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Soil Nitrogen - Experimental Design in Agriculture - Solved Past Exam, Exams of Experimental Techniques

This course addresses the needs of the student preparing for a career in agricultural research or consultation and is intended to assist the scientist in the design, plot layout, analysis and interpretation of field and greenhouse experiments. This solved past exam includes: Soil Nitrogen, North Valley, Potential Differences, Power of a Test, Experimental Technique, Experimental Error, Null Hypothesis, Relative Efficiency, Sums of Squares, Experimental Design

Typology: Exams

2012/2013

Uploaded on 08/20/2013

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1) A researcher wished to compare the suitability of the north Willamette Valley and the
south Willamette Valley for production of winter canola. He measured yield of a widely
grown variety in two growers’ fields in the north Valley, and in two growers’ fields in the
south Valley. He also took soil samples. He used a t test to compare the northern sites
with the southern sites, and found that both yield and soil nitrogen were significantly
higher in the north valley.
Can he conclude that the higher yields at the northern sites were due to higher nitrogen
fertility? Explain your answer.
No. This is an observational study. The effects of nitrogen cannot be separated from all
of the other potential differences between the sites. To determine the effect of nitrogen
on crop yield, he would need to conduct another experiment using several N levels, and
randomly apply nitrogen treatments to experimental units using an appropriate
experimental design.
2) a) What is meant by the Power of a test?
The power of a test is the probability of detecting differences that exist among the
treatments in the experiment.
b) List three things that a researcher can do to increase power in an experiment.
- choose treatment levels that are farther apart to increase differences among the
responses that are being measured.
- reduce experimental error by increasing the plot size or number of replications
- increase degrees of freedom by using more treatment levels or more replication
- control experimental error by refining the experimental technique
- select a higher alpha level (Type I error rate)
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1) A researcher wished to compare the suitability of the north Willamette Valley and the south Willamette Valley for production of winter canola. He measured yield of a widely grown variety in two growers’ fields in the north Valley, and in two growers’ fields in the south Valley. He also took soil samples. He used a t test to compare the northern sites with the southern sites, and found that both yield and soil nitrogen were significantly higher in the north valley. Can he conclude that the higher yields at the northern sites were due to higher nitrogen fertility? Explain your answer.

No. This is an observational study. The effects of nitrogen cannot be separated from all of the other potential differences between the sites. To determine the effect of nitrogen on crop yield, he would need to conduct another experiment using several N levels, and randomly apply nitrogen treatments to experimental units using an appropriate experimental design.

2) a) What is meant by the Power of a test? The power of a test is the probability of detecting differences that exist among the treatments in the experiment.

b) List three things that a researcher can do to increase power in an experiment.

  • choose treatment levels that are farther apart to increase differences among the responses that are being measured.
  • reduce experimental error by increasing the plot size or number of replications
  • increase degrees of freedom by using more treatment levels or more replication
  • control experimental error by refining the experimental technique
  • select a higher alpha level (Type I error rate)

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3) The response of soybeans to four seed treatments was measured in a Randomized Complete Block Design. Fill in the shaded cells to complete the ANOVA.

Source DF SS MS F Total 23 5195 Blocks 5 2345 469 9. Treatment 3 2085 695 13. Error 15 765 51

a) Do the results indicate that there were differences among the treatments? Support your statement with a significance test using the tables provided at the end of this exam. F calculated = 13. F critical (α = 0.05, 3, 15 df) = 3. 13.627 > 3.29, so we reject the null hypothesis and conclude that there are differences among the seed treatments.

b) The mean grain yield is 46 bu/acre. What is the Coefficient of Variation (CV%) for this experiment? show your work.

MSE

CV% *

Y

CV% = (sqrt(51)/46)*100 = 15.52%

c) What is the relative efficiency of this design compared to a CRD?

     

r 1 MSB r t 1 MSE RE * rt 1 MSE

    

RE = (6-1)469+6(4-1)51/[(64-1)*51] = 2.

d) Was blocking effective? What is your evidence?  The RE is considerably greater than 1  The F test for blocks is significant

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5) You have conducted an experiment to compare the tillering capacity of three varieties of wheat in a Completely Randomized Design with four replicates of each variety. The table below shows the means for tillers/plant for each variety, as well as the grand mean for the entire experiment.

a) Use the means to compute the Sums of Squares for Varieties (show your work).

