Assignment 3
Cray Lester
2023-02-23
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## Boston
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## select##Question 13
- This question should be answered using the Weekly data set, which is part of the ISLR2 package. This data is similar in nature to the Smarket data from this chapter’s lab, except that it contains 1, 089 weekly returns for 21 years, from the beginning of 1990 to the end of
- Produce some numerical and graphical summaries of the Weekly data. Do there appear to be any patterns?
summary(Weekly)## Year Lag1 Lag2 Lag3
## Min. :1990 Min. :-18.1950 Min. :-18.1950 Min. :-18.1950
## 1st Qu.:1995 1st Qu.: -1.1540 1st Qu.: -1.1540 1st Qu.: -1.1580
## Median :2000 Median : 0.2410 Median : 0.2410 Median : 0.2410
## Mean :2000 Mean : 0.1506 Mean : 0.1511 Mean : 0.1472
## 3rd Qu.:2005 3rd Qu.: 1.4050 3rd Qu.: 1.4090 3rd Qu.: 1.4090
## Max. :2010 Max. : 12.0260 Max. : 12.0260 Max. : 12.0260
## Lag4 Lag5 Volume Today
## Min. :-18.1950 Min. :-18.1950 Min. :0.08747 Min. :-18.1950
## 1st Qu.: -1.1580 1st Qu.: -1.1660 1st Qu.:0.33202 1st Qu.: -1.1540
## Median : 0.2380 Median : 0.2340 Median :1.00268 Median : 0.2410
## Mean : 0.1458 Mean : 0.1399 Mean :1.57462 Mean : 0.1499
## 3rd Qu.: 1.4090 3rd Qu.: 1.4050 3rd Qu.:2.05373 3rd Qu.: 1.4050
## Max. : 12.0260 Max. : 12.0260 Max. :9.32821 Max. : 12.0260
## Direction
## Down:484
## Up :605
##
##
##
## pairs(Weekly)
It looks like there is an exponential relationship between volume and year. Most of the Lag variables look the same. Nothing else really stands out.
- Use the full data set to perform a logistic regression with Direction as the response and the five lag variables plus Volume as predictors. Use the summary function to print the results. Do any of the predictors appear to be statistically significant? If so, which ones?
logreg <- glm(Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume, family = 'binomial', data = Weekly)
summary(logreg)##
## Call:
## glm(formula = Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 +
## Volume, family = "binomial", data = Weekly)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.6949 -1.2565 0.9913 1.0849 1.4579
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.26686 0.08593 3.106 0.0019 **
## Lag1 -0.04127 0.02641 -1.563 0.1181
## Lag2 0.05844 0.02686 2.175 0.0296 *
## Lag3 -0.01606 0.02666 -0.602 0.5469
## Lag4 -0.02779 0.02646 -1.050 0.2937
## Lag5 -0.01447 0.02638 -0.549 0.5833
## Volume -0.02274 0.03690 -0.616 0.5377
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 1496.2 on 1088 degrees of freedom
## Residual deviance: 1486.4 on 1082 degrees of freedom
## AIC: 1500.4
##
## Number of Fisher Scoring iterations: 4It looks like the Intercept and Lag2 are significant at the \(\alpha\) = .05 level.
- Compute the confusion matrix and overall fraction of correct predictions. Explain what the confusion matrix is telling you about the types of mistakes made by logistic regression.
prob <- predict(logreg, type = 'response')
confusion <- rep("Down", 1089)
confusion[prob > .5] = 'Up'
table(confusion, Weekly$Direction)##
## confusion Down Up
## Down 54 48
## Up 430 557This confusion matrix is saying that we’re classifying 56.1065% of the observations correctly. We are misclassifying 478 observations. It seems like our model most often sorts Up and very rarely sorts Down.
