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CHAPTERlO 10-1. a) 1) n.e parameter of interest is the difference in meatl.i 111 -112_. t\ot.-: that Ao = o . ~) ~:Jil-JI~=OOt#l=l/2 .)) H1 : Jt1-p2 tOoq:1 *II:~ 4) Th.e test statistic is .;) Rl.iect Ho ib.o < -Za,''J = -1.91)ono >'1-o.f'J = 1.¢fora = o .o.;. 6) X t=4.? X2 =7.8 0")=10 G'J=.S n1 =lO n:~ =l.; z = 0 (4.7 - 7 .8 ) (10) 2 ( 5)2 • l"-1-<0- +-1-5 - 0.9 7) Conclusion: &r.au.se -1.96 < ~.9 < 1.91). do not reje::t th;e null ~.ypoth.esis. There is not sufficient evickooe toocmclll& t~.s t 1M two means differ a t a= o .o.;. P-valu.e = 2(1 - ¢(0:9)) = 'J{l-0.8159 .;o) = 0..368 b) (10)2 (5)2 (10)2 (5)2 (4.7 - 7 .8 )- 1.96,F--'-+-- < JL1 - IL? < (4.7 - 7.8) + 1.96,1"----'-+--10 15 - 10 15 With:9.;~oonfidencs:. tf!.e trcediffereooc in tJt.c means is between -9.79andj..;9.1k>con:s.e:t.ero is contained: in this interval. "'-eoooclu-dt: ~.here is no significant difference hm~·een the means. We fa il to reject the nnllll)potf:.esis. 6. -- 6. 6. - 6. c) .8 = z a/2 0 -z t:J2 - 0 (12 (12. (12. !12 - ' +....1. -' +....1. "• "z "• 1)2 3 3 = 1.9 6 -r=....;.== -- 1.96 - r=....;.== (10)2 (5)2 ( 10/ ( 5)2 .1~~+-- +-- 10 15 10 15 = <I{ 1.08 )- <I{ - 2.8 3) = 0.8599 -- 0.0023 = 0.86 Po"'·e:r = 1-0.86 = 0 .14 d) Assume the samplcsi7.esareto b.-:tqual, use a= o.o.;.j~ = o.o.;.,aod6 =.) ( )2 ( 2 2) ( )2 ( 2 2) ~ Z012 +z,u1 +a2 _ 1.96 + 1.645 10 +5 , _ - -- - - 85 - 6 2 (~2 lO·j . a) 1) TJ!.e psrameter of interest is t he difference in meansp1 - 112· Note that .!l.Q = o. ::!) Ho:pl-#::!=OOt#l=f.l::! 3) Hl:JIJ -#::! > Oor pl>ll:J. 4) The teststatistic is (x1 -x?) -~0 z = ... 0 z 2 a, az - +- , , "2 5) Reject Ho if zo > 1.a = 2j!S for a= 0 .01. 6) X1=~S X2 =~lj 0)=10 02_= 5 BJ=lO ll2_ =15 z = 0 (24.5- 21.3) = 0.937 (10)2 (5)2 ,/"-'.:..!.-+- 1 0 15 7) Conclusion: lk\':acse 0.937 < 2..J"2.5. we fa il to ti'ject the null hypothesis. Tf!.ete is not sufficient f:\idence tooonclude that th.el\\"0 means differ at a= 0 .01. P·\'aJue = 1 - 41(0.94) = 1 - 0.8 :!.64 = 0 .17 j6 b) " - " >(x - x )-z ,.., ~""2 - I 2 o (10/ (5)2 fL1- fL2 > (24.5- 21.J)- 2.325,/"-'=-+--10 15 , .. - /Lz ;::: - 4.7 4 With99%oonfidence. tf!.e trued.ifferenoe in t.Jt.e means is greater than- 4 .74. Because 0 isoontaiood in this intet\oaJ. we fa il to reject ttt.c null hypothesis. c) {J = 6 2.325- 2 z = " 2 2 (10)2 ( 5/ a 1 a 2_ - +- +- "• 112 10 15 = <1{!.74) = 0.959 Po\'l(!f = l- 0:9() = ().()4 d) Assumethes.amplesizesareto bet:<rual...sea = 0 .05. ~ = o.os.and~ = 3 " "' (•., +z,tH -taiL (!.645+1.645t( lo2 +52L 339 - 6 2 (2)2 10-.;. a ) 1) Tlte pa101meter of interest is the di fference in breskingstrengths111 -112 and Uo = 10 l) fi():S~1-1'2=10 3) H1 :111 -~>10 4) n.e test statistic is .;) Reject H() if zo > :za = 1.64.; for a= o.o.; 6) X 1 =16l,s X <~=l.).5 .0 S:tO OJ=1.0 02=1.0 n1 = 10 ~= 12 7) Conclusion: B«:ause -.;.84 < 1J>45 faiJ to reject IJ!.e null hypotlt.esis. TJ!..ere is insufficient evideooe to support t!>.e use of plastic 1 a t a = o.o.;. P·\'Sine= 1-<1>( -.;.84) = 1-0 = 1 b) ( 1) 2 ( I/ - +- 10 12 c) 13= 1.645- ( !2-IO) =<I{ -3.o3)= 0.0012 ~ '{10+12 Power= 1-0.0012 = 0.9981) (z ofz +z,/(a~ +a;) ( 1.6 45+1.6 45)2(1+1) d) " = = 5.42 - 6 (tl-tli (12 -1 0)2 Yes. t!>.esamp!e sir.e is adequate 11)·7· X l = 89.6 X 1 = 92-5 ? 2 uj = 1.5 u 2 = 1.2 n1 = 1.5 n~= ~o a) 1) Tlt.e parameter of interest is tltediffereooe in mesn rood octane number Ill -11~ and llo = o 2) Ho:!Jl-112=00t}l)=/12 j.) Hl: !Jl-li:!<OOr iJl </11 4) The te.ststatistic is .5) Reject Ho if t.o < -:za = -1.645 for a= o .o.s 6) X1 =89.6 X2 =92S ? 2 n j = 1.5 n 2 = 1.2 •o = (8 9 .6 - 9 2.5) = _ 7 .25 ~ v~+w- 7) Conclusion: Be::at:.SC -7.2.5 < -1.-G.t5 reject tf>.e nn11 l!.ypotJ!.esisandoonclu.de t.Jt.e mean road octane nt:.mber for formulation 2 o:oeeds that of formula tion 1 using a= o .o.:;. P-valne• Jlz s -7:~..;) = 1- Kz ~ 7:~.5) = 1-1 "'o b) 9.s%oonfidence intrn'81: (8 9 .6 - 92.5) - 1.96 ~·~ + ~·~ ~ '"• _ ,'2 ~(89 .6 - 92.5) +1.96 - 3.684<,,, - JL2 < - 2.116 1.5 1.2 15 + 20 With 9.s%oonfidmce, the mean road octane munber for fonnnla tion 2 exceeds that of formulation 1 by between 2.116and j.l,&t. c) 9.5'% IC\<tlofoonfidence.E = l.aod7<0.02.:; = 1.96 n-[•o;25 r (u~ +ui)= [ 1.~ 6 r ( 1.5+1.2) = 10.37' 10~. x 1 = 6.5 .~-2 Ot=S n1 = 10 Catahost 1 Catalvst ::!. X~ =6842 02_ =3 ll2 = U') a) 95'%oonfidence interval onp1 -112• tl'..ediffereooe io me<~nactivecoooentration ( ) u 2 a 2 ( ) a 2 a~ X - ~ - z - 1 +_l_:::; JL - fJ• < X - X +z - 1 +- · I .... a/2 11 11 1 2 I 2- ol2 11 1 I 2 I 1 2 (6 5.22- 6 8 .42) - 1.96 (3) 2 + ( 3) 2 < P, - ,.., < (6 5.22- 68 .42) +1.96 (3)2 (3)2 - +- 10 10 - 10 I 0 - 5.8 3 ~ fL I - fL2 ~ - 0.57 Weate95%confident t~.a t tlte meaoactivecoooentra tionof cstai)'St 2 exceeds that of cataJy~t 1 by tx:tv.·reotJ.s7 and5.&Jgfl. P·\'Siu.e: (65 .22- 68 .42) 32 32 - +-10 10 - 2.38 Tlten P·valn.e = 2{0.008656) = 0 .0173. b) Yes. because tlt.e 95% oonfidenoe interva.l does not oontaio the vaJtle zero. We oonclndc that t.Jt.e mean acthoe ooooentration depends on tJt.;e ~.oi~of catalyst. 96 ( 5) 96 (5) I. -...:~::"" I - I. - ?.;.;.:.."::"" 32 32 32 32 - +- - +- 10 10 10 10 c) fi = = ~- 1.77 )-~ - 5.69 )= 0.038364- 0 = 0.038 364 Power= 1-P = 1 -O.Oj83f>4 = 0 .9(>16. d) Calculate thevaJueofn nsing a and~. (1.96 + 1.77 t (9 +9) ( 5)2 I 0.02, 'O.crd-Ore.10 isonlysJig.htly too ft'lo·ssmples. n.e samplesi1.esaread(lQ.nate to detect t lt.ediffe«:nceof 5· n.;edats from tft.;e first sample n = 15 a ppear to be oorma.llydistributOO. .. "' ., ... "' c "' ., e ., ., ., c.. .., "" ., 10 s 1 I I -;£----~----±- ____ L ____ L-:-~-= I - I I • I ' ' ' ··-+- if I ·r .--- ---/ :·.'·§---~ . - - ; / r- ' I -u I 700 75) 'O.edata from tJt.;eseoondssmplen = 8appear to benonnallydistributed 99 --.. 1-1 --·--+1 .. ·--.. -!,- .. --1 .. __ L~I .. ___ ·/-/1-+---1 90 _.. ·--+ --l./'_+1 ----1 ==_::§~-~~:.;i. .. :~. .. 1 ..- so ~ 70 !! so ., so c.. .., : =:1 -;/- ::· __ 10 -·l-- .... _ +-1 r--/ / . s =:V-~:.· :+-1 100 Plots for bothsamplesaresf!.o\' .. -n in the fol!o""in& figl!re. 99 9S 90 ~ 80 Q) 70 u 60 ill so "- "" "' 20 10 s ·--1-----~--- I i i /1 ' I ' =:1== :=t= / ' I =f-:;2/ I I ----- -·--------- .,t- --j--- - ' .y' ;. ' --·-------- ' --f.:-- =r ·--;T • ----- • . ·---------- -/ ' • --~-- /~ ·j I ·------- - - ' . / . ' --,--7- . /; ' I --1- --4- 6S 7S sectiOn 1~1 10-1 I. a) 2 ' '2- 5.28 ' Tl:~9.;% upper one-sided oonfidmoe interval: ro.o.;.26 = 1.7 06 ~l- ~!,>_ s -lJ>BSo P-vslll~ = P(r < -j.OO): o.ool.) < p.,'aJtle < u .oo.; Tlti.s i.soOMi&d test because tl>.e ~.ypotl'..ese.saf't': mu1 - m.u = o \WS'tl.S less th:ano . b) Bectn:.seO.OOl,S < P·\'tllu.e < o.oo.; tl':.e P·\'tllu.e <a= o .o.;. TbcretOrt. we reject tl".e null f!.)'J)Otltes.i.sof mlu- mt.