Midterm_1_s
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Midterm_1_s

Course Number: CSS 590, Fall 2008

College/University: Oregon State

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Experimental Design in Agriculture CSS 590 First Midterm Winter 2009 Name: KEY 6 pts 1) A new tillage practice can reduce soil erosion in soybeans by 20% compared to conventional tillage methods, but no formal experiments have yet been done to verify the superiority of the new method. An agronomist conducts an experiment to compare soil erosion using four different tillage practices, including the new and...

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Design Experimental in Agriculture CSS 590 First Midterm Winter 2009 Name: KEY 6 pts 1) A new tillage practice can reduce soil erosion in soybeans by 20% compared to conventional tillage methods, but no formal experiments have yet been done to verify the superiority of the new method. An agronomist conducts an experiment to compare soil erosion using four different tillage practices, including the new and conventional methods. The experiment is replicated four times in a Randomized Complete Block Design. The observed F value for tillage treatments from the ANOVA is 3.25. Using the F table at the end of this exam and an alpha level of 0.05, what is the outcome of this experiment? a. The null hypothesis is accepted and a Type I error is committed b. The null hypothesis is rejected and a Type I error is committed c. The null hypothesis is accepted and a Type II error is committed d. The null hypothesis is rejected and the researcher concludes that there are differences among the tillage practices 16 pts 2) You are an agronomist investigating the effects of nitrogen fertility on yield of canola. You intend to apply five rates of nitrogen fertilizer in a replicated field experiment. Describe at least four potential natural sources of experimental error in your field and steps you could take to control them. 1) border effects make sure that the plot area is surrounded by border rows to maintain uniform competition 2) stand losses overplant and thin the seedlings back to a uniform plant density; fence the experimental area to keep out rodents and other animals. 3) Seasonal variation conduct the experiment over several years to ensure that results are consistent from one year to the next. 4) Field gradients (moisture, fertility, etc.) for known patterns of variation, blocking can be used to get more precise comparisons among treatments. 5) Soil heterogeneity increase plot size (up to a point) or number of replicates or measure a concomitant variable to use as a covariate. Other answers are possible... 1 12 pts 3) You wish to compare eight varieties of dry beans in a field that was determined to have a soil variability index of b=0.4. The last time you conducted a trial in this field you obtained a mean yield of 2000 kg/ha and a standard deviation of 260 kg/ha using a plot size of 8 m2 and four replications in a randomized block design. What size plot would you need to have an 80% probability of detecting differences of 25% of the mean, using a significance level of 5%? Use the tables at the end of this exam and show your work. (hint: this is a little tricky because your estimates of s and d are not on the same scale; you will need to convert one or the other before you apply the formula) dfe = (r-1)(t-1) t(0.05, 21 df) t(0.40, 21 df) s mean CV=(260/2000)*100 r d% d in units=.25*2000 b= (4-1)(8-1) = 21 2.08 0.859 260 2000 13 4 25 500 0.4 Xb = [2*(2.08+0.859)2*132]/(4*252) OR Xb = [2*(2.08+0.859)2*2602]/(4*5002) Xb = 1.16782 (Xb)1/0.4 = (1.16782) 1/0.4 X = 1.4738 1.4738 x 8 m2 = 11.79 m2 6 pts 4) An experiment is underway to compare the effects of five weed control practices on mint production. The treatments are replicated in three randomized complete blocks. Four plants are sampled in each plot to obtain an estimate of leaf area. The researcher conducts two analyses. In the first case she calculates the mean leaf area for each plot and carries out an ANOVA on plot means. In the second case she analyzes the original data including the individual plant observations in each plot. Will she reach the same conclusions from both analyses regarding the effects of the weed control treatments on leaf area? Why or why not? 2 Yes, the F tests and Probability values will be the same using both approaches, so the conclusions will be the same. The SS and MS from the analysis using subplots will be 3 times the magnitude of the SS and MS in the analysis of means, but ratios (F values) will be the same. The df for treatments and for error (blocks x treatment) will also be the same in both cases. The researcher must be careful in the analysis using subplots to separate the experimental error (blocks x treatment) from the sampling error in the model. 