Stat 401XV Final Exam Spring 2017

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1 Stat 40XV Final Exam Spring 07 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!

I need to know how large p must be in order to have overall system reliability (probability of functioning).99. Set up an equation you could solve in order to find this p for me.

2 6 pts. A so-called " k out of n system" will function provided at least k of its n components function. Consider a "4 out of 5 system" with independent components that each have reliability (probability of functioning) p. I need to know how large p must be in order to have overall system reliability (probability of functioning).99. Set up an equation you could solve in order to find this p for me.. Customers arrive at a service counter with inter-arrival times (times between consecutive arrivals) modeled as independent exponential random variables with mean min. a) Under this model, what fraction of inter-arrival times are less than.5 min? 7 pts b) Under this model, approximate the probability that less than 50 customers arrive in a particular 60 minute period. (Hint: This is the probability that the sum of 50 inter-arrival times is larger than 60.)

3. A student project concerned measurement of resistivity of a type of copper wire at two different temperatures. Seven pieces of this were used in the study, and measured resistances at 0.

56 6 pts a) Give and interpret a 95% lower confidence bound for the mean increase in resistivity of this wire associated with an increase in temperature from 0.0 C to.8 C.

3 3. A student project concerned measurement of resistivity of a type of copper wire at two different temperatures. Seven pieces of this were used in the study, and measured resistances at 0.0 C and at 8.8 C are in the following table. (Units are 0 Ω m.) Wire C Resistivity C Resistivity pts a) Give and interpret a 95% lower confidence bound for the mean increase in resistivity of this wire associated with an increase in temperature from 0.0 C to.8 C. (PLUG IN COMPLETELY, but there is no need to simplify. Say what the "95%" means.) 5 pts b) Give a two-sided interval that you are "95% sure" will bracket 99% of measured increases in resistivity of this wire associated with an increase of temperature 0.0 C to.8 C. (PLUG IN COMPLETELY, but there is no need to simplify.) In a second study concerning resistivity of this wire, two different meters were both used in measuring resistance at.8 C for the same n = 70 specimens. For 50 of the 70 specimens/trials, meter A produced a higher reading than did meter B. 6 pts c) Give a p -vlaue for assessing whether there is clear evidence that the fraction of specimens for which meter A produces a higher reading than meter B exceeds.5. 3

3 4. Beginning on Page 8 there is R analysis of a partially replicated factorial experiment due to R. Snee treated in Engineering Statistics by Hogg and Ledolter.

04% () C-Amount of an Additive lb () vs lb () on y = percentage impurity produced by a chemical process. Use that in the following questions.

(Some of these "sample means" are based on only observation.

combinations produce the same mean purity. F = df.

4 3 4. Beginning on Page 8 there is R analysis of a partially replicated factorial experiment due to R. Snee treated in Engineering Statistics by Hogg and Ledolter. It concerned the effects of factors Factor Levels A-Polymer Type Standard () vs New (But Expensive) () B-Polymer Concentration.0% () vs.04% () C-Amount of an Additive lb () vs lb () on y = percentage impurity produced by a chemical process. Use that in the following questions. 5 pts 6 pts a) Give "margins of error" based on 95% two-sided confidence limits to associate with the 8 sample means in the study. (Some of these "sample means" are based on only observation.) Where n combination = : Where n combination = : b) Give the value of an F statistic and degrees of freedom for testing the hypothesis that all 8 experimental combinations produce the same mean purity. F = df.. =, c) Based on the last 3 runs of the lm() routine with these data, what model for y in terms of the experimental variables do you judge to be best? (Name and interpret values of detectable effects and say what other effects are not detectable.) d) For the first case, the predicted value produced in the final lm() run is.895. If it were printed out, what would be the corresponding value for the next-to-final run? If it is.895 say why. If it is not.895 say why not. 4

5. There is a dataset on the UCI Machine Learning Data Set Repository that provides -0 quality ratings by experts ( y ) for wine samples and corresponding results of chemical analyses ( x = ( x, x,,

Consider first a SLR analysis of the variable quality using the predictor variable alcohol. Below is a scatterplot for these variables and the least squares line through the data pairs.

