Solution_Review+for+Final+Exam_Summer+2016.pdf

Solution_Review+for+Final+Exam_Summer+2016.pdf - Stat 35000...

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Unformatted text preview: Stat 35000 Sample Final Exam Summer2016 Your name: K43)! - . Turn off your cell phone before the exam begins! - Show your work on all questions. Unsupported work may not receive full credit. 0 Report decimal answers to three decimal places. - You are responsible for upholding IUPUI’s standard for academic integrity. This includes protecting your work from the eyes of other students. . You are allowed the following aids: a one-page (front and back) 8.5 x 11 handwritten formula sheet, a scientificlgraphing calculator, and pencils. . When you are done, turn in both your exam and your formula sheet. Problem Points Possible Points Earned 2 40 ...... 3 40 3‘ ...... 4 37 l 5 32 B ”45 7-- .18 Total-W 256" Problem 1. (6+8+7+8+9=38points) The strength of a randomly chosen thread has mean 0.5 lbs and standard deviation 0.2 lbs. Assume that the strength of a rope is the sum of the strengths of the individual threads. There is a rope consisting of 100 threads. 1. Can you find the probability that the strength ofa thread exceeds 0.7 lbs, i.e., P(X>0.7)? If so, pEease calculate the probability; if not please explain Why. N3,- <’?r'ti'l(€3 5.2.qu Git: flit-3i: {refit/it,» iittf Cg». Vii/“é git/“i7 UV at" Leg ft 2 We are interested in the strength of the rope consisting of 100 threads, denoted by T. Specify name of the approximate distribution and parameter values for T. {D{) 1- .. \-"“r t? r'x- L ”in”? of; X< Xvi: xii/1.) C: iZi-‘I-l UV 2‘ 5“) “3“ 2") (Mini: is»? iii? to s} W -.. - t . $2 see i soft M (iii/Twiseg 3%“ng :il {Yaw- [(396 a”. 3“};- 3:25.” , , . .- .2- \ Wt ”””””” -- ti t tit .22 :5; , 2.- it 2 3. What theorem did you use to obtain the above approximate distribution? Under what condition can you use this theorem? ((2 ti "it’d-i- [,5 in ii The tied-413th amkt flfiflfi”?30 4. Find the probability that the strength ofthe rope consisting of 100 threads is greater than 54 lbs, i.e., P(T> 54). T rt“ {1-- it; .- " "if.“ -- K t E '"' fug f L n... :3 j \ . i .w =- g ‘ ’ \ "It“: , (f 2...,- EWW-m; ...... 7 wlwtwin—”WWWWWWW-w i 5. Find the 5th percentile of T distribution. That is the value that divides the T distribution into 5% and upper 95%. “W“? T . W--------_-.---..-:~. 2:6 .3. {Li - at; .- ,. ,f 2 - /i.(- t it. .- 2 \- j/ T\‘\\ to at; ““““ t i ‘ " i ““4" “’3’ gusts f it \ Xi {H 5 '1 {Ti-{‘5 NW. wt ‘ = f;.-_\ 5.5... .. .LL»---m’r// I M £0 +- £3 ‘ L i.- bi‘i-i’j “Mn-iw-Mmiww . H.-- W” \‘h. A”? 4 "Wm-,4)“ W..- -2--. WWW WWW-MWLW , gee"; 2-; ZZ _- ..... W, t {4 fl _. amt-(ix) .C a.) .. M K - 2 r_- j _ . . ' -.