- Now fit the logistic regression model using a training data period from 1990 to 2008, with Lag2 as the only predictor. Compute the confusion matrix and the overall fraction of correct predictions for the held out data (that is, the data from 2009 and 2010).
training <- Weekly[Weekly$Year<2009,]
test <- Weekly[Weekly$Year>=2009,]
nlogreg <- glm(Direction ~ Lag2, data = training, family = 'binomial')
summary(nlogreg)##
## Call:
## glm(formula = Direction ~ Lag2, family = "binomial", data = training)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.536 -1.264 1.021 1.091 1.368
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.20326 0.06428 3.162 0.00157 **
## Lag2 0.05810 0.02870 2.024 0.04298 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 1354.7 on 984 degrees of freedom
## Residual deviance: 1350.5 on 983 degrees of freedom
## AIC: 1354.5
##
## Number of Fisher Scoring iterations: 4testprob <- predict(nlogreg, type = 'response', newdata = test)
testold <- Weekly$Direction[Weekly$Year>=2009]
testprediction <- rep('Down', 104)
testprediction[testprob>.5] = "Up"
table(testprediction,testold)## testold
## testprediction Down Up
## Down 9 5
## Up 34 56We are still overestimating Up, but our accuracy increased to 62.5% by using a training data set.
- Repeat (d) using LDA.
ldafit <-lda(Direction~Lag2, data = training)
ldafit## Call:
## lda(Direction ~ Lag2, data = training)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag2
## Down -0.03568254
## Up 0.26036581
##
## Coefficients of linear discriminants:
## LD1
## Lag2 0.4414162ldapredict <- predict(ldafit, newdata = test, type = 'response')
ldaclass <- ldapredict$class
table(ldaclass, test$Direction)##
## ldaclass Down Up
## Down 9 5
## Up 34 56This is the same as we just got using logistic regression.
- Repeat (d) using QDA.
qdafit <- qda(Direction ~ Lag2, data = training)
qdafit## Call:
## qda(Direction ~ Lag2, data = training)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag2
## Down -0.03568254
## Up 0.26036581qdapredict <- predict(qdafit, newdata = test, type = 'response')
qdaclass <- qdapredict$class
table(qdaclass, test$Direction)##
## qdaclass Down Up
## Down 0 0
## Up 43 61It appears that we overestimate for Up.We correctly classify 58.65% of our data.
- Repeat (d) using KNN with K = 1.
set.seed(10)
ktrainx <- cbind(training$Lag2)
ktestx <- cbind(test$Lag2)
ktrainy <- cbind(training$Direction)
knn1predict <- knn(ktrainx, ktestx, ktrainy, k =1)
table(knn1predict, test$Direction)##
## knn1predict Down Up
## 1 21 29
## 2 22 32We have a correct classification rate of 50.96%.
- Repeat (d) using naive Bayes.
train=(Weekly$Year<2009)
weekly09 <- Weekly[!train,]
direction09 <- Weekly$Direction[!train]
nbayes=naiveBayes(Direction~Lag2 ,data=Weekly ,subset=train)
nbayes##
## Naive Bayes Classifier for Discrete Predictors
##
## Call:
## naiveBayes.default(x = X, y = Y, laplace = laplace)
##
## A-priori probabilities:
## Y
## Down Up
## 0.4477157 0.5522843
##
## Conditional probabilities:
## Lag2
## Y [,1] [,2]
## Down -0.03568254 2.199504
## Up 0.26036581 2.317485nbayesclass <- predict(nbayes, weekly09)
table(nbayesclass, direction09)## direction09
## nbayesclass Down Up
## Down 0 0
## Up 43 61This classifcation rate is the same as our qda model, around 58.65%
- Which of these methods appears to provide the best results on this data?
It appears that logistic regression is the best predictor for this data set at a 62.5% successful classification rate. This is of course without any transformations or interactions in the data.
