Name: SOLUTIONS Exam 01 (Midterm Part 2 take home, open everything)

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1 Name: SOLUTIONS Exam 01 (Midterm Part 2 take home, open everything) To help you budget your time, questions are marked with *s. One * indicates a straightforward question testing foundational knowledge. Two ** indicate a more challenging question requiring use of several key concepts. Three *** indicate a challenging question requiring a clever use of key concepts and/or techniques not covered in class yet. Four **** indicates serious fun. Show your work on all problems. If you can t solve a question completely, set it up and solve it as far as you can. If you require a laptop to get your final solution, describe in detail what you would do to solve the problem. Recall our class s rules for playing Toss Up! You should be able to skim these rules, since they are the same rules we ve been using since the first class. Step 1: Write each player's name along the top of a piece of paper. This is your score sheet. Step 2: Decide who will go first. After the first player takes his turn, play continues clockwise. Step 3: To start your turn, roll 5 dice. (Note Toss Up! is usually played with 10 dice, but we will use 5 to simplify future calculations.) When you look at the dice, think of green for go, red for stop and yellow for caution. If you rolled any greens, slide them over to the side. Each green is worth 1 point at the end of your turn. Step 4: Decide if you want to continue your turn or stop. If you want to continue, you will roll all the remaining dice that did not turn up green on your previous roll. If you roll more greens, set them aside with the others. As long as you roll at least one green die, you may continue rolling as many times as you want. If you roll enough times so all 5 dice are green, you can start all over rolling the 5 dice and adding additional points to what you already scored. The number of points you can earn on a turn is unlimited. But if you roll at least one red and no greens, you have run a "red light" and your turn ends scoring 0 points for that turn, i.e. you lose all your points for that turn. If you roll all yellows, you have just passed a "yellow light" and nothing happens. After a yellow light, you can decide if you want to continue rolling or not. You don t lose any points just for running a yellow light. Step 5: Record each player's score at the end of their turn, and keep a running total. Step 6: Win the game by having the most points when the game ends. When someone finishes their turn with 100 points or more, each of the other plays gets one last turn. After everyone has finished their final turn, the player with the highest score wins the game. 1

2 A Toss Up! die is a fair, six sided die with one red side, two yellow sides, and three green sides. *1 2 pts) Let X = the points scored for a single roll of 5 Toss Up! dice. What is the sampling space for X? Hint: both a red light and a yellow light score 0 pts. S = {0,1,2,3,4,5} *2 3 pts) Referring to X in question 1, consider X = k, where k is a member of the sampling space of X. Write the probability of X = k. Hint: The distribution of X belongs to a class that is very familiar to you. P( X = k ) = choose(5,k) * (1/2) k * (1/2) (5 k) = choose(5,k) * (1/2) 5 *3 5 pts) Both a red light and a yellow light score 0 pts, though a red light is much worse because you lose all your points for that turn and your turn is over. What is the probability of running a red light for a single roll of 5 Toss Up! dice? Show your work and write your final solution to three accurate decimal places. Solution 1: P( run red ) = P( run red OR run yellow ) P( run yellow ) = P( X=0 ) P( run yellow ) = choose(5,0) * (1/2) 5 (2/6) 5 = 1/32 1/243 = = Solution 2: P( run red ) = P(5 reds) + P(4 reds and 1 yellow) + P(3 reds and 2 yellows) + P(2 reds and 3 yellows) + P(1 red and 4 yellows) = choose(5,5)*(1/6) 5 *(1/3) 0 + choose(5,4)*(1/6) 4 *(1/3) 1 + choose(5,3)*(1/6) 3 *(1/3) 2 + choose(5,2)*(1/6) 2 *(1/3) 3 + choose(5,1)*(1/6) 1 *(1/3) 4 = choose(5,5)*(1/6)^5*(1/3)^0 + choose(5,4)*(1/6)^4*(1/3)^1 + choose(5,3)*(1/6)^3*(1/3)^2 + choose(5,2)*(1/6)^2*(1/3)^3 + choose(5,1)*(1/6)^1*(1/3)^4 = = *4 5 pts) Referring to X in question 1, what is the probability of running a red light for a single roll of 5 Toss Up! dice given that roll scored 0 points, i.e. P( rolled a red light X=0 )? Show your work and write your final solution to three accurate decimal places. P( rolled a red light X=0 ) = (1/32 1/243) / (1/2)^5 = / = =

