Stat 401B Final Exam Fall 2016

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1 Stat 40B Final Exam Fall 0 I have neither given nor received unauthorized assistance on this exam. Name Signed Date Name Printed ATTENTION! Incorrect numerical answers unaccompanied by supporting reasoning will receive NO partial credit. Correct numerical answers to difficult questions unaccompanied by supporting reasoning may not receive full credit. SHOW YOUR WORK/EXPLAIN YOURSELF!

2 . A manufacturing process produces cylindrical parts. pts a) Suppose that parts are L inches long with diameter D (also in inches). Part volume is then V = LD π /4. Suppose that L and D can be modeled as independent continuous random variables, ~ U 9.99,0.0 D ~ U.99,.00. The mean and standard deviation of part volume ( EV L ( ) and ( ) and VarV respectively) are of interest and simulation will be used to evaluate these. Provide a few lines of R code that will do this. (The R syntax for density, cdf, random variable, etc. calls associated U ab, distribution is "_unif(..,min=a,max=b,..)".) with the ( ) 8 pts b) Suppose that the parts are steel and as manufactured have weights with mean 0. oz and standard deviation. oz. Use the central limit theorem and approximate the probability that such parts have a total weight above oz. (Hint: Rephrase the question in terms of sample average weight.). The 03 International Journal of Microbiology article "Design and Optimization of a Process for Sugarcane Molasses Fermentation by Saccharomyces cerevisiae Using Response Surface Methodology" by El-Gendy, Madian, and Amr, presents results of a study made to optimize the performance of a bioethanol production process. This question employs some results in that paper. Under a first single set of process conditions, n = runs of the process produce yields (in gm/l) of bioethanol with sample mean y =.9 and sample standard deviation s =.03. a) Give 9% two-sided confidence limits for the standard deviation of process yield under these conditions. (Plug in completely, but you need not simplify.)

3 b) Provide a number, #, such that you are 9% sure that 99% of all process yields under these conditions are at least the value #. (Plug in completely, but you need not simplify.) c) Suppose that a single additional run of the process made under a second set of conditions produces a yield of y =. Assuming that the standard deviation of yields is the same for both sets of conditions, significance testing will be used to evaluate whether this new setup has a different mean yield than the first. Give the value of an appropriate test statistic and name an appropriate reference distribution to be used in finding a p -value. Value of the test statistic Reference distribution pts d) Suppose that several process conditions of interest differ only in the value of incubation period ( x x, y data pairs are needed to make confidence limits for in h). What model assumptions for a set of ( ) the rate of change of mean y with respect to x? Beginning on Page 8 there is some R code and output for a MLR analysis of n = process runs. These are potentially useful for describing yield, y (in gm/l), as a function of the process variables x = incubation period (h) x = initial ph x3 = incubation temperature ( C) x = molasses concentration (wt %) 4 3

4 e) What is estimated by the value " Period 0.43 " reported in the table on the output? (What does the value.43 represent in the context of the problem?) f) Does a model linear in the predictors x, x, x3, and x 4 provide useful ability to predict yield? Provide some quantitative support for your answer. YES or NO (circle one) As it turns out, a MLR with the (4) predictors x, x, x, x, x, x, x, x, x x, x x, x x, x x, x x, x x has R =.98 and a LOOCV RMSPE.8. The (quadratic) model fit by least squares predicts (an optimal) yield for x =, x =., x3 = 40, and x4 = 8.. This set of processing conditions has y ˆ = 9. and se = yˆ pts g) Based on the information above and the R printout, fill out the ANOVA table for computing the overall F for the quadratic model and provide s SF for the quadratic model. ANOVA Table Source SS df MS F Regression Error Total s SF = h) What do comparisons between s =.03 (from the bottom of Page ), s SF from above, and the LOOCV RMSPE of.8 suggest about this situation? s versus s SF s SF versus LOOCV RMSPE 4

5 i) Based on the (quadratic) MLR model, give 9% prediction limits for the next yield at process conditions x =, x =., x3 = 40, and x4 = 8.. (Plug in completely, but do not simplify.) j) Before recommending adoption of process conditions from part i), what steps would you take, and why? 3. The "Appendicitis data set" on the KEEL website provides measured values of medical variables for N = 0 patients and values of a 0- variable indicating whether the patient had an appendicitis. Beginning on Page 9 there is R code and output for a logistic regression analysis of these data. a) Which of the medical variables seems least helpful in predicting whether or not a patient has an appendicitis? Why? Using bestglm() in the bestglm package and cross validation (presumably on the log likelihood criterion) it is possible to identify a good "reduced" logistic regression model as one with the two predictor variables At3 and At4. There is some R code and output for this model included. b) Give two-sided 9% confidence limits for the log odds ratio of the probability that a patient with At3 =.0 and At4 = 0 has an appendicitis. (Plug in completely, but do not simplify.)

