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

d

step-by-step explanation:

hello and welcome!

so, when we see the words "proportional" along with a table, we need to know that we're looking numbers that divide by the same thing.

looking at this table, we simply have to do a trial and error, and divide to see which numbers have the same quotient. if you look at choice a & c, they both have equal quotients for three of the numbers, with one exception, so those can't be the answer. looking at b, there is no similarity, so that can't be it. then, we see d; if you divide all the numbers in the table, you'll see that they all divide by two, which is a proportional relationship, and our answer!

also, always remember that the x will always be the number that goes into y, and somewhere behind this table, is a formula. for this specific table, the formula would be:

since our x-value is being divided by two to get the y-value. and if they were to ask for more numbers in the series, this is the formula you would use to find them.

step-by-step explanation: i dont