3 **5 8 pts) Consider a batch of 100 Toss Up! dice has 5 defective dice in it. The defective dice look identical to the fair dice, but the probability of rolling a yellow on the defective dice is 1/2 instead of the usual 1/3. Suppose you randomly selected one die and rolled it four times, rolling a yellow all four times. What is the probability that you had selected a defective die? Show your work and write your final solution to three accurate decimal places. P( defective ) = 5/100 = 1/20 P( not defective ) = 95/100 = 19/20 P( roll 4 yellows defective ) = 1/2^4 P( roll 4 yellows not defective ) = 1/3^4 P( defective roll 4 yellows ) = P( roll 4 yellows defective )*P( defective ) P( roll 4 yellows defective )*P( defective ) + P( roll 4 yellows not defective )*P( not defective ) = (1/2^4 * 1/20) / (1/2^4 * 1/20 + 1/3^4 * 19/20) = / = =

4 ***6 15 pts) Referring to the batch of dice in question 5, you design a test to detect defective dice. The test works as follows. You roll a die 5 times and if the die comes up yellow 4 or 5 times, then the test classifies the die as defective. If the die comes up yellow 0 to 3 times, the test classified the die as okay. What are the sensitivity, specificity, positive predictive value, and negative predictive value of this test? Write solutions to three accurate decimal places. sensitivity = P( 4 or 5 yellows die is defective ) = choose(5,4)*(1/2)^5 + choose(5,5)*(1/2)^5 = = specificity = P( 0, 1, 2, or 3 die is not defective ) = 1 P( 4 or 5 die is not defective ) = 1 ( choose(5,4)*(1/3)^4*(2/3)^1 + choose(5,5)*(1/3)^5 ) = = = positive predictive value = P( die is defective 4 or 5 yellows ) = sensitivity*prevalence / (sensitivity*prevalence+(1 specificity)*(1 prevalence)) = *0.05 / (0.1875*0.05+( )*0.95) = / = = negative predictive value = P( die is not defective 0, 1, 2, or 3 yellows ) = specificity*(1 prevalence) / (specificity*(1 prevalence)+(1 sensitivity)*prevalence) = *0.95 / ( *0.95+( )*0.05) = / = = Check my solution in R: Ndefectives = 10^7 Nworking = 19*Ndefectives YellowDefectives = ( rbinom( Ndefectives, 5, 1/2 ) > 3 ) YellowWorking = ( rbinom( Nworking, 5, 1/3 ) > 3 ) sensitivity = mean( YellowDefectives ) specificity = mean( 1 YellowWorking ) PPV = sum(yellowdefectives) / sum( c(yellowdefectives, YellowWorking) ) NPV = sum( (1 YellowWorking) ) / sum( (1 c(yellowdefectives, YellowWorking)) ) round( c(sensitivity, specificity, PPV, NPV), 3 ) 0.187, 0.955, 0.179,

5 *7 5 pts) To prepare for the midterm exam, Uche and Robert have been playing Toss Up! one on one. They ve played 14 games so far. Uche is up 9 wins to Robert s 5 wins. Create a 95% confidence interval for the probability of Uche winning when playing against Robert using the Wilson Interval (aka the plus four method). Show your work and write your final solution to three accurate decimal places. x = 9 n = 14 xt = 11 nt = 18 p = x/n pt = xt/nt critval = qnorm(0.025, lower.tail=f) lb = pt critval * sqrt(pt*(1 pt)/nt) ub = pt + critval * sqrt(pt*(1 pt)/nt) # pt and se c( pt, sqrt(pt*(1 pt)/nt) ) , # unrounded ci c(lb, ub) ( , ) # rounded ci round( c(lb, ub), 3 ) (0.386, 0.836) **8 6 pts) Consider the following random variables. Note ~ iid = independent identically distributed. B1, B2,, B100 ~ iid Binomial( n=100, θ=0.01 ) P1, P2,, P200 ~ iid Poisson( λ=2 ) H1, H2,, H300 ~ iid Hypergeometric( N=30, n=10, C=9 ) U1, U2,, U400 ~ iid Uniform( a=3, b=5 ) X = B1+B2+ +B100+P1+P2+ +P200+H1+H2+ +H300+U1+U2+ +U400 Find the expectation of Z, i.e. E[ X ]. Show your work and write your final solution to three accurate decimal places. 100*1+200*2+300*3+400*4 =

6 **9 8 pts) Referring to the variable X defined in question 8, find the variance of X, i.e. V[ X ]. Show your work and write your final solution to three accurate decimal places. V[Z] = 100*(100*0.01*0.99)+200*(2)+300*(9*(10/30)*(20/30)*(21/29))+400*(4/12) = SD[Z] = sqrt(100*(100*0.01*0.99)+200*(2)+300*(9*(10/30)*(20/30)*(21/29))+400*(4/12)) = **10 5 pts) Referring to the variable X defined in question 8, find the probability that X > , i.e. P( X > ). Show your work and write your final solution to three accurate decimal places. This is a large sum of independent random variables with a known sum and variance. Use the CLT. P( Z = ( )/ ) = The book only takes Z to two decimals so either of the following answers are acceptable: round( pnorm( , lower.tail=f), 3 ) = round( pnorm(1.70, lower.tail=f), 3 ) =

