Math 2200 Spring 2018, Exam 2

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1 1 Math 2200 Spring 2018, Exam 2 You may use any calculator. You may use ONE cheat sheet in the form of a 4 x 6 note card (the medium size of the standard three sizes). The exam is out of 100 points. Each problem is worth 5 points. 1. Bone Mineral Density (BMD) is used as a measurement of the likelihood of bone fracture. The usual unit for BMD is gcm 2 but, for the problems found on this exam, we will use the unit mgcm 2 and round each measured value of BMD to the nearest integer. In every population (by age, gender, and race), the distribution of BMD is approximately normal. Mean BMD for 25 year old white women is 955 with standard deviation 123. Mean BMD for 25 year old black women is 1040 with standard deviation 135. Sam is 25 years old, white, and female. Her BMD is Nic is 25 years old, black, and female. The percentage of Nic s peers whose BMD is less than Nic s is the same as the percentage of Sam s peers whose BMD is less than Sam s. What is Nic s BMD? A) 1198 B) 1199 C) 1200 D) 1201 E) 1202 Solution B) 1199 Sam s BMD z-score is ( )123 or Nic has the same BMD z-score as Sam s (because they exceed the same proportion of their groups). Therefore, if x is Nic s raw BMD score, we have (x 1040)135 = , or x = ( )(135), or x = Rounding to the nearest integer, we have x = A slightly different approach is to calculate the proportion p S of Sam s peers with lesser BMD. We can do that in R with pnorm. Then we can use the quantile function to find the BMD in Nic s peer group that has the same proportion p S. Let us consider this alternative approach a verification of the first method. > pnorm(1100, mean = 955, sd = 123) # Proportion of Sam s peers with BMD < Sam s [1] > qnorm( , mean = 1040, sd = 135) # Answer: BMD for Nic s peers that corresponds to Sam s [1] > pnorm( , mean = 1040, sd = 135) # A check for a typo [1] Scores of a psychological test administered to prison inmates are normally distributed with mean 172 and standard deviation 33. Prisoners with scores less than 130 are labeled compliant. Prisoners with scores greater than 210 are labeled dangerous. Members of the in-between group are labeled latently hostile. The population of latently hostile inmates is what fraction of the total prison population? A) B) C) D) E) Solution D) The z-scores corresponding to 210 and 130 are ( )33, or and ( )33, or , respectively. Thus, the requested fraction is Φ(1.1515) Φ( ), or Φ(1.1515) (1 Φ(1.2727)) or Φ(1.1515) + Φ(1.2727) 1. Using the given table, and and therefore Φ(1.1515) = Φ(1.15) (Φ(1.16) Φ(1.15)) = ( ) = Φ(1.2727) = Φ(1.27) (Φ(1.28) Φ(1.27)) = ( ) = Proportion of latently hostile inmates = Φ(1.1515) + Φ(1.2727) 1 = =

2 2 The area under a normal density function is 1. In the figure below, the proportion that we seek is the shaded area under the curve. The code below begins with several lines that result in the displayed figure. Then a vector z of two z-scores is created: the first corresponding to the raw score 130 and the second corresponding to the raw score 210. The last line results in the requested proportion. > Raw.Scores = seq(from = 80, to = 260, by = 0.25) > Normal.Density = dnorm(raw.scores, mean = 172, sd = 33) > plot(raw.scores, Normal.Density, type = "l", lwd=2, xlim = c(80,260), ylim = c(0,0.013), + main = "Normal Density, Mean 172, Standard Deviation = 33") > lines(c(80,260),c(0,0)) > polygon( x=c(raw.scores[c(200,200:520,520)]), + y= c(0, Normal.Density[200:520], 0), col="gray90") > z = c( ( )33, ( )33 ) > z [1] > pnorm( z[2] ) - pnorm( z[1] ) # Requested fraction [1] In 1982, Robert K. Murray and Tim H. Blessing surveyed American academic historians asking them to rank U.S. presidents. The lengthy questionnaire also asked the surveyees if they leaned liberal or conservative. The two groups (190 liberals, 50 conservatives) agreed on 9 of 10 presidents in the top ten: Abraham Lincoln, Franklin D. Roosevelt, George Washington, Thomas Jefferson, Theodore Roosevelt, Woodrow Wilson, Andrew Jackson, Harry S. Truman, and John Adams (in order of the rankings assigned by the liberal historians). In the second row of the table are the rankings given by conservative historians to the same nine presidents listed in the same order. Note: Liberals ranked Lyndon Johnson 9 th and conservatives ranked Dwight Eisenhower 9 th. Both have been excluded. Calculate the Kendall tau correlation coefficient A) B) C) D) E) Solution D) The ordered pairs are (1,1), (2,3), (3,2), (4,4), (5,5), (6,8), (7,6), (8,7), (10,10). Each must be compared with all the others that follow. Before we begin, we remark that (4,4) and (5,5) are not ties. In fact neither group of historians ranked any pair of presidents on the list as being equal. There are no ties at all. we also note that the absence of a president ranked 9 th does not affect Kendall tau in any way. ANy

