1Stat 511 HW#4 Spring 2008 (corrected) 1. The data set in the following table is due originally to Chau and Kelley (Journal of Coatings Technology 1993) and has been used by many authors since. Given are values of 123coating opacity(weight) fraction of pigment 1(weight) fraction of pigment 2(weight) fraction of a polymeric binderyxxx==== for 14n = specimens of a coating used on some identification labels. 1x 2x 3x y .13 .67 .20 .710,.680 .45 .35 .20 .802,.822 .45 .21 .34 .823,.798 .13 .53 .34 .698,.711 .29 .51 .20 .772 .45 .28 .27 .818 .29 .37 .34 .772 .13 .60 .27 .700 .29 .44 .27 .861,.818 Note that 1231xxx++=. a) Argue that two equivalent analyses of this data set can be made using the regression models 11 2 2 33ii iiiyx x xβββε=+ ++ (1A) and 011 22iiiiyxxααα ε=+ + + (1B) Then argue that two equivalent analyses of this data set can be made using the regression models 11 22 33 41 2 51 3 62 3ii iiiiii iiiy x x x xx xx xxββββββε=+ ++ + + + (2A) and 22011223142512iiiiiiiiyxxxxxxααα αα α ε=+++++ + (2B) (This is a so-called "mixture study." For such a study, the first of each of these two pairs in some sense treats the linearly dependent predictors symmetrically, while the second is probably easier to think about and work with using standard software.) Use R to do the following. b) Fit model (1B) using lm(). Based on this fit, compute the least squares estimate of the parameter vector β in model (1A). Use the estimated covariance matrix for OLSα to find an2estimated covariance matrix for OLSβ . Then fit model (1A) directly using lm(). (A " 1− " in the model specification will fit a no-intercept regression.) c) Fit model (2B) to these data and normal plot standardized residuals. d) In the model (2B) test 0345H: 0ααα===. Report a p-value. Does quadratic curvature in response (as a function of the x's) appear to be statistically detectable? e) Use some multivariate calculus on the fitted quadratic equation and find the location ()12,xx of an absolute maximum. Use R matrix calculations to find 90% two-sided confidence limits for the mean response here. Then find 90% two-sided prediction limits for a new response from this set of conditions. 2. (Testing “Lack of Fit” … See Section 6.6 of Christensen) Suppose that in the usual linear model =+YXβε X is of full rank (k ). Suppose further that there aremn<distinct rows inX and that mk> . One can then make up a "cell means" model forY (where observations having the same corresponding row in X are given the same mean response) say *=+YXμε This model puts no restrictions on the means of the observations except that those with identical corresponding rows ofX are equal. It is the case that()()*CC⊂XXand it thus makes sense to test the hypothesis()0H:E C∈YXin the cell means model. This can be done using ()()()()**'/'/mkFnm−−=−−XXXYP P YYI P Y and this is usually known as testing for "lack of fit." Use R and matrix calculations to find a p-value for testing lack of fit to the quadratic regression model (2B) in Problem 1. 3. Below is a small table of fake 2-way factorial data. Enter them into R in three vectors of length 12n = . Call these vectors "y", "A", and "B". Level 1 of B Level 2 of B Level 3 of B Level 1 of A 12 13,14,15 20 Level 2 of A 8 10 6,7 Level 3 of A 10 13 7 a) Create and print out an R data frame using the commands > d<-data.frame(y,A,B) > d3 b) Turn the numerical variables A and B into variables that R will recognize as levels of qualitative factors by issuing the commands > d$A<-as.factor(d$A) > d$B<-as.factor(d$B) Then compute and print out the cell means by typing > means<-tapply(d$y,list(d$A,d$B),mean) > means You may find out more about the function tapply by typing > ?tapply c) Make a crude interaction plot by doing the following. First type > x.axis<-unique(d$B) to set up horizontal plotting positions for the sample means. Then make a "matrix plot" with lines connecting points by issuing the commands > matplot(c(1,3),c(5,25),type="n",xlab="B",ylab="Mean Response",main="y") > matlines(x.axis,means,type="b") The first of these commands sets up the axes and makes a dummy plot with invisible points "plotted" at (1,5) and (3,25). The second puts the lines and identifying A levels (as plotting symbols) on the plot. d) Set the default for the restriction used to create a full rank model matrix, run the linear models routine and find both sets of "Type I" sums of squares by issuing the following commands > options(contrasts=c("contr.sum","contr.sum")) > lm.out1<-lm(y~A*B,data=d) > summary.aov(lm.out1,ssType=1) > lm.out2<-lm(y~B*A,data=d) > summary.aov(lm.out2,ssType=1) See if anything changes if you ask R to compute "Type III" sums of squares by issuing the command > summary.aov(lm.out1,ssType=3) (In the past R has failed to respond to the request for Type III sums of squares without warning you that it is going to fail to do so.)4 e) Start over with this problem, doing the calculations "from scratch" using your basic linear models knowledge and matrix calculations in R. Compute all of Type I, Type II and Type III sums of squares here, using the sum restriction in the first two cases (and the order of factors A,B). Then compute Type I and Type II sums of squares using the SAS baseline restriction. f) Now suppose that by some misfortune, the observation from the ()1, 3 cell of this complete 33× factorial somehow gets lost and one has only 11n=observations from 8k = cells (and thus "incomplete factorial" data). Test the hypothesis that at least for the cells where one has data, there are no interactions, i.e. ()()**E||Cαβ∈Y1XX. (Note that this matrix()**||αβ1X X should be of full rank.) g) In the incomplete factorial context of part f), the function**13μαβ++ is estimable. What is the OLS estimate for it? (Note that this is the mean response for the missing cell only if the same no-interaction model used to describe the 8 cells extends to the 9th. This is the kind of assumption one makes in regression analysis when using a fitted prediction equation to estimate a mean response at a set of conditions not in one's original data set. It might well be argued, however, that the link between observed and unobserved conditions is intuitively stronger with quantitative factors than with qualitative