Past Final Test with Answers on Introduction to Statistics - 2004 | STAT 1040, Exams of Statistics

Download Past Final Test with Answers on Introduction to Statistics - 2004 | STAT 1040 and more Exams Statistics in PDF only on Docsity! Name. Stat 1040, Spring 2004 Final Test, Friday April 23", 7:30-9:20 am tod > Show your work. The test is worth 100 points and you have 110 minutes. 2 brted ablaldaon pore? G13 ¢| 1. (9 pts) In an experiment, 216 stamped, addressed letters are lost, and the rate of return is recorded. Some of the letters were addressed to Mr. M. J. Davis; some to Dandee Davis, c/o Hooters Club; some to M.J. Davis, c/o Friends of the Communist Party. Test to see if there is a relationship between the addressee and the rate of retum.. Clearly state the null and alternative hypotheses, calculate the appropriate test statistic, find the value, and state your conclusion. -27 br pocrrie jh Davis Hooters Communist Total “3 4 aml alk anwgh Returned 32001729 8 26 Vl Le Not Returned 40 55 43 138 44 ig a¢ Totals 72 72 72 216 2 thok fet mdijindlon 4 Oo ty mall: edldrwrreg ond ake of ilies ane sraoindteds 3 he. bette wr shrink aHtrrcbart elders sok it of ren te net iron | dione drab om tddrnacre\ y gt, ot dick me low in hdtv (1) QY peel: W72 | te pH , Te “e : (2 ae | \ 2. y (herd! BLU, (HP ()9-2¢) Te et pay [2 4 (10-¥6)" . (ss-46) (ya) ~~ «6 G6 ae Ads (3-H: [2-( =2 @ = 253 © 3) 472199 Lhe 5.59 ont 9.21 & % ! ad Pa anbe Latin [and 5% &) , : H+ meget Ah ral Pa rete 65%) @ 0 ret od eben fiprala om aldetans anit in oubihiady rigncferd D (ther nt matt caine | w=E7 2.(6 pts) In a random sample of 200 homes in Jefferson, Minois (population 25,000), it was found that the average number of televisions sets per home was 3.2 with a standard deviation of 0.9. a. Construct a 95% confidence interval for the average number of television set per home 24 in Jefferson, Hlinois. ~2 fet bah rier, det o 2 SD =0.9 SE sam = Jy 0.9 2 12, RR @ © 6 12, + = 0.06345. % 0, 064 SEave = se Favs, 45% cr. 3,24 ae 06% = 22 BLS 3,0} dy 3,33 g}>.T 95% of the 7 Oe m Je ferson have between (your lower limit! and ¢ upper limit) televisions. (2 pts) Fabyus We are 95% Lak Abad He, wririag failo do theres Alin stnordact- Ga Hh, hats neh ter Aha 98° of R harman oth Aetenytin Hing Kimeil _| lel oT @y. our interval in part (a) represents a 95% confidence interval for the average number of television set in the Semple. (2 pts) t tabae: Thea ISL ambit ATG. qurt dons son Hho d. G pts) Suppose that you find out that televigsons sets in Jeff rh 8 Thindis donot grtint apabiibion g | follow the nonmal curve, but have a long right tail. Can you still rely on your results in part (a)? Explain We on techy wh He CF pot Hh rertag . ince tA amp. nine (200] bn qpwille, dri we Gon irri Mob Hie desbrilatign Atk t@ pollens He warrmad ture ( wrk hicy out vtaedlr ant potedal} teen te dabss cy nck feller th merme er. 34 i 3. Roger has tossed a fair coin six times and got heads every time. His friend tells him that his chance of getting a tail on the next toss.is almost certain. Do you agree? (3 pts) () We t [h Cyt i feat th korg troy a Ail on the nash fin amine ot 50h ||  8. In Pearson’s data about fathers and sons, fathers had an average height of 68 inches with a standard deviation of 2.7 inches. (4 points cach) a. Suppose one father is 64 inches tall, express his i it as 2 percentile ourtn Sa. f-c8 a -t4 . a 27 . beta, doom “$5 LS: Te. 61% @ , Heo 14 ptf 2 9 54 ta lng tS On BEO 66 O 6 shuk wl Ad 2 [16] b. Another father is in the 78” percentile, express his height in inches. ants.