-.. .22 m, Mitzi Problem 2. (6+9+9+6+10=4O points) Let X = number obtained after rolling a fair six- sided die. 1. What is the probability of getting a six after rolling the die, Le. PCX: 6). ,,,,,,,,, i/f (3C5. >5 5 C / C 2. Let Y be the number of six that appears when rolling the die for 60 times. What is the exact distribution of Y? Please specify name of the distribution and its parameter vaiues. 4 , f3 Ci. rig- gg,C{‘gj‘-g skiff C CC; CC C C. K) ijii-tft: j; Calf CsC “COCK: 5365‘” EiiibpficCC {Cl 116:5} / ) Cf ----- “’Kfi’c I{fCC SCC) 21 ..... 222-- _W.. .... mi- 3. Find the probability that no six that appears from the 60 rolls using the exact E distribution. < — - fluflWW.-.-----_-_‘_ ““5 ”m Clip? Ct 13;: " {3"3“ IIIII ”YET: jg ‘ : mam t-it wit C ****** 5 ‘ i 2 .. g; -‘ 5;!“ E fin” {Kb it .- {fiéufl ------ CU - ----------------- ---- ~~~~~~ 4. What Is the approximate distribution of Y? Please specify name and parameter values J M Cry/117%“ 53%,} M“ Ma (Cw/a 5. Find the probability that more than 15 six appears from the 60 rolls using the ap- proximate distribution of Y. VC'TCC>'$ (W PCT§i§> _ . CM ,7 (C55 \ -.- p C -.--:..-_W- - :- - ””9113? W ------ j ,n E i‘ I ,AUE {Q‘ngg » 3:. {we FC a; fi {UK/‘3 Problem 3. (10+3+3+2+12+3+7=40 points) A biologist is studying the cellutose content of a variety of alfalfa hay Suppose the cellulose content comes from a Normal population with unknown mean but known standard deviation 0: 8mg In a sample of 25 cuttings the mean cellulose content of 145 mg.A AMA” WA “1W” """"" a) Find a 95% confidence interval for the mean cellulose content. It ‘1‘: 7%; fin @533"! 02.2.. M33059: ifffiéfii‘tf-I’KFE” Rig U53 IW SQIJVHLPEIEI (:35 III “(15:33 {3:435 ”325?": (23:, ‘3?“ ----- >5 I <4 Ila”? 03W? 5 J {a :{ 3C3” # ti (‘5 f'w “ I M w w WW Hm - "“3 I l I MK” 54 lllllll i E , 4 i A m 1:3» :{t I £33413? of“: :2“);;;; {MI 4333 f "“ ' 3% a 4, 33 W3 I MEL I I: I 4 W - "wt; ‘ 75 - if 4 4,, I! 35732, 3” gee??? v l 75 I} I444 3I33 I III4 ,NI4;4WMW;;WmmeWWM5M» b) A previous study ciaimed that the mean cellulose content was 140 mg Perform a hypothesis test to determine ifthe mean cellulose content IS greater than 140 mg. 1. State the null and alternative hypotheses. HI IIIIo II nIcIM>Wa> 2. What does the parameter p in the hypotheses represent? {5% 3 [email protected][email protected]’3n (I; {M .EO‘ -5 I‘jaI’ITlgéfI/‘j‘ 3 Which method would you use? 4 M “\gfm’e III-I2 {enema 37:22:44 kwméliflw .w [’4 {"L M Li) LIL it. { 1&3" tulip “If H I, 3; 5 III: II”'~nMW-?‘} M 7 ffémwigfi mmmmmmmm m E: I g ’3 (3 5. State your conclusion in the context of the problem, i.e. whether mean cellulose content is greater than 140 mg. (assume a = 0 05) $555653 {5...}! pat-{55; :TQ emf <1. Xguufl hint He ft? {eve/t; $3“ U CU shit? twee wirttgwmfig é; x’rrttaww Effie» ‘ I 531 WILLQ ffltfiif’ct‘t (£1 Jay/535‘“ f/{fltt‘e} ii“ ’53.. [LEE/rt.” ‘I‘g‘ffi u ‘2‘”‘hfl :35 5 it 2;; 4/5 U ., 6. Someone might argue that the above testing procedure (calculating a p— v—aiue etc) is unnecessary. Their reasoning goes as follows: the sample produced an average cellulose content of 145 mg, which is more than 140mg, so conclude mean cetluiose content is greater than 140mg. Do you agree with this fine of reasoning? Explain. N90» X ..... (Qt: '5]; {555 0‘6““ [if ”gs/{£9 WEI}- ('3? the” {X if} {_’_:1f55’{ju~gtf3’3{jf__ ( 0%. {1,5 tier}; {71:5 £552 «tats-mt sift t” ["t? t’t’fatfitt’L It 51 5*- Cttt 55-in W5 out av (55. 