Stat 231 Final Key F11

Stat 231 Final Key F11 - Stat 231 Final Exam Fall 2011 l...

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Unformatted text preview: Stat 231 Final Exam Fall 2011 l have neither given nor received unauthorized assistance on this exam. KEY Name Signed Date Name Printed 10 pts 15 pts (9 @ 1. An experiment was run to compare the fracture toughness of high purity 18% Ni-maraging steel to that of commercial purity steel of the same type. Summary statistics for the measured toughness of several specimens of each type are below. (Units are MPadm .) High Purity Commercial Purity 111 :10 specimens tested n2 : 8 specimens tested 35, 265.6 35, 259.8 31 = .8 s2 : l.l 21) Give a one-sided lower 95% confidence bound for the ratio of standard deviations of toughness for specimens of this type, 01 /02. (Plug in completely, but you need not simplify.) Vista 3' “ J— rflM-g is V/ L. “Pink 5'4 5 Fuel“ " affl/ 7’? 5,7 b) Because it is expensive, an engineer wants to specify High Purity steel for use in a project only if its mean fracture toughness exceeds that of Commercial Purity steel by more than 4 MPadm . With a 2.10 do the data summarized above force the conclusion that High Purity steel should be specified? (5km! 4x1— Wlmlé 7-— 542? 1w» «5: l 3 ® {Lug Ihl’MEl—W @ ’flm 5MPL¢5 7N! rs Mrm. M41124?" LW ’— H‘lslA MlM-S downy-M614) I : 11M“— SWi‘W-s fig): 0%; H7 ® Ha: lul— M7,: +0 : 3- Ha: nil-Maia a Rug HM WW is 'fih'V; arr/(14w: TIM—l— % Ml 5mm. writ b4 L @ 5—; _4D I kl k 'Fuwlv‘ $124 1‘; T: '121' mil/'4 _l- \ “3“” M WW I»! “A; S‘hkl . 11 H7, will 1:4 MJ‘d‘JQ'L (“14 1 ,bsa‘N-M vulm of T, (Zr. .91 ,l #D'Il‘t7 flQh (.445 10 pts c) Give 95% two-sided confidence limits for the difference in mean fracture toughness for specimens of these two types of steel (High Purity minus Commercial Purity). (Plug in completely, but you need not simplify.) —— \ \ [ML 74:12 4—“ ’17 7:] *1: film “Wis 15 ID (65.b~59.6 132.545 911.6%). 10 pts (1) Assuming that fracture toughness measurements of specimens of Commercial Purity steel are normally distributed, give two-sided limits that you are 95% sure contain 99% of such measurements. (Plug in completely, but you need not simplify.) M ii— Tsz l‘l‘flo (his is 59.51" 415910.!) 174 z z 2. Company officials at a meat processor are concerned with the fraction of packages of sausage filled by a particular machine that contain more than 16.1 oz of sausage. They plan to weigh the net contents of a sample of packages filled by the machine. 10 pts 21) If 15 packages are to be weighed and roughly 30% of all packages actually contain more than 16.1 oz of sausage, find the probability that 8 or more in the sample hold more than 16.1 oz of x: 1. m mist. mm mu m 14.10% .54 Ma” — — + Ms L Blw‘ml Am w Ill—l5 MA P—.S I N; WM ’— ’l—(o)+qcll)+---+—l—{73 Wwa .950 =* -05 DNW‘R I—t/ (9 l — ml. 7 pts b) Find a sample size that would guarantee that a 95% confidence interval for the fraction of all packages filled to more than 16.1 oz will have width no more than .02. (The interval will be of form fairA for A no larger than .01.) W Wm— [39 (I); ) E .01 . (.91, J. 60 W’me 1-99 Lil/Gd— 4 .01 c.{. (v? —>— 7|- (L) V\ a 9,M‘l- 3 Pts c) Company officials eventually settle on a sample size of 400 and find 150 (of 400) packages inspected contain more than 16.1 oz of sausage. Give 95% two-sided confidence limits for the fraction of all packages filled by this machine that contain more than 16.1 oz of meat (assuming that it is behaving in a physically stable manner). 