1 Quiz Practice Questions 7 (Attendance 14) for Statistics 514 ...

1Quiz Practice Questions 7 (Attendance 14) for Statistics 514Design of ExperimentsChapter 32 Neter et al. and KuhnThese are practice questions for the quiz. The quiz (not the practice questions) isworth 5% and marked out of 5 points. One or more questions is closely, but notnecessarily exactly, related to one or more of these questions will appear on the quiz.These practice questions are not to be handed in. Quizzes are to be done using Vistaon the Internet before 4am (West Lafayette time!) of the date of the quiz. Vistawill not allow any quiz to be done late. It is highly recommended that you completethis practice quiz, by hand, before logging onto Vista. The quiz is an individualone which means that each student does this quiz by themselves without help fromothers.Applied Linear Statistical Models (Neter et al.) Questions.Chapter Problem(s) hints32, pages 1306–1309 32.732.13 Whipped topping

Practice Quiz Questions 7 (Attendance 14), 32.7, pp 1306–1309 2(32.7) Five–factor response surface(a) Number of regression coefficientsThe response function is given byY i = β 0 X 0 + β 1 X i1 + β 2 X i2 + β 3 X i3 + β 4 X i4 + β 5 X i5+ β 11 Xi1 2 + β 22 Xi2 2 + β 33 Xi3 2 + β 44 Xi4 2 + β 55 Xi52+ β 12 X i1 X i2 + β 13 X i1 X i3 + β 14 X i1 X i4 + β 15 X i1 X i5+ β 23 X i2 X i3 + β 24 X i2 X i4 + β 25 X i2 X i5+ β 34 X i3 X i4 + β 35 X i3 X i5 + β 45 X i4 X i5 + ε ijkand so there are ? regression coefficients.(b) Linear, quadratic and two–factor terms?There are? linear main effects? quadratic main effects? two–factor effects(c) Number of design pointsNeed at least 21 design points to estimate the 21 regression coefficients.Need at least ? design points for a five–factor central composite designthat is based on a 2 5−1V fractional factorial design becausen T = 2 k−f n c + 2kc s + n 0 = ?where there are n c = c s = n 0 = 1 replications at the corner points, starpoints and center points, respectively.

Practice Quiz Questions 7 (Attendance 14), 32.13, pp 1306–1309 3(32.13) Whipped topping: qz7-32-13-topping-response(a) Two–factor rotatable central composite designThe second–order response function isY i = β 0 X 0 + β 1 X i1 + β 2 X i2 + β 12 X i1 X i2+ β 11 Xi1 2 + β 22 Xi2 2 + ε ijkwhere, from SAS, the estimated coefficients arecoefficient b q p–valueb 0 189.75 < 0.0001b 1 28.247 0.0008b 2 -0.772 0.87b 12 ? 0.0748b 11 -18.128 0.0115b 22 -6.875 0.223It looks like an appropriate hierarchical model would beY i = β 0 X 0 + β 1 X i1 + β 11 X 2 i1 + ε ijksince b 1 is significant (active), but b 2 is not significant.(b) ResidualsLook at the SAS output.

Practice Quiz Questions 7 (Attendance 14), 32.13, pp 1306–1309 4(c) Lack of fit testFrom SAS, the ANOVA table isSource Sum Of Squares Degrees of Freedom Mean SquaresRegression 9316.05569 5 1863.2114Error 978.860975 6 163.1435Lack of Fit ? 3 ?Pure Error ? 3 ?Total 10295 11H 0 : µ = β 0 X 0 + β 1 X i1 + β 2 X i2 + β 12 X i1 X i2 + β 11 Xi1 2 + β 22 Xi2 2 versusH a : µ ≠ β 0 X 0 + β 1 X i1 + β 2 X i2 + β 12 X i1 X i2 + β 11 Xi1 2 + β 22 Xi22(In other words, H o : no lack of fit versus H a : lack of fit)The test statistic isF ∗ =SSE − SSPEdf E − df P E÷ SSPEdf P E= SSLFdf LF= ? 3 ÷ ? 3= 3.24÷ SSPEdf P EThe critical value is F (1 − α; df LF , df P E ) = F (0.99; 3, 3) = 29.5since F ∗ = 3.84 < F = 29.5accept null; that is, there is not lack of fit(In other words, the model appears to be a good one.)(d) Three–dimensional plot and contour plot for response surfaceLook at the SAS output.(e) Calculation of maximum for the response surfaceB =[−18.128 13.75/213.75/2 −6.875], b ∗ =[28.247−0.772]and soX s = − 1 [?2 B−1 b ∗ =?]

Practice Quiz Questions 7 (Attendance 14), 32.13, pp 1306–1309 5(f) Confidence interval for mean responseFrom SAS,Ŷ h = ?s{Ŷh} = 13.70t(1 − α/2; n − p) = t(1 − 0.05/2; 12 − 6) = 2.447and so the 95% confidence interval isŶ h ± t(1 − α/2; n − p)s{Ŷh} = ? ± 13.70(2.447) = (173.1, 240.1)