% more efficient than a CRD
4

F Distribution 5% Points
Denominator Numerator
Student's t Distribution
(2-tailed probability) df 1 2 3 4 5 6 7 df 0.4 0.2 0.05
1 161.45 199.50 215.71 224.58 230.16 233.99 236.77 1 1.376 3.078 12.706
2 18.51 19.00 19.16 19.25 19.30 19.33 19.35 2 1.061 1.886 4.303
3 10.13 9.55 9.28 9.12 9.01 8.94 8.89 3 0.978 1.638 3.182
4 7.71 6.94 6.59 6.39 6.26 6.16 6.09 4 0.941 1.533 2.776
5 6.61 5.79 5.41 5.19 5.05 4.95 4.88 5 0.920 1.476 2.571
6 5.99 5.14 4.76 4.53 4.39 4.28 4.21 6 0.906 1.440 2.447
7 5.59 4.74 4.35 4.12 3.97 3.87 3.79 7 0.896 1.415 2.365
8 5.32 4.46 4.07 3.84 3.69 3.58 3.50 8 0.889 1.397 2.306
9 5.12 4.26 3.86 3.63 3.48 3.37 3.29 9 0.883 1.383 2.262
10 4.96 4.10 3.71 3.48 3.33 3.22 3.14 10 0.879 1.372 2.228
11 4.84 3.98 3.59 3.36 3.20 3.09 3.01 11 0.876 1.363 2.201
12 4.75 3.89 3.49 3.26 3.11 3.00 2.91 12 0.873 1.356 2.179
13 4.67 3.81 3.41 3.18 3.03 2.92 2.83 13 0.870 1.350 2.160
14 4.60 3.74 3.34 3.11 2.96 2.85 2.76 14 0.868 1.345 2.145
15 4.54 3.68 3.29 3.06 2.90 2.79 2.71 15 0.866 1.341 2.131
16 4.49 3.63 3.24 3.01 2.85 2.74 2.66 16 0.865 1.337 2.120
17 4.45 3.59 3.20 2.96 2.81 2.70 2.61 17 0.863 1.333 2.110
18 4.41 3.55 3.16 2.93 2.77 2.66 2.58 18 0.862 1.330 2.101
19 4.38 3.52 3.13 2.90 2.74 2.63 2.54 19 0.861 1.328 2.093
20 4.35 3.49 3.10 2.87 2.71 2.60 2.51 20 0.860 1.325 2.086
21 4.32 3.47 3.07 2.84 2.68 2.57 2.49 21 0.859 1.323 2.080
22 4.30 3.44 3.05 2.82 2.66 2.55 2.46 22 0.858 1.321 2.074
23 4.28 3.42 3.03 2.80 2.64 2.53 2.44 23 0.858 1.319 2.069
24 4.26 3.40 3.01 2.78 2.62 2.51 2.42 24 0.857 1.318 2.064
25 4.24 3.39 2.99 2.76 2.60 2.49 2.40 25 0.856 1.316 2.060
26 4.23 3.37 2.98 2.74 2.59 2.47 2.39 26 0.856 1.315 2.056
27 4.21 3.35 2.96 2.73 2.57 2.46 2.37 27 0.855 1.314 2.052
28 4.20 3.34 2.95 2.71 2.56 2.45 2.36 28 0.855 1.313 2.048
29 4.18 3.33 2.93 2.70 2.55 2.43 2.35 29 0.854 1.311 2.045
30 4.17 3.32 2.92 2.69 2.53 2.42 2.33 30 0.854 1.310 2.042
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