3 3 transformation that preserves order preserves Kendall tau. If you like, you can re-rank John Adams as 9 th on each list. Whether he is 9 th or 10 th, the order is preserved: he is last on the list. There are 8 concordant pairs associated with (1,1): Concordant: (2,3), (3,2), (4,4), (5,5), (6,8), (7,6), (8,7), (10,10). There are 6 concordant pairs and 1 discordant pair associated with (2,3). Concordant: (4,4), (5,5), (6,8), (7,6), (8,7), (10,10); Discordant: (3,2). There are 6 concordant pairs associated with (3,2). Concordant: (4,4), (5,5), (6,8), (7,6), (8,7), (10,10). There are 5 concordant pairs associated with (4,4). Concordant: (5,5), (6,8), (7,6), (8,7), (10,10). There are 4 concordant pairs associated with (5,5). Concordant: (6,8), (7,6), (8,7), (10,10). There is 1 concordant pair and 2 discordant pairs associated with (6,8). Concordant: (10,10); Discordant: (7,6), (8,7). There are 2 concordant pairs associated with (7,6). Concordant: (8,7), (10,10). There is 1 concordant pair associated with (8,7). Concordant: (10,10). Altogether, there are , or 33 concordant pairs and 3 discordant pairs. Thus τ = 33 ( 3 9 ) = = 5 6 = > liberal = c(1,2,3,4,5,6,7,8,9) > conservative = c(1,3,2,4,5,8,6,7,9) > cor(liberal,conservative,method = "kendall") [1] For the dataset in the preceding problem, find the Spearman rho correlation coefficient. A) B) C) D) E) Solution E) For Spearman rho we do change the values of John Adams from (10,10) to (9,9), although in the calculation we perform, a calculation that uses the Strange Formula for Spearman rho, the change happens to have no effect: the column difference is 0 either way. The column differences are 0, 1, 1, 0, 0, 2, 1, 1, 0 and their squares are 0, 1, 1, 0, 0, 4, 1, 1, 0. The sum of these squares is 8. Thus ρ = (9 2 1) 9 j=1 d 2 j = (80) (8) = = = > cor(liberal,conservative,method = "spearman") [1]

4 4 5. The dataset below consists of X = mean BMD (White women) and Y = mean BMD (Black women) at ages 25 to 85 in 10 year increments. Age X Y Summary statistics are X = , Y = , s X = , s Y = , and r X Y = What is the slope of the linear model of Y X? A) B) C) D) E) Solution A) If m is the slope of the regression line, then we have m = r X Y s Y = (0.9742) s X = > ww.bmd = c( 955, 945, 920, 876, 809, 740, 679) > bw.bmd = c(1040, 1017, 1034, 990, 969, 928, 859) > bmd.lm = lm(bw.bmd ~ ww.bmd) > coef(bmd.lm) (Intercept) ww.bmd We reuse the dataset of the preceding problem but we have eliminated one variable because it plays no role in this problem. The current dataset involves Age and Y = mean BMD (Black women). The table also contains the z-scores of the Y observations. What is the Pearson correlation coefficient between Age and Y? (Note: Problems 7, 9, and 10 will also depend on this data and possibly calculations done for this problem.) Age Y z Y A) B) C) D) E) Solution C) The mean age is the middle value, 55. Thus, the deviations are 30, 20, 10, 0, 10, 20, 30 and the sum of the square deviations is 2 ( ), or The standard deviation of Age is s Age = = It follows that ( (0.9731)( 30) + (0.6195)( 20) + (0.8809)( 10) r Age Y = 1 6 ) +(0.2043)(0) + ( )(10) + ( )(20) + ( )(30) ( ) ( ) 1 1 = ( ) =