5 5. There is a dataset on the UCI Machine Learning Data Set Repository that provides -0 quality ratings by experts ( y ) for wine samples and corresponding results of chemical analyses ( x = ( x, x,, x) ). This problem concerns data analysis for 599 red wine samples. Beginning on Page 0 there is relevant R code and output. Consider first a SLR analysis of the variable quality using the predictor variable alcohol. Below is a scatterplot for these variables and the least squares line through the data pairs. (The plotting locations have ben randomly "jittered" slightly to minimize the visual effects of over-plotting.) a) Say what the plot suggests about the appropriateness of the Gaussian simple linear regression model (particularly the modeling of "errors" ε i ). b) Would you be willing to use a 95% prediction interval for the expert quality rating, y, of a new specimen with alcohol content x = based on these data and the Gaussian SLR model? Explain. 5 pts c) Is there definitive evidence that average quality rating increases with alcohol content? Provide quantitative support for your answer based on the R output. 5

Suppose that one suspends any concerns about model assumptions and adopts the usual MLR model y = β0 + βx+ βx + + βx+ ε for

d) Interpret the fitted regression coefficient for x = grams of acetic acid per cubic decimeter.

the other 0 chemical analysis results add detectably to one's ability to predict quality rating. F = df.

(Leave behind f) Below is a table of some summaries for several linear predictors fit by least squares.

6 Suppose that one suspends any concerns about model assumptions and adopts the usual MLR model y = β0 + βx+ βx + + βx+ ε for quality rating as a function of the chemical analysis results. d) Interpret the fitted regression coefficient for x = grams of acetic acid per cubic decimeter. 8 pts e) Give the value of an F statistic and degrees of freedom for judging whether after accounting for x = alcohol content, the other 0 chemical analysis results add detectably to one's ability to predict quality rating. F = df.. =, Consider now the only issue of building an effective predictor of Gaussian model assumptions.) y = quality rating. (Leave behind f) Below is a table of some summaries for several linear predictors fit by least squares. Which linear predictor (set of chemical analysis terms) is most attractive and why? Chemical Analysis Terms R MSE CV -RMSPE through ,3,5,6,7,9,0, ,5,6,7,9,0, ,5,7,9,0, ,5,7,0, ,7,0, ,0, ,

g) Searching for an elastic net predictor for y = quality rating based on the predictors, the best CV- RMSPE available seems to be about.650 for α.00and λ.03.

h) There is code and output from train() in caret for k -nearest-neighbor and random forest predictors for y = quality rating based on the predictors.

(Give numerical support for your latter answer.) i) The printout presents a scatterplot matrix and correlations between y and MLR, knn, and random forest predictions.

6 pts j) Rather than predict y, one could instead use a classification tree to identify chemical analysis x, x,, x that produce y 7. The tree below was fit using rpart (and cp =.0).

7 g) Searching for an elastic net predictor for y = quality rating based on the predictors, the best CV- RMSPE available seems to be about.650 for α.00and λ.03. The predictions it produces are not much different from ordinary MLR. Why is this not surprising given the elastic net parameters and what you know about the MLR model from part f)? h) There is code and output from train() in caret for k -nearest-neighbor and random forest predictors for y = quality rating based on the predictors. What value of " k " is best for the former and what value of " mtry " is best for the latter? How do these predictors compare to each other and to MLR predictors in terms of performance? (Give numerical support for your latter answer.) i) The printout presents a scatterplot matrix and correlations between y and MLR, knn, and random forest predictions. It seems impossible to improve much on the best of these predictors using a linear combination of them. Based on the information available to you, give rationale for this happening. 6 pts j) Rather than predict y, one could instead use a classification tree to identify chemical analysis x, x,, x that produce y 7. The tree below was fit using rpart (and cp =.0). vectors ( ) What is the misclassification rate for this tree on the training set? Describe in simple terms what chemical analysis results it associates with a quality score of 7 or more. ("" is the " y 7 " class and "to the left" is "the condition holds" circumstance.) 7

