chapter 9

Prévia do material em texto

CHAPTER9 
9·1 a ) H0 : p = ~. H1: p" i,; Yes. becauset.J!.e bypotft.esis is sUited intennsoh)..e paramcter ofinterest.ioequ.ality is in t.bealternathoe 
~.ypothesis.and t~.e,'8 lue in the null andaJteroothoe hypnt~.eses mat cites. 
b) Ho : o > 10, H1 : o = 10 No, becau.se tl'.e inequality is in th:e null hypothesis. 
c) H0 : X = 50~ H 1 : X ;;= SO No. OO:.atse t.J!.e 1\)pof.i!.'i".Sis i.sstated in termsofthestatis tic l'ilt~"'tthan t.hi:parameter. 
d ) Ho : p = 0 .1, H1 : p = o~ No, the \'8lue; in tr.e null aodaltemati\tt hypotheses do oot match and botboftr.e hypotheses are equality 
sUitements. 
e) H0 : s = jO. H1 : .~ > 30 No. OO:.atst tr.e 1\)poth'i".Sis is stated in terms of the statis tic l'ilt~"'t than t.hi: parameter. 
9•3 a) Ho : o=20nm,H1 : o <~onm 
b ) n.is rest!lt dots not provide strong evick:noe that the standard deviation has not ~"'CC red need. Tltae is inst:fficient eo.id"eoce to r-ejtd t.hi: 
ouJI h~pothesis. but this is ootstrong support for the null hypotlt'i".Sis. 
9·.; a) a= P(r-ejo::t flo when Ho is true) 
- [X- 11- 11 5-12) =P(X~ 1 15 whenlt=1 2) = P r.: < , 17 =P(Z ~ -2) u l -..tn 0.) / -..t4 
= 0.02275. 
Th.e probability of u;jecting. th:e null h.~poiltesis wlten it is true is0 .0~27.; . 
b) 
1> = P(accept Ho when 11 = 11.25) = P(X > 11.51!1- = 11.25) 
= p[·){ -p > 115 - 11.25] = P(Z > l.O) 
u i .Jn 051../4 
= 1 -P(Z ~ 1.0) = 1 - 0.84134 = 0.15866 
Tf'.e probability of accepting the null !'typotltesis when it is taJ.se is0.1.;866. 
c) ~ = P(acccpt Ho wlten 11 = u .1_;) = 
= P(X > 1LSf,u.=1LS) = P(xJn> 1LS_-~5) 
u / n 0. ~ 1 4 
= P(Z> 0) = 1- P(Z ~ 0) = 1- 0.5 = 0.5 
Th.e probability of accepting tlbe: null ~.ypnt.lt.esi s wJ!.eoit is faJse is o,s. 
9·7 T)..ecritical val1:e for t.hi:one-sid"ed test is 
X s 12 - Zc.o.s r!n 
a ) a= 0.01. n = 4. from Table Ill - ~.33 =7.aa.ndX s l-!42 
b) a= o .o.;.. n = 4. from Table lll -1.6.; = 7-aa.odX s ll-59 
c) a=O.Ol. n=l6, fromTable lll - 2.3;3=taandX ::: n .71 
d) a=O.O.;. n= 16. ftomTable m -1.6.; =Zaaod.X s u .95 
9-9 a) x =1 1.25, thenP-va1ue = P(z < 1 ~:~~:j~_2) = P(Z < -3) = 0.001 35 
b) x = 11.0, then P-value = P( Z < 1;.~;J;f) = P(Z < - 4) < 0.000033 
c) x = 11.75, then P-va1ue = P(z < 1175-:}/] = P(Z < -1) = 0.158655 
- 0.51 4 -
()-u Use n = ,s. e\'f:l}tlting else ~.eldoonstant {from tl'.e\ -alnes in E!l:etcise 9- 6): 
a) P(X~98.5) +P(X > 1015) 
= P(x -1oo < 98s- 1oo] + P[x -1oo > 1o1.s -1oo] 
21$ - 2/ $ 21$ 21$ 
=P(Z ~ -168) + P(Z > 168) 
= P(Z ~ -168) + (1-P(Z~ 1.68)) 
= 0.04648 + (1 - 0.95352) = 0.09296 
b) i> =P(98.5~X~ 1015whenlt= 103) 
= P(98.5- 103 < x -103 < 1015 -103] 
2 /~ - 2/~ - 2 /~ 
= P(- 5.03 ~ Z ~ - 1.68) 
= P(Z ~ - 1.68) - P(Z ~ - 5.03) 
= 0.04648- 0 = 0.04648 
c) i> =P(98.5~x ~101.5 whenll = 105) 
= P(98.5- 105 < x -105 < 101.5- 105] 
2 /~ - 2 /~ - 21-15 
= P(-7.27~ Z ~ - 3.91) 
= P(Z ~ - 3.91) - P(Z ~ - 7.27) 
= 0.00005 - 0 = 0.00005 
[tis smaller becal!Si! it is oot likelytoaooept tltie product w~.eo W.e true mean i.sas J!.igh as u).;. 