252. /\__l_5_°_. ~=_\_"3-.—_ 352 l-N=“‘ =_6238 T_ +00 “ 375 P 4% '7 F +04 _~3 .3792.) mm) um $1M: ESL-9%" M- .S’ISi 1.9L (-fip“ M .0475 3. One measure of air pollution is the amount of beta radioactivity in the air (e. g. measured in ,uCi / m3 ). Below are summary statistics from 5 pollution measurement stations (all located within 10 feet of each other at a given site) at 4 different Ames locations on a single summer day. Location 1 Location 2 Location 3 Location 4 111:5 n1: 113:5 114:5 $123.1 3 22.9 $3 23.4 324 23.0 $12.1 $22.3 $32.1 $42.2 13 pts 21) Give 2-sided 95% confidence limits for the standard deviation of pollution measurements at a fixed location in Ames, based on the values above (and the assumption that measurements at a given location are normal with a standard deviation that is common across locations). Use information from ALL 4 locations in the calculation. 2 2 _ (5—0(-i3>+(6—0(-5)‘ a—s—our 4 (5—002) _ 0375 S?°'l"’l — (S—ID—Hs—I) +(5—1)+— (5-D _ ‘ ‘ Mr 4 m—v— so = /W\5 m1" 5fulA-il Y1. MAJ SfaDW( E . [20—4 ‘ZOA. 12 pts b) Locations 1 and 2 are at schools in residential areas, and locations 3 and 4 are in commercial . . . . l l . . d1str1cts. What 1s a "marg1n of error" for 3 (x3 + x4) — 5 (x1 + x2) based on 95% confidence 11m1ts . l . . . . for the quant1ty £013 + ,u4) — ,u1 + ,uz) (that m1ght be taken as a measure of 1ncrease 1n pollutlon in commercial areas over that in residential ones)? —L C, CmfiAh“ Mfrs film i," J7: (Mal/4+) ’ 2. (MW/4;) m Ci )5 $50M 2%. L1. + 1_ 2. Bl3.4+3.0)-Ji(5_l+23) :K i 2JZOQ93B)W 4. A QC inspector periodically inspects 9 in X 9 in floor tiles produced by a particular production line in the person's facility, looking for pock marks above a critical size. The inspector has come to the conclusion that the number of such blemishes on a single tile can be modeled as Poisson with mean A : .5 blemishes. 10 pts 21) Evaluate the probability that a given tile has at least one pock mark that is above the critical size. X: it [F Tiak ers aY\ l {S RlSSM wl‘l'K A: __5 a 1>[X>,Q=l—?[K=D]=l—10{6)= l— = .3335 10 pts b) Approximate the probability that 100 tiles have from 45 to 55 pock marks total using the central limit theorem. (This is the probability of an average of .45 to .55 pock marks per tile.) \Nd’h Xi: ll: ol' erS on ’Hth' , Wake mlous'l'M m flu nv‘ 7: J—ZXL mi PEAS é if _55:] (or loch ’iZlM'W] 190 Huma— Dl. W gaff]an zmrm rm] 4. Rigsng . ._ __ T . NW M7=Mx“ *3M VV’A=E=_07OW W W MA ‘01 W 4LT, >( is kpfmwm‘l‘x'h/vao'bmn . 5° W‘- “”l _ .555 .s _ q::.070'1{ 2" .070'7l H x 445:5 M 2-2: _070’7| : —‘75 655 AM ?[—.73<2<.7gj= 71%23-I2l77 Ms '9 : .Sé‘l—é 5. Attached at the end of this exam is some JMP output that concerns the analysis of a classical data set (of NH. Prater) concerned with the yield of gasoline in the refining of crude oil. The response variable is y : yield expressed as a percent of crude oil converted to gasoline after distillation and fractionation and predictors include the process variable x1 : the "endpoint" or temperature (OF) at which all the gasoline is vaporized and crude oil properties x2 : 10% point ASTM (the temperature at which 10% of the crude oil has become vapor) x3 : the crude oil gravity (degrees API), and x4 : the crude oil vapor pressure (lbf/inz) First consider a simple linear regression analysis of y based on x1 alone. 5 pts a) What fraction of the raw variability in y is accounted for by a linear equation in x1? (Report a number.) 