8 R Code and Output for Chemical Process Analyses Type<-c(,,,3,4,5,5,6,7,7,8) PolyType<-c(,,,,,,,,,,) PolyConc<-c(,,,,,,,,,,) AddAmount<-c(rep(,5),rep(,6)) Impurity<-c(,,.,.,.5,.9,.7,.,.,.3,.5) cbind(type,polytype,polyconc,addamount,impurity) Type PolyType PolyConc AddAmount Impurity [,].0 [,].0 [3,] [4,] [5,] [6,] [7,] [8,] 6. [9,] [0,] [,] aggregate(impurity,by=list(type),fun="mean") Group. x aggregate(impurity,by=list(type),fun="sd") Group. x NA NA NA NA NA Type<-as.factor(Type) PolyType<-as.factor(PolyType) PolyConc<-as.factor(PolyConc) AddAmount<-as.factor(AddAmount) Snee<-data.frame(Type,PolyType,PolyConc,AddAmount,Impurity) summary(snee) Type PolyType PolyConc AddAmount Impurity : :6 :6 :5 Min. : : :5 :5 :6 st Qu.: : Median : : Mean : : 3rd Qu.: : Max. :.000 (Other): options(contrasts = rep("contr.sum", )) Snee.out<-lm(Impurity~Type,data=Snee) summary(snee.out) 8

Call: lm(formula = Impurity ~ Type, data = Snee) Residuals: 3 4 5 6 7-3.643e-7 -.000e-0.000e-0.343e-9 -.058e-7.000e-0 -.000e-0 8 9 0-4.40e-7-5.000e-0 5.000e-0-8.439e-8 Coefficients: Estimate Std.

9 Call: lm(formula = Impurity ~ Type, data = Snee) Residuals: e e-0.000e-0.343e e-7.000e e e e e e-8 Coefficients: Estimate Std. Error t value Pr(t) (Intercept) *** Type Type Type * * Type Type5 Type * Type * --- Signif. codes: 0 *** 0.00 ** 0.0 * Residual standard error: 0.5 on 3 degrees of freedom Multiple R-squared: 0.967, Adjusted R-squared: F-statistic:.6 on 7 and 3 DF, p-value: Snee.out<-lm(Impurity~PolyType*PolyConc*AddAmount,data=Snee) summary(snee.out) Call: lm(formula = Impurity ~ PolyType * PolyConc * AddAmount, data = Snee) Residuals: e e-0.000e e e-8.000e e e e e-0.6e-7 Coefficients: Estimate Std. Error t value Pr(t) (Intercept) *** PolyType PolyConc AddAmount ** PolyType:PolyConc PolyType:AddAmount PolyConc:AddAmount PolyType:PolyConc:AddAmount Signif. codes: 0 *** 0.00 ** 0.0 * Residual standard error: 0.5 on 3 degrees of freedom Multiple R-squared: 0.967, Adjusted R-squared: F-statistic:.6 on 7 and 3 DF, p-value: Snee.out3<-lm(Impurity~PolyType+PolyConc,data=Snee) summary(snee.out3) Call: lm(formula = Impurity ~ PolyType + PolyConc, data = Snee) Residuals: Min Q Median 3Q Max

10 Coefficients: Estimate Std. Error t value Pr(t) (Intercept) e-08 *** PolyType ** PolyConc e-06 *** Signif. codes: 0 *** 0.00 ** 0.0 * Residual standard error: on 8 degrees of freedom Multiple R-squared: , Adjusted R-squared: F-statistic: on and 8 DF, p-value: 8.569e-06 predict(snee.out3) R Code and Output for the Wines Data wines<-read.clipboard(header=true,sep=",") wines$quality<-as.numeric(wines$quality) Good<-rep(0,599) for (i in :599) if (wines$quality[i] 6) Good[i]<- GoodF<-as.factor(Good) Wines<-data.frame(wines,GoodF) summary(wines) fixed.acidity volatile.acidity citric.acid residual.sugar chlorides Min. : 4.60 Min. :0.00 Min. :0.000 Min. : Min. :0.000 st Qu.: 7.0 st Qu.: st Qu.:0.090 st Qu.:.900 st Qu.: Median : 7.90 Median :0.500 Median :0.60 Median :.00 Median : Mean : 8.3 Mean :0.578 Mean :0.7 Mean :.539 Mean : rd Qu.: 9.0 3rd Qu.: rd Qu.:0.40 3rd Qu.:.600 3rd Qu.: Max. :5.90 Max. :.5800 Max. :.000 Max. :5.500 Max. :0.600 free.sulfur.dioxide total.sulfur.dioxide density ph sulphates Min. :.00 Min. : 6.00 Min. :0.990 Min. :.740 Min. : st Qu.: 7.00 st Qu.:.00 st Qu.: st Qu.:3.0 st Qu.: Median :4.00 Median : Median : Median :3.30 Median :0.600 Mean :5.87 Mean : Mean : Mean :3.3 Mean : rd Qu.:.00 3rd Qu.: rd Qu.: rd Qu.: rd Qu.: Max. :7.00 Max. :89.00 Max. :.0037 Max. :4.00 Max. :.0000 alcohol quality GoodF Min. : 8.40 Min. : :38 st Qu.: 9.50 st Qu.:5.000 : 7 Median :0.0 Median :6.000 Mean :0.4 Mean : rd Qu.:.0 3rd Qu.:6.000 Max. :4.90 Max. :8.000 Wines<-Wines[,:] summary(lm(quality~alcohol,data=wines)) Call: lm(formula = quality ~ alcohol, data = Wines) Residuals: Min Q Median 3Q Max Coefficients: Estimate Std. Error t value Pr(t) (Intercept) <e-6 *** alcohol <e-6 *** --- Signif. codes: 0 *** 0.00 ** 0.0 * Residual standard error: on 597 degrees of freedom Multiple R-squared: 0.67, Adjusted R-squared: 0.63 F-statistic: on and 597 DF, p-value: <.e-6 0

