Lab 8 - Continuation of Simulation Let us investigate one of the problems from last week regarding a possible discrimination suit.

Transcription

1 Lab 8 - Continuation of Simulation Let us investigate one of the problems from last week regarding a possible discrimination suit. A company with a large sales staff announces openings for 3 positions as regional managers. 22 employees apply, 10 men and 12 women. After interviewing all, the 3 positions are given to 3 women. The men complain of job discrimination. Do they have a case? Let us use simulation in R to find out how many times, on average, a 3 woman managerial staff is picked from 10 men and 12 women, when chosen by chance (i.e., when every candidate has an equal chance of being chosen). The code and graphs are shown below. # 3 women picked from 10 men 12 women # n <- c(50) for (j in 1:50) { n[j] <- 0 for (i in 1:100) { candidates1 <- c(1:22) samp1 <- sample(candidates1,3, replace=false) y <- (samp1> 10) y1 <- as.numeric(y) if(sum(y1)>2) n[j] <- n[j]+1 n summary(n/100) hist(n, main="dotted is red mean, dashed is median") abline(v=mean(n), lty=3, lwd=3, col="red") abline(v=median(n),lty=2) y ; y1 Output is shown below. Next -1-

2 In this code we have a nested for() loop program. In the i for() loop, we randomly pick 3 numbers from a list of 22 (with NO REPLACEMENT in the sample() command). We did not have to say replace=false in the sample() command, because that is the default argument (i.e., so leaving it out means do not replace when sampling). Also, we did not include a prob= argument, so each of the 22 numbers has an equal chance of being chosen as one of the 3. I am specifying in my simulation that men are numbers 1 through 10 and women 11 through 22. After we pick 3 numbers randomly, storing them in samp1, we then see if any of the 3 picks are women (numbers greater than 10). We will get a TRUE TRUE FALSE, for example, kind of display for y, as shown above. Our next statement makes y into a numeric statement (y1), where TRUE becomes 1 and FALSE becomes 0. If() the sum() of y1 is greater than 2 (therefore 3), then we have picked an all woman group of managers. When we have this sum of 3, we then add 1 to our counter, n. We repeat picking 3 from the 22 a total of 100 times in the inner loop, and end up with a total count of all women picked groups stored in n. We then repeat this same simulation of 100 picks of a 3 person group again, storing results in a reinitialized n, repeating a total of 50 times (performed by the outer loop). We will end up with 50 n's, one for each simulation of 100 picks. To store these 50 n's, we had to make n a 50 element vector. We do that by first letting R know that we want n to be a vector with the first line in the program. Then with each loop of the j for() loops we increment the n index, and at the end we have a 50 element vector. Our first command after exiting the loops lists n and then finds the summary() statistics of the 50 element n/100, as output above. We next make a histogram of n, with abline() showing mean and median, as shown above. Notice that we can put more than one command on the same line in R, if we separate them with a semicolon ( ; ). -2-

3 You see that in my simulation I got all women picked from about 7/100 = 7% to 22/100 = 22% of the time this is not rare, but it is not frequent. If I was running this study for the men, I would say that they did not have a discrimination case, because picking a 3 woman group is not a rare event (less than, say 5% of the time), which is traditionally needed for evidence of discrimination. My median (which is close to the mean) is about 15/100 = 15% of the time an all woman group will be randomly picked infrequent, but not rare. Homework [1]: Run another 50 simulations of 100 trials, where you have less than 10 women in the group of 22 i.e., start with 9 women, 13 men, then 8 women, 14 men, etc. Find out, using your simulations, how many women out of 22 would it take to only have a complete woman group of 3 picked less than 5% (middle value) of the time. Report that number of women and show your histogram/summary statistics showing your evidence. Electronics Store Promotion problem-- An electronics store holds a contest to attract shoppers. When a potential customer enters the store, he/she is given a deck of 5 cards (an ace and 4 other non-ace cards). He/she is asked to well-shuffle the cards (face down), and then he/she turns over the top card, then second, etc. until all 5 are chosen. If the ace appears on the first card, the customer gets $100 of free CD's or DVD's from the store. If the ace is the second card, the customer gets $50, if 3rd card $20, if 4th card $10, and if 5th card $5. The store owner needs to know what is the average amount of merchandise value he will give away per customer. Below is the code and an output simulating solutions to this problem. I probably added a bunch of features you wouldn't use in this simulation, but I wanted to show you some of the features of R you can use in simulations, for your inspection. # Electronics Store Problem # pay1 <- c(100,50,20,10,5) win1 <- c(50) for(j in 1:50) { count1 <- rep(0,5) for (i in 1:100) { cards1 <-sample(1:5, 5) vec1 <-(cards1==1) vec1 <- as.numeric(vec1) count1 <- count1 + vec1 win1[j] <- sum(count1 * pay1) win1 summary(win1) hist(win1, main="vertical line is mean") abline(v=mean(win1), lty=2) cat("winnings per customer are about", mean(win1)/100, "dollars","\n") -3-

4 next In this program we wanted to randomly place the ace into one of 5 places, then pick/place another card, etc., until we have picked 100 ace placements. We store these placements in vector count1, which is a 5 element vector which has initially been made (0,0,0,0,0), but has been increased by the number of times the ace showed up in the first, second, etc. place. When we have completed a simulation of 100 cards, count1 might look like this. This shows count1, after a single run in the inner for() loop, containing 19 times the ace showed up first, 20 times second, 17 times third, 24 times fourth, and 20 times last, for a total of 100 times. We then store the total winnings of those 100 customers in win1, where we multiply count1 by pay1, coming up with the sum() of total winnings for that one simulation of 100 customers as the sum of 100*19+50*20+20*17+10*24+5*20 dollars. You see in the output the total winnings of 100 customers simulated 50 times in the vector win1. We then make a summary, histogram, and statement of the average winnings in our