~ =oat th.eo:o.; or tl':.e 0.01 I£'\'CI of significance. c) Yes.. t.f!.e .sample sumda rd&viations are quite different. Consequ.ently, one would not w·ant to assume that t~.e pnpulation varia noES are equal. d) lftltultemathoe hypothesis were changed tomw- ml!2 ~ o. thent.J!.e P·\'altle = 2 .. P{r < -j.OO)ando.oo.; < P-vshte < (LOl. Because the P-valtle <a= o .o.;. wereject the null hypot.J!.esi.sofmn:l - mm. = o attlt.e o.o.; le>.-d of significance. 10·13. a) 1) The parameterofinterest is t~.edifferenoe in mean.s.111 -If.:!. with A() = o 2) f-lo:#1 -~:!=0or~1 =~2 3) H1 :p1 -~2 <0orp1 <Jb.! 4) T~.e testsUttistic is 6) x, = 6.2 2 s, =4 n, = 15 (x1 - x,) - ~0 I = • 0 I I s - +-p 111 IJ2 • s; = 6.25 n,. I 5 s = p = to= (6.2 - 7.8) =- 1.94 2.26~ I + I I 5 15 ( 111- l)st +(112 - l)s; n l +n2 - 2 14(4) +14(6.25) = 2.26 28 7) Cooch:sion: Ba':ausc- 1.94 < - 1.701 reject tf!.e null ~.ypotlwsis at tlteo.o.; le>.el of significance. p.,oaJtle = P(r > 1.94) O.Ol,S < P·\'tllu.e < o.o.; b) 9.;9Gconfidt':n0o'! interval: ro.o.;-28 = 1.701 Because7.ero is not contained in this interval. weare9s9ticonfid"ent thatv1 < WJ. c) .:1= j Usespasane.stimateofo: d= p.2- J•, 3 = 0.66 2sp 2(2.26) • Usiog Chart Vtr {g)witlld = o .66aodn = n1 = ''12 we get n = !.'.n - 1 = :19and_ = o.os. T~.erefore. ~ = o .o.; and t~.epnweris 1 -IS= 0 .9.; d 25 0 55 • "*+1 21 d) ~=o.os, = = . .T~.ereforen •40a.ndl1 = :: .Titus.n =n1 =ltlt = 21. 2(2.26) 2 • lO·lS. a) 1) The paramcter of interest is t.f!.ediffereoce in mean toddiamcter .p1 - 112· 2) Ho : #l-~~ =oorpl=#1! 6) J) Hl:#l-#2l<OOt Jll t#2 4) TUt;>.Ststatisticis ( x - x ) -tl f = I 2 0 0 I I s - +- p n, " 2 .5) Reject t ile null }!.ypofl!.es.is if to < -ra{~.nr ... •~-2 where - ro .02.),30 = -2.042 or to > ta{~.n1 ..a. ,~-2 where ro .O~jO = 2.04!!- fora= o.os x, = 8.73 2 03' s, = . ) n = LS I si = 0.40 n,= 17 s = p = t = (8 .73- 8 .68) 0.230 0 I I 0.6 14 - +- 1 5 J7 ? ? ( n1- l)sj +(n2 - l)s; n1 +n2 - 2 14(0.35) + 16(0.40) = 0.6 14 30 7) Conclusion: Because. -2.042 < 0 .2;30 < '1.041!. fail t o r-eject t.J!..e null ~.ypothais. n.ereis insufficient t'Videoce tooonch:<k that tlte two ma<:hint.S prodncedifferent mesndiamctcrsat a= o .o.;. P·vsln.e : 2P(r > cUjO) > l (O.t$0 ). P·t 'Sin.e > 0.80 b) 9s%oonfidence inter\'al: to.Ol.j.JO = 2.04~ (8.73- 8 .68) - 2.042(0.6 14)1 1 + I :0; p1- p2 <(8 .73- 8 .68 )+2.042(0.643)1 I + I 15 17 IS 17 - 0.394 < '"•- '"' s; 0.494 &r.au.se zero isoontained in this interval. t~.ere is insufficient evidence tooonclu.® th:a t th.e two machines produ~ rods '~i tbdiffermt mean diameters. 10·17. a) 1) Tt:i' parameter of'interest is the difference inmeancataJyst )ield,p1 - ~~. '~itb 6o = o. ::!) Ho:pl-#1!=0 0t#l=#::! j) H1 :p~-~~~ < OOt#l <Jli 4) n.eteststatisticis 6) ' -' ' 2 - ( x1- x,) -tl0 I = -0 I I s - +- p n, "2 (86 -89) t 0 = -==--~.:.....- 1 I 2.4899 12 + 15 - 3. I I s -p - , ? ( n1- l)sj +(n2 -l)s; n1 +n1 - 2 I 1(3)' + 14(2)' • 2.4899 25 7) Conch:sion: Because <Pl < - 248.;, reject tt.e. nllll hypotJt.esisand oooch<.de that tlu:mean)ield of catalyst 2 e.xcoods tf!.at of catalyst Jata=OiH. p1 - ~~~ s -0.60,lor t:quivaJentlyp1 "'0.60j s II-:! Weare(}:)?f.oonfident tl>..at the mean yield of catalyst 2 o:C«'ds th.at of rutaJyst 1 by a t least o .60,l units. .999 i!:- ·99 ;.; .9S .c r: .20 .OS .Q1 .00 10-19. a ) A.coording to th.e oormaJ probabilityplots,lh.eassumptionof oormaJitya ppears to reasonable bccasll(: the data from both tlte samples fa ll a pproximatelyaJong a straight lioe. Tlbeeq11:alityofvariancesdoes not a ppear to bese\l!tcl)'\iolatcci cithetsince the slopes aN' a pproximately tltesame for both samples. Normal Probability Plot 9.5 10.0 10.5 solution Ard~ HCIII'!uhy l e-:n ............ ., .. P.V.11uc 0.595 .99 ' ~ ' .~ .99 .95 -.c .eo .. .c .so ~ .zo .OS .01 .oo 10.0 A'oWage<: 10 •• StDe'.r: 0..230940 .. 10 Normal Probability Plot 10.1 HU tQl tO.~ tO.S 10.6 10.7 solution Nd~~r.~C~n<~ Homu llty f~ A>SoJ.weii: 0 ..2 11 P.VAte:: 0.8011 b) 1) The parameter of interest is t.J>.edifference in mean etch rate,Jl1 - Jl~. with:6() = o. 2) H():Jll-W~=OOrJll"'ll2 :J) H1 :Jll -#:! "*OOrJ11t:IJ2 4) The test statisticis (x1 -x,)-~0 t = -0 I I s - +- p tt l 112 5) Reject t1:e nu.JI hypotJ!.esis iff() < - ra,lvt1.a.n2- 2 wltere -ro.02.5,!8 = - :1..101 or ro > taj~Jtl'-:~-2 wl'teri! ro.o2.5.18 = 2.101 for a= o.os. 6) x, = 9.97 . , - 0.422 X = 10.4 2 s = p 2 2 ( 111 - l)s1 +(u2 - l)s2 111 +n2 - 2 - 9(0.422)2 +9(0.23 1)2 -0.340 I 8 I = (9 .97- 10.4) =-2.83 0 I I 0.340 - + - 10 I 0 7) ConclttSion: Becau:se -:1..83 < -2.101 reject t~.e null hypothes.isaoclocmcludet.!'..e ""'O machines mean etch rates differ a t a= 0.05. P·\'alue = ~F(r < -2.8 3) ~{0 .005) < p., oaJne < :l{O.O!O) = 0.010 < P·\'3lti.e < o.oao c) 95%oonfidence: inten11l: to.o2.5.18 = ~1..101 Weare95%oonfident that tl'te mean etch rate for solution i exceeds t!".e mean etch rnte for solntion 1 byb.."""·eeno.ll05 aod 0.7495· 10-:h. a ) 1) Tlte: parameter of interest is the difference: in mean melting point.#1 - #1· '"ith~ = o 1) Ho :#!-P'1=0orpl=P2 6) b) 3) Hl:Jil-#2~oor#l'*#2 4) Ttu: test .statistic is .;) Reject tl'tf: nu.lll;ypntt.es.i.s if to < -raj2.•t1.a.r:2-1 wltere -ro.oo~4o = -2::011 or to > tof2.n1 ... n2:-2 where ro .o~4o = 1.021 for a = o.o.; x = 420 I s = 4 I n, = 21 X = 426 2 ' s = .) 2 n,= 21 s = p = 10 = (420-426) =-S.498 3.536{ I + I 21 21 2 2 ( 111- l )s1 +( 112 - l )s2 n1 +u2 - 2 20( 4)2 + 20(3)2 = 3.536 40 7) Conclnsioo: Because -.;498 < -2.011 tejet'l the nulll;ypot.he:.si.s. TheaJioysdiffer .significantly in mean melting point a t a= o.o.;. P·\'Sh.-e = 2-P(r < -.5498) P·\'SJne < 0 .0010 ' _ J_= 0.37 5 2( 4) • Using tlt.eappropriatech:an in tf!.eAppeodix. v.i ttt ~ = 0 .10 aod a= o.o.; we haw: n = 1.; . • 11 + I Tf!.erefore:,11 = 38. n1 = n2: = .).8 2 u)~. a) l) Tr.ep<~tilmeterofintcrest is t.bediffereooein mesnwear amou.nt.#1 - w1• withUQ = o :!) Ho: Jll- Jt:!. =(I Or #1 = jl:!_ :}) Hl:Pl-Jl:!+.OoriJJ!#J/2 4) Tltet e&tstatis ticis 6) X 1=:!0 X 2=15 SJ=2 S2_:8 n1 = :i5 02= :i.5 [}, +s; l2 'Jt "z 1/ = ---'7---''-'--- [}, ]2 [s; l , , + 112 n -1 n -1 I 2 1/ ~ 26 (tJUncated) 26.98 3.03 7) Conclusion: Beca~3.03 > :!..056 reject thencll hypotf!.esis. TIM: data support ilteclaim tf!.a t tlte twooompanie& prodl!Ct material ~ithsignificantlydiftertnt wear a t tlt.e:0.05 IC\l:l of significance. P·valu.e = 2P(t > 3.0J). 2{0.00:i,;) < P-vah:.e < 2{0.005). 0 .005 < P· vaJu.e < 0.010 h) 1) Tltep<~rameterofinterf:.St is tf!.ediffereoce in mesnwear amollnt,J11 - ~· ::!.) fi(.l : ,:l-J/2=0 .J) Hl:JI!