chapter 8

Prévia do material em texto

CHAPTERS 
8-1 a ) Theocmfi denoe lew:l tor X- 2.14u I Jii < f1- < X + 2.14c I Jn isdetennined hy tlte \'alt>eof'1-o which is ::!.14. From 
Table lll. ¢(::!.14) = ffL<2.14) = 0 .9838 and theoonfidenc:e lt\'tl is .2(0.9838- o .s) = 96.76%. 
b) Tlt.eoonfidenc:e lt\'tl for X - 2. 49a I Jii < 11 < X + 2.49a I Jii isdctermined bytf!.e byt.ft.evaln.cott.o which is 2.14. From 
Table lll.<ll(:L49) = P("L< 249) = 0 .99.)6andtheoonfideooe le".'ti is.:l{0 .99~S) = 98.7.2%. 
c) TJ!.eoontidmce lt'.>el tbr X -l.SSo- / .Jii < Jl.· < X+ l.SSu I ..[ii is<k."'ermined bytlt(: bytf!.eva!t>eof'l.(). wl\ichis ::!.14. From 
Table lll.¢(1.8.5) = ffL< 1.85) = o.9(>78andtheoonfid~ la'tl is93..;6%. 
8•3 a ) A 1.a = 1.29wocldgjve res-ul t ina 9Q%one-si<k:d confidence intm'<ll. 
b) A 1<J = 1.6.; wouldgh\:' result in a 9s5;';one-sidedoonfidcooe: intcn:al. 
c) A 1-(l = ::!.xt wouldgi~ result ina <)95Wione-sid00oonfid.enoe inten'al. 
8-5 a) S:lmp!emeanfrom tltet'irstoonfi denoe inter\'al = 38.0::! •{6l.()8-38.M)/.:l =.so 
Sample mean from tl:'f:seoondoonfidenci!intervaJ = 39.9.5 4-{60.o.s-39.9.S)/::! =s o 
b) The9.:;%Cl is('\B.M. 61.98)aod tf!.e QO%Cl is('\9.9.;. M .o.:;). Tlt.e high.o."f' tf!.eoontidmce lt\'tl. tf!.e\,ider theCl. 
8-7 a) Findnfor tf!.e lengtboftJ!.ey.:;WiCl tobe40.7"8/ 2 = 1.¢ 
>/ 2 length = (>.<)6X20)/.[ri = 20 
392= 20.fri 
n = (3~~2t = 3.84 
Tt.ercl'ore. n = 4 . 
b) Find n for tf!.e le~th oftf!.e '!)!)~Cl to be40. 7"'/2 = 2.:;8 
1/ 2 length. = {t .<)6X20)/..Jn = 20 
39' = 20.fri 
n = ( 5~06r = 6.66 
n .erefore. n = 7. 
8-9 a ) The99W.Cl on tt.cmeanealcit!m ooncentrationwocld be longer. 
b) No. th.'t t is oot tf!.eoorrcct interpretation of a oonfidenoe intmoal. The probability thst .11 is between 049and0.82 iscitJ!.er o or 1. 
c) \' es. this is theoorrect ioterpri!'ta tionof a oonfideooe: interval. Tlteuppcr and lower limitsoft~.eoonfid.enoe limits are Nndom \ 'ariab!cs. 
8-11 95%Two-sided:Cl ont~.etruemeanyield: where.?= 90480,0 = 3 . n = 5 and"Lo.O:l,5 = 1.¢ . 
X - z0.<Yb-u I ../n < 11· < X + z0.025o I .{ri 
90.480 - 1 96(3)1.[5 < 11·< 90.480 + 196(3) 1.[5 
87.85 < ,u< 93 .11 
8-13 a) '!)9%'[\.\.·o-sidedCI on tr..e trn.e mean piston ring, diameter 
fOr a= O.Ol,7.a,f::!. = 1-o.oo.; = 2,s8 .and .Y = 74.03"6. o = 0 .001. n = 1.5 
x - zo.oos(~) < 11< x + zooos(Jn) 
74.036- 2.58(0·001) < < 74.036 ° 2.58(0·001] 
.Jf5 - 11 - T .Jf5 
74.0353 5, 115,74.0367 
b) '!)9%0ne-sid.edCt ooth.etn~meanpistonring diameter 
fOr a= O.OJ.ZQ = 1.o.o 1 = ~,3.3and .'i = 74.0:}6. o = 0 .001. n = 15 
- 0 < X - Zo.OJ .{ri - 11 
74.036 - 2.33( 0~1) < 11 
74.03545,11 
Tlte lower boundoftheone-sictedoonficteoce intm'al is less tf!.an tr.e lo,,•er hound oftlte two-sided ooofideoce. This is because the 'f) opel prob.!lbilityof 
9¢\Sonesided oonfi deooe intm'SI {or a= o .Ol) in the left tail tor int~.e loh·er bound) is greater than Type [ probabilityof 99% f\.\.·o-sid:i:doonfidenoe 
interval (oraf!! = o .oo.s) int.f!..e left tail. 
8-15 a ) 95~two-sidedCI o~ themeanoompresshutrength 
f.a/2 = 1.0.0 ::!.,5 = 1.¢.aod.f = .)-2.;0,o = 1000. n = 12 
x - Zoms (~) < 11< x + Zo o2s[ ~) 
3250 - 1 96(31 62] < < 3250+ 1 96(31 62] Jfi - 11 - Jfi 
3232.11 < 11 < 3267.89 
b) 9!)%Two-sidedCl on the true meanoompressh't strf:Dgth 
f.a/2 = w.oo.s = 2,:;8 
x - Zooos(~) < 11 < x + Zooos(Jn) 
3250 -2.58(3~~_2) < 11-< 3250+ 2.58[ 3~2) 
3226.4 < .u< 3273.6 
8-17 Set tlt.ewidth to6 ~.ol!rs witho = :!:.),ZO.O:i,s = l.9f>soh'£'for n. 