( s.Jn] 8 > Otben ,B= .P z.12 -~ , whereu = 2 
a) ~ = P(98.69< X < 101.31 1~1 = 103) = P(- 6.47 < Z < - 2.54) = 0.0055 
b) ~ = P(98.2S < X < 101751~1 = 103) = P(- 5.31 < Z < - 140) = 0.0808 
a) Asnioc:rease.s.fjdecreases 
9-15 a) ct =P(X > l85wben ~1 =175) 
=P(x - t 75 > t85 - m ) 
20 I .JIO 20 I .JIO 
= P(Z > !58) 
= 1-P(Z S: 158) 
= I - 0.94295 = 0.05705 
b) ~ =P(XS: 185wben p = 185) 
_ P[x -t85 < t85 -I85J 
- 20 I.JIO - 20 I .JIO 
= P(Z S: 0) = 0.5 
c) ~ = P( X S: 185 when ' I = 195) 
_ P[x -t95 < t85 -I95J 
- 20 I .Ji0 - 20 I.JIO 
= P(Z S: - 158) = 0.05705 
- (20) 9·>7 X > 175+ z • .Jn 
a) a= 0.01. n = 10, tt..eoz.a = -:! ,'\'2 aDd critical \ 'Sine is !89.67 
b) a= o.o.:;. n = u), thenza = 1.{>4aodcritical value is 185.93 
c) a= 0 .01. n = 16. t.ltenZ<z = 2.~ and critical valu.e is 186.6 
d) a= 0 .05, n = 16. tlumZa = 1.64 and critical value is 18.3.:!' 
X - J.Io 
9-19 p.,'alu.e = 1 -~))'"'J! . .ere Z0 = .J1i 
u I n 
a) - 180 -175 X = 180 then Zv = = = 0.79 
201 vlO 
P-value = 1 - .P(0.79} = 1-0.7852= 0.2 148 
- 190 - 175 X = 190 then Z0 = = = 2.37 201 vlO 
b) 
P-value = 1 - .P(2.37) = 1- 0.991106 = 0.008894 
c) - 170- 175 X = 170 then Z0 = 201.JIO = - 0.79 
P-value = 1 - .P(- 0.79) =1- 0.21 4764 = 0.785236 
9~1 Using n = 16: 
a) ct = P( X S: 4 85 1 ,I = 5) + P( X > 5.15 I ,I = 5) 
= P( Jt - 5 < 4.85 - 5) + Pf .X - 5 > 5.15- 5] 
0.25 / .Jl6 - 0.25 / .Jl6 . 0.25/ .Jl6 0.2S / .Jl6 
= P(Z S: - 2.4) + P(Z > 2.4) 
= P(Z S: - 2.4} + (1 - P(Z S: 24)) 
= 2(1 - P(Z S: 2.4)} 
= 2(1 - 0.99180) = 2(0.0082) = 0.0164 
b) ~ = P(4.85 S: X S: 5.15 1 p = 5.1) 
= P (4.85- 5.1 < x - 5.1 < 5.15- 5.1) 
0.25 / Tt6 - 0.25 / .,Jf6 - 0.25 / .,Jf6 
= P( -4 S: Z S: 0.8) = P(Z S: 0.8) - P(Z S: -4) 
= 0.78814 - 0 = 0.78814 
1 - ~ = 021186 
c) With. larger sample size, the value of a decreased t"rom approximatelyo.os, to 0.016. The power declined modestly from 0 .287 to 0211 
while the \'SI~e for a declined sttbstaotially. lfth.ctest ''ith. n = 16\'l'er-eoondt!.cted at thea \'alll.eof 0 .1l89. then it would !'..'t\'f:greater pm,·cr 
than the test ~itb n = 8. 
x - Po 9.~ P-value = 2(1 - <P(IZol)) where z0 = .Jn 
u l n 
a) 
b) 
c) 
-. - - ? h • - 5.2 - 5 ? 26 x - ) .- t en •o - Ts -· 
.25 / 8 
- 47 b 4·7 - 5 339 x = . t en z0 = _25 1.J8 =- . 