-#2>0 4) Tlte te&tstatis ticis 5) Reject tf!.e nl!ll hypothesis iff() > r0 .05.27 w~.ere r0 .o5.26 = 1.706 for a= 0 .05 since 6) x 1 ='2-0 x -2. =15 SJ=2 S;i_=8 t = 0 7) Conch:sion: Because 3.03 > 1.706 r-eja.'1 the null hypothesis. Tltedata support lite claim that tf!.e material from oompany 1 bass higha meanwesr than tf!.e material from oompany:! a t a 0 .05 IC\'tl of sig:nificaooe. c) For part {a) use a 95'% two-sided oonfidenoe intenal: to.02.5;:!-6 = ~.056 ., ., ., ., (x - x )- t I 2 Cr.v s· r s" r - 1- +...1. < JL - JL < (x - X ) +t _I +...1. Ill ft 2 - I 2 - I 2 (.).v 111 IJ 2 (2)2 (8 i (20 - I 5) - 2.056 25+25 ~ JL1 - l"z <(20 - 15)+ 2.056 1.6 09 ~ /LI - JL2 < 8 .39 I FOr part {b)use a 95'% lower one-sided oonfidenct inten'31: r0 .05.-26 = 1.706 (x - x )- t I 2. a.v (2t (s t (20-15)-1.706 - + - < JL - JL. 25 25 - I " 2. I 8 6 < 1'1 - 1Lz (2)2 (8 )2 - +- 25 25 For part a )we are95'%oonfident the meanabtt~sh-ewesr from oompany 1 exceeds the meanabrashoe wear from oompany2 by betwf!f:n l.f)()9and 8.391 mgjulOO. For part b)wesre95%oonfident the meanabrasl\oewesr from oompany 1 e.xcteds t.J!.e mean abrssh't"-'ear from oomp<~tl)':!. by a t lesst :u86 mg/1000. 10·:!.5· a ) 1) Tl>~ parameter ofintcrest is tJ!.;ediffercnoe in mean \\idthoft.J!.e back.sided.ip~uts for fl!.e sing.lespiDdlesaw process \l!l"SllS tJ!.;e duaJ spindle saw pt()()('.SS,p1 -IJ'J_ i) f-fo:pl-Pi=OOt lfl=P1 j) H1 :p1 -IJ.i 1:0orp1 *~ 4) Tlte teststatistic is .5) Reject tlte null hypotl:'i'Sis if I() < - raf2,n1 ... n1-!l: w~.ere -ro.o~.:!8 = - 1.048or to > taju11 ... n1-:l wf!.ere: ro.ol.:;,18 = ~.048 fora= o.o.; 6) x, = 66.385 x2 = 45.278 = 14(7 .895)2 + 14(8 .6 12)2 = 8 .26 28 n = 15 1 n,. 15 t = (66.385 -45.278) = 7 .00 0 ~ I I 8.26 - +- 15 15 7) Conclusion: Becsc.se 7.00 > ~L048, we reject tt.e. nllll llypotft.esisat a= 0 .0.:;. P·va.lue• «) b) 95%oonfi~ intmoaJ: to.0:!..,:;.:!8 = 2.048 · HI HI (6 6 .J8 5- 45.27 8)- 2.048(8 .26) - +- :::; ' '•- /.'2 <(6 6 .38 5- 45.27 8 )+2.048(8 .26) - +- 15 15 15 15 Because r.ero is not oontainodintlt.is interval. weart9.;9Goonfident that the means are different. 15 • 15+1 c) For ~<O.ouoo d = 2 {8 .26 ) = 0 .9 I,,.,.itlt:a =o.o.;thenus.ingCh3rtVtl{e)wefindn >l.:;.Tft.enn> 2 = 8 10·27. a) 1) Tlte parameter of interest is t~.edifference in mean rmmber of periods in a sample of :!00 tra ins for twodiffetf:nt lt\~sofooise wltage.lOOmvand 150mv. 111-111-v.ith~ = 0 i) f-io:pl-~=00tl)l=IJ2 $) H1: 111 -112 > o or #1 > 112 4) Tlte test statistic is s, = 2.6 n, = 100 n,. I 00 to= (7 .9 - 6.9) 2 , I I .) - +- 100 100 = 2.8 2 99(2.6)2 +99(2.4)2 = 2.5 198 7) Conch:.sion: Because 2.8:! > l.S-:!6 reject th;e nuJJ hypothes.is at tft.eO.ol l t~.oel of significanct. P·\'8Jtle = P(t > 1.8 2) P·\'altle• O.OOl.:; b) 9!}%oonfidenoeinte:rvsl: to.o1,198 =1~26 ,,,- ,,, >(x1- x2)- t . _2(s) - tr.lll~! p "• _I +-1 "2 ''•- ''2 ;::: 0.1 7 8 Because7.ero is not oontained in this intervaJ. weare99%oonfid"ent that mean 1 o:cood:s mean i . 10·29. a) Th.edata appe3r to be oonn.allydi.stributodaOO t.lte\oariaooesappear to be approximately equal. Tlteslopesoftlte line.; on tt,e normal proba hi !ity plots are aJmo.st tlt.e .same. Normal Probability Plot for Brand 1 ... Brand 2 ML Estimates • .. 95 1 l "' -1-· ., -c "' "' ., v ~ "' so CL. .., "' "/"+- ., • 10 s • i r 244 254 '" 274 ,., Data b) 1) Tlte parameter ofintcre.st is t~.edifference in meano\'f!taJl distanct.J11 - #1. with!l()-= o. 6) 2) f-to:J11-1J.2=0orp1=#2 3) Ht :pl -#2 1o0orp1 *lt1 4) Tltete.st statistic i.s x-, = 275.7 s, = 8.03 n, = IO x2 = 265.3 s2 = 10.04 n, = 10 ( x - x ) - 6. t = I 2 0 0 Rl s - +- p n 1 n 2 s = p 2 2 ( 111 - l)s1 + ( n2 - l)s2 n1 +n2 - 2 = 9( 8 .03l +9( I0.04l = 9 .09 20 10 = (27 5.7 - 26 5.3) 2_558 9.09 - 1 +-1 10 I 0 7) Concll!Sion: Eltt..!tu.se 2,s.;8 > 2.101 reject tile null }!.ypotf!.esis. Tt.e.data support tft.eclaim that tt.cmeansdiffer at a= o .o.;. P-vsJu~ = 2P(r > 2,s.;8) p.,•ahte" 2:(0.<H) = 0.02 C) ( - _) I , I < <(- _) I I .Y-X -( S -T- JI.-J_I.. X-X +t S -+-1 2 cu•p - 12 - 1 2 u.v p ~ ~ ~ ~ (275.7 - 265.3) - 2. 10 1(9 .09)~ I + I <Jlt - 1'2 ~ (275.7 - 265.3)+2. 10 1(9.09)~ I + I 10 10 10 10 5 d) d = 0.27 5 2(9 .09) 3 e) ~ = o.~ d = = 0.16 5n• = 100 Tlterefore.n = .;1. 2(9 .09) U)-31. a) 1. TJ!.e_pa.tilmctersofinterest aret lt.e mean current {note: set circu.it uquaJ to sample:! so tt-.a t Table X can be u.sed. Tlterd'oro:p1 = mean of circuit~ aDd Jl:! = mean of cir..~i t 1). :!. Ho : ~~ = #1! 3· H1 : #1 > #1! ( n1 +112)(n1 +n? + I) 4. The test statistic is W2_ = 2 - w1 • • 5· We rtjo::t Ho ifl~ ~ w 0.02.S = 51, because a= o.o~ and: n1 = 8andn2 = 9.Appendix A. Table Xgi...-s tl.ecritieaJ vaJne. 6. w1 = 78and W:! = 75 and because. 75 is less th:an51 fa il to reje::t Ho 7. \.onclusion, fa il to reject H(J . Tt.ere is oot t!OOllgh evidence tooonc:lcdeth.a t the meso of' circuit 1 exceeds the mean of circuit 2. b) 1. 'O.e pal'ametersofinterest are the mean image brightnes.softJ!.etwo t.-bcs. :!. Ho : #1 = 112 :). H1 : 111>~ H~ - P.w 4. n.e testsUttis ticis Z0 = -'----'L (f ... ' 5· We rtjo::t Ho if Zo > ~.0:!.5= 1.96 for a= 0 .0-:15. 6. w1=78.#w1 =7:!anda!. = 108 78 - 72 ' z = 0.58 0 I 0.39 Because Zo < 1.¢ . fa il to reje::t HtJ. 7. \.onclusion: fa il to reje::t H(J. Tlt..."re is not a significsnt difference in t.J!.e f!.ea t gain tOr the heating units at a= 0 .05. P-valn.e= :!h- P(Z<O,s8)) =0,S6l9 ul·JJ. a ) 1. Tlte para.mctersofintere&t are lite me<~n ~!.ea t gains for ~!.ea ting units :!. Ho : 111 = #:! 3· Hl : #!'t.-#1! ( 111 +112)(111 +n2 + I) 4 . n.e test statis tic is w2 = 2 w, . • 6 . We rtjo::t Ho ifW ~ ·w O.OJ = 78, because a= 0 .01 andn1 = 10 andJlfl = lO,Appe:odixA. TableXgh oe.s tbecriticaJ vaJu . 7. w1 = n aDd W:! = 1,33and becal!SC77 is less t itan 78. wecanrejo::t Ho 8. \.onclusion: ri:je::t HtJ andoonclndetl>.at fl!.ere is a significantdiftCr-mce intl>.e heating units at a= 0 .05. b) 1. Tlteparamctersof interest are the mean ".('..'I t gain for heating units. :!. H1 !#1•#2 3. Ho :#1 =#2 w,-1" - ...:..__.:."'cc• 4. n.e testsUttis ticis Z0 = (f ... ' 5 · RO!jo::t Ho if!Zol > Zo.02.:;= 1.¢ for a= 0 .05 2 6. wl=77.#w1=l05and 0"w = 175 z = 77 -105=-2.12 0 I ' 2' J . J B<x:allS< IJ;,I > >.¢.reject H0 ' 7. \.onclusion: ri:je::t HtJ andooncludetl>.at fl!.ere is a difference in IJ!.e hestgain for fl!.e f. .eating llnitsa t a= 0 .05. P-valn.e = 211 - P( Z < 2.19)) = 0.034 10-:).5. a) 1. Tlte parameters of interest an: the mean temperatures. :!. Ho : 111 = #:! 3. H1 !JIJ tf.I:! ( 11 +11 )(11 +n + I) 4. n.e testsUttis ticiSl-1'2 = I 2 I 2 W I' 2 5· Wereject Ho if W ~ w ~.OS = 185. becansi:a = 0.05and n1 = 15 andn::! = 15.Appendix A. TableXgh~ t.J!.ecriticaJ vaJuo:. 