1/2 width = (1 .96)(25) t ..fii = 3 
49 = 3../ii 
(49)
2 n= 3 = 266.78 
Therefore, n = 267. 
• 8-19 To decrease tf!..e leog_thoftJ!..e Cl byooe half. the sample size must be increased by4 times (2 ). 
Z0 12u I .Jii = 0.51 
Now, to decrease by half. divide hotlts:ides by 2. 
(zar~ I .Jii) I 2 = (1 / 2) /2 
(zal~ 12../ii) = 1/ 4 
(Zar2u I~)= // 4 
• Th.o:refore. t lt.es.amplesitt must be increased by :i . 
8·2.1 a) '!}9% f\.,rosid.OOCl on the mt.'ln temperature 
x - zooos( _fi; J < 11< x +zooos( ./n) 
13.77 - 2.57(~) < J.t < 13.77 + 2 57(~) 
13.383 < J.t<l4.157 
b) 9.5% Jowc:r~nfidenoe bou.odon IJ!.e mean temperature 
fora= o.o.;.7.a = 7-o.o.; = 1.6.; and.t' = lj.n .o = os. n = u 
c) 95%oonfidmoet ltat tlt.e error of estimsti~ tf!.e mt.'ln temperature for wheat grown is less than 2 degf"fleS Celsius. 
Fora= o.o.;.'l.a/2 = ~.o~ = 1.¢,ando = o,s, E =- 2 
n=(z•tr = [L9625)r = 0.2401 
Always round tzp to the De>: t nonbcr, tlterefore n = 1. 
d) Set t lt.e width to 1-5 degrees Celsius \\itho = o,s.zo.o'2..5 = l.¢sol\t:for n. 
1/2 width = (1 96)(0 5) t .f,i = 0. 75 
0.98 = 0.75../ii 
' 
n = (0·98)- = 1707 
0.75 
Therefore, n = 2. 
8·13 a) ro .o~.l~ = 2.179 b) ro.02.,5.24 = 2.064 c) ro.oo.;.13 = 3.012 d) ro.ooo.;,t.; = 4.013 
sum 251 848 _ 
H; a)i\;ean =
2
N = 10 = 2).1848 
Varianoe = = (.srl.>ev) -= 1.60.; = 2,5760 
b) 9.;%oonfidenoe intc:rvalonmean 
n =H) .Y. = 2_;.1848 S = l .M.; i0 .02..;.9 =- ~L:i(t.l. 
_ [ sj , - (s) x - 1oms,9 --:,;; < w::, x + Ioms,9 .Jii 
25.1848 - 2262(1:05] < 11 < 25.1848+ 2262[1:0) 
24.037 < 11 < 26.333 
8·:17 9.;%confideooe intenosl on mean tire life 
n = 16 .1:' = 60.139·7 s = 364.5.94 ro.o2..;.1.; = 2.131 
X - 10.025,15( :};;) < 11 < X+ Fo.015,l 5 [:};; l 
60139.7 - 2.131[361/4) < J.t < 60139.7 + 2.131[361/4) 
5819733 < J.t < 62082 07 
8-~9 .i = uo s = 0 .015 n = 2.5 
9.;%CI on tlt-c mesn \-alone of S)TUpdispens.ed 
For a = o .o5and n = 2.5.1aj:to-1 = t0 .02,5.~ = ~.064 
X - I O.OH,24 [ Jn) < Jk < X + 10.025,24 [ Jn) 
1.10 - 2.064(0·015) < .< 1.10+ 2064(0015) = _ Jl _ rr 
"L,J "L,) 
1094 < Jk < 1.106 
8-31 99~ upper oonfidenceintm'al oo mesoSBP 
n = 14 .f = n8.:} s = 9.9 ro .o1.1$ = 2.650 
J1 < x + 1o.oos,n lJnJ 
11 < 118.3 + 2 6so(~ J 
jk < 125.312 
By examining tltt' oormal probabiJity pJot. it a ppears that t lte data are normally distributed. Tlt..."N! does oot a ppear to be m::mgb f:'i denoe to reject the 
t;ypot~.-es.is that tr..e frnqnenc:ie.s are normally distributed. 
.. 
.. 
.. 
§: 
m .. 
Q. .. 
.. 
.. ,. 
10 
• 
,,. 
Normal Probability Plot for Frequencies 
M..r......... -~a 
• 
• 
• 
• 
• 
.,. 
8·j3 Tlwdata a ppear to be oonnaJiydisttib~>tOO b3s00on euminationoftlte normal probability plot below. 
TbcretOre. tlttte i seo.ide~ to support that the annual rainfall is ncmnaJiydistribnte<t 
9s9tioonfid"eoce intrn'8l on meanannllAI rainfall 
n = 20 .? = 485.8 s = 90.34 ro .o~.:P9 = 2.093. 
Probability Plot of Rainfall 
N:lnMI· 9S¥.0 
x - toms,l9( Jn) < J1 < x + lo.oa,I9(Jn] 
485.8- 2.093(~4)< J1 < 485.8 + 2.093(9~] 
443.520 < Jk < 528.080 
8·35 99%oonfide:ooe intcrval onmesncurrcnt required 
Assmne that tlt-cdsta area random sample from a oonnal distribu.tion. 
n = 10 .? = 317.2 s = 15.7 ro.oo.;.19 = j .2,50 
317.2 - 325o(Jd) < 11< 317.2 +3 25o(Jd) 
30!06 < J1 < 33334 
·- 46.9 SUiu to$ 
. ., 
AO ... 
