Quiz 2 in ST512: Statistical Analysis of Experimental Data, Fall Semester 2008 - Prof. Jas, Quizzes of Statistics

Download Quiz 2 in ST512: Statistical Analysis of Experimental Data, Fall Semester 2008 - Prof. Jas and more Quizzes Statistics in PDF only on Docsity! ST512 Fall Semester, 2008 Quiz 2 Name: Directions: Answer questions as directed. Please show work. Partial credit may be awarded for correct expressions given in incomplete answers. To save time, it is ok to give answers as numeric expressions without carrying out every last operation. If the question asked for the standard error of the estimated mean of a population at x = 14 in a simple linear regression, the following response would receive full credit: ŜE(β̂0 + 14β̂1) = √√√√16.4 ( 1 38 + (14 − 10.8)2 190 ) You may assume a level of significance of α = .05 and a confidence level of 95% for any tests or confidence intervals. There is a blank page at the end for extra space. You may also use the back of the page if necessary. 1. Consider a balanced experiment with a completely randomized design to test for the effect of a factor with t = 4 levels. Assume the usual model for the response on experimental unit j (j = 1, 2, . . . , n) randomized to treatment i: Yij = µ + τi + Eij where Eij are assumed i.i.d. N(0, σ 2). Circle T for true or F for false, or provide a short answer: (a) T or    F : the expectation of the error mean square is E[MS(E)] = σ 2 + n ∑ τ 2i (b)    T or F: As n increases, the type II error rate goes down. (c)    T or F: As σ increases, the type II error rate goes up. (d) T or    F : For simultaneous 95% confidence intervals for the six differences among the four treatment means, the Bonferroni procedure will yield more narrow con- fidence intervals than the Tukey procedure. (e) Suppose that interest lies in tests of k = 5 contrasts involving the four treatment means. i. Give the definition of the familywise error rate in this context. Pr( at least one type I error) ii. Suppose n = 5, which procedure will lead to more narrow simultaneous 95% confidence intervals, Scheffe or Bonferroni? Since t(.025/k) < √ (4 − 1)F (.05, 3, 16), the intervals based on the Bonferroni procedure will be more narrow. 3. An experiment investigates the effect of seeding rates (90,110,130,150 and 170 lb/acre) on wheat yield. A field is divided into four rows of five plots each. Within each row, the five seeding rates are randomized to the five plots. In the summations given below, yij denotes the yield for treatment i assigned to row j: 5∑ i=1 4∑ j=1 (yij − ȳ++) 2 = 543.7 5∑ i=1 4∑ j=1 (ȳi+ − ȳ++) 2 = 416.8 5∑ i=1 4∑ j=1 (ȳ+j − ȳ++) 2 = 74.4 5∑ i=1 4∑ j=1 (yij − ȳi+ − ȳ+j + ȳ++) 2 = 52.6 with treatment means and standard deviations Treatment (i) Seeding rate rows mean (ȳi+) std. dev. 1 90 4 47.7 3.9 2 110 4 53.1 3.9 3 130 4 55.8 1.9 4 150 4 60.8 0.6 5 170 4 58.8 2.7 (a) This is a randomized complete block design. What are the “blocks”? rows (b) What is the block sum of squares? SS[block] =? 74.4 (c) Estimate the mean difference in yield between levels 150 and 170 lb/acre. Report a standard error. ȳ5+ − ȳ4+ = −2 ŜE(ȳ5+ − ȳ4+) = √ MS(E) 2 4 = √ 52.6 12 2 4 = √ 4.83(2/4) = 1.48 (d) Estimated linear and quadratic polynomial contrasts are given below. θ̂L = −2ȳ1+ − ȳ2+ + 0ȳ3+ + ȳ4+ + 2ȳ5+ = 29.7 θ̂Q = 2ȳ1+ − ȳ2+ + −2ȳ3+ − ȳ4+ + 2ȳ5+ = −12.5 i. Are these estimated contrasts orthogonal? What is their covariance, Cov(θ̂L, θ̂Q)? Yes they are orthogonal, and therefore have covariance 0. ii. Test for lack of fit in a model in which mean yield depends linearly on seeding rate, (with different intercepts for different rows). Is the linear model ok? SS(θ̂L) = θ̂2L 10/4 = 352.8 FLOF = SS[LOF ]/(5 − 1 − 1) MS(E) = (416.8 − 352.8)/3 52.6/12 = 64/3 4.38 = 4.86 F (.05, 3, 12) = 3.49 Since FLOF > F (.05, 3, 12), we reject the simple linear regression model in favor of a more complex polynomial. (e) A reduced model is fit (SAS output below) that includes parameters for the effects of rows 1,2 and 3 ({ρj}), and only two parameters for a quadratic effect of seeding rate (xi): yij = µ4 + ρj + β1xi + β2x 2 i + Eij. Sum of Source DF Squares Mean Square F Value Pr > F Model 5 471.8648571 94.3729714 18.39 <.0001 Error 14 71.8371429 5.1312245 Corrected Total 19 543.7020000 Standard Parameter Estimate Error t Value Pr > |t| Intercept 2.297500000 B 12.44551412 0.18 0.8562 row 1 -3.060000000 B 1.43265132 -2.14 0.0508 row 2 -1.120000000 B 1.43265132 -0.78 0.4474 row 3 -5.060000000 B 1.43265132 -3.53 0.0033 row 4 0.000000000 B . . . seedrate 0.728857143 0.19757013 3.69 0.0024 seedrate2 -0.002232143 0.00075676 -2.95 0.0106 i. For a given row, estimate the difference between seeding rate x = 150 and seeding rate x = 170 under the reduced model. β̂1 ∗ 150 + β̂2 ∗ 150 2 − β̂1 ∗ 170 + β̂2 ∗ 170 2 = −0.3 ii. Test for lack of fit of the quadratic model. FLOF = SS[LOF ]/(5 − 2 − 1) MS(E) = 416.8 + 74.4 − 471.86)/2 4.38 = 19.3/2 4.38 = 2.20 F (.05, 2, 12) = 3.89 Since F does not exceed the critical value, do not reject H0. We may conclude that there is no significant lack of fit of the quadratic model.