2": _ 5%3 5 pts b) What are 95% confidence limits for the increase in mean yield that accompanies a l°F increase in "endpoint" in this model? (Report numerical limits.) Mac b, :l: ’17 95b, H¢r4 M‘s is .109 : 2-049402") {wax/L, 2.$79 0'? 17—51; A5“ 7 pts c) What are 95% prediction limits for the next yield of this process run at an endpoint of 300°F ? G— D 5: (Plug in completely, but you need not simplify.) Ulsz q: tsug‘, lat—Jr, Elem 4- 300(.\09)] it 2.0426199 I poo—332.03)" ‘ HE; J“ 3653.7le Now consider analyses of y as potentially depending upon x1, x2, x3, and x4 represented in the JMP reports. 10 pts (1) After accounting for endpoint (x1), do the crude oil properties (x2, x3, and x4) add detectable ability to explain/model/predict the yield (y)? Provide an observed value of an F test statistic, degrees of freedom, and an answer for a : .05. Fall MAO/ll -' w l , I; ,363 , x4, uwtx i Z: M? _ flak”)- SSZMJMAJ / 14-? (347,93 - mandala—13 ' M$€(€M\\ 4-993 ll 7 F : l 06 d. f . : 3 /No (circle the correct one) 5 pts e) If you had to choose a model for y based on some or all of the predictors using only the information provided on the JMP reports, which predictor(s) would you employ? WHY? (Explain your choice!) predictor(s) you would employ: 25' (MT/1W) lbw—X 17c; (ASTRA) Explanation: KL [5 N g [M A u 14,1 rrLL “ADA! MX KMSE rs W-M Minxub. ’2. 41,1 '9“! 44. W i”! llec,‘ butl— 5+1 ,L [o’l' W M fir» A snw‘u pa t‘ul‘o'b\ M Mull/)4 Milli. ’“N‘l Mw/lA bnflar‘ ’nwa XD-H :3 5 pts f) In a model including all predictors, give 95% two-sided confidence limits for the increase in mean yield that accompanies a l°F increase in "endpoint" with properties of the crude oil held fixed. (Plug in completely, but you need not simplify.) We btitfi‘c’bl Hm ms is .l551‘2.osz(069‘f) 1? 25% [Sl— W‘Tl' 432’] AS” 8 pts g) In a model including all predictors, a standard error of prediction for the conditions of the first set of predictors in the data is 1.193 . Give 95% confidence limits for the mean yield under that set of conditions. (Plug in completely, but you need not simplify.) um Qircsalx um (rm r: E 912+ (.l56)(206)+ (—.|60)(l90)+ (.227)(30.9)4—(.se4)(8.é:l]t 2 .092 (1.193) 6. Bert and Ernie are both applying for summer work at the Large Corporation. There are 3 positions open and a total of 10 applicants including Bert and Ernie. Bert is not eligible for job A as his uncle is the supervisor for that position. Otherwise, all applicants are eligible for all 3 jobs. 7 pts 21) In how many different ways can the 3 jobs AB, and C be filled by the 10 applicants? WNW oi A 'Du-c E 4%.,“ C, . TN MAVwa 9+ Yo$$ll°lllfias N4 M / CalMA‘B 9 - 9-5 s 446 A Balm/X 7 pts b) Suppose that all applicants are equally qualified, so that hiring is done essentially at random (except for the restriction that Bert cannot have job A). What is the probability that neither Bert nor Ernie is hired? o'l' WW" VLo'l' \‘M/Dlw‘ Bm+16(ni¢ arc it MS /3‘ “A '56 iii “’3 W4 half/LA‘B 334: __ So ?(V\DBW+M&V 'VLO Ernie): W —' _S/ZS 6 Pts 7. Events A,B are independent. Suppose P(A) : .3 andP(B) 2.4. What is P(not(A or (In set notation this is P((AUB)C) .) y? P(Af) PlB)= .l'2. 8. A manufacturing process produces axels whose diameters may be modeled as continuous random variables. The diameter (in inches) of a single axel, X , may be described by a distribution with probability density (pictured below) 1— for .99<x<l.01 f(x) = .02 0 otherwise X 2 Axels are (exactly) 20 inches long, so the volume of an axel is V : 20[7Z'[ 2 j J: 572'X2 . 