Introduction to Statistics - Spring 2005 Final Exam with Answers | STAT 1040, Exams of Statistics

Download Introduction to Statistics - Spring 2005 Final Exam with Answers | STAT 1040 and more Exams Statistics in PDF only on Docsity! Stat 1040, Spring 2005 Name: Final Test, May 3, 1:30pm-—3:20pm leo > Show your work. The test is out of 100 points and you have 110 minutes to finish. Questions 1 to 5 below are based on survey results related to the Terri Schiavo case, as published in Time, April 4, 2005 (pp. 22-30). Terri Schiavo suffered severe brain damage during cardiac arrest in 1990. Over the next 15 years, Terri was kept alive through a feeding tube. For 10 years, Terri’s husband, Michael Schiavo and Terri’s parents, the Schindlers, battled in court over the issue of whether or not to remove Terri’s feeding tube. After a final court order in March 2005, Terri’s feeding tube was removed and Terri was allowed to die in early April 2005. 16>) yo 1. A simple random sample of 1010 Americans was polled by Time on the following question. Question A: “Do you agree with the decision to remove Schiavo’s feeding tube?” Time reports that of the 1010 people, 595 (about 59%) agree 349 (about 35%) disagree 66 (about 7%) don’t know. {l 2 ] (a) (3 points) Assume that Terri’s parents, who strongly opposed the removal of Terri’s feeding tube, initiated an alternative poll with the following question: Question B: “Do you agree with the decision to remove Terri’s feeding tube, thus leiting her starve to a slow and painful death?” Other things being caval the percentage of people that would have answered. agree to question would have been (a) higher than 59%, (b) about 59%, or Oo (c) much less than 59%. ircle your answer and explain clearly. iets pea get ape mnartlinty (ohome a alent ond nainfol destd*) Aardiy wlan Hebcnbe “tml de wesk bo Kon st ie, @ Tarr s Arartnts ong far vtmane Ast dioding fi] (b) (4 volts) Time reports that “the margin of error fuadng the bing a e who agree is £3 percentage points.” Is this (a) our usual standard error?’ b) about twice ~~ our usual standard error) or (c) about one half of our usual standard error. Circle your answer and show the necessary calculations. wie, bakahroy : [SE % * nt + lok = 54% W hin Sde \o5¥ 0.41 = O44 « Vo59-0.4T) ao. 3% 2595 x2, aity Awe SE sane (ite. 0,49 = 15.59 (51% fiz] (c) (3 points) Time further presents outeomes for different subgroups (such as for church goers/non church goers and for different political affilations) and states that “the margin of error is higher for subgroups”. Assume that. we are looking at a subgroup that consists of about 253 people. For these 253 people, the standard error. for the percentage who agree would be i. about four times as big AGG: 253. 5 > ( 2) = ‘ ~ oie ii, about twice as big ; iii, about 2 as big ax: SE iin oderwh THe = hime no hi, 00 He prt SE as the standard error for the sample of 1010 people. Just circle your answer — no calculation is necessary. 4/26) 2. A newspaper reporter wants to do some follow-up interviews with some of the par- ticipants in the survey and randomly selects 3 different people from the original 1010 participants. Recall that 595 answered agree, 349 answered disagree, and 66 an- swered don’t know to the Question (A). ~) fer tak Colleton wre? . sb% av (a) (3 points) The chance that the 3 randomly selected participants for the follow-up interview all answered agree to question (A) is about 20.4 — %. 5357. i ms Lak anpeto, PE Bae apt, errs tare gion lS Ilene? ~Tpog Rad ayrtio, sat gs gqy @ 55 4 —_ . Co - o tphnin fiat nogtes > Top @ |uts are: Toca ae = 20,1042 20.4% ia ] (b) (3 points) The chance that the 3 randomly selected participants for the follow-up interview all answered disagree to question (A) is about yl %, UA seneen: SLE QD | Both disaets, 342 ajate (A 2 Ud, coop “1008 fod dinuugtiny == 3S . 49 © sy © 3% _ goyts 4% Baba lot cliseeprttr s (00% At 3 chongyer, 3 tof ~ ' {12 ] (c} (3 points) The chance that at least 1 of the 3 randomly selected participants for the follow-up interview answered agree to question (A) is about 93.1%. 3 lob docoact appee: BILE . AS Gy) 3 deca nat cope, qinta fot clr dead cqte ES wie” wie Pe | alt Blok peg, WO aw Owe 00926,9% rad desta uy Wid ey 109” Toad * Tent dito age hg D| ah ben | apneic; 10% Oey - I31% le Hp. Use the fact that about 59% of the 1010 Americans in the Time sae / agree to Question (A). . ‘ 2 for Calckibion, yrs . 