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+Assignment 7 - Answers
+Leona Zhu
+11/30/2019
+ +In the following assignment you will be looking at data from an one level of an online geography tutoring system used by 5th grade students. The game involves a pre-test of geography knowledge (pre.test), a series of assignments for which you have the average score (av.assignment.score), the number of messages sent by each student to other students about the assignments (messages), the number of forum posts students posted asking questions about the assignment (forum.posts), a post test at the end of the level (post.test) and whether or not the system allowed the students to go on to the next level (level.up).
+
+
+Part I
+#Upload data
+library(purrr)
+library(tidyr)
+library(dplyr)##
+## Attaching package: 'dplyr'## The following objects are masked from 'package:stats':
+##
+## filter, lag## The following objects are masked from 'package:base':
+##
+## intersect, setdiff, setequal, unionlibrary(ggplot2)
+#install.packages("GGally")
+library(GGally)## Registered S3 method overwritten by 'GGally':
+## method from
+## +.gg ggplot2##
+## Attaching package: 'GGally'## The following object is masked from 'package:dplyr':
+##
+## nasalibrary(rpart)
+D1 <- read.csv("online.data.csv")#Visualization
+#Start by creating histograms of the distributions for all variables (#HINT: look up "facet" in the ggplot documentation)
+D1 %>%keep(is.numeric)%>% gather() %>%
+ ggplot(aes(value)) +
+ facet_wrap(~ key, scales = "free") +
+ geom_histogram()## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
#Then visualize the relationships between variables
+ggpairs(select(D1, -id))## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
+## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
+## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
+## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
+## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
#Try to capture an intution about the data and the relationships
+
+#messages have strong correlation with post test score and average assignment score. It shows that student send more messages talk about the assiggnment with other students get better score on post test and assignments.#Classification tree
+#Create a classification tree that predicts whether a student "levels up" in the online course using three variables of your choice (As we did last time, set all controls to their minimums)
+
+ctree_1 <- rpart(level.up ~ pre.test.score + messages + forum.posts,
+ data = D1)
+
+#Plot and generate a CP table for your tree
+printcp(ctree_1)##
+## Classification tree:
+## rpart(formula = level.up ~ pre.test.score + messages + forum.posts,
+## data = D1)
+##
+## Variables actually used in tree construction:
+## [1] messages pre.test.score
+##
+## Root node error: 400/1000 = 0.4
+##
+## n= 1000
+##
+## CP nsplit rel error xerror xstd
+## 1 0.54250 0 1.0000 1.0000 0.038730
+## 2 0.01125 1 0.4575 0.4650 0.030762
+## 3 0.01000 3 0.4350 0.4875 0.031322#Generate a probability value that represents the probability that a student levels up based your classification tree
+ctree_2<- prune.rpart(ctree_1, cp = 0.01125)
+printcp(ctree_2)##
+## Classification tree:
+## rpart(formula = level.up ~ pre.test.score + messages + forum.posts,
+## data = D1)
+##
+## Variables actually used in tree construction:
+## [1] messages
+##
+## Root node error: 400/1000 = 0.4
+##
+## n= 1000
+##
+## CP nsplit rel error xerror xstd
+## 1 0.54250 0 1.0000 1.000 0.038730
+## 2 0.01125 1 0.4575 0.465 0.030762D1$pred <- predict(ctree_1, type = "prob")[,2]#Last class we used type = "class" which predicted the classification for us, this time we are using type = "prob" to see the probability that our classififcation is based on.
+
+Part II
+#Now you can generate the ROC curve for your model. You will need to install the package ROCR to do this.
+library(ROCR)## Loading required package: gplots##
+## Attaching package: 'gplots'## The following object is masked from 'package:stats':
+##
+## lowess#Plot the curve
+pred.detail <- prediction(D1$pred, D1$level.up)
+plot(performance(pred.detail, "tpr", "fpr"))+ abline(0, 1, lty = 2)
## integer(0)#Calculate the Area Under the Curve
+unlist(slot(performance(pred.detail,"auc"), "y.values"))#Unlist liberates the AUC value from the "performance" object created by ROCR## [1] 0.8825125#Now repeat this process, but using the variables you did not use for the previous model and compare the plots & results of your two models. Which one do you think was the better model? Why?