5.55 weave [rim/45553 We. 555553-45 (i’ctteté’é‘tt .45... ‘IW somatic”; meanest“ i“"’?w 52055127Warmer/t HQ whet 0.15»an wigs/t» £95; at V54 [55% [5255 “UL/5:55,“; ‘QCE‘M’f Problem 4. (6+4+5+10+6+6= —37 points) An investigator theorizes that peopte who participate in a regutar program of exercise will have levels of systolic blood pressure that are sgfi fIca YIXI deereanrom that of people who dommndeartipapate in a regularmprogra exeét‘seé‘ WW5 Ideatlfi“ InvestIgator randomly” assigns 21 subjects to” “ash exercise program for 10 weeks and 21 subjects to a non exercise comparison group. After ten weeks the sample mean systolic blood pressure and sampie standard deviation for the two groups are given below. r group I Sample Size Sample Mean ‘ Sample Standard deviation ‘ exercise 21 137 t 10 ‘ Non-exercise 21 127' t 9 1. State the null and alternative hypotheses. a 1 (it; ”(Lt/3:“ O t} C. H, ; it; H2 '75 O 2. What do the parametets a and 512 in the hypotheses represent. Fl 5” gt” {‘4' , ‘7 _ m c 5 {a >UL’51‘Q 73v 555ij 755% w“ a, \ {ti/(t «4 m‘” ‘1 E { Ii. 5‘ _ 3. Which method would you use? W“‘”\. (a) 1-sarnp1e T (b) 1—sample Z /(c) 2 sampte T ‘5 (d) 2-samp1e Z .2 . f 3". t- . . . .. r“ " SEXY 9 wit}? {it/x; V?- ifif hit-0‘42 it“; “3““ J: ‘32 4. Assuming the population standard deviations are unknown but equal calculate the 955% _ _,.. ....~ confidence interval for the mean difference. -‘ (”55151 (:1; {151/ 1511 My ié {<1 1:. 555 ,, 25.1525 - 1 mi ,: W. 1 .4... w — M-.. M... m “M w1f~§w1:i2fli11nil5?5.€5 $4131 5 >1 1; 23445; . .~‘ :7. 5.1.213. ”j" l 0 j” “5:“ (5 i <15 1““‘1115. :: “ 1151’ 5‘” 1:3 1%. :12». 1121:“ ,1 . [I . gay"; 2/ 1 "111 'f‘lh? 5? “"' “(3‘43 141-1111 311:. 51 - 35331111141111 1111.} II t3 {2: E. 5. Use the calculated confidence interval above to test hypotheses given in pang. Does \33 exercise have an effect on the systolic blood pressure? 13“.." 1111-); 15511. {1‘23 1?”? 110111 11111“ 1 1 1151111119,. Cw 1511536§35 {#15131 127171. 111.31”: “15111,, Mjfi3ffv1111175 131115191“? :2: w— (“)3 CF12 E (x/ ‘ 13131;- 31131115 5151,. 1!”) .955 3331115- 7517-1511135 {3 K11,“ (>11... 511111.151: (.3111. {riff-{3:135 3:). 1’1 55 1“: 0511121! 5 32115551355151.- 6 How would you g1aph1caliy check the assumption that the two populations are normal? / ( 55155011351111.1141 01121125 {5111“ 11111.15 3.2112111? 5‘ 513513“ 3155055 Problem 5. (8+3+3+2+2+11+3=32 points) Twelve cars were equipped with radiai tires and driven over a test course. Then the same 12 cars (with the same drivers) wereequipped with reguiar beited tires and driven over the same course. After each run, the cars' gas economy (in kmfi) was measured. is there evidence that radial tires produce better fuel economy? (Assume normality of data.) Data can be found in the table below. Gaseco.