- Experiment with different combinations of predictors, including possible transformations and interactions, for each of the methods. Report the variables, method, and associated confusion matrix that appears to provide the best results on the held out data. Note that you should also experiment with values for K in the KNN classifier.
log2 <- glm(Direction ~ Lag1*Lag2, data = Weekly, family = binomial, subset = train)
summary(log2)##
## Call:
## glm(formula = Direction ~ Lag1 * Lag2, family = binomial, data = Weekly,
## subset = train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.573 -1.259 1.003 1.086 1.596
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.211419 0.064589 3.273 0.00106 **
## Lag1 -0.051505 0.030727 -1.676 0.09370 .
## Lag2 0.053471 0.029193 1.832 0.06700 .
## Lag1:Lag2 0.001921 0.007460 0.257 0.79680
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 1354.7 on 984 degrees of freedom
## Residual deviance: 1346.9 on 981 degrees of freedom
## AIC: 1354.9
##
## Number of Fisher Scoring iterations: 4log2prob <- predict(log2, weekly09, type = 'response')
log2pred <- rep("Down", length(log2prob))
log2pred[log2prob>.5] = 'Up'
table(log2pred, direction09)## direction09
## log2pred Down Up
## Down 7 8
## Up 36 53I added in the Lag1 variable and the interaction between Lag1 and Lag2 in the logistic regression model. It seems like it classifies 57.69% correct, which is worse than if we just used the Lag2 variable by itself.
lda2 <- lda(Direction ~ Lag1*Lag2, data = Weekly, family=binomial, subset=train)
lda2predict <- predict(lda2, weekly09)
table(lda2predict$class, direction09)## direction09
## Down Up
## Down 7 8
## Up 36 53This predicted the exact same as our logistic regression problem.
qda2 <- qda(Direction ~poly(Lag2,2), data =Weekly, subset = train)
gda2predict <- predict(qda2, weekly09)$class
table(gda2predict, direction09)## direction09
## gda2predict Down Up
## Down 7 3
## Up 36 58By using a squared Lag2 in the model, we improve our accuracy to 62.5% where our all time best currenty resides.
knn5predict <- knn(ktrainx, ktestx, ktrainy, k =5)
knn10predict <- knn(ktrainx, ktestx, ktrainy, k =10)
table(knn5predict, test$Direction)##
## knn5predict Down Up
## 1 16 21
## 2 27 40table(knn10predict, test$Direction)##
## knn10predict Down Up
## 1 17 19
## 2 26 42Having a k=5 gives us an accuracy of 53.85% and k=10 gave us an accuracy of 56.73%. Overall, our inital logisitic regression model gave us the best accuracy of 62.5% We were able to replicate that by squaring our Lag2 term in qda.
##Question 14
- In this problem, you will develop a model to predict whether a given car gets high or low gas mileage based on the Auto data set.
- Create a binary variable, mpg01, that contains a 1 if mpg contains a value above its median, and a 0 if mpg contains a value below its median. You can compute the median using the median() function. Note you may find it helpful to use the data.frame() function to create a single data set containing both mpg01 and the other Auto variables.
mpg01 <- ifelse( Auto$mpg > median(Auto$mpg), yes = 1, no = 0)
Auto <- data.frame(Auto, mpg01)
head(Auto)- Explore the data graphically in order to investigate the association between mpg01 and the other features. Which of the other features seem most likely to be useful in predicting mpg01? Scatterplots and boxplots may be useful tools to answer this question. Describe your findings.