P-value = 2(1 - <I-(3.39)) = 2(1 - 0.99965) = 0.0007 
x = 5.1 then z0 = 
5
·
1
-
5 
= 1.1313 
25 / :rs 
P-value = 2(1 - .P(L1313)) = 2(1 - 0.870762) = 0.2585 
9·2.5 X- bin(l.;.04)Hu: p = U4aodH1: p" O.tl 
P1 = 4/15-= 0 .267 P2 = 8/15 = 0.,;3;3 
.-\.o:lept Region: 0 .267 ~ jJ ~ 0.,:;33 
Rcj~Region: jJ <tJ.267ot jJ >0.$33 
UsetJt.;e nonnaJ approximation (or p<~rtsa)and b) 
a) Wit-en p = 04.a = P(j; < 0 .26]) ... P{p > O.,;.jj) 
= p z < 0.267 - 0.4 +P z > 0.533 - 0.4 
0.4(0.6) l4(0 6) 
15 15 
= P(Z < - 1.05) + P(Z > LOS) 
= P(Z < - !.OS) + (1 - P(Z < 1.05)) 
= 0.14686 + 0.14686 
= 0.29372 
b) Wbeop=O.l 
" = P(0_267 < 0 < 0533) = p 0.267 - 0 2 < z < 0.533 - 02 
" -. - 0.2(0.8) - - 0.2(0.8) 
secuon'9·2 
= P(0.6S < Z < 3.22) 
= P(Z < 3.22) - P(Z < 0.65) 
= 0.99936 - 0.74215 
= 0.2572 
15 15 
9·27 n.cprob!emstatement implieiHo: p = o.6,H1: p > o.6aDd&l'in~Ss:naoceptanoe: regiooas f; < ~~~ = 0.80and rejection region 
as jJ >o.SO 
a) a =P( .f; > 0.80 i p = 0.60) =P Z > 0·80 - 0·60 
0.6(0.4) 
500 
= P(Z > 9.13) =I- P(Z:S: 9.13) "' 0 
9 ·29 a) Ho :11= 10.H1 :11> 10 
b) H0 :Jt=7,H1:wt7 
c) Ho :11-=5.H1 :Jl<5 
9<P a) <t=O.Ol.tr.ena=z.a•2.;')3 
b) a=o.o.;,tltcna =Za•l.6.$ 
c) <t=O.l.lf!.ena =:t:awl.29 
9-3<\ a) P·vaiU<: = >{1 - <l>~t !;,Ill= ' {1 - .p~,.OS)) •0.04 
b) P-val.e= ' {1-<l>{I J;,ll= 2{1-1>{1&!))•0.066 
c) P-value = 2{1- <I>~tl;,IJ) = 2(1- <l>{04ll• 0.69 
9·35 a) P-valne=W~l=.P{2 .0S) "o.91) 
b) P-valne = <1>(7.()) = <l>(-1.84) •0.03 
c) P-vaht.e = <P(7.()) = <1>{04) w0.65 
9·37 a) StOev= /N<SEMean)=0.7495 
19.889 - 20 
z0 - - 0.468 
- 0 75 I -./[6 
P-va1ue = 1-<I>(Z0) = 1- <I>( - 0.468) = 1- 0.3199 = 0.6801 
Because P-vaht.e >a= o.o.; wefail to rej~ the null 1\ypo~is tJ!.at \1 = ~>Oat tlt.eo.os leo.oel of significanoe:. 
b) A one...sidi'd test because tle.aJtemative ltypotl'.csis is mu > 20. 
- (j - q 
c) 9.;'%Cl ofthemesnis X - Zo.025 r < Ji.·< X+ 20.025 r 
-vn " " 
19.889- (1 .96) o;;; < !1 < 19.889+(1.96) o;;; 
-v10 -v10 
19.4242 < ,u< 20.3539 
d) p . .,.aJlle = 2h - <l>(Zo)J = ::!h- <1>{0468)) = 2h- 0 .6801) = 0.6398 
s 2.365 
9-39 a) SEMeanfromtiwsamp!estsndarddeviatioo = JN = .Jf2 = 0.6827 
b) Aone-s.idOO test because t~.ealte:rna tive hypotltesis is mu > <)9. 
