Homework--2

(2b) Write a new data.frame object that has three columns. Two columns, X and Y , jointly describe the values that X and Y can take on together. The third column, probs , describes the probability each of these pairs of events occurs jointly. Print your data.frame so that it shows in your report. x2 <- c( 0 , 1 , 2 , 3 ) y2 <- c( 0 , 0 , 0 , 1 ) joint_prob <- c( 1 / 8 , 3 / 8 , 3 / 8 , 1 / 8 ) new_df <- data.frame(x2,y2,joint_prob) print(new_df) ## x2 y2 joint_prob ## 1 0 0 0.125 ## 2 1 0 0.375 ## 3 2 0 0.375 ## 4 3 1 0.125 (2c) Report the conditional mean of X given that Y equals 0. Recall that conditional probability can be written as: P [ A | B ] = P [ AB ] P [ B ] #mean of x given y=0 #when y=0, x can be 0,1, or 2 x_interest <- sum(new_df$x2[which(new_df$y2== 0 )]*new_df$joint_prob[which(new_df$y2== 0 )]) total_prob <- sum(new_df$joint_prob[which(new_df$y2== 0 )]) print(x_interest/total_prob) ## [1] 1.285714 (3) Assume there is a drug that is used by 0.5% of the population. We also know that a blood test #’ for the drug has a 3% positive rate. Let’s further assume that if a person uses the drug, the test will return a positive result 95% of the time. If a person tests positive for the drug, what is the likelihood that she is actually a user of that drug. (Referring to the Bayes Rule example that we discussed in class will help you complete this problem.) #usage = 0.5% #positive rate = 3% #true positive = 95% #Using Baye ' s theorem we have the following logic: # Pr{ + | user } = Pr{ user | positive}*Pr{user}/Pr{positive} prob <- 0.95 * 0.005 / 0.03 print(prob) ## [1] 0.1583333 3