This preview shows pages 1–13. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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% Nimaraging 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 onesided 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— rﬂMg 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 MlMS downyM614) I :
11M“— SWi‘Ws ﬁg): 0%;
H7
® Ha: lul— M7,: +0 : 3 Ha: nilMaia 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‘NM vulm of T, (Zr. .91 ,l #D'Il‘t7 ﬂQh
(.445 10 pts c) Give 95% twosided 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 twosided 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 199 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% twosided 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 lN=“‘ =_6238
T_ +00 “ 375 P 4% '7 F +04
_~3 .3792.) mm)
um $1M: ESL9%" 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 2sided 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)+— (5D _ ‘ ‘ Mr 4 m—v—
so = /W\5 m1" 5fulAil 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 22: _070’7 : —‘75 655 AM ?[—.73<2<.7gj= 71%23I2l77 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 : 2049402")
{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\\ 4993 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 WM 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 XDH :3 5 pts f) In a model including all predictors, give 95% twosided 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 'Duc E 4%.,“ C, . TN MAVwa 9+
Yo$$ll°lllﬁas N4 M / CalMA‘B
9  95 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 Hodl _
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 madeI5 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 ﬁmmaryifﬁ _ _ _ —— __'
_ “_ ' ' = ' RSquare 5.55213?
" F Raquarrmﬁ 5.5555“ \
evaluations done I 5 55 :I v 1: 55515555 Square Errrrr 3334414 ( M
Mean MR5555555 19.55533 5:55 arr555m {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 ...
View
Full
Document
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.
 Fall '08
 Staff

Click to edit the document details