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E2Afall10key - PubH6450 NAME l(E\JERsloA(4 Exam 2 Lecture...

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Unformatted text preview: PubH6450 NAME; l/(E \JERsloA/ (4+ Exam 2 (11/11/10) Lecture Instructor (please circle): Susan Telke Andy Mugglin Directions: This is an open-book, open-notes exam; however, sharing of books, notes, handouts, homework or lab papers, calculators, or verbal comments is not permitted. You may use a calculator of your choosing, but laptop computers and any kind of intemet connection are not permitted. For multiple choice questions, please clearly indicate your answer(s). For short answer problems, please show all relevant work necessary to arrive at a solution, as partial credit will be awarded. Correct answers without appropriate justification will generally not receive full credit. DO NOT LOOK AT OR BEGIN THE EXAM UNTIL THE INSTRUCTOR ANNOUNCES THAT THE EXAM IS STARTING. You will have from 1:25pm to 3:20pm to complete this exam, so do not spend too much time on any one problem. With such a big class, there is always someone who needs to take the exam early or late. After the exam is over, DO NOT DISCUSS THIS EXAM with anyone who was not in this same room taking it with you. We anticipate returning the exams in class on Tuesday. Advice on rounding: We suggest that you round calculations (mean, s.d., etc.) to 2 decimal places (if the data are whole numbers), 3 decimal places (if the data are known to the nearest 0.1), etc. Probabilities are usually rounded to 3 or 4 decimal places, but no more. Good luck! I (1) Problem 1 (20 points): A doctor treating patients with high cholesterol prescribes medicine to lower the patients’ cholesterol. Her patients’ cholesterol levels are measured at each visit. She notices that, in some of her patients, their cholesterol levels are not changing in subsequent measurements, and she suspects that this is because these patients are forgetting to take their medication. She plans to conduct an experiment on a randomly selected sample of 25 patients. She calls her patients between the next two visits to remind them to take their medication. Then she calculates the mean difference in cholesterol levels between these visits. Previous studies have shown that the changes (in measurement between visits) are normally distributed with a standard deviation of G = 80, so she will proceed with that assumption. She wants to perform a statistical test (using a = 035) to see if there is statistical evidence of a change in cholesterol that might be attributed to this intervention of calling patients between visits. a) State the hypotheses (both null and alternative) that she is interested in testing. Be sure to define any symbols. (Ll) \‘XO"MCJ :0 US. HAIMA'fiO MA :3) The PoOMLa—UDN memo CHANéE ml CHOLésTZQOL- BrwN \Hblrs F041 THas-e WHO Rgcelvé‘ A “REMWOER” PHONE CALL. b) In the context of this problem, what is a type I error? What is the probability that she will commit one? PqueIB :: 01:: 570:1)? L”) A «\{PE’ 1 ERROR. WOULD OCCUK. [F' 7745’ FwEflE6/gK CDAJCLUDEJ 71/5 KEMI/va/Q CALLED FEsqus ”2/ fl CflPM/Gé //\/ mfg/V Cb/DLZJ‘I‘ERDL ppé Tr Pop/A/LfiT/D/V/ wflzn/ //V mwx/ /f fl/i; ”or c) In the context of this problem, what is a type 11 error? (a) H "TyPE J graze/L Wo/I/Lao 044 Me. At ‘77/2 flEriffflIZél/E/9 600’de51 7% fEm/n/pa< M7 KE$V47 //V 19’ Wgfl/l/ C flfl/éé Dy/Mfl’f/OM 5' ”WISE/€047 Wflé/V /f' /f W5. _ (2) d) She is worried that she may have inadequate power to detect a clinically meaningful change with n=25, and she only has adequate power to detect a large difference D. If she wants to have the same level of power to detect a difference that is half as big as D, what should her new sample size be? WOMEN)“; fl: ' D9. ll 25 W Q (ohm crL : $01,!st (gr 13, “EN“: Devan —:. an\:100 e) Suppose that she makes this change to the sample size, and at the end of her experiment, she conducts the planned hypothesis test and achieves a p-value p = 0.013. She concludes that the intervention has made a difference that should not be attributed to chance alone. In making this conclusion, the experiment has possibly made a: type I error (circle one). @ or No (Why/Why not?) rd axial \‘L (3) 4:?