6. w1 = :!.58 and ILI2 = :!07 and brr.au.se both 2.:;8and 207 are greater than 185. we fail to reje::t Ho. 7. \.onclusion: fa il to reje::t f-lo . Th.o."f'eis not a significsnt difference in t.J!.e mean pipeddlectiontemperatllteata = 0 .05. b) 1. 'O.e pal'ametersofinterest are the mean etch l'il tes. :!. Ho :111 =112 :). H1:111•~ w,-1" - ...:..__.:."cc• 4. n.e testsUttis ticis Z0 = u,., 5· We rtjo::t Ho ifl7"()1 > ~.o::!5 = 1.¢fora = 0 .05. 2 6. w1=73.#w1=l05and O"w1 = 175 7 3-105 z0 = - 2.42 13.23 Because !Zol > 1.96. ri:je::t Ho. 7. \.onclusion: ri:je::t H(J. Tlt.ere is a sig.nificaot <iiffercooc betwetn tlt~ meanrtcb rates. P-valu.e = 0 .0155 SEctiOA 1~ 10-37. s) d =O.::qj8 Sd=1).1j.5l.n=9 9.;%confi denoe in1m'al: d _ f [ 5 d l < L < d +t [ Sd l ,,n,n-1 T, - 1 d - '<>i2fl-l T, 0.27 38 - 2,306 [ 0-~ I] < I'd ::; 0.27 38 + 2.306 [ 0·~ I] 0.1699::; l l, ::; 0.377 6 With 95% confin'enoe, tJ!.e mean s J!.ea r strength of Karls.TUJ!.e mcthod exceeds tt-.e. mean sho:.u strtngtf!. of tft.e l~igh mrtttod by between 0 .lf,c'J9 a 00 OJ:n6. Becsuse7.ero is oot incln<i.."'<l in this intervaJ. tJ!.e intervaJ i.soons.istent \\ith rejecting the nuJJ hypotr.es.is. f.t!.at the means a.reequ.al. Tlte9.;'%confidence intm'al is directly related to a testofl\ypotl'.es.is \\it.ho.o.; !ew:l of signifieanoe and thecoodusions rescftedare identieaJ. b) [tis only oeoess.sry(or the differences to be nonnallydistributed for tJ!.e paired r-test to be a ppropriate and riliab!c. Tbetefore, t.J!.;e t-test is appropriate. Normal Probability Plot ' 99 ' ' • ' ' . ··:· ---· ~-- ·---~--- ·--~ ------~-1 .99 . -:·- --- •• -:---.--. -r ---.- --:··----. n c .95 ·-.--------.--------,--------.--·-:.;.;;·1 = ' '. , __ .--, :0 .80 ·-;~------...-------,----~-~-· ----,-1 E ' • . J.!-- I I 2 .so --:---.- ---~~-:-------:-------:-1 ~ .20 --.~;------···------,-------, ] .OS • -lt · --- ·- -:-- · · -- · -:--- · · • -~- • · --- · ~-~ • ' + ' ' .Ot ·-····--·--.------- .. --------.-··---- .. -~ ' ' ' ' . . 00 ·-·-----·--1··----- .. ·------ ... ------ 1• I I ! ! ! 0 .12 0 .22 0 .32 diff 0.42 0 .52 10-39. d =868,l75 S(i=l.29fl,n=8wheredj =hranch-brand :l 99~ confid-ence intm+a!:d- t <>12 . .-1 [ J,} fLd < d+tan ... -l [ 1] 8 68 .J7 5 - 3.499 [ lls-O ] < /Ld < 868.37 5 + 3.499[ Jls-O l - 727.46 ::; ! ,, ::;2464.21 Because this confidence intrn'8l oontains 7.ero. there is oosignificant difference bctv.·ren the two braodsoftireata 1%significance le.,.el. 10-41. a) 1) Tl'oe parilmeter ofirrt::rest is t.r.edifference: in bloodc~.olesterol level, Pd. wh.eredi = Before-After. -2) fi() : IJd=O 3) H1 :Jid> O 4) Tlte t<Ststatistic is .;) Reject the null hypothesis if to> ro.o,s.14 wl:iTe ro .o.;.14 = 1.761 fOra= 0.05 6) d = ~6.867 t = 26.867 = 5.46 5 0 J9 .04 / .JI5 7) Conclusion: Becaus.e.;;46.; > 1.761 rej«1 t.t'..c null 1\ypothesis. The data support tlteclaim that t~.e mean difference inc~.ol ei>terol IC\'els is significantly less after diet and anaerobic exercise progr-am at t.r.eo.os IC\'el of significan::e. P·\'ahte= P{t > .:;..;6.;) • o h) 9.;~confidence interva l: d -lo . .-1[-t,} fLd 26 .867 - 1.76 1['~] </Ld I 8.20::; 11, Bee:.<~ use IJ!.e lower bound is pos:iti\'e, "'ith9.;%oonfidence IJ!.e mcandifferenoe in bloodc~nle.stcroiiC\'f:l is significantly less after the diet and aerobic exercise program. 10-43. a) 1) The parameter ofintct'f$t is tltediffere:oce in meanweig.ht ,Pd w~dj = Weight Before- We:igltt After. 2) ff,) : IJd=O 3) Ht : lld. >O 4) Tlte test statis tic is .;) Reject t~.e nuJ I I;ypot~.-es.is if to > ro.os.9 wit-ere ro.o.;;~ = 1.83$ for a= o.o.; (,) d =17 17 t0 = r::- = 8 .387 6 .4 1/ v JO 7) Conclusion: Because 8.387 > 1.8$;3 rej0::t the mill hypothesis and oooclnde th:at the: mean weigltt loss issignificantlyg.reater thsn 1.ero. Tbat is, tJ!.cdata support tt.eclaim that t llis particular diet modifica tion program iseffecthl!: in ri.'dlt.cing. weigbt a t th:eo.o.; leo.'f:l o( signifies nee. b) 1) Tlte parameter of in1erest is t ltedi ffcrenoe in meanwcigltt !oss,J.Id wlt.cN J}j = Bet ore- After. 2) Ho :#d= !0 '3) H1 : Jfd> tO 4) Th:e teststatisticis .;) Reject tJ!.e null hypothesis if~ > r0 .o5,9 \''lter'f! r0 .o5 ,9 = 1.8$;3 for a= o .o.; 6) d =" 17 -10 to = 6 .41 / JiO 3.45 7) Conclusion: Because345 > 1.833 reject the null hypothesis. There iseo.'icteoce to support tlteclsim that t"-is particular diet modifica tion program iseffecthl: inpmdlt.cinga mean \,·eight loss of a t least U') !bsat tlt.eo.o.; leo.'CI of significance. c) UseSdasanestimateforo: Yes. the sample sit.<: of 10 isadt'quate for this test. 10-45. a ) Tfte probabilitypJot below show that nonnalityassumption is reasonable. Normal P robabll tiy plot of the difference of I Q Normal 991r------.-----.------------------~-- r---~~ ·0.015 ' I "' ' J ' - ~-. ' ' / _1, -- .J_ -I. ~+-+ . .-! ., -- ... -. ~.. -~ I I 1,/ I I •• ~· 1 - T- ' - ---_,-, ... ~ -1- I -I- -~ -~- - - --!-----i---£ -·~>--- - -1--- . ' . . .. ' . . ~- __..,_ -~ - ~ iv- --l- ~ ~-1 -1-- b) d = -0.015 Sd = 0,S093 n= lO ro .o 2.,5.9 = 2.:!-M 9s%oonfidenoe in1mal: ' ' ----1- - --- --..... - -4 --~ --.;-- -.--- •. ;· ' - .~ ... :( --; T -r 1 , r •,/~--- ---1 r--- - T-- · --.. ,;- -f- --, --1 - -~ -1 ~- i' , •' I ' -- -t- --oo~- -~-----~- -t-- -- • ;- -- ! : ! --.,- -r---- • I .,.-; --+ - · 1.0 · O.S 0.0 Dill o.s 1.0 d - t o/2.n-l [ t.l ~ l"d ~ d +t n12.n-ll t.l - 0.37 9 < }Jd < 0.349 3 ~~ StOe-.· 0.509-4 N 10 1(S 0.1<19 P.V~lle :~o0. 1 !iO Becacseuro isoontaioodin lhi:oonfideoce intervaJ there is oot sufficient eo.ideooe thst tt.e mean lQ depe:odson birthor<k"'t. c) I~= J-0.9=0.1 d= d = t.J=-1 = 1.96 " sd Thus 6 ~ n wollld be eno&gb. 10-47. l) Parameters of interest are tl:-e mOOiancMicsterollew:ls for twoacthities. 2) H0 : l'o = 0 2) or Ho : fi - J'l = O 3) H1 : p0 > 0 3) H 1 : 1; - ~>0 . 4) r .;) BecaU:S('a = o .o.; and n = t.;.Appendix A. Table Vlll gh\'S th:ectitical valneof r ;OS = 3. We reject Ho in fa ..-or of H1 if r ~!:: 3. ·= 6) Tlt.e tl:St statisticis ,. '· Observation Before After Difference 1 265 229 36 2 240 231 9 3 258 227 31 4 295 240 55 5 251 238 13 6 245 241 4 7 287 234 53 8 314 256 58 9 260 247 13 10 279 239 40 11 283 246 37 12 240 218 22 13 238 219 19 14 225 226 - 1 15 247 233 14 • • ~['5)(0.5)'( 0.5)20-r = 0.00049 P-vaJne = P(R ~ J' = l4 l p = o..;) = L.., r r = l3 Sign + + + + + + + + + + + + + + 7) Conclusion: Because the P-vaJne = 0.00049 is less thana= o.o.;. reject tr..e nu.IJ hypothesis. Tf!.ere is a significant d.ifferrn::e io the mediancf!.ole.st~olle~o'clsafter diet andv:erciseat a.= o.o.;. 104 9. a lf.,.,., = 1.59 b • -??8 ~ t e.ao,!i1,9 - - · ... c) r.,, .. , - 2.64 lO·.;l. l) Tf!.e parameters of interest are t.J!.esta.nda.rdde\iatiomol~ 2) H . 