{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Stat 231 Final F10 Key

Stat 231 Final F10 Key - Stat 231 Final Exam Fall 2010 I...

Info iconThis preview shows pages 1–11. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 8
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 10
Background image of page 11
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Stat 231 Final Exam Fall 2010 I have neither given nor received unauthorized assistance on this exam. g 9K Name Signed Date Name Printed 1. In a 1992 Journal of Quality Technology paper, Eibl, Kess, and Pukelsheim studied a number of different set-ups for a painting process and their impacts on the variable y : measured painting coat thickness (mm) (several pieces of material were painted and evaluated for each set-up considered). Some summary statistics for pieces from each of 5 particular set-ups included in the study are below. Set-u #l Set-u #2 Set-u #3 Set-u #4 Set-u #5 11,24 11,24 113:4 114:4 115:4 y1 : .980 y2 : 1.525 y3 : 1.130 y4 :1.735 y5 21490 S1 : .146 S2 : .100 S3 :.207 s4 : .078 S5 : .080 Initially, consider ONLY the information from the study about Set-up #1, and assume that for this set-up measured paint thickness is normally distributed. 5 pts a) Give two-sided 95% prediction limits for measured paint thickness on a single additional item painted under Set-up #1. (Plug in completely, but you need not simplify.) M64 Wits,‘ Hi (All. z “45 .9504: 3.024144% 4+5;— (wtfrs ML MM) 5 pts b) Give endpoints of a two-sided interval that you are 95% sure contains at least 99% of all thicknesses for items painted under Set-up #1. (Plug in completely, but you need not simplify.) his; ”if/P s 7/ . 950 i: 5.299 (-149) (Ml? m mm) Now consider ONLY Set-ups #1 and #2, under the assumption that thicknesses of paint on items painted under either set-up are normally distributed. 5 pts c) Give endpoints of a two-sided 90% confidence interval that compares how consistent Set-ups #1 and #2 are in terms of measured paint thickness produced. (Plug in completely, but you need not simplify.) 51 5| l md /_,,__. l/léé S; l? S1 W film“ MFR 440'- rcvcrsal NW .100 9.2?) .100 4/ 9.18 2 ( mugs) 5 pts (1) Give 95% two-sided confidence limits for the difference in mean measured paint thicknesses produced by Set-ups #1 and #2. (Plug in completely, but you need not simplify.) M56 (ll-'fl-q—li‘tviflfkif' w‘l’b Ill-pm: wn (Irwin-15 Now consider the results from ALL of Set-ups #1 through #5. 5 pts e) For yi]. : measured thickness for piece j from Set-up 1', below is a normal plot of all 20 values _ f [—‘l \ (ya — yl. ) / if? a: jl. Say what it indicates to you about the appropriateness of analyzing these experimental results based on the one-way normal model. *1% ”Ms K anwl fld’ If (1.95 Wadi—265i ftsiAI/mls M 11.9 W5 ZavCl'bX—t “{le filfl' 1‘ g ‘1“ am “511,471+ 1w. AM “ms E 11.5 “mics 1M) MW‘Jl—I‘MS 5+ E rpvulolms with W 0Y1!— N“! E a: ”-1 w( “Milli I (0%! ’FMEE “.115 w ,wf- MW A Algmfif‘sbvw ) -15-2-15 -1 415 D L15 1 1.5 2 Shldlnflzad maid Threw-lass J As it turns out, in this problem SSTr : 1.525 and SSE : .260 . 