2

SST  iri Y iY

SST = 4[(3.65-4.42)^2+(4.43-4.42)^2+(5.18-4.42)^2] =

b) The MSE in this experiment was 0.15. Calculate a 95% confidence interval for the mean of the ‘Karl’ variety.

t (alpha=0.05, 9 df) = 2. half-width = 2.262 * sqrt(0.15/4) = 0. lower end = 5.18-0.438 = 4. upper end = 5.18 + 0.438 = 5.

c) One of your colleagues suggests that you should have used a Randomized Block Design rather than a CRD, because the RBD is the most common type of experimental design in field trials. What justifications can you provide for using the CRD in this experiment?

 It’s a small experiment, and there are no known gradients in the field, so there is no justification for blocking.  The CRD will give you the greatest number of degrees of freedom in the error, which increases the power of the test.  Missing plots do not complicate the analysis of a CRD.

Varieties Custer Ike Karl Grand Mean Means 3.65 4.43 5.18 4.

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F Distribution 5% Points Student's t Distribution

  • df 1 2 3 4 5 6 df 0.4 0.5 0.05 0. Denominator Numerator (2-tailed probability) - 1 161.45 199.5 215.71 224.58 230.16 233.99 1 1.376 1.000 12.706 63. - 2 18.51 19 19.16 19.25 19.30 19.33 2 1.061 0.816 4.303 9. - 3 10.13 9.55 9.28 9.12 9.01 8.94 3 0.978 0.765 3.182 5. - 4 7.71 6.94 6.59 6.39 6.26 6.16 4 0.941 0.741 2.776 4. - 5 6.61 5.79 5.41 5.19 5.05 4.95 5 0.920 0.727 2.571 4. - 6 5.99 5.14 4.76 4.53 4.39 4.28 6 0.906 0.718 2.447 3. - 7 5.59 4.74 4.35 4.12 3.97 3.87 7 0.896 0.711 2.365 3. - 8 5.32 4.46 4.07 3.84 3.69 3.58 8 0.889 0.706 2.306 3. - 9 5.12 4.26 3.86 3.63 3.48 3.37 9 0.883 0.703 2.262 3.
    • 10 4.96 4.1 3.71 3.48 3.32 3.22 10 0.879 0.700 2.228 3.
    • 11 4.84 3.98 3.59 3.36 3.20 3.09 11 0.876 0.697 2.201 3.
    • 12 4.75 3.88 3.49 3.26 3.10 3.00 12 0.873 0.695 2.179 3.
    • 13 4.67 3.8 3.41 3.18 3.02 2.92 13 0.870 0.694 2.160 3.
    • 14 4.60 3.74 3.34 3.11 2.96 2.85 14 0.868 0.692 2.145 2.
    • 15 4.54 3.68 3.29 3.06 2.90 2.79 15 0.866 0.691 2.131 2.
    • 16 4.49 3.63 3.24 3.01 2.85 2.74 16 0.865 0.690 2.120 2.
    • 17 4.45 3.59 3.20 2.96 2.81 2.70 17 0.863 0.689 2.110 2.
    • 18 4.41 3.55 3.16 2.93 2.77 2.66 18 0.862 0.688 2.101 2.
    • 19 4.38 3.52 3.13 2.90 2.74 2.63 19 0.861 0.688 2.093 2.
    • 20 4.35 3.49 3.10 2.87 2.71 2.60 20 0.860 0.687 2.086 2.
    • 21 4.32 3.47 3.07 2.84 2.68 2.57 21 0.859 0.686 2.080 2.
    • 22 4.30 3.44 3.05 2.82 2.66 2.55 22 0.858 0.686 2.074 2.
    • 23 4.28 3.42 3.03 2.80 2.64 2.53 23 0.858 0.685 2.069 2.
    • 24 4.26 3.40 3.00 2.78 2.62 2.51 24 0.857 0.685 2.064 2.
    • 25 4.24 3.38 2.99 2.76 2.60 2.49 25 0.856 0.684 2.060 2.
    • 26 26 0.856 0.684 2.056 2.
    • 27 27 0.855 0.684 2.052 2.
    • 28 28 0.855 0.683 2.048 2.
    • 29 29 0.854 0.683 2.045 2.
    • 30 30 0.854 0.683 2.042 2.