6 4. Beginning on Page 0 there is R code and output for a factorial analysis of some experimental data on the charge lives of batteries made of 3 materials at 3 different temperatures taken from an experimental design book of Montgomery. (Though the temperature is clearly quantitative, here treat both factors as qualitative.) The data for this study comprise a balanced 3 3 factorial data set. a) Are there statistically detectable interactions between Material and Temperature? Explain. YES or NO (circle one) b) Give 9% two-sided confidence limits for the difference between the Material and Material main effects. (Plug in completely, but you need not simplify.) c) Find the fitted/predicted value of battery life for a battery of Material under Temperature for a "main effects only" model of life. (If this is not possible based on the given information, say why.). There is a famous "Boston Housing" data set on the UCI ML Data Repository. It concerns the median home price in counties around Boston in the late 90's as predicted by 3 measures of community composition. This question concerns use of data from 4 counties with complete records and prediction of "MEDV". There is R code and output provided (in pretty much the same format as for Lab #) beginning on Page. (As a baseline, MLR of MEDV on 3 predictors produces ssf = 4.3 and R =.404.) Use the output to answer the following questions. a) Which of the predictors of MEDV do you like the best, and why?

7 b) What (if anything) convinces you that you can do better here than MLR for prediction purposes? c) Why/how is it obvious that the ordinary MLR (OLS) predictor differs very little from the elastic net predictor in this case? What about the particular elastic net fit (chosen by repeated cross-validation) makes this similarity unsurprising? c) What is the origin of the "vertical stripes" appearance of the plots in the "Tree" column of the matrix of scatterplots of predicted values? d) Below is a schematic of the tree predictor (plotting is "condition TRUE to the LEFT"). Give a simple description of the conditions (values of the predictors) producing the largest predicted MEDV. e) If you were going to "stack" two of the predictors here, which two would you consider and why?

8 R Code and Output for Bioethanol Analyses Biofuels Period InitialpH Temp Conc Yield summary(lm(yield~.,data=biofuels)) Call: lm(formula = Yield ~., data = Biofuels) Residuals: Min Q Median 3Q Max Coefficients: Estimate Std. Error t value Pr( t ) (Intercept) Period InitialpH Temp Conc Residual standard error: 8.8 on degrees of freedom Multiple R-squared: 0.009, Adjusted R-squared: -0.3 F-statistic: on 4 and DF, p-value: 0.99 anova(lm(yield~.,data=biofuels)) Analysis of Variance Table Response: Yield Df Sum Sq Mean Sq F value Pr(F) Period InitialpH Temp Conc Residuals

9 R Code and Output for the Appendicitis Data Analyses Appendicitis[:0,] At At At3 At4 At At At Class summary(glm(class~.,data=appendicitis)) Call: glm(formula = Class ~., data = Appendicitis) Deviance Residuals: Min Q Median 3Q Max Coefficients: Estimate Std. Error t value Pr( t ) (Intercept) At e-08 *** ** At At3 At At At At Signif. codes: 0 *** 0.00 ** 0.0 * (Dispersion parameter for gaussian family taken to be 0.038) Null deviance:.840 on 0 degrees of freedom Residual deviance: 0.09 on 98 degrees of freedom AIC: 3.8 Number of Fisher Scoring iterations: summary(glm(class~at3+at4,data=appendicitis)) Call: glm(formula = Class ~ At3 + At4, data = Appendicitis) Deviance Residuals: Min Q Median 3Q Max Coefficients: Estimate Std. Error t value Pr( t ) (Intercept) e- *** At e-0 *** At ** --- Signif. codes: 0 *** 0.00 ** 0.0 * (Dispersion parameter for gaussian family taken to be 0.883) Null deviance:.84 on 0 degrees of freedom Residual deviance:.4 on 03 degrees of freedom AIC: 9.98 Number of Fisher Scoring iterations: 9

10 Predlogit<-predict(glm(Class~At3+At4,data=Appendicitis),se.fit=TRUE)$fit SEPredlogit<-predict(glm(Class~At3+At4,data=Appendicitis), + se.fit=true)$se.fit cbind(appendicitis$at3,appendicitis$at4,appendicitis$class, + round(predlogit,3),round(sepredlogit,3))[:0,] [,] [,] [,3] [,4] [,] R Code and Output for Battery Life Study Batteries Life MaterialA TempB