cor(Auto[,-9])## mpg cylinders displacement horsepower weight
## mpg 1.0000000 -0.7776175 -0.8051269 -0.7784268 -0.8322442
## cylinders -0.7776175 1.0000000 0.9508233 0.8429834 0.8975273
## displacement -0.8051269 0.9508233 1.0000000 0.8972570 0.9329944
## horsepower -0.7784268 0.8429834 0.8972570 1.0000000 0.8645377
## weight -0.8322442 0.8975273 0.9329944 0.8645377 1.0000000
## acceleration 0.4233285 -0.5046834 -0.5438005 -0.6891955 -0.4168392
## year 0.5805410 -0.3456474 -0.3698552 -0.4163615 -0.3091199
## origin 0.5652088 -0.5689316 -0.6145351 -0.4551715 -0.5850054
## mpg01 0.8369392 -0.7591939 -0.7534766 -0.6670526 -0.7577566
## acceleration year origin mpg01
## mpg 0.4233285 0.5805410 0.5652088 0.8369392
## cylinders -0.5046834 -0.3456474 -0.5689316 -0.7591939
## displacement -0.5438005 -0.3698552 -0.6145351 -0.7534766
## horsepower -0.6891955 -0.4163615 -0.4551715 -0.6670526
## weight -0.4168392 -0.3091199 -0.5850054 -0.7577566
## acceleration 1.0000000 0.2903161 0.2127458 0.3468215
## year 0.2903161 1.0000000 0.1815277 0.4299042
## origin 0.2127458 0.1815277 1.0000000 0.5136984
## mpg01 0.3468215 0.4299042 0.5136984 1.0000000pairs(Auto)
- Split the data into a training set and a test set.
trainb <- (Auto$year %% 2 == 0) #This picks an even year
testb <- !trainb
Atrain <- Auto[trainb,]
Atest <- Auto[testb,]
mpg01test <- mpg01[testb]- Perform LDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
ldampg <- lda(mpg01 ~ cylinders + weight + horsepower + displacement + year, data = Auto, subset = trainb)
ldapredictmpg <- predict(ldampg, Atest)
table(ldapredictmpg$class, mpg01test)## mpg01test
## 0 1
## 0 87 6
## 1 13 76mean(ldapredictmpg$class != mpg01test)## [1] 0.1043956- Perform QDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
qdampg <- qda(mpg01 ~ cylinders + weight + horsepower + displacement + year, data = Auto, subset = trainb)
qdapredict <- predict(qdampg, Atest)
table(qdapredict$class, mpg01test)## mpg01test
## 0 1
## 0 89 12
## 1 11 70mean(qdapredict$class != mpg01test)## [1] 0.1263736- Perform logistic regression on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
mpglog <- glm(mpg01~ cylinders + weight + horsepower + displacement + year, data = Atrain, family =binomial)
summary(mpglog)##
## Call:
## glm(formula = mpg01 ~ cylinders + weight + horsepower + displacement +
## year, family = binomial, data = Atrain)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.38487 -0.00601 0.01455 0.13620 1.76182
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -14.183575 7.621940 -1.861 0.0628 .
## cylinders -0.021014 0.767203 -0.027 0.9781
## weight -0.005155 0.001602 -3.217 0.0013 **
## horsepower -0.050225 0.027943 -1.797 0.0723 .
## displacement -0.018794 0.018025 -1.043 0.2971
## year 0.478279 0.122101 3.917 8.96e-05 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 289.577 on 209 degrees of freedom
## Residual deviance: 58.285 on 204 degrees of freedom
## AIC: 70.285
##
## Number of Fisher Scoring iterations: 9logpredict <- predict(mpglog, Atest, type = 'response')
predmpg <- rep(0, length(logpredict))
predmpg[logpredict>.05] <- 1
table(predmpg, mpg01test)## mpg01test
## predmpg 0 1
## 0 79 3
## 1 21 79mean(predmpg != mpg01test)## [1] 0.1318681- Perform naive Bayes on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
n.bayes<- naiveBayes(mpg01 ~ cylinders + displacement + horsepower + weight + year, data = Atrain)
n.bayesclass <- predict(n.bayes, Atest)
table(n.bayesclass, mpg01test)## mpg01test
## n.bayesclass 0 1
## 0 88 11
## 1 12 71mean(n.bayesclass != mpg01test)## [1] 0.1263736- Perform KNN on the training data, with several values of K, in order to predict mpg01. Use only the variables that seemed most associated with mpg01 in (b). What test errors do you obtain? Which value of K seems to perform the best on this data set?