+ctree_3<-rpart((level.up)~post.test.score + av.assignment.score,method="class",data=D1)
+printcp(ctree_3)##
+## Classification tree:
+## rpart(formula = (level.up) ~ post.test.score + av.assignment.score,
+## data = D1, method = "class")
+##
+## Variables actually used in tree construction:
+## [1] av.assignment.score post.test.score
+##
+## Root node error: 400/1000 = 0.4
+##
+## n= 1000
+##
+## CP nsplit rel error xerror xstd
+## 1 0.93 0 1.00 1.00 0.038730
+## 2 0.07 1 0.07 0.07 0.013042
+## 3 0.01 2 0.00 0.00 0.000000D1$pred1 <- predict(ctree_3, type = "prob")[,2] #what is [,2]?
+pred.detail1<-prediction(D1$pred1, D1$level.up)
+
+plot(performance(pred.detail1, "tpr", "fpr"))+abline(0, 1, lty = 2,)
## integer(0)unlist(slot(performance(pred.detail1,"auc"), "y.values"))## [1] 1##The auc of first model is 0.88;The auc of second model is 1.0 and the xerror of second model is 0. Which means the second model is perfect. Therefore the second model is better.
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+
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+
+
+Part III
+#Thresholds
+#Look at the ROC plot for your first model. Based on this plot choose a probability threshold that balances capturing the most correct predictions against false positives. Then generate a new variable in your data set that classifies each student according to your chosen threshold.
+
+#I use 0.7 as my threshold.
+
+D1$threshold.pred1 <- ifelse(D1$pred >= 0.7, 1, 0)
+
+#Now generate three diagnostics:
+
+D1$accuracy.model1 <- mean(ifelse(D1$level.up == D1$threshold.pred1, 1, 0))
+
+##True positive,false positive, false negative
+D1$truepos<-ifelse(D1$level.up == 1 & D1$threshold.pred1 == 1,1,0)
+D1$falsepos<-ifelse(D1$level.up == 0 & D1$threshold.pred1 == 1,1,0)
+D1$falseneg<-ifelse(D1$level.up == 1 & D1$threshold.pred1 == 0,1,0)
+
+
+D1$precision.model1 <- sum(D1$truepos.model1)/(sum(D1$truepos.model1) + sum(D1$falsepos.model1))
+
+D1$recall.model1 <- sum(D1$truepos.model1)/(sum(D1$truepos.model1) + sum(D1$falseneg.model1))
+
+#Finally, calculate Kappa for your model according to:
+
+#First generate the table of comparisons
+table1 <- table(D1$level.up, D1$threshold.pred1)
+
+#Convert to matrix
+matrix1 <- as.matrix(table1)
+
+#Calculate kappa
+kappa(matrix1, exact = TRUE)/kappa(matrix1)## [1] 1.040758#Now choose a different threshold value and repeat these diagnostics. What conclusions can you draw about your two thresholds?
+
+#I use 0.5 this time.
+
+D1$threshold.pred2 <- ifelse(D1$pred1 > 0.5, 1,0)
+D1$accuracy.model2 <- mean(ifelse(D1$level.up == D1$threshold.pred2, 1, 0))
+
+
+D1$truepos.model2 <- ifelse(D1$level.up == "1" & D1$threshold.pred2 == "1", 1, 0)
+D1$falsepos.model2 <- ifelse(D1$level.up == "0" & D1$threshold.pred2 == "1", 1,0)
+D1$falseneg.model2 <- ifelse(D1$level.up == "1" & D1$threshold.pred2 == "0", 1,0)
+D1$precision.model2 <- sum(D1$truepos.model2)/(sum(D1$truepos.model2) + sum(D1$falsepos.model2))
+D1$recall.model2 <- sum(D1$truepos.model2)/(sum(D1$truepos.model2) + sum(D1$falseneg.model2))
+
+#Table
+table2 <- table(D1$level.up, D1$threshold.pred2)
+#Matrix
+matrix2 <- as.matrix(table2)
+#Kappa
+kappa(matrix2, exact = TRUE)/kappa(matrix2)## [1] 1.153846matrix1##
+## 0 1
+## no 561 39
+## yes 189 211matrix2##
+## 0 1
+## no 600 0
+## yes 0 400##Model 1 kappa value is 1.04; Model 2 kappa value is 1.15. Theredore model 2 is better.
+
+To Submit Your Assignment
+Please submit your assignment by first “knitting” your RMarkdown document into an html file and then commit, push and pull request both the RMarkdown file and the html file.
+