"-'ill~... .o..ilt 
8-37 a ) Tlt-cdsta appear to be nortlU'lilydistribnted based onexantinationoftm: nonnaJ prob..~bilityplot belo~·. 
b) 9.5% Mo-sidOOoonfid'I':OOI'! intervaJ on meanoomprehens.iw:strength 
Normal Probability Plot for Strength 
ML ES>:Itro:ttes • 95% Cl 
99 
9$ 
•• 
•• 
-
70 
c: 
•• ~ so 
a> 40 
"- 30 
20 
10 
s 
1 
, .. 
n = 12 .'i-= 22.59-9 s = j..;h ro .02.),ll = 2.201 
2250 
Data 
n = 12 x = 2259.9 s = 35.6 10.025,11 = 2.201 
X - r0.02S,llS};; < 11 < X+ 1o.02S,ll [};; l 
2259.9 - 2.20 1[ 3~) < 11·<2259.9 +2 201(3~) 
'1112 - - '1112 
2237.3 < 11 < 2282.5 
c) 95'% lovrer<Onfidenoe bound on mesnstrengtll 
X - lo.OS,ll ( .!,;) < 11 
22599-1796(~)<11 
2241 4 < 11 
2350 
Ml Estimates 
Mu n 2259 .92 
StD£-V 34.0550 
8-39 a) Tlt-edata appear to be normallydistrihnted basOO:on examination of the nonnaJ probability plot be!O\Io·. 
TJ>.ereforc, t.i'.ercisevide~n to support t.i'.at thespeed-upofCNNis oorm.allydistribntod. 
b) <).;%confidence irrt::rvsl on mcans]X't'd-up 
n = 13 X= 4.313 s = 04,3.-28 ro .02.).l2 = 2.179 
PfoNbllity Plot of Spe4:d 
tt:.f«ooll · ~ 0 
X - TomS,ll [.!,; l < 11 < X+ 1o.025,11 [.!,; l 
4.313 - 2.179[0·4328) < < 4.313 2.179(0·4328) 
.J13 - 11 - + .J13 
4.051 < 11< 4.575 
c) 95'% lovrer oonfidence bouodon meanspood-up 
n = 13 .:r = 4.313 s = 043-28 r0 .o5.12 = 1.782 
x - ro.os,12 ( J,;) < 11 
4.313 - 1782[0.4328) < :.;IT - 11 
4.099 < 11 
- · UIJ ~f)M 0411 
. " o\D on1 
......... 0.~\ 
8-41 a ) TJ!.edata appear to be oorm.allydistMbnted. There is oot strong eo.idence th.a t the percentage of enrichment deviates from oorm.ality. 
b) 99% f\.\.·o~idOOoonfideooe: interval on meanperoentsgeenrichment 
For a= 0.01 and n = l!l:.taj2.n-1 = to.oo.;.,u = 3.106 . . t = -2.9017 s = o.om 
X - /0 005,11 [:}; l < jl < X + 10.005,11 (},;) 
2.902- 3 106(0; 3) < jl < 2.902 + 3.106[0; 3) 
2.813 < ji < 2.991 
2 2 • 
8-<13 a) 9slill upperClanddf=24 X! - o ,<lf = X0.9$,24 = 13.8) 
,2 - 2 - ') 1 67 b) 9')% 1ovrerClanddf=9 XaAf - Xo.Ol,9 - - · 
2 2 2 2 
cl <)Olii\Cl anddf= , 9 >-a / 2,df = Xo.05,1 9 = 30-14 and XJ-12,</f = xo.95,19 = 10.12 
8-4.; 299% lower oonfidmoc bou.OO foro from the previons E'.'\ercise is 
0 .0000307.510 
o .oo.;.;45 s a 
One may t3ke t.f!.e square root oft~.e vuiaflC.'e bound to obtain t~.e oonfickoce bot:.nd for the stao&ard de\iation. 
8-47 9.;%oonfidence intrr-'tll tor o: gi\l!n n = 5J.s = 0.37. 
First find tlteoonfideoot intervaJ for a . 
Fora=o.o.;andn = S l~x~n,n-1 = X~ms.so = 7 1.42 andxf-on,n-1 = X~.975,50 = 32.36. 
50(0.37)2 < <72 < 50(0.37)2 
71.4~ - - 32.36 
0.0(}6 s 0 ... $0.211.5 
Taking t lt.esquare root oftlt.eendpointsofthis intcrv31 we obtain, Ojl < a < 0 46. 
8-49 n.edata appe.'lt to be oonnallydistribntcd based onexaminationoftt.e. normal probability plot below. 
Therefore. there ise-,.ideooc to support that tltemean tempm1tu.reis oonnallydistribt:.tOO. 
9.;'%oonfidmoe inten'SJ for a 
n=8s=0.94&J 
17 
Probability Plot of Mean Te.Ol) 
Normal· 9S9bCI 
2l 2* 2S 
He.a.n Te~ 
x;,21,- J = X~.o25,7 = 16.01 and x L 12,n- 1 =x~.91s,; = 1.69 
7(0 9463? < 0'2 < 7(0.9463)2 
16.01 - - 1.69 
0.392 < u2 < 3.709 
0.626 < u < 1.926 
,._ 2:1.4! 