#I was having trouble binding the data for KNN here for some reason, so I found some
#code online that took a different approach.
data = scale(Auto[,-c(9,10)])
set.seed(1)
train <- sample(1:dim(Auto)[1], 392*.7, rep=FALSE)
test <- -train
training_data = data[train,c("cylinders","horsepower","weight","acceleration")]
testing_data = data[test, c("cylinders", "horsepower","weight","acceleration")]
train.mpg01 = Auto$mpg01[train]
test.mpg01= Auto$mpg01[test]set.seed(1)
knn_pred_y = knn(training_data, testing_data, train.mpg01, k = 1)
table(knn_pred_y, test.mpg01)## test.mpg01
## knn_pred_y 0 1
## 0 52 10
## 1 9 47mean(knn_pred_y != test.mpg01)## [1] 0.1610169set.seed(1)
knn_pred_y = knn(training_data, testing_data, train.mpg01, k = 5)
table(knn_pred_y, test.mpg01)## test.mpg01
## knn_pred_y 0 1
## 0 51 4
## 1 10 53mean(knn_pred_y != test.mpg01)## [1] 0.1186441set.seed(1)
knn_pred_y = knn(training_data, testing_data, train.mpg01, k = 15)
table(knn_pred_y, test.mpg01)## test.mpg01
## knn_pred_y 0 1
## 0 53 4
## 1 8 53mean(knn_pred_y != test.mpg01)## [1] 0.1016949set.seed(1)
knn_pred_y = knn(training_data, testing_data, train.mpg01, k = 20)
table(knn_pred_y, test.mpg01)## test.mpg01
## knn_pred_y 0 1
## 0 53 4
## 1 8 53mean(knn_pred_y != test.mpg01)## [1] 0.1016949It appears that the error rate stabilizes somewhere around 15. Increasing K>15 doesn’t seem to have a significant impact.
##Question 16
- Using the Boston data set, fit classification models in order to predict whether a given census tract has a crime rate above or below the median. Explore logistic regression, LDA, naive Bayes, and KNN models using various subsets of the predictors. Describe your findings. Hint: You will have to create the response variable yourself, using the variables that are contained in the Boston data set.
crime <- rep(0, length(Boston$crim))
crime[Boston$crim > median(Boston$crim)] <- 1
Boston<- data.frame(Boston,crime)
summary(Boston)## crim zn indus chas
## Min. : 0.00632 Min. : 0.00 Min. : 0.46 Min. :0.00000
## 1st Qu.: 0.08205 1st Qu.: 0.00 1st Qu.: 5.19 1st Qu.:0.00000
## Median : 0.25651 Median : 0.00 Median : 9.69 Median :0.00000
## Mean : 3.61352 Mean : 11.36 Mean :11.14 Mean :0.06917
## 3rd Qu.: 3.67708 3rd Qu.: 12.50 3rd Qu.:18.10 3rd Qu.:0.00000
## Max. :88.97620 Max. :100.00 Max. :27.74 Max. :1.00000
## nox rm age dis
## Min. :0.3850 Min. :3.561 Min. : 2.90 Min. : 1.130
## 1st Qu.:0.4490 1st Qu.:5.886 1st Qu.: 45.02 1st Qu.: 2.100
## Median :0.5380 Median :6.208 Median : 77.50 Median : 3.207
## Mean :0.5547 Mean :6.285 Mean : 68.57 Mean : 3.795
## 3rd Qu.:0.6240 3rd Qu.:6.623 3rd Qu.: 94.08 3rd Qu.: 5.188
## Max. :0.8710 Max. :8.780 Max. :100.00 Max. :12.127
## rad tax ptratio black
## Min. : 1.000 Min. :187.0 Min. :12.60 Min. : 0.32
## 1st Qu.: 4.000 1st Qu.:279.0 1st Qu.:17.40 1st Qu.:375.38