(km/I) 1 2 3 4 5 6 7 3 9 10 in 12 Mean|s.0. gelted 4.1 4.9 6.2 6.9 6.8 14.4 5.7 ‘58 6.9 4.7 16.0 4.9 550810394 Madmgelted 0.1 —0.2 0.4 0.1 0110.1 0.0102 0.5 0.2 0.1 03 1114210193 1 Radial 4.2 4.7 6617.0 6.7 14.5 5.7 6.0 7.4 4916.1 5.2 5.750l1.053l a) Find a 90% confidence interval for the mean gas economy difference between cars equipped with radial tires and cars with belted tires. 55‘55‘75 5 555 55555” 55555675555515 M {161111 5.51.11’1-131’9151‘73’. F Q3553}: 3 f, 53;: 35251 “/5956, 1 C; - "11' a. 1...- , I (3.11:5 :5 A f." (3 5518 $51 5 E} 5? .1. -13?“ 313,313 ......... 1... 1:: ---------------- ~/ 5 1 5:5 .. 1.1.513; *5 11.51513“ """"" "i g ,, 3 3 11151.1 1 1 55 i Z . .... m 41.1 W .5 ”:1 ,,,,,,,,, 3 I fl M 4 a... ~. 3 _. .. l ........ 1.21142: .1... i 1’31; .3113” 1 33.33. 33 1.33 -5 % -1. 1.13 11 Mg 15 “ ’ “‘1‘ --------- ~12 / .:6.1111 "’4: 51111.5?» 56111-5555 H1? 3 .. L12» {1 5 1 3 9.151%-l * b) Based on the above confidence interval, do we find any evidence that radial tires, on average, produce better fuel economy? 5% are (<5: 555a aux.) 5.53; newt-6545?"? 5 63-5553? [as 55515555. 555555.55..- {133,55 {35g [fig/U Lflisflfiimi ”55.55;? 55755515655 "5"??? ‘3 {5WD 65595;; Attittjr’ “5055/5 53: diet"? «ifi’t’i- Z5 3 t; 5/1. (3% artifafifi “ 0) Next, we perform the hypothesis test to determine whether radial tires, on average produce better fuel economy. 1. State the null and alternative hypotheses. it. fitrwtitfirc v a. n: it Mum 2. What doesthe arameterintheh otheses re resent? . x K D W p W 5551 Varfifi‘wfi. “fry/€15; . 5953 55955 35g“!- :: 5’35? €0.55 9 act; (it {metres} 59w {TQMS‘ fig 5:: R . .......... 3. Which method would you use? ( (a) 1 sampleT (b) 1 sample Z (c) 2-sample T (d) 2-sample Z ”hm-A M 4. Calculate the test statistic and find the Rejection Region, using 0: = 0.01. 7-1.. wwwwwww t- a a t a 5. State your conclusion and interpret the result in the context for the researchers. . ,_. u . 77 r. . . ’ 5)" (“a .- :5 .1“. .. “5“} g s" % (Fifi... .,,N.. A{§F‘a{ Wu"? LE]: 53‘;- L? .51; (2:: 1’? g {a} [5 «Er {“ng is gr} 5% Tat/5:1: f 153% 5’; fit... {:1 V W ,/s. . . . f w pm M i u " 3 15555 tile 5 5;, t? e 5.5“ n...) 55% 05 e L. "t 5;} [315-353; (”A at Vijtgf dig; {7 t3% 5151559 194552 (New)? {3 .595 55515 (ti-:2 9555;522:2355" 1555:1355“ \‘tkcégatf 55555“; 5* ,- 5?) t“: (Leciéfij 5’ - g}? affix/{(lfl 54715—3/ 554/ iii/Lt? {i {2” {n.- €155 {.45}? k} Problem 6. (5+10+5+5+10+10=45 points) The owner of a towing and garage service wants to know how the lowest temperature on a winter day affects the number of emergency road service calls he receives. He records the lowest temperature (in 0F) and the number of emergency road service calls he received on 30 randomly selected winter days. Summary statistics of data are: E2l4, SJ. 216.6, fi=18.6, s), 28.3, ”:30 1. In view of the study the owner wants to answer, what is the independent variable, and what is the response variable? X‘ 2:: 1512222 3% 1’2: 2<’2--<2{2.2222:2t%2-232 X” *1“?