- . .., 
• • 
.. ""' 
... ~..,. cua 
8-.:P 95~oonfidenci: interva l foro 
n = lS s = 0 .008.31 
.X~/2/t-l = X~.o25,7 = 16.01 and xL n,n-l =xl 975,7 = 1.69 
7(0 9463i < q2 < 7(0.9463)1 
16.01 - - 1.69 
0.392 < u2 < 3.709 
0.626 < q < 1.926 
The data do not appesr to be: normally distributed based on an examination of the normal probability plot be!.oh'. TbcretOre. the<).;% confidmce inten:al 
foro is oot valid. 
Probability Plot of Gauge Cap 
N:!r!NII . 95¥.. Cl 
8-53 a) 9s9rioonfid.enoe inten'al on the fr.actiondefective procbtood witltthis tool. 
p=~=004333 
300 
11 = 300 
p- z, ,2 ~ <p <p +:an p() - p) v-----;;- n 
" - ~411 
- ·-. " .~oo c.m 
,...... .. G.Ct& 
==~= 0.04333(0.95667) < < 0.04333 .L 1.96 0.04333(0.95667) 0.04333 - 1.96 300 _ p _ ' 300 
0.02029 < p < 0.06637 
b) 95% upper oonfid.m:le bowxha = l.O.os = 1.65 
o < n-'- z ~ 
.-Y • o /2 
n 
"""-:-===~ p < 0.04333+ 1.650 0.04333(0.95667) 
300 
p < 0.06273 
8-55 a) 95~oonfidence interval for the proportionofoollegegrnduates in Ohio tf!..a t \ 'Otod for George Bush . 
• _ 412 _ 0 '36 D---- .) 
- 768 11 = 768 za/2 = 1.96 
ft - z.,2 p (1- p) <p <ft+ z. ,2po p) 
n n 
~~~ ~~~ 
0.536 - 1.96 053~8464) < p < 0.536+ 1.96 0.53~~~464) 
0.501 < p < 0.571 
b) 95~ lower oonfidmoe bcmndont.f!.e proportion of oollegegradtU~tes inOitio that \'OtOO for George Buslt. 
p-z.~<p 
0.536 - 1.64 0.536(0.464) < 768 _ p 
0.506 < p 
8-.;1 a) 9.;~oonfidmoe interval ontbe proportion of rats that art: uDdf:l'-wcig.ht 
P• - z p(l p) < p < o+z a /2 n - - · a /2 11 p(! - p) 
04- 1.96l4{0 6) < < 0.4 1.96l 4(06) 30 _ p - + 30 
0.225 < p < 0575 
b) E = 0.0:!-, a= o.o.;.l.af~ '""'lo.0:!..5 = 1.<}6and P = 04 as the initial estimate of p. 
2 2 
n=[2at2) jJ(l - p) = [ 1 96) 0.4{1 - 04) = 2304.96. E 0.02 . 
11 " 2305. 
c) E = o .M ,a = o.o.;. r.aj'J = 7<1.o~s = 1.9()at Jeast:9.;%oonfid-ent 
n = [z"n)\o25) = (196 j'
2 
(0.25) = 2401. 
E 002 
8·.'}9 Theworstcasewollld: be for p = o.:;. tiles with E.= O.Os and a= OA'H.'la/2 = 7.0 .00.; = i.:;Sweobtaina sample size of: 
J/ = [ 2~2r p(l-p) = [~:~~r 05 (1-0.5) = 665.64 , n;;;; 666. 
8~>1 95'% prediction intm'<IJ on the lifeoft lw: nv:t tire 
ghl!llj=60l.:W.7 &=,3(>4,5.94 0=16 
-tOr a= o.o.; 1cf2.0-l = to.02.;.,15 = 'J.131 
- . g~ - g~ X - lo.02S,ISS + - < Xn+l < X + to.02S,l5S + -
11 n 
601397 - 2.131(3645 94)JI + 1~ < x•+l < 60139 7+2.131(3645 94)J 1+ /6 
52131.1 < Xn+l < 68148.3 
The prediction interval is considerabi)'.,..'ider than the9.;% confidence interval {58,197,3 s J.t s 62,082.07}. This is expected boca~ the pro:fiction intervaJ 
ooeds to include t~.e\'8riabilityin th.e parameter u.timatcsas wcll as the \"atiabilityin a future observation. 
S-63 9.5911 Prf:diction lntervaJ on th.e,olumeofs)'l"'i;pofthe DE".Xt be\oeragedispens.ed 
.'i = uo s = 0.015 n = 'J.S fa/:!.,n-1 = to.02,5,'J4 = -2.064 
x - fo.oo;,24sJI + : < Xn~l < x + fo.025,24sJI + : 
1.10 - 2064{0.015)J1+ ; 5 < x,_1 < U0 - 2064(0015)J1+ ; 5 
1.068 < Xn~l <113 
The prfdictionintcn'9l is \~icier thant!teoonfid.."''Oe i.nten•aJ:1.094 ~ ll s 1.106. 
S-6.5 9.:;911 prOOiction intmal on thewlmneof syrupofth.e next be\'E!tagedispens.ed 
n = :!.1) .f = 48,s.8s = 90.$4taj'J,n-t = l().o~.; .l9 = :!..093 
x - to.025.19s) l + 
1 
< x,+l < x + to.025.l~JI + 1 
· n · n 
485.8 - 2.093(90.34))1 + 
2
1
0 
< x,+! < 485 8 - 2.093(90.34))1+ 
2
1
0 
292 049 < Xn+l < 679.551 
Tl\e9.;% prediction intm•aJ is wi<k:r than th.e95'%oonfidenoe interval . 
...,. R· I < . <- . R· I x - to.oos 9s , - - -\n+l - x • to.oos.9s • -
' n · 11 
317 2 - 3250(15.7)) 1+ I~ < x•+l < 317.2- 3250(15.7))1+ I~ 
263.7< xn+l < 370.7 
The length of tlt.e prediction inter."SJ is longer. 