7 An elite Rubik s cube solver can solve a Rubik s cube in under 1 minute. Robert s personal goal is to average under 9 minutes. He s saved his last 30 solving times. They are as follows (sorted and in minutes). 3.9, 6.1, 6.3, 6.5, 6.5, 6.7, 6.7, 6.9, 7.1, 7.1, 7.3, 7.5, 7.9, 7.9, 8.5, 8.9, 8.9, 8.9, 9.1, 9.1, 9.3, 9.5, 9.9, 10.1, 10.3, 10.3, 11.5, 11.7, 12.1, 12.3 sample size, n = 30 sample mean, x_bar = sample standard deviation, s = *11 2 pts) Based on the Rubik s cube solving times above, write in standard mathematical notation the null hypothesis for a statistical test of whether Robert s true average cube solving time is different from his personal goal of 9 minutes. Ho: μ = 9 min *12 2 pts) Based on the Rubik s cube solving times above, write in standard mathematical notation the alternative hypothesis for a statistical test of whether Robert s true average cube solving time is different from his personal goal of 9 minutes. Ho: μ 9 min *13 3 pts) Referring to questions 11 and 12, write in standard mathematical notation the rejection region for the test with a Type I error rate, α = TS = xbar 9 / (s/sqrt(30)) > t 0.05,29 = qt( 0.05, 29, lower.tail=f ) = **14 6 pts) Referring to questions 11 and 12, calculate the appropriate test statistic. Show your work and write your final solution to three accurate decimal places. TS = / ( /sqrt(30)) = / = = Check my solution in R: x = c(3.9, 6.1, 6.3, 6.5, 6.5, 6.7, 6.7, 6.9, 7.1, 7.1, 7.3, 7.5, 7.9, 7.9, 8.5, 8.9, 8.9, 8.9, 9.1, 9.1, 9.3, 9.5, 9.9, 10.1, 10.3, 10.3, 11.5, 11.7, 12.1, 12.3) t.test( x, mu=9, conf.level=0.90 ) *15 3 pts) Referring to questions 11 through 14, write a brief conclusion for the hypothesis test. (For sake of time, do not include a confidence interval during the part 1 in class version of this exam.) While the observed average solution time is less than 9, the evidence is not sufficiently strong to conclude that the true average time is less than 9, i.e. we fail to reject the null hypothesis. A 90% confidence interval for the true average time is (7.87, 9.11). 7

8 To improve his cube solving times, Robert turns to performance enhancing substances. The following are his last 10 cube solving times under the influence of Red Bull (sorted and in minutes). 0.4, 3.3, 3.4, 4.6, 4.6, 6.5, 7.8, 10.1, 11.5, 12.8 sample size, n = 10 sample mean, x_bar = 6.50 sample standard deviation, s = *16 2 pts) Based on the substance enhanced times above and the unenhanced times from questions 11 15, write in standard mathematical notation the null hypothesis for a statistical test of whether Robert s true average enhanced time is different from his true average unenhanced time. Ho: μ enhanced μ unenhanced = 0 *17 2 pts) Referring to question 16, what would be the appropriate degrees of freedom for a t test that assumed equal variances for the two samples? df = = 38 **18 3 pts) Based specifically on the data referenced in questions 11 16, what would be the likely consequence of assuming equal variances for the t test? The unenhanced sample size is three times bigger, but it has a sample standard deviation that is roughly half as large as the enhanced sample. Thus if the equal variances assumption is false, it is likely this sample will be underestimating the true standard error of the difference in sample means. **19 6 pts) Referring to question 16, calculate the appropriate test statistic without assuming equal variances. Show your work and write your final solution to three accurate decimal places. TS = x bar.un x bar.en / sqrt( s un 2 /n un + s en 2 /n en ) = / sqrt( ^2/ ^2/10) = / = = **20 3 pts) Referring to question 16, estimate the appropriate p value from question 19. For the part 1 in class exam, you may write your estimate as an interval (lower bound < p value < upper bound). For the part 2 take home, solve for the p value to three decimal places. In class: (0.10 < p value < 0.20) Take home: x = c(3.9, 6.1, 6.3, 6.5, 6.5, 6.7, 6.7, 6.9, 7.1, 7.1, 7.3, 7.5, 7.9, 7.9, 8.5, 8.9, 8.9, 8.9, 9.1, 9.1, 9.3, 9.5, 9.9, 10.1, 10.3, 10.3, 11.5, 11.7, 12.1, 12.3) y = c(0.4, 3.3, 3.4, 4.6, 4.6, 6.5, 7.8, 10.1, 11.5, 12.8) t.test( x, y, var.equal=f ) p value =