8 if, ng fined e's gird ia g] (a) (7 points) Find-an approximate 99% confidence interval for the percentage of all Americans that would have answered agree to Question (A). ata hath ateag: ‘ Mi %, ae mh Hed Hy ln joo the come Skyy 8 Sd= \/a.s95-0.4f 04g (p or . i, y i SomeVootay. sy OD vin UY ook setts mir (154 deel» = Suan tO O49. {S. ; ‘ 4 wD Go = BSL jooms 191% @D | We E. 9% x 100 -190% (vio 259% + 1%, 2 = 55% We 63% Dif Kity vt dd, 4Yy »[Ee]v. Leonardo da Vinci (1452-1519) theorized that if you put your arms out hae and “+ Ti€asured from the fingertip of one hand to the fingertip of the other, this ‘wingspan’ distance would approximately equal your height. A group of fourth-grade students measured their height and wingspan and found <2 pera average height = 49.5 inches with an SD of 1.8 inches average wingdpan = 48.9 inches with an SD of 2.1 inches The scatter diagram was football-shaped and the correlation coefficient: was 0.8. (201 (a) (5. points) Find the equation of the regression line for predicting wingspan from height. Xz hight alae 9s, 5 21-9933 oe wy =v Sx 48 , : : - ° = 4S 8~ 0,933 °45,5 = 2.72 weet? oO OX 2 muptasion syuehin: ts = 2,424 0,533) or fryers 2.92 40,533 hist | W {b) (2 points) Does the intercept make sense? Explain. : 4 Yes does, Ta intbragt rye Hh piled winypcen fot = abl Med on felt, Ms Mok Osrche 5 E225 SO Adina Me ertmne healt 00 The ann Le sarkromt daub pel tion Wihdeyer He indkrtecl i, st mantel male ainer hed, Vi] (c) (3 points) Predict the wingspan of a fourth~grader who is 52 inches tall. few bandh = 2, ninepgen? 1.214 0,533-52 = [S24 rho] QD ig] (d) (2 points) Find the rms error for your answer in part (b). EMS, Let = bo? + SD, = Vi -o.8* 2.1 = SRo@. 2.) = Vex ape ad-2y = [1.16 anukes hs | (e) (2 points) One of these fourth-graders is 52 inches tall. Is it likely that he or she would have a wingspan of 45 inches? Explain cldarly using your answers to (c) and (d). gdh ainapyen flor 52 tre SH.24 Hel (prom (cl) cu WSS G1 yas @ 1.26 26 “dey Bopadichal manyon ar ef SI, 4 umole. Hore Oo of hilton de unt 52 inthis tall awadd hag ctrch, alt (ot en adett®) mytinggen, 5  i2>fi2]s. (12 points) The following table comes from a simple random sample of university } students, each of whom was asked whether they agree or disagree with-the statement “Cell phone use while driving should be prohibited”. ~) for td talolibion ariet Male Female bed * Agree 13 16 29 is4 Be ~36 fer wncorrcit, AA Disagree | 240 206 | 446 234.0 208, ae 53-299 475 an) = ~ 1 if all ok omaper Test to: see whether male and female university students differ in their responses. You. should clearly state the null and the alternative hypothesis, find a test statistic and an Pe adgeadbne P-value, and state your conclusions in everyday language. Aa foh fer wrdepninee: © eapeamend sloyndioh? e mera Viordene golinenk ope re wt inlet Bho A basher ter io dffeot OD Dy Aatgeobick: 2LIS? = 15.4 wh, (oe Able tage clae\ - 2 Yo I cm A ete - (eal + (usted t (at) - Roenotsl -o@ Me 2--Q-)21 @ uy edoack nejuh Me rll (Probar 54) @ 3) 4% 0.85 Lehnan 0.46 tod 1.0% v girdit ak sapeernit ert inlipndiod C ~ Pe valet dibwtin 58% tnd 37, tt dng © 9. (12 points) An English exam is taken by 2000 students. The exam scores are known a fiz] to follow the normal curve. The teacher says that the average of all 2000 test scores is 75, but one of the students thinks the average is actually lower. She takes a age lum random sample of 9 students and finds they got the following scores: ~ 2 for te 63, 53, 84, 82, 35, 50, 68, 73, 92 36 fot aon 1 ok Drunpyeh Test to determine whether the average really is 75, against the alterthtthe that the student is correct. You should clearly state the null and the alternative hypothesis, find a test statistic and an approximate P-value, and state your conclusions in everyday deft: fetes nonvrhanin cw sot foot 32= 183) @ ih rode Shame Var? 5S. © . Ssil oj wll: ane, lt KA nth bb i frstiokicl , @) Soy a £42 C) £-ej hoki = 75 VD pe 6S .-83% 0 @ atari: ery of dick sere bo on Monsroiohd (3) Gi GAR Aty betug <35 0 A= 319k @ 3¢): y = CBE SBSH L435 Sot HE TBEIL [3 Leah og Act 36 (mel mt =1 | seme ey # a foi 3h bin 0H ek LLP =O x9 F-vabea Ldaern ake onl (0% aiompde SD = (aol. +r66a ly). ad mtd ol { Perdue >5%) @ all feud tll de 22 = fe ‘ert. (2.32 ) 6 eg fat ( he, ay cable 75) Y