5 pts 1') Give the values of spooled, and of an F statistic for testing H0:.,u1— 2,43: A421”; along with its associated degrees of freedom 3mm: \IMsE >1] ssé)/—<nr aim/(205) .132 Mg? _ $SW/(r-‘3 _ 1.675/(5—0 ._/ =ZZ.O F : MSE — éSE/(nr‘f) ' 190/120—9) s =fl FzL'O df— Ar [5 Pooled _ Henceforth if you couldn't do part f) you may use the incorrect value of s — 2.0 .) Pooled— d/ 39 /b 5 pts g) Give two-side nf1dence limits for the difference in mean measured paint thicknesses produced by Set-ups #l and #2 under the one-way model assumptions (Plug in completely, but you need not simplify.) Vlea V4‘11i’tswflul 0—1—15); 611‘“ Amos n-V\ @$0~1525 r 2131(.132)\l fir (”His art mg 5 pts h) Set-ups #l and #3 involve a high belt speed while the others involve a low belt speed. So it is possible that the linear combination of means 3 might be of interest. What is an appropriate margin of error for estimating this with 95% confidence? (Plug in completely, but you need not simplify.) We tam [2% Z-l3l (132) (9.1%.); + 36%)")??— lwflk Mb WA) 2. Below is a pdf, f (x), for a simple continuous distribution. f(x)=# x if—[<7L<l L 0 otherwise Suppose that a random variable X has pdf f (x) given above. s a) Evaluate P[—5< ‘ <5]. 5 5pt X P[—,54X4.§j: J LEAXVC 4% <5 " mm at [5“- Nr“ ’4- A —| 1 7‘ _ flgyéh :(W: .7325 5 pts b) What are the values of EX and VarX ? I > 0 =. we b1 am ( lax—um 4 mm o .. I 0 law: axe—(60% E><2 w o I :2;+ D 76+ 1‘ S 1_ 50 \ltrX: JasLGZEX-LFSDZZQDXZ -‘= “’1'- i_| : p Z —\ EX: 0 VarX: 'g 5 pts c) Suppose that )7 is the sample mean of n : 25 independent variables, each with pdf f (x). Approximate P _-_ S ¢7 < 5— . (If you were unable to do part b) you may use the incorrect value VarX=-7here-> 'X [g WWWMI mm wfl W #1fo a M «I: “RAE—5 : E 50 NL WA ‘Z'Vfl-lug 5 .5—0 H 2: =55:- 3,93 AS“ 1 J3; E ’5— 22-:‘3-g3 [ So 'l><‘.54><<.$\ 73 l-0 ,5 6 .5 3. Nine one-inch holes are to be drilled in a steel hydraulic cylinder barrel. This part is fixtured once on a CNC machining center and all nine holes are drilled one after another. Of interest is the random variable W : the number of holes that fail to meet engineering tolerances for radial position 5 pts a) What feature of the "Bernoulli trials" model seems least appropriate here? Explain. :Ethl wo‘l’Mnlfi ham gang/41‘) (flunk 91L Th}- 5 M55 as memwl {Moswfltfl' (mills. If A Farr ’5 MooWofil'L/k olnmfwl WW AXll/LKL Tbil— WMl/l PVL$MMILHII¥ “VI/Ml ’l‘v Mb. All 9 holes +1) lac posi’hwcl m+sfa<¢ a Lmeriw‘ Norma/s. W vas hula; hit-M7 nfl‘ bdm/L libtl t/QWAIMT‘VOM $315,“\ Pail/H14 Mob 3% MAI/Pmma b “Wink" 1‘5 W woblm. For the next part of the question, ignore any misgivings raised in part a) about the usefulness of a Bernoulli trials model here. 5 pts 5 pts 5 pts b) If one judges that the chance that any single one of the nines holes on a barrel fails to meet engineering tolerances for radial position is 5%, find P [W S 2] based on a Bernoulli trials model. «(fill— W 4’5 Blwmwl n=9 mi T: .og‘ ? \Ne = mum M») t 1: 99)" + 9(.9€)”( 09) Jr (2)(.95)7(.os)2 = (35V (Q95)"+ 9(35)(.05)+ 