2 2 0 . a 1 = a2 ' ) H . 2<' 2 -' I ' 0'1 ·~ 0'2 4) Tlte t<Ststatistic is I d) f'""·" = -.--''--= -= 0.529 ·~.2~105 I.S 9 e) r • .,,.,. = --'-- f0.10,9 .2-1 I - = 0.525 1.9 I I) f. - _;..._ _ _!_-0.311 O.~AIS f , 2, o.o~ · ~· J . - .;) Reject the null hypothesis if to <fo.9.;4 .9 = 1/fo.o.;,9.4 = 1j 6 = 0 .1666 tbr a= o .o.;. 6) n = 5 n, = 10 I • ? sj = 23.2 ..; = 28.8 /, = ( 232) 0.806 0 ( 28 .8 ) 7) Conclusion: Becanse 0.1666 < o.806do oot reject t lbc null ~.ypotltesis. t;.;'%oonfideooe: int~'a l : a 2 [52] ~:s; --;- ft-('r.n - l .n -1 a- s- : ' 2 2 2 2' 2 2 a, J . · a l a l -~(--)• /,0 059 4 Where / 05 • 4 = 6 -~ 4.83 or - < 2.20 2 28 8 · · · · r· 2 0'2 • 0'2 °2 Becatse t.J!.e value one is contained v.;thin this interval, t.J!£te is no significant difference in the vsriances. 10-.;3. a) 1) TILe paramctersof interest .arefl!.estandarddeviatiom.o1.o2.. 2) H . 2 2 0 . a 1 = a 2 3) H . 2 2 l . a ,=n2 4) TlbetestsUitisticis s2 J: =-1 0 2 s2 .;) Reject tlte null hypotbesis i ffo </ .0.975.l4.14 = OJ,Sorfo > /o.o2.!;,14.14 = 3 for a= 0.05 6) n = 15 I n = 15 2 f. = 2.3 = 1.21 0 1.9 7) Cooclusion: BecauscO.J;33 < 1.21 < 3 fail to rejt'CI the nuJI hypott.esis. Th.ere is oot sufficient e-.i&oce that tt.ere is a difftreooe intf!..o: standard de\iation. 9.;%ocmfidmce iote:n'al: [ 2] 2 [ 2] ~ a 1 s1 ? J;- a12Jl -I JJ -1 S - 2 < - 2 fun.,; -l.r~ -1 S - !I Cf S !I 2 2 2 ( 1.21)0.333< "t <(1.21)3 .,.- 2 2 0.403<a~ 5 3.63 a · 2 Because tlte vaJue one is oontained ,.,;thin this inten'al, tJ!.cre is oo significant differeooe: in the \'ftrianoes . .,. b) >. - 1 = 2 ltl =J~ = 15 0=0.05 a2 COOrt VU (o)wdiod e = 0.3.5 tltenfl!.epowen- ~ = 0 .6.; a l <:) ~=O.O.;ando:~.=otf 2sotJ!.at~= 2 andn=31 2 10·5.5·· s) 1) The parametersofirrt:'!rest are the time to assemble standard de\iatiom,o1P 2 2) H0 : a~ =a; 2 2 3) H, : a 1 = a 2 4) T!be test statis tic is 6) 11 = 25 I ? s· /, =-1 0 ? s· 2 n = 21 2 /, - (0.98 )2 -0.9 23 0 ( 1.02)2 sl = 0.98 s~ = 1.02 7) Conclusion: Because 0.365 < 0.9~ < ~.86 (sil to reject tf!.e nullllypotlt.c.s.is. Tf!.-ere is not sufficient evick:noe to support t~.-eclaim that men and womendifrer in repeatability for tJ!.isassemb!ytaskat W.e 0.02 Jt'\oel of significance. ASSUMPTlONS: Assum-e thst tf!.e two populations are i~t and normally distributed. b) 98?f,oontidmce ioten'<l l: .,.2 (0.923)0.36 55 -+ < (0.923)2.8 6 a · 2 a 2 0.336 9 5 -+ < 2.6 40 "2 Because the vall!.(' one is ocmtained "-ithin this i ntenosl, there is no significant differet:~Ce betv;am th-e \'arianoe of t~.e repeatability of men and ,~·omen for tlt-e assembly task at a !!'%s.ignificanoe le-.oel. 10 ·.')7. a ) 90% oonfidence i nter\'aJ for the r <~tio of varia DOE$! [(OJ5) J0.412 ~ .,.r ~ [(0.35) ]2.33 (0.40) "'2 (0.40) (1 0.6 004 ~ - 1 < 1.428 b) 95% oonfi<ieoct i nter\'al: [( 0.3 S) ]0.342 < 0'~ < [<0·35)]2.8 2 (0.40) - ,.2 - (0.40) " 0.5468 ~-1 < 1.5710 2 The 95'% oonfidence i nten:a.J is wider th.an tf!.e 90'% oonfid.cnce interva!. c) 90'% loh'er-~Sidedoonfidence intet\'al: [;, 1 "~ s; ~-ct.n,-1.11~-1 <a; [(OJS>)o.soo <a~ (0.40) - u; 0'2 (1 0.438 <--';- 0.6 6 I ~ - 1 a- a 2 2 10·59· 1) Tt.c parameters of interest arett.c melting \'S.riances. n; ,a;. 2) H . 2 2 o · a l= u2 'J) H · 2 2 •. a, ::& 0'2 4) Th.e teststatistic is (12 0'2 5) Rejt\."1 the null llypotf!.esis iffo < fo .9].;..20;:l:O wherefo.9].;..20;:l:O = 0 40.;8or fo > fo .o :i.;.l0.-20 w~.erefo .02,5.20 .:l:O = :l:46fora= o .o.;. 7) Condusion: Because0.<~0.;8 < 1.78 < :l:46fail to reject the null hypothes.is. Thepopulation\osriancesdonotdiftCr a t tf:.eo.o.; IE'\~ of significance. 10-61. 1) The parametersofint::rest are theo\W.all distance standard df!'\iations.o1.o-2 2) H . 2 2 o· "'I = a2 3) 4) Tlteteststatisticis i f =-1 0 2 s2 .;) Reject t~.e null hypotf!..('Sis iffo </.0/~7.;,<).9 = o.~Sor fo > !0.02,5.9.9 = 4.03 (or a= o.o.; 6) n 1• 10 "2 - 10 s,- 8.03 ' f. - (S .OJ)" -0.640 0 (10.04)2 s2 - 10.04 7) Conclusion: Because 0.-248 < 0 .6.10 < 4.fJ4 fa il to reject the null hypotlbes.is. There is oot sufficient E'\idenct t!tat t lbestand:ard <ieviationsoftt.e.over.aJI dista1Xl€Softho: two brands differ a t tt.eo.o.; level of significaDCe. 95'%oonfidenoe interval: [s~ ]l,...,n.n-l.n -I < "221 < [ s~ ]/<J/2~ -l.n -1 s· ':n s- · ~ 2 2 2 (12 (0.640)0.248 < -t < (0.6 40)4.03 0'2 .,.2 0.159 < - 1 < 2.57 9 - ? - u - 2 " 0.399 < - 1 < 1.606 A 95% Joweroonfidenci: bound on tf!.e ratio of standardde\ia tio"O$ is gi~o by a 2 Be::ause lf!.e \'aJue one is oontai ned "'ithin this interval. there is oo significant difference in lf!..o: \'aria noes of the dista ooc.; a t a 596 significa I'A."f: level. 2) 3) 4) n.eteststatistic is S) Reject tf!..e m!.ll ~.)pGIM.s.is iffo <f..o .9J5,14.14 = 0.3$ or fo > fo .O:tj.l4.l4 = 3 for a= o.o.; 6) 111 - 15 n2 - 15 s12 - 0.008312 j~- 1.35 ' ' s2• -0.00714' 7) Cooclu:s.ion: Eklcsuseo.,m < 1.3.5 < 3 fail to reject t~.e null hypothesis. There is not su.fficient e~.idenceth:at there is a difference in the \"aria noes of the weig Itt measurements ~·eeo tr.e l\..-o slt.ects of paper. 9.;%oonfideooe intmoal: Be::ali:Se t.hevah;,eone isoontainedv.ithinthis interval. th.ert is no significant difference in the variances. 2 2 10-6.;. a) 1) Tlte paNimetersofinterest aret.het:tch-ratevariant'('.$, U 1 ,a2. 2) H . 2 2 • . 17,= 172 'J). H · z 2 I ' q l = tT2 4) Tlte teststatistic is 6) n = 10 I s, = 0.422 n2 = 10 s2 = 0.23 1 /, = (0.422l 3.337 0 (0.231)2 7) Conclusion: Because 0.248 < 3.337 < 4.03 fa iJ to reject t.f!.e ncJJ hypothesis. Tf!.ereis oot sufficient e-.idence that th-e etch rate \'arianoesdiffer a t tlw!o.o.; la'E:I of significance. b) With).= .J2 = 14/1 = 0 .10 aoda = O.b.;, we find from CJ!.artVll {o) tt::atll1 • = n2 • = I 00. T~uerefore. t.hesample;ofs.ize: 10 would not beadec!.uste. b) • = 188 = 0.752 P, 250 . . p = 245 = 0.7 2 350 p, - p2 Test statistic is z 0 =-;=..:.b~'===~ • . ) [ I I l where p(l - p - +- /Jl "2 0.052 zo = --,--------~--~ (0.7217 )(1-0. 72 17 >[-1- +- 1-) 250 350 1.4012 P-\'Sh.lf: =ll - fV-< l40l.2)J .. l - 0 .9194 = 0 .080(, 95'% IO\'fet oonfidmoe interval oothed:ifference: (p - p) - z I 2 a (0.052) -1.65 0.7 52(~~00.7 52)+ 0.7(~~00.7) 5. P,- p2 - 0.008 55. p1- p2 • 188+245 p = 250+350 0.7217 c) Tlte P-vsluo: = 0.0806 is less thana= 0.10. Therefore, we reject t!"~£ null hypotl",£S.isthat p1 - P::l: = 0 s t tJ!.e 0.1le11el of significance. (fa= o.o.;. t.be P-vsh.lf: = 0.0806 isgre.1ter tt..sna = 0A)5 and we f.s iJ to reject tlte null hypot.besis. H)-6!). a) 1) The paNmctersof interest .srethe proportionofmters in favor ofBush\'S t lto:se in f.s,-cr of Kerry. Pl and P::l: :!:) Ho :pl : }):): 3) H1:P1iM_ 4) Teststatisticis 5) RejilCl tt.e null ~.ypot.J!. . .