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276 lines
12 KiB
Plaintext
276 lines
12 KiB
Plaintext
Part B: Choose one of Questions 10 or 11
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5. We now examine the differences between LDA and QDA.
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(a) If the Bayes decision boundary is linear, do we expect LDA
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or QDA to perform better on the training set? On the test set?
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The QDA has more flexibility, so it will match the training
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set more closely than the LDA. The LDA will perform better
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on the test set because the real relationship is linear, so
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the QDA would have additional bias.
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(b) If the Bayes decision boundary is non-linear, do we expect
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LDA or QDA to perform better on the training set? On the test
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set?
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QDA will still perform better on the training set, but now
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should also perform better than LDA on the test set, since
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QDA will account for the additional degree of freedom in the
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real relationship.
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(c) In general, as the sample size n increases, do we expect the
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test prediction accuracy of QDA relative to LDA to improve,
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decline, or be unchanged? Why?
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Definitely increase. The LDA has an advantage when the
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training set is small because it is less sensitive to the
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fluctuations of those few data. As the size of the training
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set grows, the QDA is able to optimize its coefficients
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well, and assuming the real relationship is not linear, the
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QDA should eventually out-perform the LDA.
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(d) True or False: Even if the Bayes decision boundary for a
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given problem is linear, we will probably achieve a superior
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test error rate using QDA rather than LDA because QDA is
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flexible enough to model a linear decision boundary. Justify
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your answer.
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False. The QDA will likely be biased because it will fit to
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training data that don't completely represent the
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relationship that will be observed in test data.
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8. Suppose that we take a data set, divide it into equally-sized
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training and test sets, and then try out two different
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classification procedures. First we use logistic regression and
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get an error rate of 20 % on the training data and 30 % on the
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test data. Next we use 1-nearest neighbors (i.e. K = 1) and get
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an average error rate (averaged over both test and training data
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sets) of 18 %. Based on these results, which method should we
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prefer to use for classification of new observations? Why?
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Definitely 1-nearest neighbor. The logistic regression
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performed more poorly on the training data, to which it has
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been optimized as much as possible, and yet the nearest
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neighbor model performs better over the entire dataset.
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Considering the logistic regression performed even worse on
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the test data, the average error rate of the logistic
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regression over the training and test data is 25%. This
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suggests that the relationship may not even be linear, and
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the nearest-neighbor is a very solid method for modeling
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non-linear classifications, so if the real relationship is
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not linear, it easily explains why the nearest-neighbor
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method would do so much better. Everything here seems to
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point at using the nearest-neighbor.
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9. This problem has to do with odds.
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(a) On average, what fraction of people with an odds of 0.37 of
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defaulting on their credit card payment will in fact default?
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.27
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(b) Suppose that an individual has a 16 % chance of defaulting
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on her credit card payment. What are the odds that she will de-
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fault?
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.19
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11. In this problem, you will develop a model to predict whether a
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given car gets high or low gas mileage based on the Auto data
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set.
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──────────────────────────────────────────────────────────────────────────
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(a) Create a binary variable, mpg01 , that contains a 1 if mpg
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contains a value above its median, and a 0 if mpg contains a
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value below its median. You can compute the median using the
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median() function. Note you may find it helpful to use the
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data.frame() function to create a single data set containing
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both mpg01 and the other Auto variables.
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> auto$mpg01=rep(0,397)
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> auto$mpg01[auto$mpg>median(auto$mpg)]=1
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> auto$mpg01
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[1] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 1 1 1 1 1 0 0 0 0 0 1 1 1 1 0 0 0 0
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[38] 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
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[75] 0 0 0 0 0 1 0 1 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0
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[112] 0 0 0 1 0 0 1 1 0 0 0 1 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1
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[149] 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 1 1 0 1 0 0 0 0 1 1 1 1 1
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[186] 1 1 0 0 0 0 0 0 1 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0 1 1 1 1 1 0
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[223] 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 1 1 1 1 1 0 0 0 0 0 0 1 0 0 0
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[260] 0 0 0 0 0 0 0 1 1 1 1 0 1 1 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1
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[297] 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
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[334] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 0 0 1 1 1
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[371] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1
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──────────────────────────────────────────────────────────────────────────
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(b) Explore the data graphically in order to investigate the
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associ- ation between mpg01 and the other features. Which of the
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other features seem most likely to be useful in predicting mpg01
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? Scat- terplots and boxplots may be useful tools to answer this
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ques- tion. Describe your findings.
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Horsepower clearly has the best relationship from the
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scatter plots. All of the mpgs over the median are on one
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side of the plot. Weight and acceleration are alright, but
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there is significant overlap between middle values.
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Displacement is on the cusp and the other variables don't
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have a terribly useful relationship with this median.
