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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 setups 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 setup considered). Some summary
statistics for pieces from each of 5 particular setups included in the study are below. Setu #l Setu #2 Setu #3 Setu #4 Setu #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 Setup #1, and assume that for
this setup measured paint thickness is normally distributed. 5 pts a) Give twosided 95% prediction limits for measured paint thickness on a single additional item
painted under Setup #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 twosided interval that you are 95% sure contains at least 99% of all
thicknesses for items painted under Setup #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 Setups #1 and #2, under the assumption that thicknesses of paint on
items painted under either setup are normally distributed. 5 pts c) Give endpoints of a twosided 90% confidence interval that compares how consistent Setups #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% twosided conﬁdence limits for the difference in mean measured paint thicknesses
produced by Setups #1 and #2. (Plug in completely, but you need not simplify.) M56 (ll'ﬂq—li‘tviﬂfkif' w‘l’b Illpm: wn (Irwin15 Now consider the results from ALL of Setups #1 through #5. 5 pts e) For yi]. : measured thickness for piece j from Setup 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 oneway 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 ) 15215 1 415 D L15 1 1.5 2 Shldlnﬂzad maid Threwlass 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 twoside nf1dence limits for the difference in mean measured paint thicknesses
produced by Setups #l and #2 under the oneway model assumptions (Plug in completely, but you
need not simplify.) Vlea V4‘11i’tswﬂul 0—1—15); 611‘“ Amos nV\ @$0~1525 r 2131(.132)\l fir (”His art mg 5 pts h) Setups #l and #3 involve a high belt speed while the others involve a low belt speed. So it is
possible that the linear combination of means 3
might be of interest. What is an appropriate margin of error for estimating this with 95%
confidence? (Plug in completely, but you need not simplify.) We tam [2%
Zl3l (132) (9.1%.); + 36%)")??—
lwﬂk Mb WA) 2. Below is a pdf, f (x), for a simple continuous distribution. f(x)=# x if—[<7L<l L 0 otherwise Suppose that a random variable X has pdf f (x) given above. s a) Evaluate P[—5< ‘ <5]. 5
5pt X P[—,54X4.§j: J LEAXVC 4%
<5
" mm at [5“ Nr“ ’4 A
— 1 7‘ _ ﬂgyéh :(W:
.7325 5 pts b) What are the values of EX and VarX ? I > 0 =.
we b1 am ( lax—um 4 mm o .. I 0 law: axe—(60% E><2 w o I :2;+ D 76+ 1‘ S 1_
50 \ltrX: JasLGZEXLFSDZZQDXZ ‘= “’1' i_ : p Z
—\
EX: 0 VarX: 'g 5 pts c) Suppose that )7 is the sample mean of n : 25 independent variables, each with pdf f (x). Approximate P __ S ¢7 < 5— . (If you were unable to do part b) you may use the incorrect value VarX=7here> 'X [g WWWMI mm wﬂ W #1fo a
M «I: “RAE—5 : E 50 NL WA ‘Z'Vﬂlug 5
.5—0
H 2: =55: 3,93
AS“ 1 J3;
E
’5— 22:‘3g3 [ So 'l><‘.54><<.$\ 73 l0
,5 6 .5 3. Nine oneinch 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; hitM7 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% twosided
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 1830. 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 1830 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
«FwyIndium “#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 Ill5
143 _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
1830 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 llwxll: in Fl \‘1 5'” “”5 birtSE—bl Lat. (25+ (914.er
w 5.(,a(l.2£)t 5.bD(l9é)(.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 zmslgl'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. flug 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 YMMVml 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 3predictor model that has the highest R2 among all possible 3predictor 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 545? WM “mm. 123—) 1L gm L¢=3 MAW
altAwe
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 2313 E E 131131.933: 5 1 5315.331 33534 13313 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 13345 7’
El E1 11313131131 3 1 11353.33 233.333 321133 [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 23323 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}
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9.9.591r 2.3139 13.2559 531
9.9599 2.3399 15.5155 1.5
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9.9522 22491 5.317? 6
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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 31:11111131 . . 9.954? 21954 3.2174115
11115911151191.51133495115911591935W91M13131.551111m31131.91151.111511119159111119.1131anrﬂ'IHIfMLhﬂlﬂﬂ 19 9.9545 2.2991 5.5493 6
dlsﬁtuhEthuaJiamﬁhuuldar 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.ﬂMWSLMWd1Wh.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'IlfDM
13133133313531.3931. E15131313193'31 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 .r1':_'—11'.'.1."11'11:I'II:'.':.‘ ''1 _'.I'.'.'..1r 19135. 91351311531951. E15191531Erl11.133131.E151.1115. 91151 .11119'1. 91151. 15153113. 95111. 1:311E1’311.h319111
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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 Plainill
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 30943
H9911 ulespunsa T?D31T2
{IIDaemon: {or Sum W919]: 145
9] Analyst of VINE“
Sum of
Emma El: Swan: Inn Summ Fltﬁn
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 illDU?
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. III111
RSEFCL IERHSE=25321 Raqum 11915033
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“ Elfnah 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 ...
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