Randomly split the messages into a training set d1 (80% of messages) and a testing set d2 (20% of messages). calculate the testing accuracy, confusion matrix, precision, recall, and f-score of the na¨ıve bayes classifier in determining whether a message is spam or ham. submit your source code. note: let’s assume that spam is the positive class

In the step-by-step

Step-by-step explanation:

This is the code i created using the R software and the packages "caret" and "e0171".

The script was supposed to work in all general cases.

library(caret)

library(e1071)

# Categorical vector

spam <- c("spam","not_spam")

spam_vec <- sample(spam,60, replace = T)

# Supposing two independent variables so that the kappa will be close to 0.

x1 <- rnorm(60)

x2 <- rnorm(60)

# Creating the dataset

data1 <- cbind(spam_vec,x1,x2)

data1 <- as.data.frame(data1)

names(data1) <- make.names(c("spamvec","x","y"))

# Creating the partition

index <- createDataPartition(data1$spamvec,

                            p=0.8, list=FALSE)

training_data <- data1[index,]

testing_data <- data1[-index,]

fitControl <- trainControl(method = "cv",

                         number = 5,

                         savePred = TRUE,

                         classProb = TRUE)

tune.grid <- expand.grid(C = seq(0, 10, .1))

# Scaling the predictors

preProcess_cs <- preProcess(training_data[, -1],

                                 method = c("center", "scale"))

spam_training_cs      <- predict(preProcess_cs, training_data)

spam_testing_cs       <- predict(preProcess_cs, testing_data)

# Training a Naive Bayes to predict binary outcome

Naive_Bayes_Model=naiveBayes(spamvec ~.,

                            data=spam_training_cs,

                            tuneGrid = tune.grid,

                            trControl = fitControl)

# Confusion matrix

prediction <- predict(Naive_Bayes_Model, spam_testing_cs)

confusionMatrix(prediction, spam_testing_cs$spamvec, positive = "spam")

confM <- confusionMatrix(prediction, spam_testing_cs$spamvec, positive = "spam")

accuracy <- confM$overall[1]

accuracy

c-4 is the final answer! : )

we know that

applying the law of sines

°

so

∠j=18°

∠k=29°

∠l=180-(18+29)=133°

the answer part a) is

the measure of angle k is  ≈29°

part b)  the measure of angle k could also be

(180-29)=151°

so

∠j=18°

∠k=151°

∠l=180-(18+151)=11°

the answer part b) is

the measure of angle k could also be 151°

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