5 5 > Age = seq(from = 25, to = 85, by = 10) > bw.bmd = c(1040, 1017, 1034, 990, 969, 928, 859) > cor(age, bw.bmd) # Answer for this problem [1] > # > # We continue the session to obtain data that answers problems to come. > # > lm(bw.bmd ~ Age) # The answer to the next problem is the slope; it appears under Age. Call: lm(formula = bw.bmd ~ Age) Coefficients: (Intercept) Age > resid( lm(bw.bmd ~ Age) ) # The answer to problem 9 is the 5 th entry of the response > sum( (lm(bw.bmd ~ Age)$fitted.values - mean(y))^2 ) # Answer to the 10 th problem. [1] Recent testing of 67 female cadavers at the Mayo Clinic resulted in a linear model for Women s BMD as a function of Age. According to the linear model, women lose BMD at the rate of mgcm 2 per year as they age. (Race was not considered in the study. Source: Femoral Strength..., Journal of Bone and Mineral Research, Volume 30, Issue 12, 2015, p ) According to the linear model that is based on the data of the preceding problem, at what rate per year do Black women lose BMD? A) B) C) D) E) Solution A) Using values that have already been given or calculated, we find that the slope m of the Y Age regression line is ( ) s m = r Y Age Y = = s Age We answer with because the problem asks for loss. The verification of using R can be found in the solution to the 6 th problem. We note that different studies have led to rather different linear models. The loss per year requested in this problem refers to Black women, who lose BMD less quickly than White women. The Mayo Clinic study refers to all women. The age range we have used is In the Mayo Clinic study, the age range is There is evidence that a linear model over both large ranges is inaccurate: yearly loss seems to accelerate with age. 8. The Mayo Clinic study that is discussed in the preceding problem reported a coefficient of determination equal to In that study, what was the correlation between BMD and Age? A) B) C) D) E) Solution A) We have r = 0.43 =

6 6 9. For the linear model Y Age of Problems 6 and 7 where Y = mean BMD (Black women) and the observations are Age Y what is the residual associated with the observation (65,969)? A) 4.13 B) 1.99 C) 8.11 D) E) Solution E) In problem 7 we calculated the slope m = of the regression line. We now calculate Y = and Age = 55. The Y-intercept of the regression line is b = Y m Age = ( )(55) = The regression line is Y = Age, and the fitted value for Age = 65 is Ŷ = ( )(65) = The requested residual is , or The verification in R is in the solution to the 6 th problem. 10. For the linear model Y Age of Problems 6, 7, and 9, where Y = mean BMD (Black women) and the observations are Age Y The sum of the squares of the errors is What is the sum of the squares explained by the regression line? A) B) C) D) E) Solution C) We have SST = (7-1) Var(Y), or SST = 6 ( ) 2 (from Problem 5), or Thus SSR = SST - SSE = = The verification in R can be found in the solution to the 6 th problem. 11. Let W be the distribution of mean BMD for Black men at ages 25, 35, 45, 55, 65, 75, 85. There is a strong positive correlation between W and the mean BMD Y of black women at the same ages (although the gender correlation discussed in Problem 5 is stronger than the race correlation just mentioned). If W is treated as a response variable and Y as an explanatory variable, then the linear model W Y results in SSE = and SSR = What is the linear correlation coefficient r YW? A) B) C) D) E) Solution B) We calculate SST = SSE + SSR = = It follows that SSR r Y W = SST = = = In the table below, X is a diameter measurement and Y a volume measurement for shortleaf pine trees the observations given here have been taken from a dataset of 70 trees that Donald Bruce and F.X. Schumacher published in Mean Sd X Y ln(x) ln(y)