S-6.;> 9£1% prediction intervaJ on th.e next specimen of oonc:rctetcstOO 
gh'etl.i = ~60 s= 3SS7 o = lHora = o .o,s andn = 1.2, fa/ i.n-l = to.o.;,u = 1.796 
8-71 9Q%predictionintervsl forenricltmentdllta given."( = ~.9 s = 0.099 n = 1.2: fora= 0.10 aodn = ~~. laj :tn-1 = to.o5,11 = 1.796 
-. Rl - R·l x - to.os,ns +- < x.,+l < x + to.os,ns • -
n n 
2 9 - 1.796(0.099))1+ I~ < x.,+1 < 2 9 + 1.796(0.099))1 : I~ 
2.71< x.,+1 < 3.09 
Tlt.e9f)~oonfidenceintet\'8l is 
x - to.os,ns (1 < /.1 < x + to.os,12s (1 v-;; v;, 
2.9 - 1.796(0 099)ji; < /1· < 2.9 - 1.796(0.099)ji; 
2.85 < /.1 < 2.95 
Tt.e prediction intmosl is wider than thect on tiM: population mean ''itlt tl!esameoonfideoce. T!to!99%oonfideooe intmosl is 
x - to.oos,nsJ! < J.i < x + io.oos,nsJ! 
2.9 - 3.106(0.099)& < /.1 < 2.9 + 3.106(0.099)& 
2.81 < J.i < 2.99 
Tlbe prediction intm'a.l ise\~ ~ider iliant lt.eCl on the population mean,,ith.greater oonfidm:e. 
8·73 95%toleraooeintervsl on the life of t ho: tires tltat h:asa 95%Cl. 
ghm.Y = 60139.7 s = 3645.94 o = 16wefindk = :l.9Q.'J 
.i · ks .. 't ~ ks 
60 >39·7 ·· ·90 3!364.; ·94 ), (>() >39·7 • ' ·903!3'>45 ·94) 
(49555-54. 70723.86) 
95%oonfictence intervsl {58.197 .$ $ p $ fn.08:l.07) is shorter thsn th.e95'% tolerance intervsl. 
8-75 95% tolerance interval on thesyrup..-olrune that ltas9Q%oonfideoce JC\'cl 
.'t =uO s=0.015 n =2.5and k =24 7<t 
.Y- ks .. ? • ks 
1.10- :l474{0.0 15),1.10 ~ 2474(0.015) 
(1.06,1.14) 
8-77 95% toleranoe interval on the rainf.aJI tr.st has a oonfidmoe IC\otl of 95% 
n = 20 X= 485.8 s = 90.34 k = 2.152 
.'f-ks.X ~ks 
48.5.8- 2 .. 75-i<90J4),485.8 ~ 2.75:l(9QJ4) 
{2j?.184. 7'J44l6) 
Tiw: 95% tolerance intervsJ is much wider than tlt.e95%oonfidence intervsl on t.J!.e population mean(44!J.S2- $ p s 5.2-8.08). 
8-79 99% tolerance intervsl ontf!.e bMg,htnessofteiC\ision hlbes tf',s t ~.asa 95'%Cl. 
gh~tt~f=j.17.~ s= 15.7 n = JOweli.ndk=4433 
.'f- ks .. f • ks 
j.l7 .-2- - 4433(15.7).317 .21 ~443;.l{l5 .7) 
(:>47 .60' 386.80) 
Th.e99% toler11nce-intenosJ is mcch"'id.er tl>.an t1te95%oonfidenoe interval on t.f!..e poptilation mean:)01l)6 $ p $ .33:).j4. 
8-81 99% tolet3nctintmoal on rod-enrichment data that ha\-ea 95%CL 
gh'f:n .'i' = 2.9 s = 0.0~ o = ~~we fiodk = 4.150 
.?-ks .. ~ ·ks 
2.9- 4.150{0.099). 2-.9 .. 4.150(0.099) 
('49-3~>) 
Tlt.e99'% toleraoct intervsl is mccl\ "';der tJ!..an the95%Cl on tf!.e population mean{~.84 $ p s 2.¢). 
supplementa I Exercls.ts 
S.Sj Wherea1 "'<1::!_ =a. leta= o.o.; 
lnten>al for a1 = ~ = af :l = o.o~ 
n.e con.'idt':'lce i E:\~ fo: X - 1.96o I Jn < J1 <X+ 1.96o- I Jn is detennined b)•tbeby the wthte of Zo· \\'hich is 1.96. From Table Il l, ~·e find 
<()(1.'!)6) = P{Z < 1.9(,) = o .<p.;.and tlteoonfidenc:e IE:\oel is9.;'%. 
fnterval for 
The oon.fidentt intt.'!\'31 is X - 2.33o I .Jii < J1 <X + 1. 75q I .[ii • the con:idence leo.'cl is the same because a= 0 .05. The ~mrnctric inten'al 
does oot affect the JE:\'el of significance; ~.owe>.~, it does affect the length. Thes}mmetric intm-al issOOrter in length . 
• 8-8.; p = so.o =.; 
• • a) lt=l6findR.s ~ 744)or ,tts 2,s6) 
P(S2 > 7.44) = P(xrs > 1 5~~44) )= oos < P~··ds > 22.32) < o 10 
• Using Minitab J~S 2: 744) =0.0997 
P(S1 < 256) = P~·ds < 1 5(~S6) l = 0 05 < P(xfs < 7.68) < 0. 10 
• Using Minitab RS :; 2,s6) = 0.064 
• • b) fb r n=jO fi ndRS ~ 744)or RS s2,s6) 
P(S2 > 7.44) = P(xi9 > 29(~ 44)) = 0 025 < P(x~9 > 4315) < 0 05 
• Using Minihlb RS ~ 744) =0.044 
P(S1 < 2 56) = P(xi9 < 29(~.56) ) = 0 01 < P(xi9 < 1485) < 0 025 
• UsingMinitabff.s s2,s6) =0.0l4. 