34.05)") 93, .95” c) Suppose that in fact production records show that among the last 100 barrels inspected, 7 had at least one hole failing to meet engineering tolerances for radial position. Give 95% two-sided confidence limits for the fraction of all barrels produced on this machine that have at least one hole failing to meet specification for radial position. (Plug in completely, but you need not simplify.) l/l66 $1 2. {HE—Lara) 4. Attached at the end of this exam are some pages of J MP reports useful in the analysis of some data of Heinz, Peterson, Johnson, and Kerk concerning body dimensions measured on n : 145 males ages 18-30. We'll suppose here that one would like to quantify how y : subject weight (kg) varies with the 22 other variables (all measured in cm) for such men. a) What single predictor variable is most effective at explaining the observed variation in y ? Explain. What is the sample correlation between y and this predictor? (Make a reasonable assumption about the sign of this correlation.) Most effective single predictor: WP - film Why: (E? Mg’nu [Mid— Po$$ll0lfl Ta— 001/ A. SLR There is a Fit Y by X output for inference based on the model y = £8 +fi£ Mryflmg included in the J MP reports. Use it as you answer the questions b) through d). 5 pts b) Give 95% confidence limits for the standard deviation of weights (in kg) of males 18-30 that have a particular waist girth. (Plug in completely, but you need not simplify.) {at M a 1 MM g 7/0.” A»! “l;— MJC ”Dale? will]: ‘43 «Fwy-Indium “#7” XL /0 Tl? fipflv “Wye ’l'l,‘ Ml 96:13 1430' ; +1.99 00% (+33)) M4 XLQ5H'3( "W5)_l'aL l/;%z,3‘) W 9045; 3173.0 Ill-5 14-3 _I___+3 gavel/r150 “A 5'09 9 111% (Mil—'5 ML 1433 5 pts c) Remember that in the original data, the units of weight are kg and length has units cm. There are roughly 2.54 cm per inch and 2.205 lbs force per kg force. Give 95% confidence limits for the increase in mean subject weight in lbs that accompanies a 1 inch increase in waist girth for males 18-30 years of age. fix ZJOSHQS M25:(M=5é01-%' _ 51041, fil cm awx l 44:3 Mu (”l1 fl—a L N! Imiwrfl— wwtg’l' irfk MW [LA/6M J SNh—f- “’4 _ MA is abmk . SQ zm'fimllll qu Ill/Z, 5'40 "hvwks llwxl-l: in Fl \‘1- 5'” “”5 birtSE—bl Lat. (25+ (914.er w- 5.(,a(l.2£)t 5.bD(l-9é)(.OéD (in [LS ,r Hos/mak) 5 pts (1) On the plot on the Fit Y by X report, there are several ( y, waist. girth) points that plot outside the dotted lines. Is this a concern? If so, why, and if not why not? (Circle the correct response below and explain.) This IS a concern. < This IS NOT a concern. > Why: ”Wat Mal "“3 Spatfi 557 Fax «hum 1W1: VJ afl' M6“ xNW . WA. Ab MIN— 4>¢F¢o All Ala/7‘ Plrfi +0 ‘Cfi/Fglllvmgirk ’l’levx. film?“ 5/ ,J— 143 IS AMT 7 ”X 474‘s ‘5 zmslg-l'tfil" Wm" (flu “MW ”{— rFaM'l'S m’l‘swé ’0“ 9:76 ffW/llillm lWV‘TlE . There are two Fit Model outputs included in the J MP reports. Use them as appropriate as you answer the questions e) and f). 