7 7 Pearson s linear correlation coefficients for X and Y, for X and ln(y), for ln(x) and Y, and for ln(x) and ln(y) are: r X Y = , r X ln(y) = , r ln(x) Y = , and r ln(x) ln(y) = By using a linear model for an appropriate transformation of the observed values, find a model of the form Y = C X p where C and p are constants. What is p? (The next problem continues with this higher parabola model.) A) B) C) D) E) Solution C) By applying ln( ) to each side of Y = C X p, we obtain ln(y) = ln(c) + p ln(x). The linear model for this equation is Y) = ln(y) m ln(x) + m ln(x) where Thus m = r ln(x) ln(y) s ln(y) s ln(x) = ( ) = Y) = ( ) ( ) ln(x) = ln(x). Applying exp( ) to each side, we obtain Y = exp( ) X = ( ) X > X = c(5.1, 5.9, 8.6, 11.1, 12.2) > Y = c(3.0, 5.6, 16.6, 32.8, 28.2) > lm( log(y) ~ log(x) ) Call: lm(formula = log(y) ~ log(x)) Coefficients: (Intercept) log(x) > exp( ) [1] In the model Y = C X p of the preceding problem, what is C? A) B) C) D) E) Solution B) See the solution for the 12 th problem.

8 8 14. Weetabix Cereal has a new series of collectible cards inserted in its boxes of delicious food product. John Adams, John Quincy Adams, John Tyler, and John Kennedy are the collectible cards in the company s President Johns of the White House collection. Each box of Weetabix cereal contains one card, and the cards are inserted so that they are found in boxes with equal likelihoods. What is the probability p that the purchase of five cereal boxes will result in a complete set of the four Johns? Estimate p using the following table of random digits drawn from 1, 2, 3, and 4, and grouped in 30 blocks of 5 digits. Use each group of five digits in the table to simulate the purchase of 5 cereal boxes What estimate of p results from the simulation? A) B) C) D) E) Solution C) The collections that have 1, 2, 3, and 4 are: In the first row: In the second row: In the third row: 31412, In the fourth row: 44321, 23214, In the fifth row: Nada. Thus, p 730 = The composition of the sophomore class of Moose Jaw University is 712 female. Among all sophomores, the proportion with GPAs greater than or equal to 3.6 is 15. Gender and GPA are independent. If a sophomore is randomly selected, what is the probability that the student is male and has a GPA less than 3.6? A) B) C) D) E) Solution A) When the problem was conceived, the solution was intended to go as follows (but we advise the busy executive that a much shorter solution is given afterwards). Let F be the event of selecting a female, M = F c the event of selecting a male. Let HGPA be the event of selecting a student with a GPA greater than or equal to 3.6. Let LGPA = HGPA c be the event of choosing a student with a GPA less than 3.6. Then, in view of the independence of gender and GPA, 7 12 = P(F) ( ) = P F LGPA) (F HGPA) ( ) ( ) = P F LGPA + P F HGPA = P ( F LGPA ) + P(F)P(HGPA) = P ( F LGPA ) ( ) ( ) = P ( F LGPA ) It follows that P ( F LGPA ) = = = 7 15

9 9 Again using we have P ( F HGPA ) = P ( F ) P ( HGPA ) = ( ) ( ) 7 1 = , P ( M HGPA ) = P ( HGPA ) P ( F HGPA ) = = 5 60 = Finally, we have the partition of the entire sample space into four disjoint events: Ω = ( F LGPA ) ( F HGPA ) ( M LGPA ) ( M HGPA ). On applying P( ) to each side and solving for the probability of the specified event, we have P ( M LGPA ) ( 7 = ) 60 ( ) = = = A shorter solution using the independence of M and LGPA: P ( M LGPA ) = P ( M ) P ( LGPA ) ( = ) ( 1 1 ) ( 5 = 5 12 ) ( ) 4 = In Moose Jaw, days are either sunny xor cloudy. Weather patterns show that if it is sunny one day, then the probability of rain the next day is 217. If it is cloudy one day, then the probability of rain the next day is 25. On Monday, the CTV weatherman in Moose Jaw said, The probability that it will be sunny tomorrow is 14. What is the probability that it will rain in Moose Jaw on Wednesday? A) B) C) D) E) Solution B) Using the following tree diagram, we see that the requested probability is W Rain 14 TS ---W Dry TC ---W Rain 35 W Dry P(W R) = P(WR Ω) = P(WR (TS TC)) = P(WR TS) + P (WR TC) = P(WR TS)P(TS) + P(WR TC)P(TC) ( ) ( ) ( ) ( ) = =

