1Stat 511 Exam II April 7, 2004 Prof. Vardeman 1. Consider a 33× factorial analysis with factors A and B under an ordinary (fixed effects) linear model. The effects model under the R baseline restriction has parameter vector for mean responses ()***** * * * *2 3 2 3 22 23 32 33,,,,, , , ,µααββαβαβαβαβ′=γ a) Write out all 9 cell means in terms of the entries of γ in the table below. Let rows correspond to levels of A (1 to 3 top to bottom) and columns correspond to levels of B (1 to 3 left to right). b) Give below a matrix C so that the testable hypothesis 0H:=Cγ0 is the hypothesis 0H: 0 jjβ=∀. (As always, ...jjβµµ=−.) c) Give below a matrix C so that the testable hypothesis 0H:=Cγ0 is the hypothesis that considering only levels 2 and 3 of A and levels 2 and 3 of B (the lower right 4 cells in the table) “there are no interactions” (considering only these 4 cells an interaction plot of means would have the “parallelism” property).2d) Suppose that all cell sample sizes are 1 and write ()11 12 13 21 31 22 23 32 33,,,,,,,,yyyyyy yyy′=Y . The hypothesis in c) is that E Y is in some subspace of 9ℜ of dimension 8. Write below a 9 8× matrix 0X so that ()0C X is this subspace. (Hint: Note that this hypothesis says nothing at all about 5 of the 9 mean responses, and be careful to pay attention to the order of the entries of Y listed above.) 2. A data set of Kaplan, et al. (1972) on the metabolism of sulfisoxazole can be found on page 273 of Bates and Watts. This substance was administered intravenously to a subject and blood samples were taken over time. Concentration of sulfisoxazole in the plasma, y (in g/mlµ), was measured. Bates and Watts suggest an analysis of the data based on a model for concentration as a function of time, t (in minutes from injection), ()()1234exp expii iiyt tθθθ θε=−+−+ (*) Attached to this exam is an R printout that you should use to answer the following questions. a) There are actually two different non-linear regression models represented on the R printout. The first is model (*) and the second is the less complex model ()12expiiiytθθε=−+. What quantitative support does the printout provide for the use of model (*) rather than the simpler second model? b) In fitting model (*), what one gets as parameter estimates depends upon one’s starting values for ()1234,,,θθθθ. In fact, it is possible to produce exactly the same predicted values and SSE as shown on the printout for a different set of parameter estimates ()****1234ˆˆˆˆ,,,θθθθ. How is this already obvious from (*)?3Henceforth base your responses on model (*). c) Based on model (*), an initial concentration is 13θθ+. Give approximate 95% confidence limits for this value. (You don’t need to do arithmetic, but YOU MUST PLUG IN NUMBERS.) d) It is potentially of interest to know the value of ().5 1 3the time at which concentration is .5tθθ=+ .5t is some function of ()1234,,,θθθθ, say ()1234,,,τθθθθ. It is possible to show (don’t try to do so) that here ()()()12341234ˆˆ ˆˆ,,,ˆˆˆˆ, , , 2.06 and .01664, .4352,.08311, 240.4iθθθθττθθ θθ τθ∂=∇= =−−∂ Give approximate 95% confidence limits for .5t. (Again, don’t do arithmetic, BUT DO PLUG IN.) 3. Consider a scenario in which 12 samples of a large lot of material are sent 3 apiece to 4 different labs for hardness testing. At the labs, each specimen is tested twice. For the hardness measured on the th test of the th specimen at the th labijkykji= where 1,2,3,4 and 1,2,3 and 1,2ijk===, suppose that ijk i ij ijkyµφε=++ (**) for constants 12 34,,,µµµµ, the ()2 iid N 0,ijφφσ independent of the iid ()2N0,σ random variables ijkε. a) For the ijky written in dictionary order in Y, what are , , , and X β Zu so the model can be written in standard mixed linear model form?4The second R printout attached to this exam gives an analysis for 24 hypothetical hardness values under model (**). Use it to answer the following questions. b) Under the mixed linear model used in this problem, measurements made on the same specimen are correlated. What is an estimate of that correlation? (PLUG IN.) c) Give approximate 95% confidence limits for both φσ and σ. For φσ: For σ: | | | | | | | d) Give a sensible point (single number) prediction and standard error for the sample mean of two hardness tests made on a 13th specimen sent to Lab 1 for hardness testing. (PLUG IN.) e) R reports “8 degree of freedom” standard errors for its estimates of each of 123 4, , , and µµµ µ. Those estimates are 1.. 2.. 3.. 4.., , , and yyy y. Consider the values .ijy and say why “8” makes sense and how you think the standard error(s) could be computed from these sample means using simple computations.5Stat 511 Exam II Spring 2004 Printouts Problem 2 > time<-c(.25,.50,.75,1.00,1.50,2.00,3.00,4.00,6.00,12.00,24.00,48.00) > conc<-c(215.6,189.2,176.0,162.8,138.6,121.0,101.2,88.0,61.6,22.0,4.4,.01) > blood.fm1<-nls(formula=conc~(theta1*exp(-theta2*time)+theta3*exp(-theta4*time)),start=c(theta1=200,theta2=5,theta3=70,theta4=.1),trace=T) 55161.36 : 200.0 5.0 70.0 0.1 19963.2 : 52.4343646 2.3213531 120.7827395 0.1637390 7107.025 : 66.8313947 0.2547202 168.7018220 0.1692898 7100.616 : 62.4548374 0.2588085 173.0702195 0.1700836 7093.994 : 58.4335743 0.2630338 177.0836029 0.1708158 7093.283 : 51.0640427 0.2717472 184.4379396 0.1721668 7090.37 : 44.8794949 0.2810323 190.6084239 0.1733210 7084.891 : 39.7259944 0.2908481 195.7490068 0.1743080 7076.77 : 35.4470720 0.3011472 200.0161628 0.1751539 7066.12 : 31.8979404 0.3118771 203.5546740 0.1758805 7053.153 : 28.9518490 0.3229817 206.4912762 0.1765066 7046.573 : 24.0509948 0.3458234 211.3753644 0.1775888 7027.508 : 20.6117921 0.3700385 214.8019914 0.1784065 7018.247 : 15.7654693 0.4202012 219.6310298 0.1796486 7005.513 : 10.8169763 0.5269206 224.5724600 0.1811158 6793.44 : 8.5514174 0.7518429 226.9114329 0.1821330 5964.866 : 11.2819924 1.1022659 224.5584007 0.1818996 4472.041 : 19.9510024 1.2816124 216.8710602 0.1799516 2510.098 : 35.5643008