100 039- 98 
c) l f tt.e null h\pott.esis iscltanged to tJ!.ep = 98. z0 = f1'7\ = ~ .82.5.3 
. 2.51 "12 
BecatseW{~.S~,l) is close to 1. tf!.e P-vahze = 1 - q,(~.8~) = O.DO:!. is ' 't':f'Ysmall and close too. Ttns. tlte P-vaJue <a= o.o.;.and we rejectfl!.e null 
~.ypothcsisat lf!.e o.o.:; I~ of s.ignitlcance. 
d) 95% !owerL't of the me.'lo is X - Zo.os...;.,. < }.1 
"11 
100.039 - (1 645) 2~ < 1./. 
v 12 -
98.8518< I' 
100.039- 99 
e) tftheaJternativehypoth'i".S.isischangedtotbemul<9¢, Zo = ~ r;;; = 1.4397 
b l " 12 
P-vaJue = 211-~V"1>)J = 2h-q.{l4$97l1 = '211-0.92..;0) = 0 .15 
Becali:SC tt.e P-value >a= o .o.; we fa iJ to reject tt.e null l<ypothes.isat tt.eo.o.; levcl of significaooe. 
9"4l a) 1) T~.e -p.'l rameterofinteret.t is t~.e tree mean crankshaft wear, p. 
>l fl, : p=S 
3) H1 : p 11j 
x- p 
4l Zo = u I .Jii 
.;) Reject Ho if zo < -1.fJj'J. where a= o.os aod -1.().02.5 = - 1.¢or zo > 1<0f'J. where a= o .o.; andl().O'J.S = 1.96 
6) X ='2.78.a=0.9 
z = 2.78- 3 =-0.95 
0 0.9 1.J15 
7) Because -0:95 > - 1.9(> fa il to reject tJ!.e null hypotltesis. Ther-e is not sufficient e'\ideooe to support thecla.im tlt-c mesncrankshaft 
wear differs from;3a ta = 0 .05. 
b) 
6- <I>[- + 3- 3.25 ) <~>f 3- 3.25) 
, - "o.o2s 0.9 1 Ji5 - - zo.o:o + 0.9 1 Ji5 
= 4>(1.96 + -LOS) - <P(-196 + - 1.08) 
= <l><o.SS) - <l>(-3.04) 
= 0.8lO.'j7- (0 .00ll8) 
= 0.809j9 
(z.n + zl "2 _ (zo.o25 + zo.10? u2 (1.96 + 129f(0.9f 
c) l1 = o2 - (3 75 - 3f = (0 75)2 15.21, 11 ~16 
9-43 a ) l ) The parameter o( interest is t~.e true mean battery life in OOurs,p.. 
>) H, :p=40 
;3) H1 :,11>40 
- _ X- 1./. 
4l "o - u l .fii 
5) Reject Ho if11l > Za wherea = 0 .05 and1'0.o5 = 1.65 
6) X =40,S.O= l .~ 
z = 40.5- 40 = 126 
0 1.25 1.JIO 
7) Because l .i6 < l .65 filii to reject H.o and oonc:lcdett.e battery life is oot significantlygrester t~.an40 a t 
a= o.os. 
b) P·\'S i ll~ = l-~{1.::! 6) = l- 0 .8961 = (UOJ8 
= <~>( 7 + 40- 42 ) 
cl !3 -o.os 1.25 1.JIO 
= ~{1.&5- • -5.06) 
=<I><- ~41) 
•0.000;3:!:.5 
( z. + z~ )2 u2 ( z0_05 + z0_10) 2 u2 (1.65 + 1.29)2(1.25f __ 0 844 d) 11 = = = 
82 (40 -44)2 (4? . , 
X - Zo.o;" I .Jii < 1./. 
40.5- L65(1.25)1.Ji0 < 1./. 
39.85 < I' 
11 ~ 1 
TP.e lo\...-er bound ofthe9()%oonfidi:ooe intervaJ must be greater than40 to \wifyf.t!.at tr.e true mean excoods40 hours. 
9-45 a) 1) Tit..: parameter ofioter('St is th:etnt.e meanspood. Jl. 
~) f-1() : p = 100 
3) H1 : p< 100 
X - 11 
4l 2o = 1 r o vn 
5) Reject Ho ifzo < -2Q\<o'~.erea = o .o.;a.00-2<,.o5 = -lJ>.; 
6) X = lO!L:!-,0=4 
102.2-100 
Zo= 4 1,JS 1.56 
7) Be::ause 1..56 > -1.65 faiJ to reject the nniJ llypot~ls. n.ere is imnfficicnt evideooe tooonclcde tf!.at t!tetrue meanspood is less th:an 
ulOat a=o.o.:;. 
b) .r.o = 1-56 then P-valm: = q,V.0)•0.9<1 
. [ (95-IOO),JS] -
c) .8 = 1- <I> - zo.o; -
4 
= I - <P(-16) - 3.54) = 1 - 4>(189) = 0.02938 
Poh·er:: 1- ~ = l- O.M!)j8:: (L97062 
( z • . + z.c-)2 u2 (z0.0, + zo.J<)2 o2 (I 6,- + I 03)2 (4)2 
" ' ' . . = 4 60 -- -
dl n = 82 = (95 - 100f = (Sf . , U :: ) 
e) 95%Confidoenoe: lnter\'a l 
~t<x+zo{{n) 
11·< 1022+ 1 65( Js) 
~t< 104.53 
Because 100 is inclcdro in the Cl there is not sufficient E'\idmoe to reject t~.-e null hypotf!..('S,js, 
9-47 a) l) The paramcter of interest is t.toe tnteawrrage b.a t1cry life.jl. 