• • c) For n = 71 Its ~ 744)or ffs s 2..;6) 
P(S1 > 7.44) = P~·do > ?0(~.44) }= 0 005 < P(xfo > 104 16) < 0 01 
• UsingMinitabRS 2: 744) =o.oo.;J 
P(S1 < 2.56) = P~·d0 < ?0(~ 56)) = P(x?o < 35.84)< 0.005 
• Using Minitab ltS s 2..;6) < 0.001 
d) TJ!.e probabilities get sma ller as o inc:rease.s.As n increases. the sample \'ariance s~.ou.ldapproach the popu.Jation\'arianoe: tf!.erd"ore, tf!.e 
likeliltoodof obtaining a sample variance much larger th:aotJ!.e populatioo\'atiaooe,,ill dec:rea.se. 
e) The probabilith·s get smaller as n inc:reastS.As n inc:rease.s. tf!.e sample\'lttianoesh.ouldapproacll the population\>atiafX'i:: tlterefore. the 
likd iltood:of obtaining a sampJe variance mc.cllsm.sl!cr than tl:e population\'atianoe v.ill d(l(!fea.se. 
8-87 a) Tit.: probability plot s.f!..()\11'$ that thed:ata appear to be normally distributed. Tlberefore. tlbere: is no evideoceoonclc.dethat ilw 
oompret.msive strength d.<lta are oormaJ ly distributed. 
b) 99% 1ov..-eroonfidmceboundont~.e meao .Y = !5.1~. s = 84.2, 1t = 9 r0 .01,8 = ~.8:>6 
x + Tom,s[:};) < 11 
25 12+ 2 896(8; ) < 11 
16.99 < 11 
'O.e lo\.\o·er bouodon t.Jt.e99%oonfidenoe interval sJ!.ov..'S th:at tltie meanoompret..:nsl\oestrength is most likely be greater than 16.99 Mega pascals. 
c) 98% th·o-sided oonfide:nct inten'31 on the: mean .Y = ~.l.:l. s = 841, n = 9 to.o1.8 = 2.8:)6 
x -lom,s(J; J < 11 < :X -lom,s(J; J 
25.12 - 2.896(8; ) < 11 < 25.12 - 2.896(J§ l 
16.99 < J1 < 33 25 
The bounds on tJ:-e 98% two-sided oonfi<k:ooe: intcn:a I s.ho,,'S that t~.e mean oompre:bensi \'e strengtb v..il 1 mOISt likely be greilter tha o 16.99 Megapas.-:.a Is 
aod Jess tf!.an3J.!!..5 Mega-pascals. The lower boundofllw9:>%ooes.idOO. C[ is the sameas the lower bound ofthe:98% two-sided C[ {t.rJs is because of the 
value: of a). 
• • d) 99~ ooe-sidOO: upper bound on tltie oonfidenoe inte:rva I on o oompre~.enshoe strength 
? 
s = 8.42, s- = 70.90 ·d99s = 1.65 
' 
(]1 < 8(8.42)2 
- 1.65 
o 2 < 343.74 
Tr.e upper bo~ndon the:99%confid'eoce intrnoal on the,oarianoes.OO''I'S that tlt.e\'3rianotoftheoompre~ . ..ms.h'e str-ength is most likely less tf!-.anj43.74 
Megapasca.ls . 
• e) 9&~ 1\~o-sided:oonfi<k:oce intervaJ ono of comprcl'.ensiwstre:ngth 
s = 8.42, s2 = 70.90 x5.o~9 = 20.09 x5.99,& = 1.65 
8(8 42)2 <o2 < 8(8 42)2 
20.09 - - 1.65 
28.23 < o 2 < 343.74 
Tlte bmmdson the~ 1'\'lo-sidOOoonfidet'lt'e·interval on tf!~varianoe sltows that tf!.e\'3riaooeoft~.-eoompteltms.hotstrqth is most likely Jess than 
34j..74 MegapascaJs and greater than :l-8.~ Mf%apascaJs . Tlt.e upper bound oft~.e9!)?6one-sid:ed <.."1 is t.Jt.es.ameas the upper bound oft}!.e: 98% two· 
sided Cl because va h.o.e: of a for tf!.e ooe-sidOO example is ooo-h:alft~.e vaJti.e for t~.e f\~·o-si<ted exam pie. 
0 98% tv..·o-sidedoonfideoce intenoal on tlt-e mean X= 2:}, s6,31. n = 9 ro.o1.8 = 2..81}6 
x -10_01,8( Jn l < 11 < x + lo.ot,s( Jn) 
23- 2.896(631) < u.< 23-'-2.896(631) 
.f9 _,._ ' .f9 
16 91 < J1 < 29 09 
• 98% th·o-sided oontlde:nct inten'3J ono oompreltmshoe strength 
s = 6.31, s2 = 39.8 x&m,9 = 20.09 x6.99,& = 1.65 
8(39.8) < 0'2 < 8(39.8) 
2009 - - 1.65 
? 15.85 < u - < 192.97 
Fixing. tf!.e mistake dec:ressOO: tr..e val 1:es of the sample mes nand tf!.e sample standard deviation. Be::auu the sample staor'...atd deviation was dec:reasOO 
tf!.e l'<'idths of tr.e oonfideooe i otervaJs were also decreased. 