9 **21 3 pts) The following two questions follow up on an interesting question posed during our class. Consider the two sample test where X and Y are normal with known variances: X ~ N( μ X, σ X = 1 ) and Y ~ N( μ Y, σ Y = 1 ) The researchers have funding to collect 100 samples from each of X and Y, i.e. N X = 100 and N Y = 100. They have the option of two testing procedures to test the hypothesis: Ho: μ X μ Y = 0 vs. Ha: μ X μ Y 0. Procedure A: Use a single Normal Z test with a Type I error rate = α A. Procedure B: Use two Normal Z tests based on the first half and the second half of the 100 data point collected, i.e. N X1 = N Y1 = 50 and N X2 = N Y2 = 50. To reject the null hypothesis in procedure B, both tests need to be rejected, with each test being rejected at a Type I error rate of λ. What is the Type I error rate of procedure B in terms of λ, i.e. solve for α B in terms of λ? Hint: this is actually pretty simple; it only gets two stars because you had to read through all the math speak. α B = λ 2 ***22 15 pts) Referring to question 21, let α A = α B = Let the smallest clinically interesting difference in means equal μ X μ Y = Compare the relative power of procedure A to procedure B, i.e. solve for (1 β A ) / (1 β B ). Power for procedure A = P( reject Ho Ho false ) = P( Z > μ X μ Y =0.20 ) = P( Z > μ X μ Y =0.20 ) = P( x bar y bar /sqrt(1/100+1/100) > μ X μ Y =0.20 ) Without loss of generality, = P( x bar y bar /sqrt(1/100+1/100) > μ X μ Y =0.20 ), i.e. call X the variable with the greater mean = P( (x bar y bar )/sqrt(1/50) > ) + P( (x bar y bar )/sqrt(1/50) > ), still given μ X μ Y =0.20 = P( (x bar y bar )/sqrt(1/50) > ) + P( (x bar y bar )/sqrt(1/50) < ), given μ X μ Y =0.20 = P( (x bar y bar 0.20)/sqrt(1/50) > ( *sqrt(1/50) 0.20)/sqrt(1/50) ) + P( (x bar y bar 0.20)/sqrt(1/50) < ( *sqrt(1/50) 0.20)/sqrt(1/50) ), given μ X μ Y =0.20 = P( Z > ) + P( Z < ) = pnorm( ,lower.tail=f)+pnorm( ,lower.tail=T) =

10 Power for one of the tests in procedure B = P( reject Ho Ho false ) = P( Z > μ X μ Y =0.20 ), Note λ = sqrt(0.1) and qnorm(sqrt(0.1)/2) = = P( Z > μ X μ Y =0.20 ) = P( x bar y bar /sqrt(1/50+1/50) > μ X μ Y =0.20 ) Without loss of generality, = P( x bar y bar /sqrt(1/50+1/50) > μ X μ Y =0.20 ), i.e. call X the variable with the greater mean = P( (x bar y bar )/sqrt(1/25) > ) + P( (x bar y bar )/sqrt(1/25) > ), still given μ X μ Y =0.20 = P( (x bar y bar )/sqrt(1/25) > ) + P( (x bar y bar )/sqrt(1/25) < ), given μ X μ Y =0.20 = P( (x bar y bar 0.20)/sqrt(1/25) > ( *sqrt(1/25) 0.20)/sqrt(1/25) ) + P( (x bar y bar 0.20)/sqrt(1/25) < ( *sqrt(1/25) 0.20)/sqrt(1/25) ), given μ X μ Y =0.20 = P( Z > ) + P( Z < ) = pnorm( ,lower.tail=f)+pnorm( ,lower.tail=T) = Power of procedure B = (Power of one test in procedure B) 2 = (pnorm( ,lower.tail=f)+pnorm( ,lower.tail=T))^2 = = Power of procedure A / Power of procedure B = / = = Aside 1: How does the power compare over a range of differences in means? If the difference in means = 0, the relative power should be 1 because the two procedures have the same Type I error rate. As the difference in means goes to infinity, the relative power should go to 1 because the power of both procedures will go to 100%. So we should see a upside down U shaped curve. Let delta = the true difference in means and without loss of generality, delta > 0. Power procedure A = P( (x bar y bar delta)/sqrt(1/50) > ( *sqrt(1/50) delta)/sqrt(1/50) ) + P( (x bar y bar delta)/sqrt(1/50) < ( *sqrt(1/50) delta)/sqrt(1/50) ), given μ X μ Y =delta = pnorm(( *sqrt(1/50) delta)/sqrt(1/50),lower.tail=f) + pnorm(( *sqrt(1/50) delta)/sqrt(1/50),lower.tail=t) 10