1 Jill .02 | | | .99 lfll 3 1.0! 10 ptS a) Find EV,the expected volume of the axel. Z“ :_ _ 0200‘ Lo] ,_ o 250 I] ( - EV: S‘ITEX‘=5TF l 3:161) Au — 25'"— 3 w .97 : 5,002— 7]— 10 Pts b) Find VarV, the variance of the volume of the axel. You must set up completely a correct expression in terms of definite integrals, but you need not evaluate. (Hint: VarV : EV2 — (E V)2 .) \lmr \l= EVQF" (EV);— : E @156 — LEV y. 1.0\ '2. = Ml 95‘ l 2‘53 Aw — cm) .‘W 10 n. w w .3:wa m Fug—T— m nuns—Sm. an . .3323 mm . . $922. 9. . . “Emfi mi | | am mt. mum man can . aEEmm ‘ aw Enmb‘ can . own Ei ‘ . :E‘ 31 an“ a :NN EEa—Em ‘ . Earp‘ in | . 5 an in SE anEaD m. m 3 En N |.|| En m “ Eisalflaummu Em P Eh M 33.". Emma. «mama- nmmwd- 3.3..” San: . um? f Emma. ESE Macks- tame- Mqud _n_...__ . .m ngmd. NEH? 33+ wNFd Erma- Em... . 3:. 0 Emma. 32...... “31..” 3am... mind EESEw w Find $35 35.9. wrhd 3a.: 22... m2..an n. n . R _E w: ii 5.3 EE Em. Eat. . .nEuEfiEH. P EEK—abuflw nu . n I i .. I - -. Em .. _ Eta..in E iuflafififii .|III EwL finfiEfi. 2.2.... 20.”; "REE unnamed. .._J_u 2n.-. “2.34 $.22 :3 m...— 11 4199mm Fit :9 Yillrl 9y Endpoint [ EJ—Limar Fl! 19 Linear Fit ‘fiefd = —1 6.66266 + 6.1693?1*Endnoirrt Al Eu I11l11aryr of Fit | 2. Esouaro 2 9.5995 ’E RSquara M] 9.49 9912 Root Mean 39 new Error TE SETS Mean of Response 19.55939 Dos amt! ons {or Burn wars] 32 A[ Lack Of Flt 1 Sum or F Rnflo Source Df 51:99:99 lean Smart! 6136‘! Lack or Fr! 29 14119199 59.9999 Prob :- 9 Pure Error 4 299.919? 92.9942 9.999? TotaJ Error 39 11599999 Ho: 95:: 9.9192 A|_Analysls of Variance | Sum o1 Source DF Squares Hear = - a '- F Ratio Modal 1 1994.99 9o.7919 Error 36 . 58.66 Prob ::- F c. Total 91 9594.9??2 «.9991- A|_Paranutlr Estimatu I from Eati'nate 9111 Error 1 Ratio Prom-[11 Inioroopt 46.6 6266 6.66T2DE 4.49 66166“ Enoooint o.1oo 5.95 {9991* b. b 5| Shpwiso Fit for Yilid Ailvlfiflpnnn Yield M 555qu Regmslon Central I A Whull Hod-l _ 555 555 51555 5.5mm 555555555] 5;: 4AM" by Fruit”? H“ 3554.537: 31 15.32341? 5.5555 5.5555 55355354 ‘ 50 HCIJI'T'I‘M Estimatns | : w T>i Slip Hislury | ‘ E _— — 35 Alhll Pfiiflbll Models | a? — t. ordered 5515 55512 made-I5 up to 41511115 551' 515551. I . 2“ Handel Number 5mm 55:55 55 1“ . 1 5.5553 3.5555 324.4555 u Vapnr 1 5.1435 15.5533 555.5515 CI D 1“ 2n 3” 4D 50 " i ' ' 132m Predicted P5 5551 ' a ' ' I" 555:5.55 5555322344 3 - - - ' A fimmary-iffi _ _ _ —— __'| _ “_ ' ' = ' RSquare 5.55213? " F Raquarrmfi 5.5555“ \ evaluations done I 5- 55 :I v 1-: 55515555 Square Errrrr 33344-14 ( M Mean MR5555555 19.55533 5:55 arr-555m {5: 55m 35915} 32 1?‘l.?130 Prat:F Ap- ‘WW .5551! ll Tum Eslimata StdErmr 15555 5555515 \> Intercept 4.020?“ 10.12315 —0.0? 0.5052 b .r. 5.1555 «1313335 53.55 1.55511 x rirw ’- 5.529229 6.12 50001“ ’ '1' 0.23245 0.00903? 2.2? 0.0311' gbbl 05:35: I 5.5533252 5.355352 155 5.1555 5| Effect Tests| Al Residual by Prodluhd Flat | 5 4 E 3 E 2 E ‘1 E 5 5—1 *5 -3 .4 5 15 25 35 45 55 Yield Pralide 45an | Frans PmnRHSE 1?4.B0?05001 2.332523? 13 ...
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This note was uploaded on 02/11/2012 for the course STAT 231 taught by Professor Staff during the Fall '08 term at Iowa State.

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Stat 231 Final Key F11 - Stat 231 Final Exam Fall 2011 l...

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