g) Tl:e: exercise pro,ictess = 841 {instead ofthessmple:,oariance).A 98% 1'\~o-sidOO oonfidmoe interval on the mean .Y. = ~. s = 841. n = 
9 f().Ol.S = -2..896 
x - loot,s(Jn l < 11 < x + 7oo~s[ Jn] 
25 - 2 896(s;1) < J1 < 25+ 2 896(s;1) 
1688 < J1 < 33.12 
• 98% th·o-sided oontlde:nct inten'3J on o of oompr-ef!..enshoe strmgttl 
s = 8.41, s2 = 70.73 x6.ot,9 = 20.09 x5.99,s = 1.65 
? ' 8(8.41)- <o' < 8(8.41)-
2009 - - 1.65 
' 28.16 < CT- < 342.94 
Fixing the mistake did oot haw an affect on the sample meaoor thesamplestandsrd de>.ia tion. Thl(!)•ate\'E:ryclose to t.f!.eorigiool \'alu~. TJt.ewi<it.hsof 
f.l!.e oonfidmoe ioterva lsare aJso ''el'J' simi! a r. 
h) When a mistake:nvaJne: is O"..at tt.e sample mean, tJt.e mistake \\oil! not a.ffect tt.e sample mean. staodsrd<W.iationoroonfidmoe iotenoaJs 
greatly. Howe\'et, wlt.eo the mistake is not near tl.e:sample mean. the\oaJne:cangre.atlyaffect tf!..es.ample mean.standsrdde>.iationand 
oonfid-cnce inter\'als. TJ!.e: fa rther from tlte mean. tltegreilter the effect. 
8-89 X=l5-3JS=-0.6 2. fl.:: :l:O k= :l:,Sf)(! 
a) 95'% tolmmoe interval of ltemoglobinvalnes \\i th 91')96confidenoe 
.? - ks .. 'i- ks 
15-13- :l:,s64(0.6:l:), l5 .33 .._ 'i,s64(0 .62) 
{lj.74, 16:9:!:) 
b) 99% tolerance intenal ofh.emoglobio, oaJu.e.; witl't 9Q%oonfidencek = jj6S 
."f- ks,f ... ks 
l5J.l- 3.368(0 .6:l:), l5~- 3,368{0 .~) 
{1,3.:!:4.17 4.:!) 
8-91 a ) Tlw"e is ooe-.i&oce to reject tt.eass-umption t.J!..st the data are normaJiydistribnted. 
Nonnal Probaility Plot for Foam Height 
Ml htlm~~ .... ·WJ~.O 
.. 
.. 
" 
-
.. 
c 
" "' ~ 
" 
"' 
., 
~ 
" 
I I I 
• . 
• 
" 
.. 
, 
" 
' 
y 
/ . 
I y I 
.. 
"' '" Data 
X - Io.o2S,9 ( Jn) < J1 < X - Io.025,9 ( Jn) 
203.2- 2 262(~) < J1 < 2032+ 2.262[ ~) 
197.84 < J1 < 208.56 
c) 9.;%ptedictioninter.oal on a futnr-eS.!ImpJe 
I/ 
• 
• 
I 
-
/ 
2032-2.262(7.50)~1+ 1~ < J1<203 2 + 2262(7S0)~1 : 110 
185.41 < }1< 22099 
d) 9.;96 toleranoe inten 'al on foam tt.eight '~i t~99%confideooe k = 4 .2.6.; 
.f - ks •. 'l ... ks 
203.:!: - 4.:!:6.;(7 ,s). :!:03.~ .._ 4.:!:6,5{7 ,s) 
{l7l.:ll' 23.5 .19) 
I 
(:) Tlt.e9,s%Cl on the popdation mean is the narrowest interval. For t f!...eCl. 9.5%ot sn.ch intervals contain tit~ population mean. For t f!.e 
prediction inter.>al. 9.5%of s-uch inten oals \\i ll OO\'ef' a futuredsta \ 'alue. T~is inten oal is quite a bit wid-er th:an theCl ontl'.e mean. Tl:~ 
tolerance interval is tf!.-ev.idf'.St ilrt:!rval of a ll. For t.J!.e tolerance intervsJ. 99%of s t:cb intervals v.ill iocln.de9.;%oftf!.-e truedistribl!tionof 
foam ".ci&J!t. 
8-1)3 9.;~oonfid.enoeinten 'al onU..e proportiooofbaseballs \\i th a ooefficient of resti tution that exce00s0.6.J.5. 
f> = :o = 0.2 11 = 40 
fP(Y) fr - =av----;;- < P 
02- L6sl·2<0·8) < p 40 -
0.0956 < p 
8-95 a) Th-cr(! is ooe-.idence to support that tJ!.cdata a re oot oorma llydistributed. T~data points appear to fa ll along tJ!.e normal 
probability line. 
" 
Normal Probability Plot for Tar Content 
M. ~t!IXf'.AISSbO 
:/ 
' 
.. 
., 
" i: A~ 
.. 