-sis ifzo < -Z.O.O!!..S or t.o '> .1.0.025 wkere .1.0.025 = 1.<)6 for a= o .o.:;. 6) n1 = :x:t20 x 1 =tOn Xi= 9:}0 i>, = 0.53 • = 107 1+930 = 0.49 5 p 2020+2020 z 0 = r====o;.;;. 5;.;;3....;0.;..4;.;;6===7 = 4.45 0.49 5(1 - 0.49 5)[ 2~20 + 2~20 l 7) Coochz.sion: Becatse:445 > 1.96 reject th.e nu.ll hypntlt.esis andocmc:h<.de)\"S there is a significant differmce in the proportions a t th.e 011.) Jeo.'el of signific.snce. P·\'<lln<: = 2h- P(Z < 445)) ., o b) 95'%confid-enoe intervaJ on tlt.i!differm::e: '( I ') ' ( I ' ) P1 - P1 + P2 - Pz < _ <( · _ • ) +z - P1 P2 - P, Pz ,n 0.039 < p1- p2 5. 0. 1 lkcat:M! this inter\'sJ does oot oontain t~..e\'ah:.e wo. \<,;esrey,;%oonfident there is a difference in th.e proportions. JO·i1. a) 1) The p:~.rnmetersof interest are tlt.e proportion of sati.sftlctory lcns.e.;,p1 andp2_. :!:) Ho :pl : }):): 3) H1 : Pl 'IJ':!: 4) Test statistic is 6) ~ =jOO X:):: 19{) i>, = 0.843 where p, = 0.653 p = 253+196 = 0.748 - 300+300 0.8 43- 0.6 53 z = -,---=,;;_...;.;.;.;.;;,___~ = 5.3 6 0 0.748 ( 1- 0.748 )[ 3~0 +3~0 ] 7) Conclusion: Becstl.Sil$.36 > :!:,sS reject t.h.e null hypothes.isandooncludc ~u t.h . .erc is .a significant difference in th.e fNictionof polisHI~·indll.ttddd'ects produced byt.J!.e h-.·o polishing soll!tionsat t.J!.ediJl lt'\d of significance. P·\'Sloo= .:!:!1- P(Z < 5,16)J = 0 b) Byeonstructi~ a 99%oonfideooe interval ontt.edifference in proportions, tltesamcquestioncan beallS\'o·erOO bywbet~.cr or not zero is oontaioed in Ow intcr\'al. s, 2.23 U)-J.'j. a) SE Mean1 = / = ----r':" = 0.50 ""' v20 x 1 = I 1.87 x2 = 12.73 s~ = 2.23 ' n = 20 I n,= 20 Oegrcesoffrtedom = n1 • 't:l- :l =.20-~ = 38 s = p 2 2 ( " ' - l)s, +( "z - l)s2 = (20-1)2.232 +{20-1)3.192 = 2.7522 20+20- 2 (x1- x?) - ~. t = -0 I I s - +- p n, "2 ( - 0.8 6) = - 0.9 8 8 I 2.7 522~ 1 + 1 20 20 P·\'8llle = 'liP(r < -0.9881)1and.:t(O.lO) <P-vaJu < 2{0.::!.:;) = o.~o <P-vallle < os The9.:;% n\·o-sidcl confidence inten 'tl l: Iaj:!..ltr• n2-2 = ro.02,5,38 = '2.0~ (x, - xz)- Cz. . -2s -1 +-1- < JL,-1'2 < (x ,- xz) +t s _I +- 1- {l/ ·'~~: P n 1 n 2 9~-: P n 1 n 2 (- 0.8 6 ) - (2.024)(2.7 522)~ ; 0 + ; 0 <J.I, - JJ.z <( - 0.86) +(2.024)(2.7 522)~ ; 0 + 2~ - 2.622< JL,- !Lz <0.902 b) This is l\'fo-sided test breause t.Maltemati\'E: hypotltes.is is p:1 - Jt2 oot = o. c) Bcc:au.se: tft.e0.20 < P-vsll:e < o,s and: t~.-e P-vah.~i: >a= o.o.; we fa il to reject t~.e null hypotf!..('Sisat ttoeo.o.; l£\'tl of signifiestn. rf a= 0.01. weaJso fai l to reject t~.e nllll hypot.hes.is. 10 ·7 5. a) Assumptions tf!..at must bt met are nonnaJity. ~-caJity of variaoce, and i rxiependeooe of the obsetvatiol'6. Nonnality and t!qu.aJity of varianofS appear to be re.'lson.'lble from the oonnaJ probability plots. TJ!.edata appear to fall a long straight lim and t.J!.es.lopesappear to be th-e s.ame. r ndepeodmoe of t~.-e observations for each sample is obtained if ra nd.om sam pits are selected. Normal Probabilit y Plot ' ' ' ' ' ' ' .99 ---~~-. -:--- --:--~- -;- ----:-----:-- --~· ' . ' ' ' ' ' t:;- .99 ---.. ~--- .. --~~ .. -~--.----~.-----.---- ... 1 :a .ss . -:----:--~--:----·:-~-~~:-----:--~-: ·l ~ .80 ---~~---~--~-~-~--·-----, ·-:···--: £ :~ ~-~r~ ~ ~~~~ ~~~-~-~:~~~~~~~~~~r~~~~:l .OS · ' -~~ • • .;. •• ~~~-~ • -:- ·-- ~:- • ---:-- --~ · .Ql - --~~-. _;. __ --~-~ •• ;. --- .: •• --.:-- -- ... .001 ' ' ' ' ' ' ' ~- -.. ~-- .,. .. ~-.. -~ ···- --- ., .. ---.-- --.. ' ' ' ' ! ' ' . . . . . . 14 IS 16 17 18 19 20 9-hout A'ltr~ lG.lSSii StOtv: l..Oil9C9 "" b) X1 =16-36 SJ = -2.07 n1 =9 X::! =ll48;l ~ = 2,37 n2=6 .m "' .99 ·" .9S = .0 .80 rn .0 0 .so ~ a. .20 .OS .o> .00 NormalProbability Plot I I I I I I + I ~-:~- ~~r---~---~----~- -- f- -~ -:--- -~ · - ~- ~ --~---~---~----~. -- ~. -- ~- ---~ .II ~-:~-~~r---~---~----~---f--~~- --~·~ I o I + o o L- • o · -.· ~ · -'" --- r--- ,--- -r ·-• t~ .._......,., -- ·r · ' • ' ' •..---r ' ' -~-~--~---~---~~-~---'---~----~· ~ -:~- J_,~+::-~=-~~~--~. --f.--~--- ~:- j Oo • o I I I I ! I · -.· · · ~r-··r---,----r · · · r · -~ -.· -~ -.. 0 I I I I o o ! ~-.--~-r---r---,----r~ -r~ ·-.· ··r I I ' I i I I I - ~-- ~~~--- .. ---~·----..... t ••• --i• •• -~. I I I I I I I I . . . . . . . . . . 8 g 10 11 12 13 14 l S 1-hour s = 8(2.07)2 +5(2.37/ = 2.19 p JJ (x1- x?) - t_,. +n - 2 (s ).-1 +-1 </L1- Jt2 <(x1- x?)+t .. n • _?(s) _I +-1 - w - JI, . ! p 1. n, 112 _. w•.n, -t~~! - . p , , 112, (1 6.36 - I I .48 3)- 3.012(2.19 )~ ~ + ~ < JL,- I"z <(16 .36 - I 1.48 3)+3.012(2.19 )~~ + ~ 1.40< ,.,-,,2 < 8.36 <:) Yes, weare9!)%oonfident the results from t.J!.e first test oooditioocxcood the res-c.!tsoftlteseoond testoortiJtion btcau.setf!.eoontldettt inten'al contains only positive\ 'Slues. 2 / 2 d) 9.s%oonfid-cnceinterva!tortT1 a2 ? "- -'. 9.s%oonfidenoe in1mal on ., . a; I I frm 5JJ 5 = = --= 0.207 5,/002,$ 5 = 6.7 6 . . .. fo.oos52 4. 8 2 . -· .. 2 2 2-s, . u, s, - f. < - < - ,. 2 o.9'7 s.s.s - ., - ., Jo.o2s..s .s s; u; s; [ 4.283 J(o.2075)s"~ s[4.283 ]<6 .76J 5.6 17 a; 5.6 17 2 a, 0.1582 < - <5.157 - ? - a- 2 e) 8ecsuse IJ!.e\'Silleone isoontainedwithin tl·.is intervaJ, U..e population,>ari.anoesdo oot diffo:rata .;;%sig:nificanoe le".t:l. 10·77· a ) 1) Tf!.eparometerofinterestis theme<~oweiglt.tl oss.lld where lli = Initia l Weig.ltt - Final Weiglt.t. 2) fl<> : Jid=3 3) Hl:#d>3 4) T}!.etest s tatis tic is - d -6.. to - sd l ..[,; .S) Reject H(J iff() > lo.n- 1 wlt.-m: 1<.1.0.';.7 = l.B<},sfor a= 0.05. 6) d = 4·12.5 t = 4. 125 - 3 = 2.554 0 1.246 / Ts 7) Conclusion: Booat:Se ~;554 > 1.89.5. reject tft.e m:ll hypotltesisand oonclu<k:tJt.e 8\'f:Nigeweig~.t loss issignificantlygr-eater than3at Q = o.o.s. b) 2) fl<> :Jid=3 3) H1 :1Jd>3 4) The test statistic is 5 ) Reject H.., iff() > la.o -1 ,.,.lter-e to.01,7 =- ~.'!)98 for a= 0 .01. 6) d = •. !>.; I = 4.125- J = 2.554 0 1.246 / ../8 7) Conclusion: Booat:Se ~;554 <~.998. fai l to reject t~.e nllll hypot.be.s.is. 'O.e 8\ 'f:tageweigM loss is oot significaotlygrester t~.an3at a= 0.0!. c) :l) Ho :1/d=.S ;l) H1: #d>.S 4) The test statistic is 5) Reject H.., iff() > fa.o - 1 wlter-e to.o5•7 = 1.8<}.5 for a= o.o.s 6) d = •. !>.; t = 4.125- 5 =- 1.986 0 1.246/ Ts 7) Conclusion: Booat:Se -1.986 < 1.89.;. fa iJ to reject W.e null hypotJt.;esisandoonclndethat tlte&\'tl'age \~·cight Joss is not significantly gri".ii tcr than.;ata = o .o.;. d) 2) fl<> 'Jid= 5 ;l) H1:#d>.S 4) Tr.e test statis tic is .S) Rcjo::t Ho if to> 'a.o-l w~.erefo .o1 ,7 = ::!.~fora= 0.01. 6) d = •. !>.; t = 4.125- 5 =- 1.986 0 1.246 / Ts 7) Conclt:s.ion: Bec:au.se -1.986 < ~.99$, fa il to reject tlbe null ~•)'J)Otlte.sisaodoonchtde th:ea\..