Hypothesis Testing - Experimental Design in Agriculture - Solved Past Exam, Exams of Experimental Techniques

Download Hypothesis Testing - Experimental Design in Agriculture - Solved Past Exam and more Exams Experimental Techniques in PDF only on Docsity! 1 1) a) Define what is meant by a Type II error in hypothesis testing, and describe the relationship between Type II error and the power of a statistical test. Type II error () is the probability of accepting the null hypothesis when it is false i.e., not detecting differences among treatments that actually exist. The power of a test = 1-. Power is the probability of detecting real differences among treatments. b) Power is greatest when (circle the correct answers) i) differences among means are a) small b) large ii) experimental error is a) small b) large iii) error degrees of freedom are a) small b) large iii) Type I error level for the trial is a) small b) large 2) The budget for your research project this year is limited and you wish to make the best use of resources. You are planning to conduct an experiment on a particular field at a research station that is known to have a soil variability index of 0.40 for a wheat crop, using a standard plot size of 1.5m x 5m. It costs $3.00 to plant and harvest a plot when you consider fixed costs that do not vary with plot size. Variable costs for a standard plot are $2.00 for chemical inputs plus about 50 cents for labor. What is the optimum plot size? b=0.4 K1=3.00 k2=2.5 Xopt=0.8 Optimum plot size = 0.8*1.5*5 = 6 m2 xopt = bK1 (1-b)K2 8 pts 8 pts 10 pts docsity.com 2 3) You wish to conduct an experiment to compare eight varieties of green beans. The last time you conducted a trial in this field, the coefficient of variation was 14.5% using a standard plot size of 10 m2 and four replications. You have decided to increase the number of replications to five this year, to gain better precision. Using the standard plot size in a Randomized Complete Block Design, what is the magnitude of the difference that you could expect to detect 75% of the time, using a significance level of 5%? Show your work. 4) You asked your assistant to prepare a field plan for an experiment with 8 treatments in a Randomized Complete Block Design. When he returns with the plan, you notice that he has used the same randomization in all of the blocks. Why is randomization needed to meet the assumptions for a valid ANOVA? How would you explain the importance of randomization to your assistant? Randomization is needed to ensure that the errors (residuals) are independent. Adjacent plots will tend to have similar residuals, but randomization ensures that all treatments are equally likely to appear next to each other. Randomization prevents bias in estimating treatment means and permits valid comparisons to be made among treatments. If the assignment of treatments to the experimental units is systematic, then a gradient in the field would tend to favor the treatments that consistently occur where conditions are most favorable. dfe =( t-1)(r-1) 28 t(0.05, 28 df) = 2.04840944 t(0.50, 28 df) = 0.68335282 CV 14.5 r 5 X 1 b 0.5 d 25.0519133 10 pts 2 2 2 1 2 b 2(t t ) CV d r *X   The detectable difference is 25.05% of the mean. 10 pts docsity.com 5 1) Your boss provides you with the SAS output below from a barley variety trial and asks for your help to interpret the results. The ANOVA Procedure Dependent Variable: yield Sum of Source DF Squares Mean Square F Value Pr > F Model 8 55284 6910 2.64 0.0497 Error 15 39192 2613 Corrected Total 23 94476 R-Square Coeff Var Root MSE yield Mean 0.5852 11.94 51.1 454 Source DF Anova SS Mean Square F Value Pr > F Block 3 8046 2682 1.03 0.4089 Variety 5 47238 9448 3.62 0.0029 a) What does the Pr > F for varieties indicate? It indicates the probability of making a Type I error (rejecting the null hypothesis when it is true). Another way of saying this is that it is the probability of obtaining an F value greater than the computed F value when H0 is true. Since the value is less than 0.05, we conclude that there were significant differences among the varieties. b) Without doing any additional calculations, do these results suggest that blocking was effective? No, because the F test for blocks is not significant. c) What is the standard error of a treatment mean in this experiment? se = sqrt(2613/4)=25.56 Y MSE s r  4 pts 4 pts 6 pts docsity.com 6 F Distribution 5% Points Student's t Distribution Denominator Numerator (2-tailed probability) df 1 2 3 4 5 6 df 0.4 0.5 0.05 0.01 1 161.45 199.5 215.71 224.58 230.16 233.99 1 1.376 1.000 12.706 63.667 2 18.51 19 19.16 19.25 19.30 19.33 2 1.061 0.816 4.303 9.925 3 10.13 9.55 9.28 9.12 9.01 8.94 3 0.978 0.765 3.182 5.841 4 7.71 6.94 6.59 6.39 6.26 6.16 4 0.941 0.741 2.776 4.604 5 6.61 5.79 5.41 5.19 5.05 4.95 5 0.920 0.727 2.571 4.032 6 5.99 5.14 4.76 4.53 4.39 4.28 6 0.906 0.718 2.447 3.707 7 5.59 4.74 4.35 4.12 3.97 3.87 7 0.896 0.711 2.365 3.499 8 5.32 4.46 4.07 3.84 3.69 3.58 8 0.889 0.706 2.306 3.355 9 5.12 4.26 3.86 3.63 3.48 3.37 9 0.883 0.703 2.262 3.250 10 4.96 4.1 3.71 3.48 3.32 3.22 10 0.879 0.700 2.228 3.169 11 4.84 3.98 3.59 3.36 3.20 3.09 11 0.876 0.697 2.201 3.106 12 4.75 3.88 3.49 3.26 3.10 3.00 12 0.873 0.695 2.179 3.055 13 4.67 3.8 3.41 3.18 3.02 2.92 13 0.870 0.694 2.160 3.012 14 4.60 3.74 3.34 3.11 2.96 2.85 14 0.868 0.692 2.145 2.977 15 4.54 3.68 3.29 3.06 2.90 2.79 15 0.866 0.691 2.131 2.947 16 4.49 3.63 3.24 3.01 2.85 2.74 16 0.865 0.690 2.120 2.921 17 4.45 3.59 3.20 2.96 2.81 2.70 17 0.863 0.689 2.110 2.898 18 4.41 3.55 3.16 2.93 2.77 2.66 18 0.862 0.688 2.101 2.878 19 4.38 3.52 3.13 2.90 2.74 2.63 19 0.861 0.688 2.093 2.861 20 4.35 3.49 3.10 2.87 2.71 2.60 20 0.860 0.687 2.086 2.845 21 4.32 3.47 3.07 2.84 2.68 2.57 21 0.859 0.686 2.080 2.831 22 4.30 3.44 3.05 2.82 2.66 2.55 22 0.858 0.686 2.074 2.819 23 4.28 3.42 3.03 2.80 2.64 2.53 23 0.858 0.685 2.069 2.807 24 4.26 3.40 3.00 2.78 2.62 2.51 24 0.857 0.685 2.064 2.797 25 4.24 3.38 2.99 2.76 2.60 2.49 25 0.856 0.684 2.060 2.787 26 26 0.856 0.684 2.056 2.779 27 27 0.855 0.684 2.052 2.771 28 28 0.855 0.683 2.048 2.763 29 29 0.854 0.683 2.045 2.756 30 30 0.854 0.683 2.042 2.750 docsity.com