= 18% | ants, = 12%, arta frm ~0.75 f 035 S4H.67Y (bord S64) LAC) rigunl units ang ip = 700 nho Ra pn dels = 56% Bb 1-s[64} 9. For men 18 - 24 in the HANES sample the following is sean x average height = 70 inches SD = 2.7 inches -2 Ay average weight = 162 pounds SD = 30 pounds r= 0.6 x, 4 mh tabi Lb Ky omnes (aga. (6 pts) Calculate the xe eamation for predicting weight from height. “4 - SOx) cleats ag ekg 62-662 Jo= - 304.9 Neiptortim syn wei = 304,95 C6EX] oF A weight = ~ 304.94 6.63- {al b. G pts) Prete the weight of a man who is.66 inches tall. aft © fet Ligh = bb: . ment: ~304.9 ¢66%: 66 = 135.32 gaurds WD _ {aj c. 3 pts). About how. far off do you. expect. your prediction to be?. Pind. berets V/ | — at 4 oes (roe. 30 os" 30 bg 5 ~ 08-30 « 2 pand (a), d. (4 pts) Men who were 73 inches tail averaged about 182 pounds. Te o! Cite id explain, the ones who weighed 182 pounds averaged about 73 inches tall. We cinaek alrmsley Lin eer Tike repptaatem aquaton ah we Kerk @ ar ebedeke 4 oi menptanion tchon fer peadichiy inf from aah au it Yor uch abalitbon tner sfY “14 fer inter Hat if 10. (8 pts) Suppose that the ages at which children first walk alone wid ciple ol aannyed distributed with mean 11.5 months. A simple of sample of 12 children is selected from children having low birth weights. The average age these children first walked alone is 12.2 months with a standard deviation of 1.1. Can we conclude from these data that children with low birth weights walk later than the average? Clearly state the null and altemative hypotheses, calculate the appropriate test statistic, find the p-value, and state _ your conclusion. , j ~ 122-15 ‘ ke fok : aan <% “ Ae Bee eae @ AfsiQ-1 =i ® ty wl: dan nit lolitas crm og, QD |S) Az 212 debmarn Lge end 2.20 hay bt, = 16.5 oY nil 5% LS% dhemabert: h Brim nih les hotlnsiflo val the | “DPerrnbax hokutin 15% ond 5%, (4) AL Arkon >it. { : ; pe o 1 6 6@® K) «eget te mlb CP. rales <%| ® 2 SOTs JS ogy s Ode hte be 4, i Mee gag usaat © + weaall br oh entity stgifitor®) SESqm VEE [15 = SAG LOSS + hibdlrin wi Ah bans ber bhenric SEay, = 39% 20.33 DY walk foler tl g- 11. (8 pts) Many companies are experimenting with “flex-time,” which is supposed to teduce absenteeism. One company employees have averaged 6.3 days off work in the past. The company introduces “flex-time” and a year later a simple random sample of 100 employees is selected. They average 5.5 days off work with a standard deviation of 2.9. Test to determine if flex-time reduces absenteeism. Clearly state the nul] and alternative hypotheses, calculate the appropriate test statistic, find the p-value, and state your conclusion. ~2 tok dabobebon grret 2- BA: + aamh riat 2 30 aa fot imtoo oA iy oulh: flee Tine Ao fp om abrwtatrr tty by 630 3 “4 abl, bh one btnatioe: fbx Tima selaces abornkeiod? ¢.¢, brea <&3 © . = Vil 2.52 (072.5 =29 ® a 4 . ye / SE aug? Tao" . 0.24 ® Uw reed Ae alll Poole 64 © _ S573... - ze 029 7 L468 © # regndl in high @ By arta pebmdin = 2.35 one 2.75: 99,40 % dishially ni “Dnit wef rebea: (00% 99.40% 4 ay (i) | + fbx bine atlacre heaton) 2 _