(mm‘l‘ l—lo \ \‘lo JCFML> AMD $an. 3 c type 11 error (circle one). Yes o® (Why/Mung!) é: PCDo mm EEJECT Hp l “015.thng {3) N01 “PogsigLéH 9H1? QEJECTED Ho. (3) Problem 2 (16 points): A new nutritional and exercise program is designed to help people lose weight. A random sample 55 adults were enrolled into the program. All completed the program and 8 people successfully lost weight. a) Is the sample large enough that a CLT-based confidence interval for this proportion could be computed? Justify your answer. D: 55 'X: 8 /\ CL’V anuuzgé (A) fl$*% .45 so [‘19 CLT Does NOT APPLY. (ggm‘y/lgfloycs 4451? /o res éfiuzgmr ’T/Anqg NW D" ) b) Compute a 95% confidence interval for the proportion of weight—loss success for the new program. WE mus-r use“ “Puss Foam C-ZL gag-:39» __ O.H0Cil‘l’8 o,\\00)q9 “t. 0.041 513m (carats) 0.26523) (4) 0) Consider the following interpretations of the confidence interval found in (b). Which are correct interpretations? Circle T if correct, F if incorrect. T (19 There is approximately a 95% chance that the true success rate of this program is contained within the interval found in part (b). T F The interval in (b) represents all possible values that the true success rate might be. T (a The interval in (b) contains 95 % of all of the possible values that the true Qp‘r > success rate might be. 9}th T (g If the experiment was repeated an infinite number of times, the sample proportions from 95% of the experiments would fall within the interval found in (b). _, L, Cl") F If the experiment were repeated an infinite number of times, 95% of the CIs from those replicates would contain the true success rate. T (3 The interval in (b) represents the range of probabilities that a randomly chosen participant will lose 95% of his/her excess weight. T (E The interval in (b) represents the set of values p0 that would be rejected in a hypothesis test of H0: p = p0 versus HA: p i p0, at a significance level of 0.05. (5) (1%) 0: (9‘ Problem 3 (26 points): In order to quantify the effects of indoor smoking bans at hospitality establishments, researchers visited a random sample of 61 bars and restaurants twice, once prior to and once immediately following the implementation of a state- wide smoking ban. For each venue, the post—ban fine particle concentration was subtracted from the pre— —ban fine particle concentration to get the post- ban concentration reduction PBCR . ( ) ”P051” , PRE a) The literature suggests that the average fine particle concentration in hospitality establishments is not a normal distribution. This means that PBCR will have the distribution resulting from the difference of two non— —normal random variables. A histogram of PBCR shows a unimWribution with no obvious outliers Researcher’ s plan to do a hypothesis test to determine if there is a change 1n fine particle concentration, will uncertainty about the distribution of PBCR affect the validity of the hypothesis test? Why/why not? NO“) ‘0 >40 60 )CLV AWMES 3—4 N {th , 5/613 b) The sample average PBCR for all venues was 45 ug/m3 and the corresponding sample standard deviation was 15 ug/m3. A previously hypothesized mean PBCR was 32 ug/m3, so the reduction in fine-particles was bigger than anticipated. Test whether the observed PBCR is SE’n—iiariflflflfrom the We your null and alternative hypotheses, compute the appropriate test statistic and p-value, reject the null at a: 0.05, and state your scientific conclusion. Sir—45 5:15 Movsz HD"M“M0:3Z \35- “A: MiMDl—BL JL— 71:93 Lia/j}, __ (0.14 :0 3/r7‘ [ES/(— Pwm2 7.?LED7E'ZHB é.- 2[.oo\3 —.-. .001, i7~ku4o< ) QEjéLT \r\o ”The, (km ’Promdes SU'QClCleo’H: £010;an Mob Them» Paca is diQCaren‘c Hun 32 #5)”? (6) afi’d 70135 . c) Determine a 99% confidence interval for average PBCR from the sample data. H51 249m Q5/6713 ==.7 (5fi.8§)50-H) d) Based on the 99% confidence interval, would a hypothesized PBCR of 85 rig/m3 be rejected by a two—sided hypothesis test that used a = 0.01? Why/why not? \lES‘) slam/M3 l$ N9; QOMA‘NED ‘N THE (M ‘39?) C1 . (7) Problem 4 (16 points): A sociologist wanted to determine whether the age distribution at which younger women (15—29) have their first baby is associated with whether they live in rural or urban areas. She collected a simple random sample of 350 young women. The following table lists the categories separately by age: Ae Cate_or of Women Total 15-19 25-29 a- 20 2945-150 (08 50 .\ 40 08b 90 140 Test the claim that there is an association between age categories of younger mothers and the places where they live at the 0.05 level. 60 150 (a) State your null and alternative hypotheses for this test. “d- \MD&P%N02NCE‘ oP‘ PLACE Ame Ree (LAT. HA" PssSociA’UoN o):— PLAU’e AND A54. QM (a) (b) Calculate the value of the test statistic. mom“ Do 7L?” @ BibbER ’rH AN 1x 2 TABLE ‘2 (dos—up?“ 2049401 Ute—Bosebi UDO—‘lZ-SW >41: 2 (nap : 2114 306(5) 72.8% (b) 'l' (So—79.19)L £50433? Lia—12.)?