## Median : 5.000 Median :330.0 Median :19.05 Median :391.44
## Mean : 9.549 Mean :408.2 Mean :18.46 Mean :356.67
## 3rd Qu.:24.000 3rd Qu.:666.0 3rd Qu.:20.20 3rd Qu.:396.23
## Max. :24.000 Max. :711.0 Max. :22.00 Max. :396.90
## lstat medv crime
## Min. : 1.73 Min. : 5.00 Min. :0.0
## 1st Qu.: 6.95 1st Qu.:17.02 1st Qu.:0.0
## Median :11.36 Median :21.20 Median :0.5
## Mean :12.65 Mean :22.53 Mean :0.5
## 3rd Qu.:16.95 3rd Qu.:25.00 3rd Qu.:1.0
## Max. :37.97 Max. :50.00 Max. :1.0set.seed(1)
traincrime<- sample(1:dim(Boston)[1], dim(Boston)[1]*.7, rep = FALSE)
testcrime <- -traincrime
Bostontrain <- Boston[traincrime,]
Bostontest <- Boston[testcrime,]
crimetest <- crime[testcrime]pairs(Boston)
lgcrime <- glm(crime~indus + nox + age +rad + tax +lstat, data = Boston, family = binomial)lgprobs <- predict(lgcrime, Bostontest, type = 'response')
predictlg <- rep(0, length(lgprobs))
predictlg[lgprobs>.5]<-1
table(predictlg, crimetest)## crimetest
## predictlg 0 1
## 0 65 14
## 1 8 65mean(predictlg != crimetest)## [1] 0.1447368Our logistic regression has a 14.47% error rate.
ldacrime <- lda(crime~ indus + nox + age +rad + tax +lstat, data = Boston)
predictldacrime <- predict(ldacrime, Bostontest)
table(predictldacrime$class, crimetest)## crimetest
## 0 1
## 0 71 19
## 1 2 60mean(predictldacrime$class != crimetest)## [1] 0.1381579Our lda has a 13.82% error rate, which is better than our logistic regression.
crimenbayes <- naiveBayes(crime~ indus + nox + age +rad + tax +lstat, data = Boston)
crimenbayesclass <- predict(crimenbayes, Bostontest)
table(crimenbayesclass, crimetest)## crimetest
## crimenbayesclass 0 1
## 0 68 24
## 1 5 55mean(crimenbayesclass != crimetest)## [1] 0.1907895Our naive Bayes has the highest error rate so far of 19.08%.
kbostontraining <- cbind(Boston$indus, Boston$nox, Boston$age, Boston$rad,Boston$tax ,Boston$lsta)[traincrime,]
kbostontesting <- cbind(Boston$indus, Boston$nox, Boston$age, Boston$rad, Boston$tax, Boston$lsta)[testcrime,]
kcrimetrain <- crime[traincrime]
kcrimetest <-crime[testcrime]
crimeknn1<- knn(kbostontraining, kbostontesting, kcrimetrain, k=1)
table(crimeknn1, kcrimetest)## kcrimetest
## crimeknn1 0 1
## 0 64 5
## 1 9 74mean(crimeknn1 != kcrimetest)## [1] 0.09210526crimeknn3<- knn(kbostontraining, kbostontesting, kcrimetrain, k=3)
table(crimeknn3, kcrimetest)## kcrimetest
## crimeknn3 0 1
## 0 65 6
## 1 8 73mean(crimeknn3 != kcrimetest)## [1] 0.09210526crimeknn5<- knn(kbostontraining, kbostontesting, kcrimetrain, k=5)
table(crimeknn5, kcrimetest)## kcrimetest
## crimeknn5 0 1
## 0 64 6
## 1 9 73mean(crimeknn5 != kcrimetest)## [1] 0.09868421It looks like the best k is somewhere between 1 and 3, as it gives us the lowest error rate out of all our models at 9.21%. It looks like for this data set, using Knn classification will give us the best results.