5 pts e) The two models represented on the J MP reports involved respectively 2 and 4 predictor variables. They have the largest R2 values for models of their respective numbers of predictors. If it is possible to judge whether the increase in R2 for the second compared to the first is statistically significant using an F test, find the value of statistic and give degrees of freedom. If it is not possible to use an F test based on the given information, very carefully say why. flu-g IS rw’i’ paSSleA NM- 714 bad" ZIPrchava Mao/6.! [s m a sub—Wm A 4% ml 4— Ward/W mar- _ 50 E “MH- le4 of H— as A “fa/WK MAJ/CL! " dbd'mwm/ “(W «n “Mt/Lou" pvt Amy Mr L Pfc/lmfwg , s, M YMMV-ml l:— +5§l— is lawn h i?! Wievm’ifl- 5 pts f) If you were going to drop one predictor from the 2nd model (to produce a model using 3 predictors) which one would it be, and why? If you did this, would you have the 3-predictor model that has the highest R2 among all possible 3-predictor models (including those using other predictors)? Explain. Predictor to drop: 6km», i/(fl‘, 31m 1) Why: (Er \Ms’UM swllcsl— Absolw'h rt s’mhsb‘z, T; Eb D'i'JFlVk if 11‘“ny hi’flSh H01 :0 [Wu/l TS 1M ’UM [Lafi- llOVimslu‘mwfl—I‘m’l' Ji’ :ili‘ 4- "Smiths Circle one of the following: This IS the best model of size k : 3. This IS NOT the best Model of size k : 3. Explain: fl 54-5? WM “mm. 123—) 1L gm L¢=3 MAW alt-Awe 1f WMSESNTK, Fovurmd'lr'fk) Mi Mflh'l‘ (Wall is MW m olefin/l bl Mow“ shawljzrjlfl “(VIM WWWAA think =45 8 1533mm — mm Pam 531311313 :51 33 “1113113" "515331“ E 1111313351 113311241 1 3 5533 1 11'! [:1 31353111131 5 1 3121335 23352 2.4331 131 E1 33111311133911 3 1 5232123 14.513 1311 El E1 13113313133115 3 1 3513333 124.251 31321 E 311331333111 3 1 3333323 143.132 51323 E E 13133131331 3 1 1323433 113.121 41321 E E 311133313331 5 1 1323125 151.335 23-13 E E 131131.933: 5 1 5315.331 33534 133-13 1:] [1 13133133111 3 1 3153453 114.113 5323 E 331333331 3 1 1254.314 33.133 23311 E 31133133131131 3 1 11231.33 221.533 33331 E E 131331311111 3 1 1111333 253453 13333 E E 133111191131 3 1 13324.31 4313.415 133-45 7’ El E1 11313131131 3 1 11353.33 233.333 3211-33 [1 E 1115511111 3 1 14523.3 534.115 1.5351 - ’1’ ’]Z E 31131191311 3 1 12133.33 323.312 1333 '0‘ 455 TL E E 513311.913: 3 1 3333.33 131.333 1322 IL gb E 11113311119131 5 1 353223 1331113 5.1323 £1!” I» E El 1313391111 3 1 1333333 $344.11 3.43.33 E E 1211193111 5 1 1532211 112554 233-23 El [1 31113353111 3 1 1335.321 113.454 1.1313 15 1:1 111113131131 3 1 5235.335 13.115 1.4314 E E 331313 3 1 3311.434 33212 3314 _':Ir-'| !'1 'III !.-"II 11- |_'.' 11131311111111.5955. 911111 13133111151151. 113111111 1113131951111 I115.911111.131331111.911111.11319111 9.9434 2.1224 515594 E) 9.9415 2TH? 19.9495 IE) 9.9552 9.4921 29.9119 E1 9.9529 2.4932 35.9995 11:) 9.95.215 24994 95.4195 E} 3.3315 2.2153 3.5313 1131 9.9.591r 2.3139 13.2559 531 9.9599 2.3399 15.5155 1.5 9.9525 2.2391 4.9999 1E) 9.9522 22491 5.317? 6 9.953} 22542 5.9321 IE1 9.9535 2215? 3.49215 If) 33534 22213 4.5314 Eh 9.9531 2.2299 4.9711 '5 9.9542 2.9939 3.9195 6 9.9941 229T5 3.4492 E1 9.9641 22934 3.5629 6 . . :---111:I-;r'-'-.-1.'i 31:1111-113-1 . . 9.954? 