" » 
10 
' 1/ 
b) ~%Cl ontt.emean taroontent 
"' .Y 
./ 
" 
.? = l.S.~9. s = 0.0.;66, n = 30 ta/ !!..n-1 = to.oo.;.:l:9 = .!!.756 
- s - . s 
X - Fo.005,29-,- < ,U < X + lo.Co)5.29-,-
vn · vn 
I •09 _ 2 1'60.0566 < .< J _29 _,_ 2 1"60.0566 .)~ . ) ,[30 Jl _ .) • . ) ,[30 
1 501 < ,u < 1557 
I 
" 
e) 99% prediction intena l on fl!.(! tar oontent for the ntXt sample that ,.,.;u be tested .. 
x- to.oos,19s J1 + ~ < x,~, < x + fo.cosJ9s J 1 + : 
MLEs'1r0~ 
M~" 1.5.'!1) 
SOW OM,1i 
1529- 2.756(0.0566)J 1 + 
3
1
0 
< x,+1 < 1 529+ 2.756(0 0566)J1 + 3
1
0 
1.370 < x11_ 1 < 1 688 
d) 9:>% tolerance inten11l on the \'3lnesoffl!.(! tar oontent ,.,.;tba 95% IC\\':1 of oonfi<k:ooe 
(j-h-ks) 
{l-529- ;3.j,.;O{o.o.;66). lS29 .a. 3.:}.50(0.0.;66)) 
h,33l),l.7l9) 
e)Tt:i'oonfidence inten'3l in part (b) is tOr t!ti' population mean and we may interpret this to implytJ!..a t 95'%ofsc.eh intm'als v.ili 00\W tt.e 
true population mean. for tJt.e prediction intm'al, 95'%of such intervals \\i ll oow:r s fl!tnreobset\l.'d tar oontent. for tJt.e tolerntxlo"! interval, 
99% of snclt intervals will 00\W 95% of tlt.;e true distribution 
S-97 9l)~oonfideooe interval on the population proportion 
n= 1600 x =8 P =o.oo.; '1.o.f~ =t.o.oo.; = ~.:;8 
- Jp(l p) < < - + p(1- p) p - z.n n - p - p Zat l II 
= :-=-::---,:-;:-::= 
0.005- 2.58 0.005(116-000.005) < p < 0.005 + 258 0.005(1- 0.005) 
1600 
0.0004505 < p < 0.009549 
b) & = o.oos,a = o.Ol . 'la,'~ = zo.oo.; = .:.>.:;8 
? ? 
Jl = ( 2«12)-p(l - p) = ( 2·58 )- 0.005(1 - 0.005) = 517.43 11 i! 518 
E 0.008 , 
e) E = o.oo8.a = o.01.1-af2 = zo.oo5 = 2.:;8 
2 [ ]2 n=["at2) p(l - p)= 258 0.5(1-0.5) = 26001.56 n " 26002 
E 0.008 ' 
d)A bonncionU..e trt.e popc.lation proportion rOOu.oes tlte requiraisamp!esi1.e by a su.bstantial a.mount.Assmplesiuof.;18 is mucl>. more 
reasonsble thana samplesit.eof ow:r 26.000. 
8-99 a ) TJ!.cdata appear to foliO'\'!· a normal distribution based on t1be oorm.al proOObilityplotsince iltedata fall along a straight line. 
b) lt is important tocf!.cck for norm.aJityoftJ!.edistribution uDdttl)ing thesa.mpledata sioct tbeooofideoceintervals to be constructed 
s.~.ould ~.a \l: the assumption of ootmalityfor the results to be reliable(especiallys.ioce the s.amp!esize is less than30 and ttoecentraJ limit 
tltrorem does oot a pply). 
c) 95%oonfidence intervsJ for the mt.'ln 
n = n .f = ~.73 s = 6~ t = O.O:i.S,lO = l2-2.8 
x - ro.01;,1o(:};;) < J1 < x + lo.o23,IO [:};;] 
22.73- 2 228( ~)< J1< 22.73+ 2228(~) 
18.478 < J1 < 26 982 
d) As \\ith part b. to construct a oonfidenc:e interval ontf!.e variance. tJte nonnaJityassumption must }!.old for t.J!.e results to be reliable. 
e) 95%oonfidenoe intmai tOr,a tia.noe 
n=n6j3 s=6~ 
., ., , ') 
x;;n,n- 1 = Xo.oo;,Io = 20.48 and Xi- i2,n-I = Xo.m,Io = 3.25 
' ' 10(6.33)- < o-2 < 10(6.33)-
20.48 - - 3.25 
19.565 < u 2 < 123.289 
8 ·101 a) 
P(x2 < 2>.T
1
_ <x2 ) = 1-a 
l _ g_ 21' " ?,· 2 ' 2' -
Tltena confidence interval for 
I . /1·=- IS ). 
b) n = 20. r = 10 .and t.heobs.en'Cd va.lneofT r is. l99 - 10(29) = 489. 
A9;%confid<:nodntervalfor- IS , = (28.62.101.98) I . (2( 489) 2( 489)) 
,\ 34.17 959 . 
8.103 a) n=lj2•{l :9/.lX94877/4).tt:.en•t=46 
bl (>o-sl/(94Bn/4l = h • pl/1>-pl 
p = 0 .6004 betv.·ecn 10.19omd 1041. 
8·!05 We would expllct that 950 oftt.eoonfidence intmoals would include tlte\'8Jueot p. Ttis isdne to9t.Jt.edefinitionofs oontidmce ioten'al. 
l£1: X bct the-number of intervals. that contain the true mean (!J). Wecao US(: the large sample approximation to dcterminetlte probability that P(l)30 < X< 
970). 
L · 950 o9 -o 930 0930 970 o9-o etp = 10oo =. ' PI =woo = . and Pz = 10oo=. ' 
Tr~ ,.,;,""' ;se,sti,.t<d by p(1 - p) = 0. 950(0. 050) 
II 1000 
P(0.930 < p<0.9JO) = P z < (0.970 - 0.950) - P z < (0.930 - 0.950) 
0.950(0.050) 0.950(0.050) 
1000 ~ 1000 
= P(z < 0 ~~~92)-P(z < O~g~~~2 ] = P(Z < 2.90) - P(Z < - 2.90) = 0.9963 
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