11 summary(lm(life~materiala*tempb,data=batteries)) Call: lm(formula = Life ~ MaterialA * TempB, data = Batteries) Residuals: Min Q Median 3Q Max Coefficients: Estimate Std. Error t value Pr( t ) (Intercept) < e- *** MaterialA ** MaterialA TempB e-0 *** TempB MaterialA:TempB MaterialA:TempB MaterialA:TempB ** MaterialA:TempB Signif. codes: 0 *** 0.00 ** 0.0 * Residual standard error:.98 on degrees of freedom Multiple R-squared: 0., Adjusted R-squared: 0.9 F-statistic: on 8 and DF, p-value: 9.4e-0 anova(lm(life~materiala*tempb,data=batteries)) Analysis of Variance Table Response: Life Df Sum Sq Mean Sq F value Pr(F) MaterialA ** TempB e-0 *** MaterialA:TempB * Residuals Signif. codes: 0 *** 0.00 ** 0.0 * R Code and Output for Boston Housing Data Analysis Boston[:,] CRIM ZN INDUS CHAS NOX RM AGE DIS RAD TAX PTRATIO B LSAT MEDV summary(boston) CRIM ZN INDUS CHAS NOX Min. :0.003 Min. : 0.00 Min. : 0.4 Min. : Min. :0.380 st Qu.: st Qu.: 0.00 st Qu.: 4.93 st Qu.: st Qu.:0.440 Median :0.903 Median : 0.00 Median : 8.4 Median : Median :0.90 Mean :.4083 Mean :. Mean :0. Mean :0.043 Mean : rd Qu.:.4 3rd Qu.: rd Qu.:8.0 3rd Qu.: rd Qu.:0.00 Max. :9.94 Max. :00.00 Max. :.4 Max. : Max. :0.80 RM AGE DIS RAD TAX Min. :3. Min. :.90 Min. :. Min. :.000 Min. :8.0 st Qu.:.9 st Qu.: 40.9 st Qu.:.3 st Qu.: st Qu.:.8 Median :.9 Median :.80 Median : 3.0 Median :.000 Median :.0 Mean :.344 Mean :. Mean : Mean :.83 Mean :3.4 3rd Qu.:.3 3rd Qu.: 9. 3rd Qu.:.40 3rd Qu.:.000 3rd Qu.:4.0 Max. :8.80 Max. :00.00 Max. :. Max. :4.000 Max. :.0 PTRATIO B LSAT MEDV Min. :.0 Min. : 0.3 Min. :. Min. :. st Qu.:.80 st Qu.:3. st Qu.:.88 st Qu.:8.0 Median :8.0 Median :39.08 Median :0.0 Median :.9 Mean :8. Mean :39.83 Mean :.44 Mean :3. 3rd Qu.:0.0 3rd Qu.:39. 3rd Qu.:.0 3rd Qu.:.0 Max. :.00 Max. :39.90 Max. :34.40 Max. :0.00

12 #k= for knn prediction is chosen by repeated CV sqrt(knn.reg(train=boston,y=boston[,4],k=)$press/4) [] 4.3 knnpred<-knn.reg(train=boston,y=boston[,4],k=)$pred cbind(boston$medv[:0],knnpred[:0]) [,] [,] [,] 4.0. [,] [3,] [4,] [,] [,] 8..4 [,].9. [8,]. 8.3 [9,]..0 [0,] #alpha=.00 and lambda=.3 for the elastic net are chosen by repeated CV #producing CV RMSPE x<-as.matrix(boston[,:3]) y<-as.matrix(boston[,4]) BostonNet<-glmnet(x,y,family="gaussian",alpha=.00,lambda=.3) ENetPred<-predict(BostonNet,newx=x) cbind(boston$medv[:0],enetpred[:0]) [,] [,] [,] [,] [3,] [4,] [,] [,] [,].9.49 [8,] [9,] [0,] #cp=.003 is a good choice of regression tree complexity parameter #chosen by CV and producing RMSPE BestTree<-rpart(MEDV~.,data=Boston,method="anova", control=rpart.control(cp=.003)) cbind(boston$medv[:0],predict(besttree)[:0]) [,] [,] #mtry= is a good choice for random forest parameter, chosen by CV on OOB error BostonRf<-randomForest(MEDV~.,data=Boston, + type="regression",ntree=000,mtry=) sqrt(bostonrf$mse[000]) [] 3.009

colnames(comppred)<-c("medv","ols","nn","enet","tree","rf") pairs(comppred,panel=function(x,y,.

13 comppred<-cbind(y,lm(medv~.,data=boston)$fitted.values,knnpred,enetpred, + predict(besttree),bostonrf$predicted) colnames(comppred)<-c("medv","ols","nn","enet","tree","rf") pairs(comppred,panel=function(x,y,...){ + points(x,y) + abline(0,)},xlim=c(0,0),ylim=c(0,0)) round(cor(as.matrix(comppred)),) MEDV OLS NN ENET Tree RF MEDV OLS NN ENET Tree RF

Stat 401B Final Exam Fall 2015

Stat 401B Final Exam Fall 2015 Stat 401B Final Exam Fall 015 I have neither given nor received unauthorized assistance on this exam. Name Signed Date Name Printed ATTENTION! Incorrect numerical answers unaccompanied by supporting reasoning