3 5) You have been hired as a consultant to critique a small grains research program. On a 16 pts trip to the field station you observe an experiment designed to investigate the effect of three planting densities 300, (200, and 400 plants/m2) on grain yield of oats. You know from a soil map that the clay content of the soil is higher at the northern end and lower at the southern end of the field. The experiment has been divided into three blocks. What is your assessment of the layout of this experiment? What would you do differently? What features of the design do you like? How would you explain your answers to the scientist who is conducting the experiment? N Higher clay contentm 15 200 Block 1 300 400 200 Block 2 300 400 200 25 m Block 3 300 400 Lower clay content Problems: 1) Plots are not randomized in each block which could bias results 2) Plots within each block should be oriented north to south so that all plots within a block are exposed to similar soil conditions. Good points: 1) Blocks arranged along the gradient 2) Border rows 3) Long, narrow plot shape and square blocks 4 6) A plant breeder in Zambia would like to compare three varieties of maize for grain yield: a hybrid, and open-pollinated variety (OPV), and a local (landrace) variety. The experiment is conducted as a Completely Randomized Design with 4 replications. The mean grain yields for the varieties in Mg/ha are as follows: Mean=2.8 Varieties 6 pts Hybrid Means 3.35 OPV 2.90 Local 2.15 SSvarieties=4*[(3.35-2.8)2+(2.90-2.8)2+(2.152.8)2]= 4*[(0.55)2+(0.1)2+(-0.65)2]=2.94 Fill in the shaded cells to complete the following ANOVA. Source 16 pts Total Varieties Error DF 11 2 9 SS 4.02 2.94 1.08 1.47 0.12 12.25 MS F 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 = 12.25 F critical ( = 0.05, 2, 9 df) = 4.26 4 pts 12.25 >4.26, so we reject the null hypothesis and conclude that there are differences among the varieties b) What is the standard error of a treatment mean? 4 pts seX = MSE / r sqrt(0.12/4)=0.1732 5 6 pts c) The LSD (alpha=0.05) is 0.5541 for this experiment. Is the yield of the hybrid significantly different from the OPV? Is the yield of the OPV significantly different from the Local? Explain your answer. Hybrid vs OPV OPV vs Local 3.35-2.9 = 0.45 Less than LSD so no difference 2.90-2.15 = 0.75 Greater than LSD so they are different 6 pts 7) An experiment is conducted to compare the effects of 5 irrigation treatments in oranges (trickle, basin, spray, sprinkler, and flood). Are these treatments likely to be considered to be fixed effects or random effects? Justify your answer. They are fixed effects because we are interested in this particular set of irrigation treatments per se. They do not represent a larger population of treatments. 6 F Distribution 5% Points Denominator df 1 1 161.45 2 18.51 3 10.13 4 7.71 5 6.61 6 5.99 7 5.59 8 5.32 9 5.12 10 4.96 11 4.84 12 4.75 13 4.67 14 4.6 15 4.54 16 4.49 17 4.45 18 4.41 19 4.38 20 4.35 21 4.32 22 4.3 23 4.28 24 4.26 25 4.24 26 27 28 29 30 Numerator 2 3 4 5 6 7 199.5 215.71 224.58 230.16 233.99 236.77 19 19.16 19.25 19.3 19.33 19.36 9.55 9.28 9.12 9.01 8.94 8.89 6.94 6.59 6.39 6.26 6.16 6.08 5.79 5.41 5.19 5.05 4.95 5.88 5.14 4.76 4.53 4.39 4.28 4.21 4.74 4.35 4.12 3.97 3.87 3.79 4.46 4.07 3.84 3.69 3.58 3.5 4.26 3.86 3.63 3.48 3.37 3.29 4.1 3.71 3.48 3.32 3.22 3.13 3.98 3.59 3.36 3.2 3.09 3.01 3.88 3.49 3.26 3.1 3 2.91 3.8 3.41 3.18 3.02 2.92 2.83 3.74 3.34 3.11 2.96 2.85 2.76 3.68 3.29 3.06 2.9 2.79 2.71 3.63 3.24 3.01 2.85 2.74 2.66 3.59 3.2 2.96 2.81 2.7 2.61 3.55 3.16 2.93 2.77 2.66 2.58 3.52 3.13 2.9 2.74 2.63 2.54 3.49 3.1 2.87 2.71 2.6 2.51 3.47 3.07 2.84 2.68 2.57 2.49 3.44 3.05 2.82 2.66 2.55 2.46 3.42 3.03 2.8 2.64 2.53 2.44 3.4 3 2.78 2.62 2.51 2.42 3.38 2.99 2.76 2.6 2.49 2.4 Student's t Distribution (2-tailed probability) df 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 0.4 0.05 0.01 1.376 12.706 63.667 1.061 4.303 9.925 0.978 3.182 5.841 0.941 2.776 4.604 0.920 2.571 4.032 0.906 2.447 3.707 0.896 2.365 3.499 0.889 2.306 3.355 0.883 2.262 3.25 0.879 2.228 3.169 0.876 2.201 3.106 0.873 2.179 3.055 0.870 2.16 3.012 0.868 2.145 2.977 0.866 2.131 2.947 0.865 2.12 2.921 0.863 2.11 2.898 0.862 2.101 2.878 0.861 2.093 2.861 0.860 2.086 2.845 0.859 2.080 2.831 0.858 2.074 2.819 0.858 2.069 2.807 0.857 2.064 2.797 0.856 2.060 2.787 0.856 2.056 2.779 0.855 2.052 2.771 0.855 2.048 2.763 0.854 2.045 2.756 0.854 2.042 2.750 7

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