11 Power procedure B = P( (x bar y bar delta)/sqrt(1/25) > ( *sqrt(1/25) delta)/sqrt(1/25) ) + P( (x bar y bar delta)/sqrt(1/25) < ( *sqrt(1/25) delta)/sqrt(1/25) ), given μ X μ Y =delta = pnorm(( *sqrt(1/25) delta)/sqrt(1/25),lower.tail=f) + pnorm(( *sqrt(1/25) delta)/sqrt(1/25),lower.tail=t) Power curve for procedure A delta = seq(0,1,0.001) PowerA = pnorm(( *sqrt(1/50) delta)/sqrt(1/50),lower.tail=f) + pnorm(( *sqrt(1/50) delta)/sqrt(1/50),lower.tail=t) PowerB = ( pnorm(( *sqrt(1/25) delta)/sqrt(1/25),lower.tail=f) + pnorm(( *sqrt(1/25) delta)/sqrt(1/25),lower.tail=t) )^2 plot( delta, PowerA, col="blue", xlim=c(0,1), ylim=c(0,1.5), main="power of Procedure A (in blue) vs B (in red)\nand Relative Power (A/B in black)", ylab="power and Relative Power (A/B)" ) par(new=t) plot( delta, PowerB, col="red", xlim=c(0,1), ylim=c(0,1.5), main="", xlab="", ylab="" ) par(new=t) plot( delta, PowerA/PowerB, col="black", xlim=c(0,1), ylim=c(0,1.5), main="", xlab="", ylab="" ) lines(c( 1,2),c(0.1,0.1)) lines(c( 1,2),c(1,1)) Power of Procedure A (in blue) vs B (in red) and Relative Power (A/B in black) Power and Relative Power (A/B) delta 11

12 Aside 2: Can I confirm my arithmetic power calculations via a simulation study? # define true differences delta = seq(0,1,0.01) # number of loops to run, the precision of the estimates will depend on this loops = 10^5 # initialize matrices to save the hypothesis test results, each col will be for a given delta RejectA = matrix( NA, ncol=length(delta), nrow=loops ) RejectB = matrix( NA, ncol=length(delta), nrow=loops ) # store critical values crita = qnorm(0.1/2, lower.tail=f) critb = qnorm(sqrt(0.1)/2, lower.tail=f) for( j in 1:length(delta) ){ for( i in 1:loops ){ x1 = rnorm( 50, delta[j], 1 ) x2 = rnorm( 50, delta[j], 1 ) y1 = rnorm( 50, 0, 1 ) y2 = rnorm( 50, 0, 1 ) } } tsb1 = (mean(x1) mean(y1)) / sqrt(2/50) tsb2 = (mean(x2) mean(y2)) / sqrt(2/50) tsa = (mean(c(x1,x2)) mean(c(y1,y2))) / sqrt(2/100) if( abs(tsa) > crita ){ RejectA[i,j] = 1 } else { RejectA[i,j] = 0 } if( abs(tsb1) > critb && abs(tsb2) > critb ){ RejectB[i,j] = 1 } else { RejectB[i,j] = 0 } PowerA = apply( RejectA, 2, mean ) PowerB = apply( RejectB, 2, mean ) plot( delta, PowerA, col="blue", xlim=c(0,1), ylim=c(0,1.5), main="simulated Power of Procedure A (in blue)\nvs. B (in red) and Relative Power (A/B in black)", ylab="power and Relative Power (A/B)" ) par(new=t) plot( delta, PowerB, col="red", xlim=c(0,1), ylim=c(0,1.5), main="", xlab="", ylab="" ) par(new=t) plot( delta, PowerA/PowerB, col="black", xlim=c(0,1), ylim=c(0,1.5), main="", xlab="", ylab="" ) lines(c( 1,2),c(0.1,0.1)) lines(c( 1,2),c(1,1)) 12

13 Simulated Power of Procedure A (in blue) vs. B (in red) and Relative Power (A/B in black) Power and Relative Power (A/B) delta[21] = 0.2 PowerA[21] = PowerB[21] = PowerA[21]/PowerB[21] = delta 13