— ”fqi'lr (a? 72. ': 35.50% (c) What is the distribution of your test statistic? What is the p-value for this test? ?C7(:> 35.5014) 1— o 05* (A) Z: 5>D16TruemnoN ((1) Based on the above, what conclusion should the researcher draw? NEW \5 EUIDENCE' OF" AM Assoufi‘rmm BTw/v Ace; CHTEOOQIES OF Yeomeek WHEQS AND THE Pumas 'THEL/ U05. x2) (8) Problem 5 (16 points): Cardo et a1. (NEJ M1997) conducted a case-control study on the risk factors involved in healthcare workers contracting HIV (a process called "seroconversion) after occupational exposure. The cases were health care workers who had a documented occupational exposure to HIV—infected blood by a needle stick or cut with a sharp object, HIV seroconversion that was close in time to the exposure, and no other reported concurrent exposure to HIV. Control subjects were health care workers with a documented occupational, percutaneous exposure to HIV-infected blood who were HIV seronegative at the time of exposure and at least six months later. An important consideration was whether the exposed worker was given post-exposure zidovudine to prevent HIV. a) Why do you think the researchers conducted a case—control study and not a cross-sectional or cohort study? ”Wane NLE PE“ (1) CDNVZQLT Hw FROM E¥Po$on€ 'TO \AN maze—tel) Btodb. (n \s sMMQ P609142 wHo b) Using these data, the researchers want to estimate what might be going on in the population of all occupationally exposed health care workers. Answer the following about this conceptual population: ‘ v, _ t ‘,.‘.._,,',,‘ r , g i) What is the estimated conditional probability of a case havi taken z1dovudine‘! {1) P (km was mass} — qla3 ii) What is the estimated marginal probability of an exposed worker taking zidovudine? 253 E/THEZ LA/Fl)’ /2) ”Rifle”: 4 CC; 6 TUOL/ N (7 exposed worker contracts HIV. CASE NUT Conflch iii) What is the marginal probability th \ mlzo < 2) DO NOT CRLCU LHTG/ (Fl-HS \ Log—rung. iii) What are the estimated odds of a control hav1ng taken ZiooVuumc in L. C2) 34,—: ODDS Cowman” Toma AET (9) CREATE 35070 CI. FOR DR Cl 4 /7"7l 0233 2 014085 I [WO‘ ) E*P(‘l“3>)'Ew(os%383 / C333)?” Problem6(6 points): : [0.31) [[145] OF 1 For each of the scenarios below, choose the most appropriate test to answer the scientific question (circle one). a) Suppose we have some data regarding allergic reaction to different types of penicillin (type G and type BT). These were done by administering skin tests (one G, one ET) on each of 500 subjects. A summary of the data is: 48 patients reacted to both G and BT; there were total 52 reactions to G; there were 68 total reactions to ET. The researchers would like to test whether the types of penicillin differ in their association with allergic reaction. i) Normal approximation to the Binomial for one proportion ii) Normal approximation to the Binomial for two proportions iii) Binomial exact test for one proportion iv) T test for the difference of two independent means L2) E5 T test for a single mean or a mean of a paired difference 1 McNemar’s Chi-Square test vii) Plus four confidence interval for two proportions (10) b) A report published in the Star Tribune described a study that identified a genetic variation associated with increased risk of developing lung cancer. Assume that in the general population of smokers, 15% develop lung cancer. The study sample (n=52; found that 15% of 20 smokers with one copy of the gene variation developed lung cancer and 25% of 32 smokers with two copies of the gene variation developed lung cancer. Researchers would like to test if the population proportion of smokers with two copies of the gene variation developed lung cancer is higher than that in the general population. i) Normal approximation to the Binomial for one proportion i" Normal approximation to the Binomial for two proportions Binomial exact test for one proportion 1v) T test for the difference of two independent means v) T test for a single mean or a mean of a paired difference vi) McNemar’s Chi—Square Test Vii) Plus four confidence interval for two proportions One component of the SF—36 quality of life assessment is the physical function score which ranges from 0 — 100. The mean physical function score for 60-70 year olds in the general population is 65. The SF-36 physical function score was calculated for each patient in a random sample of 36 patients age 60—70 with chronic obstructive pulmonary disease (COPD) at a University clinic. The sample mean and sample standard deviation (mean=45, 5:65) for the physical function score were calculated for this sample of 36 COPD patients. Researchers would like to test if the population mean SF—36 score is different for COPD patients compared to the general population. i) Normal approximation to the Binomial for one proportion ii) Normal approximation to the Binomial for two proportions iii) Binomial exact test for one proportion iv) T test for the difference of two independent means < v)) T test for a single mean or a mean of a paired difference v1) McNemar’s Chi-Square test vii) Plus four confidence interval for two proportions (11) Notes: Standard Normal Probabilities P(—1.645 < Z < 1.645) = 0.90 P(—1.96 < Z < 1.96 ) = 0.95 P(—2.576 < Z < 2.576 ) = 0.99 P (z < —0.96) = 0.1685 P (z < —1.41)= 0.0793 P (z < —3.87) < 0.001 P (z < —3.74) < 0.001 P (Z < —6.11)< 0.001 Chi-Square Probabilities P(le > 3.84) = 0.05 P(X22 > 5.99) = 0.05 P(X26 > 12.59) = 0.05 Student’s t probabilities P(tg > 4.29) = 0.001 P(t60 > 2.66) = 0.005 P(t60 > 2.39) = 0.01 P(t6o > 23.43) < 0.001 P(no > 6.77) <0.001 (12) ...
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