5 simulations, with the result in this run of the program, that customers will win about $37.14 each this is what I would tell the store owner he will spend, on long term average, per customer on his advertising promotion, according to my R simulations. A final statement about the line count1 <- rep(0,5). The rep() command will make a 5 element vector, with 0 replicated for all 5 of those elements with the result stored in count1. In this program we start count1 as (0,0,0,0,0) and add to it as we march through the 50 iterations of 100 simulations. You should probably look for a while, carefully, at the code and the results, until you understand what we have done this time, using both new statements and the ones we have talked about in past labs. Homework [2]: The same electronics store owner wants to conduct another advertising promotion. Again, each customer who enters gets 3 cards, face down, with one ace in the pack. After shuffling the cards (again face down) the customer turns the cards over until the ace appears. If the ace appears first, the customer gets $75 (in electronics store coupons); if it appears 2nd, customer gets $35; and if the ace comes up in 3rd, customer gets $10. Run simulations and determine how much, on average, customers will get in store coupons. Cell Phone problem: A legislator claims that our city's new law against using cell phones while driving a car has reduced cell phone use to less than 12% of all drivers. While you wait for the bus the next day, you notice that 4 out of 10 people who drive by the bus stop are using their cell phones. Does this cast doubt on the legislator's claim of the 12% figure? The results of the code shown below casts doubt on the politician's 12% figure. # cell phone problem # count1 <- c() for(i in 1:50) { count1[i] <- rbinom(1,100, prob=.12) var1 <- as.numeric(count1 >= 40) tot1 <- sum(var1) hist(count1, main="vertical line is mean") abline(v=mean(count1), lty=2, lwd=4) var1 <- sum(as.numeric(count1 >= 40)) cat("number of samples with 40 or more: ", var1, "\n") results are -5-

6 next Note that we use the binomial function rbinom() to generate a vector (vec1) of 100 elements, containing either 1 or 0, with the probability of having a 1 being.12. Then we save the total count of 1's in the variable tot1 tot1 is a vector with total counts of the 50 loops of 100 random samples containing 40 or more 1's. This indicates the number of random samples of drivers who are using their cell phones while driving which contain 40 or more drivers per sample. None of our samples contained 40 or more drivers! So much for political promises. Lady Griz basketball: Another problem suitable for simulation and solvable by utilizing the rbinom() command is now investigated. The star guard for the Lady Griz makes 55% of her 2 point field goals, 35% of her 3 point shots, and 72% of her free throws, according to last year's statistics. She attempted about

7 15 two-point shots per game, 6 three point shots, and 13 free throws (of which 9 were 1 point attempts and 4 were 2 shot attempts). Assume for this simulation that she did not do any 1-and-1 free throw attempts. How many points will she probably make per game, assuming that the regular season is 26 games this year? See code and output below. # Lady Griz # points1 <- rbinom(1,9,.72) points2free <- rbinom(1,8,.72) points2 <- rbinom(1,15,.55) points3 <- rbinom(1,6,.35) gametot <- points1 + points2free + 2*points2 + 3*points3 totpts <- c(points1, points2free,2*points2, 3*points3, gametot) names(totpts) <- c("1free", "2free", "2pts", "3pts", "totpts") totpts next The single shots made out of 9 attempts, at a 72% percentage of success, is stored in points1 and computed by the rbinom() command 9 times a sample is taken from the 0,1 binomial, where a 1 is obtained about 72% of the time. The other rbinom() commands are similar in construction to the points1 construction. Note that all of the points... labels represent shots made out of those attempted, so we had to multiply by 2 and 3 for the field goals made, in order to arrive at total points made per simulation. Also note that for the points2free we had to make 8 shots (for 4-2 shot attempted free throws). Notice that totpts is a 5 element vector and we gave each of the elements appropriate names, using the names() command. We put this approach to the simulation in a code with a for() loop to find the average number of points made in a 26 game regular season. See code and output below. -7-

8 # Lady Griz season average # totpts <- c() for (i in 1:26) { totpts[i] <- (rbinom(1,9,.72) + rbinom(1,8,.72) + 2*rbinom(1,15,.55) + 3*rbinom(1,6,.35)) totpts cat("season pts=",sum(totpts),"av pts/game=", mean(totpts), "\n") output is I probably should have spaced out the cat() command text a bit better, for clarity sake, but I will leave that for now. Homework [3]: Another Lady Griz attempts 10 two point shots per game at 56% average, 4 three point shots at 25% average, and gets to try 12 free throws (assume all single point tries), at 70% average. Using R simulation, find her 26 game regular season average and report your code and result. Extra Lab Information, not required for this lab, but interesting: I developed a way to count, randomly, how a player does on a 1 and 1 free throw opportunity, WITHOUT using high real estate computer memory with for() loops. Say our 72% free throw shooter takes 10 1-and-1 free throw attempts in a typical game. How many points would she make, using our simulation. See below. # 1 and 1 free throws # first.shot <- sample(0:1,10, replace=true, prob=c(.28,.72)) second.shot <- sample(0:1,10, replace=true, prob=c(.28,.72)) points <- first.shot + first.shot * second.shot first.shot second.shot points -8-

9 output She made 10 points with her 10 1 and 1 free throws, with her 72% free throw percentage, in my simulation. There are probably more clever ways to do this, which you might come up with. -9-