-:ragewcig.ltt loss is oot significantly grcster titans a t a= 0.01. 6) 10·79· a ) l) T)..,e p<~tilmetersofintcrest are flt.e proportions of children wlt.ooontract polio.p1.p:!. •l H,:p,=p· 3) Hl:P1~P~ 4) Tf',,e_ teststatisticis -r===p~' bl ~pd2~==7 zo = p(J-p)[-' +-' l 111 IJ2 ' = .3_ = I I O = 0.00055 p 1 11 1 201299 (Placebo) x., 33 p =~= = 0.00016 2 "2 2007 45 (Vaccine) 0.00055- 0.00016 z = --,----.....;;==__;==----..,.= 6.55 0 0.000356(1- 0.000356)[ I + I l 201299 2007 45 7) Becatse 6-55 > 1.9()l'f'ject Ho andoonclnde the proportion of childrenwltooonttilcted-polio is significantly different a t a= 0.05. b) a= O.Ol Reject H() ifzo < -Zaj:!OT .l.c) > Za/2 W~Zaj2 =~._'iS. Still Z0 = 6,s.;. Bf:eause(,,s.;. > ~,sB. reject Ho and oonc:lude the proportion of cbildrenwlt.ooontracted polioissigniJicantlydifferentat a= OiJl. c) Th.eoonclcsion;a.reflt.e same OOcause:r.o is so large it exceeds 'l.a/2 in OOtbcases. I 0-81. X '8 7 p =-1 =-J-=0.258 X 310 P? = ....1. =--=0.2583 - n2 I 200 I n1 1500 ( 0.258 - 0.258 3) ±1.9 6 0.258(0.742) + 0.2583(0.7 41 7) 1500 1200 - 0.0335 < p1- p2 :::; 0.0329 lk<:ausezero isoontained intllis inten 'tll, t!ttte is nosignificantdi:ffcri!llOC bctwem the proportions of unlisted numbers in tlt.e 1\~·oci tiesat a s% significaooe le>.l:l. b) zat~=zo.o5 =1.6S (0. 2 58 - 0.25 8 3) ± I. 6 5· . 1..::0 ·=.2 5:...:8'-'.( 0.:..:._7 4.:.::2:!.) +-'-0 ·=.2:...:58'-"3"-'( 0"-. 7_4c;.l 7....!..) 1500 1200 - 0.028 2 :::; p 1 - p2 :::; 0.027 6 Tile proportions of unlisted numbers in tf!.e two cities do not significantlydiftf:r a t a s%significa-oc-e le\l:l. c) X 77 4 p =-1 =--=0.258 I 11 1 3000 X 620 p = ....1. =--=0.2583 2 n2 2400 95Wi oonfidence: interval: ( 0.258 - 0.258 3) ± 1.9 6 0.258(0.7 42) + 0.258 3(0.7 417) 3000 2400 - 0.0238 < p1- p2 < 0.0232 90~oonfide~ interval: ( 0.258 - 0.258 3) ± 1.6 5 0.258(0.7 42) + 0.258 3(0.7 417) 3000 2400 - 0.020 I :::; p1 - p2 :::; 0.0 I 9 5 ( nc:rcssi ng tJ!.e sample siu: dec:reased tr.e '~dth of tf!.e oonfideooe i otervals. but did not change t lte ooDclllSion;. dr-awn. The ooDclllSion remain; tltat t lt.o:re is no significant differeoce. tO..Sj. a) Yes. tMreoould be some bias in tr.e resultsdtr.e to tf!.etelephonesurvcy. b) (f itoould bes.hown that tlt.ese populatiorGaresimiJar to tlte respondents. the results maybe extendEd. 10-8$. X 1 = 3'1).87 X 2 = j0.6S 01 = 0 .10 ~ = 0 .1,5 01=1:! 0:)_ =10 :a) 9()% Mo..sidi'doonfidenoe inter\'al: (0 10)2 (0 15)2 (0. 10)2 (0.15/ (30.8 7 - 30.6 8) - 1.6 45 " 1 2 + " 1 O < 1•, - 1":! S ( 30.8 7 - 30.6 8) + 1.6 45,p . .:.:..: 1 2 ..:.!._ + I O 0.0987 SJ; - ~<0.28 1 3 Wear-e91)5Wioonfident t~.at the me<~o fill volume for machioe 1 exooods that of machine 2 by between 0.0987 and 0.:!813 fl .m. b) 9.5% Mo-s.idOO oonfid(!OOC interval: (30.87 - 30.6 8) - 1.96 (O. II~/ + (O.i:)2 < 1•, _ ,,2 < (30.8 7 - 30.6 8 )+ 1.96 0.08 12 s "'' - '":! < 0.29 9 (0.10)2 + (0.15i 12 10 Wear-e9,s'%oonfident that tlt-c mesn fill \ 'Oiwnc for machine 1 o:coods t.r.at of machine:! by between 0.0812 aod 0.:!99 O.oz. Compari.sonofparu (a)s.nd(b): As U..e levcl ot oonfideooe ioc:reasr.s. IJ!.einterval "'idthalso io...~(~ith all otlter \'a lues J>.eldoonstant). <:) 9.5% upper..s.id.."C! oonfideoce interval: (0.10/ (0.15) 2 ,.., - ~-'2 < (30.87 - 30.6 8 )+ 1.6 45,/"-~'-- +-'------'- 12 10 1•,- p2 < 0.28 I 3 With 9s'%oonfideooe. the fill \'Oiume for machine 1 o:ooods th:e fill \t>lum-cofmacJtJoe:! by no moro: than 0.2813 fl .oz. d) 1) Tlto: patt~mcter of interest is IJ!.edifferenoe in mean fjU \ 'Oiumev1 -vz 2) Ho : v1-112 =Oorvl =112 j) Hl:P1-W.!:OOt #l"'#2 4) The test statisti<:is (x1 - x.,)-~0 z = ... 0 2 ., (! a" -' +_1. n l n2 ,s) Rejtd Ho ifzo < -Zo.f-'1. = -1.9()or1{) > Za/'l = 1.¢for a = o .o,s 6) X1 =$0.87 X 'J.=j0.68 OJ= 0 .10 0:)_ = 0 .1,5 n1 = 1:! na: .. 10 z = 0 (30.87 - 30.68) = 3.42 (0.10)2 + (0.15)2 I 2 10 7) Elecilusej¢ > 1.96 reject IJ!.e null hypotlw.s.iss.nciocmc:lude IJ!.e mean fill \'Oiumt.iofmachioe Janel macJt.ine i differ significantly a t a=O.O,s. p.,'SJne = 211.¢(34:!)] = :!{1-0.99969) = o .ooo6:! e) Assumethes.:tmplesiusareto bef!G.U::8l, use a= o .o,s. ll = 0 .10.aod<1 = 0 .20 8 .53, n = 9, use o, = n, = 9 c) 10-87. a) l) Tlte parameters of interest are: tft.e proportion of lenses that are unsatisfactory after tnmble·polishing,..p1,p~. l ) fi():pl=M_ 3) Hl:pli.J)l 4) Th.e test statistic is 5) Reject H() if~ < -Z<zj2 orzo > Z<z/2 WltereZaf2 = 2.)8 for a= O.OJ. 6) x1 = numbcrofdefec:ti\'elenses X 47 p =-1 =-= 0.1567 I 11 1 300 x +x p= I 2 0.2517n1 +n2 X 104 p = ...1.= -= 0.3467 2 "2 300 0.1567 - 0.3467 z = .--------:----~= - 5.36 0 0.25 1 7( 1- 0.25 1 7) [ 3~0 +3~0] 7) Conclusion: Because -.:;,36 < -2.;8 reject Ho andoonclude tbere is strong. evidence to support tlteclaim IJ!.st tlbe 1\-\'0 polishing fluids are different. b) Tlbe oooc:h:sions are the same whetJ!.er we ana lyu t~.e data u.si ng IJ!.e proportion unsatistilctory or proportion satisfactory. 11 = r 2.57 5 (0·9 +0·6 ~(0. I +0.4) +1.28J0.9(0.1) +0.6(0.4) r (0.9 - 0.6)2 5.346 = 59 .4 0.09 II = 60 ;, .99 = .95 :;; .. .80 .0 E .so a.. .20 .OS .01 .001 10-89. a) No Normal Probability Plot ' ' ' ' ' ' ,-- ---------- t·----- --- ---,-·1 f-·----------- t------ ----·--i-·1 ,-- ---------- t·----- --- ---,-·1 :··----------·-:·----- -------~--1 -·--------------·---- --- ____ .. _. t ' • ' -~------------ ,--.-- .!. ---------:-· ' ---------,---------- ---~-· - t -- . -·--------------·- - ------ ---~-· ' ' ' _, ______________ , __________ ---1-· I ! I 23.9 24:4 me rcedes 24.9 Normal Probability Plot ' ' ' .99~ - -~--- --------1-----------1--------- ?;- .99- --1-----------i-----------t --- -- := .95- --~-----------l-----------1------.. -- .0 ' ' ' «:~ .80- -,-----------,-----------r---- -:-- "8 so- ·--~----------.!----------~- ... . ' ' - ' . a.. ' ·-- •• .20- -"I"--------·:~...-------- --.-r------- . --. ' OS- ·- • r-~~- ------.-----------r------ -· • T-- ' ' .01- --.----------- .. -----------.. ------- .00~ - ... --------- -~---- ------ -~---- ---- ' Awr.~~ Z4.G7 StO~ 0.30Z030 N! 10 AIY.Ie...-~Oa'lh;l NOI'l!Uhy lut A>Sq.ured: 0334 I>..YJ\:oe: 0.011 .99 ~ .95 1 :: - .20 b) The normal probabilitrplots indicate t!tat t lbcdats follow ootllUII di.stTibutio~ bec:an.s.etltedata a ppear to fa ll along a .straight line. The plots also indicate that thevsrianoe.sappe.'lr to beequaJ be<:au.se tles.lope.;appear to be tt.esame. Normal Probability Plot ' ' ' ' ' . . . ' . - ;-- -----.------.,.------,------ --; .. 1 . ' ' . ' ·r······-,·······'t·- .... , ....... ..,. ' ' ' ' ' -r····---,--·····"1·--···-,· ..... ..,. • • • • "T"- •• • ""!" • • ••• •.,-• --· •• •o••• •• • • • -~--------'-------J. .. __________ .. • • . . • ·r····-- -----{- -----:- -----~ • ------.