10 An urn contained 6 red balls and 4 green balls. A first ball was removed from the urn and not replaced. Then a second ball was removed from the urn and not replaced. Finally a third ball was removed from the urn. In each selection, the likelihood of drawing any ball was the same as the likelihood of drawing any other ball. What is the probability that exactly 2 red balls were selected. A) B) C) D) E) Solution E) R R---G R R ---G---G G ---R---R 38 G 13 G---R G We can end up with exactly 2 red balls in these three ways: RGR, RRG, GRR. These are disjoint events. Therefore: P(exactly 2 red balls) = P(RGR) + P(RRG) + P(GRR) ( ) ( ) ( ) ( ) ( = = 1 2. ) ( 1 2 ) + ( 2 5 ) ( 2 3 ) ( ) 5 8

11 The prevalence of neuromuscular disorders is about 160 per 100,000 persons. As a biomarker of muscular disease, a creatine kinase (CK) level greater than 500 IUl has sensitivity 66% and specificity 77%. What is the probability that a member of the population with CK level > 500 IUl has a neuromuscular disorder? A) B) C) D) E) Solution A) Let NMD denote has neuromuscular disease and let POS and NEG denote CK assays greater than 500 IUl and less than 500 IUl respectively. We are given P(NMD) = = , P(POS NMD) = 0.66, and P (NEG NMD c ) = Then, acording to Bayes Theorem, P(NMD POS) = P(POS NMD) P(NMD) P(POS NMD) P(NMD) + P (POS NMD c ) P (NMD c ) = sensitivity prevalence sensitivity prevalence + (1 specificity)(1 prevalence) = (0.66)(0.0016) (0.66)(0.0016) + (1 0.77)( ) = The use of CK as a diagnostic marker originated in the 1960s. Nowadays, MRI is a more commonly used diagnostic tool: it is significantly more sensitive and more specific than CK tests. The use of CK as a test for acute myocardial infarction goes back to For that condition, CK is more sensitive (90%) but still not very specific. 19. Amiloride is a medication that is used to treat complications of heart failure and cirrhosis of the liver. In a trial designed to investigate its value as a treatment of cystic fibrosis (CF), it was found to be effective in 78.57% of CF patients. If a doctor prescribed amiloride for 20 CF patients, then what is the probability that at least 18 patients saw improvements in their conditions? (Assume that the effects of amiloride on the different patients are independent.) A) B) C) D) E) Solution A) The question asks for the probability P of 18, 19, or 20 successes in 20 independent, identically distributed Bernoulli trials where the probability of success in each trial is p = and the probability of failure is q = 1 p = Therefore ) ( ) ( ) P = ( p 18 q p 19 q p 20 q 0 = (20)(19) (0.7857) 18 (0.2143) (0.7857) 19 (0.2143)+(0.7857) 20 = On paper, twelve applicants for an executive position, 5 women and 7 men, seem equally qualified for the position. Two are selected at random for an interview. The probability that no woman is chosen for an interview is 722. The probability that two women are selected for an interview is 533. Let X be the number of women selected for an interview. What is the expectation of X? A) B) C) D) E) Solution C) The set V of values that X assumes with positive probabilities is V = {0, 1, 2}. We are given f X (0) = 722 and f X (2) = 533. It follows that f X (1) = 1 ( ) = 6666 ( ) = Thus ( ) ( ) ( ) E(X) = (0) + (1) + (2) = =

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