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The boxplots indicate that acceleration really isn't a great
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predictor of mpg01, but displacement is. It also confirms
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horsepower and weight as good predictors, and cylinders also
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seems to be very strong, even though I didn't take that from
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the scatter plots.
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I will use mpg01 ~ horsepower + weight + cylinders + displacement
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──────────────────────────────────────────────────────────────────────────
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(c) Split the data into a training set and a test set.
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Seems like a 50/50 random sampling is appropriate enough.
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> training_indices = sample(nrow(auto),397/2)
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> train_bools = rep(F,length(auto$mpg))
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> train_bools[training_indices]=T
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> head(train_bools)
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[1] TRUE TRUE TRUE FALSE TRUE FALSE
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> length(train_bools)
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[1] 397
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> train_data = auto[train_bools,]
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> test_data = auto[!train_bools,]
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Actually, I changed this now, because a solution I found
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online suggested a different test split and I was having
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trouble with the KNN model, so I followed their style. I used:
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> train <- (auto$year %% 2 == 0)
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and then the rest the same
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──────────────────────────────────────────────────────────────────────────
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(d) Perform LDA on the training data in order to predict mpg01
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using the variables that seemed most associated with mpg01 in
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(b). What is the test error of the model obtained?
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> lda.fit
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Call:
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lda(mpg01 ~ horsepower + weight + cylinders + displacement, data = train_data)
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Prior probabilities of groups:
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0 1
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0.4666667 0.5333333
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Group means:
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horsepower weight cylinders displacement
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0 131.96939 3579.827 6.755102 268.4082
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1 77.96429 2313.598 4.071429 111.7188
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Coefficients of linear discriminants:
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LD1
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horsepower 0.0060634365
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weight -0.0011442212
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cylinders -0.6390942259
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displacement 0.0004517291
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***Test Data Error Rate:
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> mean(lda.pred$class!=test_data$mpg01,na.rm=T)
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[1] 0.1428571
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──────────────────────────────────────────────────────────────────────────
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(e) Perform QDA on the training data in order to predict mpg01
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using the variables that seemed most associated with mpg01 in
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(b). What is the test error of the model obtained?
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> qda.fit
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Call:
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qda(mpg01 ~ horsepower + weight + cylinders + displacement, data = train_data)
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Prior probabilities of groups:
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0 1
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0.4666667 0.5333333
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Group means:
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horsepower weight cylinders displacement
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0 131.96939 3579.827 6.755102 268.4082
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1 77.96429 2313.598 4.071429 111.7188
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> qda.pred=predict(qda.fit,test_data,na.rm=T)
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***Test Data Error Rate:
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> mean(qda.pred$class!=test_data$mpg01,na.rm=T)
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[1] 0.1483516
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──────────────────────────────────────────────────────────────────────────
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(f) Perform logistic regression on the training data in order to
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pre- dict mpg01 using the variables that seemed most associated
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with mpg01 in (b). What is the test error of the model obtained?
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> glm.fit=glm(mpg01 ~ horsepower + weight + cylinders + displacement,data=train_data,family=binomial)
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> glm.probs=predict(glm.fit,test_data,type="response")
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> glm.pred=rep(0,nrow(test_data))
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> glm.pred[glm.probs>.5]=1
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***Test Data Error Rate:
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> mean(glm.pred!=test_data$mpg01)
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[1] 0.1373626
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──────────────────────────────────────────────────────────────────────────
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(g) Perform KNN on the training data, with several values of K,
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in order to predict mpg01 . Use only the variables that seemed
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most associated with mpg01 in (b). What test errors do you
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obtain? Which value of K seems to perform the best on this data
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set?
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The knn method can't handle the NA values, so
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> set.seed(1)
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> auto <- na.omit(auto)
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> train_bools <- (auto$year %% 2 == 0)
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> train_data = auto[train_bools,]
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> test_data = auto[!train_bools,]
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> train.X = cbind(auto$horsepower,auto$displacement,auto$weight,auto$cylinders)[train_bools,]
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> test.X = cbind(auto$horsepower,auto$displacement,auto$weight,auto$cylinders)[!train_bools,]
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> train.mpg01 = auto$mpg01[train_bools]
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***Test Data Error Rates:
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k = 1
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> knn.pred = knn(train.X,test.X,train.mpg01,k=1)
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> mean(knn.pred != test_data$mpg01)
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[1] 0.1483516
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k = 2
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> knn.pred = knn(train.X,test.X,train.mpg01,k=2)
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> mean(knn.pred != test_data$mpg01)
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[1] 0.1593407
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k = 3
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> knn.pred = knn(train.X,test.X,train.mpg01,k=3)
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> mean(knn.pred != test_data$mpg01)
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[1] 0.1648352
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k = 4
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> knn.pred = knn(train.X,test.X,train.mpg01,k=4)
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> mean(knn.pred != test_data$mpg0)
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[1] 0.1923077
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k = 1 looks like the best, since the error rate increases with k.
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