- 212+ 6? 22’“: “$622213 2/222 2 21;. 2% “~22” 225‘s 21>: 27‘ <72 .2 -- - “ 2 <1 i 2. Suppose that the correlation between the number of emergency road service calls and the lowest temperature r = —0.89. Find the least squares regression equation. - “7 L 53’”- 722 ft 4-2 2334 W‘ 2.. .I’ mow-«NWM ‘""" ~ " ‘ "fl“ {5‘2 :: '1" 2 “T "”22 W ' 2 <2 “21’3- WK EA .zfr f {~22 C .~ ”j", .2: My ‘ ‘ y s .. m ‘ H "7 at “:2 2 22-2 r >2 2: 322. . g «2 {22. 422222“; (.2222) 14?: 254 S 3. What is the predicted value ofthe number of calls for a day when the lowest temperature is 16°F. ‘ ti £2 n27) :— 2222. 2’22 2- (22:2 5242:???) (7/22 ‘2 1:: {if a 7 g 4. Is the model overestimating or underestimating for that day when the lowest temperature is 16°F and the actual number of calls is 13? "N1. {3 yefi2< ek/iwlFYlCtbf?tt<§ 2 gv2ta.rsf $1 7'5 f? 5 x [.F" \. 5. Answer the following questions by filling in the blanks based on the interpretations of 0: and a . 0 When the lowest temperature increases by 1°F, on average the number of emergency alt-’22:". 2.222722. is 232. L212 £5 road service calls the owner receives will - When lowest temperature is 0°F, on average the number of emergency road service calls the owner receives is ,2 A?" g; S Find the 95% confidence interval for the slope fl ofthe regression iine. Does the 95% CI. support the claim that )3 < O? - {:1 g“ .. (Ti-Vii if?“ "g 2 ,2" 5:“ iii 21:2th (>2: C 0 f 2, 2?va "73“” I "" i M M 5:36“ i ’/ \ Mr 52) "3:: 231%? x: k , (0 O"? m 211.3 2"" W ““'"""7"':"'§" {7 U” L” "2““ (2 H > i <2 ) : “5’2 r“: 23 ....... 25:2 3“ ""“U W2 / my?” SEW-“f ‘i/hfim—Xlwzfi: g; ’2’ ‘/ W mm W 2 2 ii ,1 2 1‘” 5%: Mpgum 222 ) fawn {032?} :pc-firw> flow?) < < 0000;" Problem 7. (6+12=18 points) A random sample of 100 data points was collected to test if the population mean, u, is different from 240 at a =0.05|evel. That is, Ho:#=240 vs. H12y¢240 The sample resulted in a mean of 235. Assume the sample is from a Normal population with known standard deviation 0:40. 1. Whatdoesarepresent?Howdoyouinterpretit? _ at: PCT WEI meow r: [3: MIT-{- H l HO mime ) I‘m-Jrhispmbiew, (>2 Vepr’wasem—tL—l he [méaél‘b‘Ly of ..1‘€3€Cbn3 --H1e [mfdhfltoin mall/t H 66mg Egaef to‘zliO/ whale it t; actuallfi eeuai “Eb 314C). 2. Suppose that the true population mean is actually 236. For 1the test above, find the probability [3 of committing a Type ll error. ”a b , . x Tegt ariasivsti‘g: Z‘ W m Moi) [ARC/{52X Hi, ... "Yaw :W'erf-irO __ eO/m At- :9th “:- i 203g «"2; CW Zogg. ‘7? 22%} :EZDEE’ (aw—II (“I t; bf 205$ 71 Cfék AW )(wzllO .51“) (w MT :7 iflé :FC2s2:QB:FC :FC2<2%) "'0me ~ 2 M 2 (games; m Ci» 0 sale) 3‘0..g%0 ...
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