11 summary(lm(quality~.,data=wines)) Call: lm(formula = quality ~., data = Wines) Residuals: Min Q Median 3Q Max Coefficients: Estimate Std. Error t value Pr(t) (Intercept).97e+0.9e fixed.acidity.499e-0.595e volatile.acidity citric.acid -.084e+00.e < e-6 *** -.86e-0.47e residual.sugar.633e-0.500e chlorides free.sulfur.dioxide -.874e e e-06 *** 4.36e-03.7e * total.sulfur.dioxide -3.65e e e-06 *** density ph -.788e+0.63e e-0.96e * sulphates 9.63e-0.43e e-5 *** alcohol.76e-0.648e < e-6 *** --- Signif. codes: 0 *** 0.00 ** 0.0 * Residual standard error: on 587 degrees of freedom Multiple R-squared: , Adjusted R-squared: F-statistic: 8.35 on and 587 DF, p-value: <.e-6 lmtune<-train(y=wines[,], + x=wines[,:], + method="lm", + preprocess = c("center","scale"), + trcontrol=traincontrol(method="repeatedcv",repeats=00,number=0)) lmtune Linear Regression 599 samples predictor Pre-processing: centered (), scaled () Resampling: Cross-Validated (0 fold, repeated 00 times) Summary of sample sizes: 440, 439, 439, 438, 438, 44,... Resampling results: RMSE Rsquared Tuning parameter 'intercept' was held constant at a value of TRUE lmpred<-predict(lmtune) knntune<-train(y=wines[,], + x=wines[,:], + method="knn", + preprocess = c("center","scale"), + tunegrid=data.frame(.k=:5), + trcontrol=traincontrol(method="repeatedcv",repeats=00,number=0)) knntune k-nearest Neighbors 599 samples predictor Pre-processing: centered (), scaled () Resampling: Cross-Validated (0 fold, repeated 00 times)

12 Summary of sample sizes: 439, 440, 439, 439, 438, 438,... Resampling results across tuning parameters: k RMSE Rsquared RMSE was used to select the optimal model using the smallest value. The final value used for the model was k = 8. knnpred<-predict(knntune) ForestTune<-train(y=Wines[,], + x=wines[,:], + tunegrid=data.frame(mtry=:), + method="rf",ntree=000, + trcontrol=traincontrol(method="oob")) ForestTune Random Forest 599 samples predictor No pre-processing Resampling results across tuning parameters: mtry RMSE Rsquared RMSE was used to select the optimal model using the smallest value. The final value used for the model was mtry = 4. rfpred<-predict(foresttune) round(cor(cbind(quality,lmpred,knnpred,rfpred)),) quality lmpred knnpred rfpred quality lmpred knnpred rfpred pairs(cbind(quality,lmpred,knnpred,rfpred))

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Stat 401B Exam 2 Fall 2017

Stat 401B Exam 2 Fall 2017 Stat 0B Exam Fall 07 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