- . -.------····------..-- ---·-·· ------. Normal Probability Plot ' ' ' ' . ' . . .99 ·-:- -------:· ------- ·:· ------- ·:·- ••. 1 >. .99 --~--------! ----·-·-:··-----·-:·---·1 ~ .95 .... --------l·------- .. ----- --~-- ~ -· ~ e a. . . . . .!!0 . • r··· .. - .. •· ....... .,. ... .. . . .. • • • .so -~ -- --··!··--- .-- · --- --~---- .20 --r. --·- -l·-·------:----- ---:---- . . . .<:6 ·- --------~---------.---------.---- ' ' . . .0 1 ··r ·· ····-~---·-···-.-·-··-··--.·-·· .001 ' . . ' ' ' ' ' ' .001···--------· ...................... . .... ---- --·- ----- •4-·-- -- -·--- --- -~- l4.5 l4.6 . . ' ' . . . . 39.5 40.5 41.5 42.5 volkswag MdcJ·-.owiM 1'4onM\1y ;.u.~ A..s.c..:-.ar.t 0.440 PN!lf ~~t~: ~30 c) Byoorr«:ting th.edats points, it is more app.:~rcnt t~.edata follow normal distributions.. Note that one tmn.sn:al obs.enationeancauseao analyst to reject tlte nonlU'liitya.ss-..>mption. 2 2 d) 95'%oonfidenoe intenoal on tlte ratiooftltevariaooes. a r I aM ? s;; = 1.49 ~.9M25 = 4.03 I I 19.9».9? 5 = f. = -= 0.248 9,9,0.025 4-03 ? s4J = 0.0204 Sy (1 ._. Sy [ 2] 2 [ 2] - l <"- ' <' - /, 2 9.9.0.9'15 ·~ 2 ·~ 2 9.9.0.025 s_lt/ aM SM [ l.4 9 ]o.248 < (]~ <[ 1.49 ]4.03 0.0204 (72 0.0204 M (72 18 .114 <--f-<294.35 qM Because tlte interval does oot include thevalneone. we reject lite hypotJo..r.sis that variabilityin mileage p..."'tform.a~ is t.Jo.esame for lite two types of \~.icles. T!teN: i.sevidenoethat iJ..e variability is gr~ter tOr a Volksw-agen than for a Merood.es. ' (12 (12 e) 1) Tt:-.eparamctetsofinterestaret.lte \ariancesinmileage performance. I ' 2 H. ? ? ::!) o: a j = a;:wttereVolkswage:ni.s rq>tesentedbyvsriance!,Mereedesby\ '8rianoe::!. H. ? ? 3) 1 : a j ::ea; 4) Tlo.eteststati.stici.s (,) .S) = l .~ n1 =tO S::!_ = O.t<f3 n~ = 10 ? s· fo = -';- s" 2 I -,--'--=-0, = 0.248rorn=o.os 4. , ?2? f. = ( 1.- )" 72.78 0 (0.143) 2 7) Conclnsion: Because 7~.78 > 4.03 reject Ho aOOoooclude that there i.sa s ignificant differeooe betv;ren Volks\1.-agenand Me:roedes in terms of mileage: \ "aria bility. Tl-.e .same ooocln.sions. a-re reacl!OO i o part {d). 10--91. a) Tlt.eassumptionof normality appears to be reasonable. This is f:\idcnt byth.c fact that the data lie along a straigltt line in tJt.e normal prob.lbilityplot. .99 .99 ~ .95 .0 .80 "' .0 .so ~ .20 .OS .01 .00 Normal Probabilit y Plot ' ' ·- ;-- -----.------ -;- ------.--------:- ' ' . ' ' ·-r-------,-------,-------,-------,- 1 0 I I 0 ·- r-- -----,------ --;- ------,--------- I ' I ' I ·- .,----- --~- ------.,------- -----,- ' __ .. _______ , ____ _ -------•-------~- ' ' . -; -- ' ' ' --,------ -;- ------,-------,- ' ' ' . - ., -------·- ------ ., ---- -- -·- ------... - . ' ' . . . - .. -------,- -- -- --.. ---- ---,-- -----, - ' . - r----- --•--- --- - ~- --- -- -•--- --- -..,- ·2 ' ' ·1 0 diff 1 2 Aveuge: ..0.222222 StOev: 1.30171 Al'\dersoo~ling N-oon3l ty TesL A -$Qv3t'E!d: 0 .526 N: 9 P-V:We: 0 .128 b) 1) n..e parameter of interest is t~.e meandiffeN!oce in tip hardness. Pd ') f!,:~~,t=O j) H1:1ld+.O 4) Theteststatisticis .;) Since oosigniflcanoe le'\oel is gh-.m. \\·e \\ill calculate P-\'8lne. Reject Ho iftft.e P-\'8lne issig.nificantlysmaJI. 6) d = -O.:l~ t = - 0.222 =-0.5 12 0 1.30/ Ji P-\oaJne = :tP{r < - O,Sl.:l) = :tP{r > O,Sl.:l)and :l{O.:l..;) < P-vslne < :1{040 ). Thn.s, o,so < P-vahw: < 0.80 lid= l 7) Conc:hJS.ion: El«:ansc the P·\'alne is lsrger thanoommon lc>.oelsofsig.nificanc:e. t8il to reject Ht, andoonclnde there is no significant difference in mean tip h:atdness. c) IS= 0.10 d =-' =-' = 0.769 q d 1.3 From Chart Vtl (0,-rith. a= 0 .01. n = 30 10-9;3. a) The data from both dept.hsap))i'llrto be normaUydistribt<ted. but tt.eslopesdo oota ppcar to becqu.al. Therefore. it i.s not 2- 2 reasonable to.:tss-umetbat a 1 = a2 . Nonnal Probabilitv Plot br sur~ce. .. bctl!:m ALE rim~~ ., ,. ., H Q. "' Il "' 0 5 ~-+=t-) /. • - - • . . I -• . --tL / ~ . / -- . _Lc I I I I 5 • Data / I / I - A - ·-~-- - • - -- - - ~~~--+- .! ·~ 7 • b) 1)Tite p:~.rameter of interest is the difference in mean HCB ooncentration • .u1 - W.!· '~ilh!lo = o. :l) fio : 111 -#'J=OOt #l =#:1 3) Hl:IJJ-11-:!~0ot#liP:l 4) n.ete:st.statisticis 5) Reject tt.e null hypoth:esis iff() < -ro.o:!:.5,1.; or to > ro.0-:1.5.1.; where ro.o:!:.5,1.; = -2.131 for a= o.o.;.Aiso 6) X l=4.804 X :1=5.839 11:1=lO 11:1:10 [si + si] 2 n l nz " = _r ~-'-~· -=-r ---=r ~.f_~:-1 = 15.06 -'--'-''- + ...!....0"- ll - l /l -1 I 2 "~ 15 (nun cared) 7) Conclusion: Becat:.se -:1.74 < -1.131 reject the null 1\ypothes.i.s. ConrJtldethat the mean HCBoonoentration isdifti:rimt a t the f\~·o depths sampled a t tlt.e o.os level of significance. 2 c) Assu.metf.e,-arianoesarecqual. Then:l = z,a.= o.o.:;.n = n1 = •'t:l = 10.n• = :ln -1 = 19,sp = O,&fand d = 2 (0.S 4 ) = 1.2. From CJ>.artVIl(e)we find JS,. O:O,s.and then calcula te povrer = 1- JS = 0.9.:;. I d) Assl!melhevarianoo.sareeq-uai.Titen!l = t.a=o.o.;,n = tt1 = J't:l.n• = :m- 1. ~ = O.J.sp = 0.84,. and d = 0.6. From 2(0.8 4) 50+1 -C~.artVU (e)wefiodn•=.:;Oand ll = 2).5 .son=26. 2 2 2 _ _ a 1 a., 10-95.TJ!.e Jl( X1- X 2) =-+......::.. andsnpposetltis is to equal a oonstantk. TJ!.i:n, wearetominimit.eC1n1 +C2n::~_ snbject to n, ,.2 0'2 (12 - 1 +...1.. = k . Using a Lagrange multiplier, weminimiu by setting t1bc p<~rtial <ieri\'ati\'eSof ;;,,:::,).) = C 1111 + C21l? + A[ a~ +a; -kl, ith res....,... ton I' · a1xD oona) to zero • II II \ I"""' 1' ".! ' • • I 2 'these equ.atiom a t'f: {) Au2 - f(,1,n2,>.) = c1- -+=o i)n ,- 1 I (I) {) >.u2 - f(n n A) = C _ ____.1_ = 0 {) l ' 2' 2. ., n ,- 2 2 (2) i) 0"2 0"2 - f (n 1 ,n 2 ,\)=-1 +...l.= k {)). . (3) [ (12 (121 Uponaddi.ng ~lLII!i2~{l}and~~~""·eobtaio CJ +C2 - >.. u: + ,: = 0. Stbstimtiog fromt:g~t!~nJ.;)lenshles u.s tosol\'efor i.toobtein C, +Cz = >.. k Then.~~-tip~(l)and(::!)aresol\'ed for n1 andn2 to obtain ? , = u~(C1 +C) 1 kC I 10-CJ7. a) a= IV-> ZtJI' Z < -za-elwmrrez h:asa standard oonnal distrib.:tion. Then, a= IV-> Ze)- JlZ <-za-el = e + a -e =a. TJ!.e(l - n)oonfi denoe in1mal tbr r n(9)ron t:se the rela tions~,ip b) Tlw(1-a )oonfi deooe intenosl for Scan u..se t!bc Cl de\'elopedin part {a) where-0 = e"(ln{9)) - Z n - x • ~ Z n - x n - x, ~ ~... J''' Oe [ ~x, H "/,) 50 <Oe 4[ •,x, H •,x,) c) _ 19j(IOG-27]+[ IOG-L9]]'" 19j( IOG-27]J 100-19]11"' l .42e l 2700 1900 < 0 < l.42e 1 2100 \ 1900 ~ 0.519<0<3.887 Be::ause tlteoonfid"eoce intrn'Sl oontains the \'8lneone. weoonclude that IMte is no significant difference in th-e proportion; a t the9s%level of sig:nificaooe. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
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