21954 3.2174115 11115911151191.51133495115911591935W91M13131.551111m31131.91151.111511119159111119.1131anrfl'IHIfMLhfllflfl 19 9.9545 2.2991 5.5493 6 dlsfituhEthuaJiamfihuulda-r 911111.w‘alstE151313391.91191.hlp.91191.fi11E1.91191.1515m91m.531£9119111319111 19 9.9545 2.2919 5.545? D 31151393191151 .wasl.911111. 511911.E151.311131.911111. 1151E1'1 31151333133111.31313131131113911. 955115153131911515 31E11 .. .._ _ |.'1 |_|.:-.=' -_,:I'|1= -.-J:. 5 -__5:1'-.'. !1'|__'1 -;:|'-‘1' i-_-jl'|' ._- 1‘ _ __ 15133.153313351533591515131315'5555331131111‘1.111535911513313111 111135.dequ9H.flMWSLMWd1-Wh.flistflrfll.h31fl I. 1.“. 111.1 .-..1|.1-.- ;1|' .-_._.: -'. -|!.'. l|'| II _1.'| I 1—.-|11 -_|:'| .-.| |- I'i '1-| :51 “99.559555591391991. 93131533319551.3111. 9131 ,mm mmfll'I'Ll'IlfD-M 13133133313531.3931. E15131-313193'31 55E1g35131535.91151.331191131.1131913 1513 .9351. 31151.11 931. 51171313131511.9551. 1131131971111 119911. 911111. 13133111191151. 5311. 911111.113i9111 1.111331.955111.111133.133111;1151.11931.51151.w3131.91151.511111L911111.13133n11.E1111.5311.E151.5319111 .--r-1':-_'—11-'.'.1."11'11:I'II:'.':.‘ |'-'|-1| _'.I'.-'.'..-1||r 19135. 91351311531951. E15191531Erl11.133131.E151.1115. 91151 .11119'1. 91151. 15153113. 95111. 1:311E1’311.h319111 13133111333111.5133. 93111311331331. 53111 mumn1nmmmmmmm 51111911111331.- Dwmmhmm'd'd'dfimmtflkflulhhhwfluMNM-h-ldI -L Wm :—2T.2fifififi+1flfl35'f1‘wal5!.'_ HIE] +— aware awn Randall M {1.152154 Hunt Ham ma Ermr Ham uf Rnspmsa . I'V- 1?? (lbs Mom {at Sum “$1 145 Ella! Sultan If W Hill: “I!!! Ffllth nodal 1 13324.1”!!! 13824.1 4379.413 Error 143 HERBE- 315 PM :- F G. Tulal 144 13322.5?!- tflflm" — Tenn 53m 51:! Emur tllflin Plain-ill mm -2?. 4. 5911119 4.5.45 ”-1 .1101)? mm 0.11MB 2|}. 96 {not}? ng E. 10 walghthclual 51] ED fl] ED 91'] 10111111121] might Prlfidld P111001 HBIFUJIT RHBE=4.1309 R59uam fl.EfiTTE Rfiuum Ml MEWS Hunt H Ian Square Error 4.1 3-0943 H9911 ulespunsa T?D31T2 {II-Daemon: {or Sum W919]: 145 9] Analyst of VINE“ Sum of Emma El: Swan: Inn- Sum-m Flt-fin Modal 2 151399.632 r949.“ 455.3633 Error 142 2423.192 17.116 Prob :- F 019191 113322.314 -c M‘I‘ EMSMEITortRIth 1rl13rcnpt 36.971333 5.33TT1‘1 -1E.14 ill-DU? mmgirm 115324341 {1.1361512 Bfifi ¢.Dflfl1* hlpgil'h 1.1449055 flflTE‘IM 15.53 113101" Efl‘lct Tull Rllidlfll bf Prldiutld Plat walnut R9 sldual 50 39 m 99 99 199119129: weightF'redtted 1' 9| '53 Rllporm 1Illllvligh'l _ Mud by Prldinlld Flt! mlnhthmual E E E E 3 E a 5D El] TD ED 90 199 111]I 1213 “in?“ Haiti“ Pt. III-111 RSEFCL IERHSE=25321 Raqum 119-15033 RSuuara Ml] 0.943453 R991 Ham Square Error 215321415 Ilaan MR4“! pause TLDMH flhsanmfinns {or Sum wars} 141.5 *[Analyslsu‘l’n'fl'lm Sum 91' Elmira: Ilium 5mm F IIIIiu 17315328 4323.93 ED131212 Error 140 1DDT.145 119 Prob:- F G. Tats] 144 133.22!“ 4.9991- 5mm Modal I]: 4 Tall“ Elf-nah SHErl'or lflflfiu W111 htamapt 4949942 9951991 99.99 4.9991' shuuldangirm 9.9992997 9945925 @4999? mum 9.599%?9 9949994 115:: «c.9999 I'llghflm 11.155151 HERBS 9.99 1.9999 halflh‘l $42155“ IDDEEBJE 13.114 4999? 51} ED fl} Bil 90111111111121} WEIQ'IPTEEEM 11 ...
View Full Document

{[ snackBarMessage ]}