,) H, : ~=4 
3) H1 : p >4 
X- 11 
4l 2o = r 
u l -vn 
sl Reject Ho if .to> 7.a where a= o.o.;and7<0.o.; = 1.6.;. 
6) X =4.0.;,o=0.2 
4.05 - 4 
z0 = = 177 0 21-v50 
7) Because 1.77 > tJ>.; rtject the nuJJ h)p()tlt.es.is. T}!.ere is sufficient f!\ideoce toocmch:-de tf!..at thetruea'l:tilge banery life excoods4 
h.ours at a= o .o.;. 
b) p.,'SJne = 1 - <Pv;:,l = 1- 4>{l .n ) •0.04 
c) .8 = <I>( z005 - (45 ~.i-[50) = 4>(1.65 - 17.68) = 4>( - 16.03) = 0 
Power =I- P = l - 0 = 1 
d) ll = (z •. + zA)
2 
o
2 
= (z0.~ T' z0.1)2 o2 = (1 6,- + I ~0)2(0 2)2 
" w . · -- . = 138 
82 (4 5- 4)2 (05)2 > 
n~2 
e) x - z0 os [,fn) < ~t 
4.05-165(~)9 
4.003 < 11· 
Becanut.J!.e lm~·er limit ot'tlte Cl is just s lightlyabovt4. weoonclude th.:a t a\wage lift is gresterth.an4 lt.oursata = o .o.;. 
9-49 a) o=O.Ol,n=20,t.hecriticaivalue=-2-539 
b) a= o.o.;. n = 12-. thecriticaJ vah.lf! = 1.79(> 
c) a= 0 .1, n = 15. tt.ecritical \oaJne = lJ45 
9·51 a) 2·0.02..5 s p s 2•o.o5 theno.o5 s l':: 0 .1 
b) 2•0.'()~:: p s ~·o.o.; t.J!.eno.o.; s p s 0 .1 
c) 2•0.25 :: p s ::~•o4 dtetH).05 s p:: o.s 
9·53 a) 1-b .05 .s p s 1-0.02..5 U..en0.95 s J':: 0 .975 
b) 0.02.5 s p:: o .o.; 
c) l-b4 sp:o: l-0.2$ t.ltcno.6sp::0.75 
9·.55 a) degreesoff'l'«dom=n-1 = 10-1 =9 
s s 
bl SE Mean= r,; = = = 0.296, !hen s = 0.9360. 
v N v10 
10 
= 12.564-12 1.905 
0.296 
r0 = 1.90.; withdf = 10 - 1 = 9. The P-vaJ .. e faJJ.s between two vaJues: 1.83;3 (for a= o .o.:;)and ~.26~ (for a= 0.02.,5).so o.o.; = ~0.0:!.,5) < P· 
vsJu.e < a(o.o.;) = 0 .1. Th.e P-vaJu.e > c = o .o.; so we fai l to rej<lct tJ!.e nullltypothcsi.sat the o.o.; levtl of significance. 
c) A f\,·o-s.idfld test btc.u<Sethe alternathoe ltypat.lbes.is is mu not = 1.2. 
d) 95% two-sidedCl 
x-i002s,{ f,; ) < P < x + Foo2s,{ Jn) 
12.564- 2262[0]5°] < 11 < 12.564+ 2.262[ 0]5°) 
11.8945 < 11 < 13.2335 
e) ,!,'uppos.e that tlteaJtemath~ ltypothesis is changed top > 1!!. Be::a1:se ro = 1.90.; > ro.os.9 = 1.8~ v;e reject fl>£ nuJI ll.ypofl!.esisat tJt.e 0.05 
IC\'tl of significance. 
t) Reject t.lte nuJJ ~.ypothesis tJ!.at the p = 11-5 \WSustJ!.ealternati'.l: llypoth.esis{JI * ll,S)at the o.o.; Jt"\l!! of significance because t.r.e p = ns is 
not include intlt.e9.;'% tv.·o-sidedCI on the mean. 
9·57 a) 1) Th.e paNimeter of interest is tft.e true meanofbody,..-eigltt.p. 
~) f-lo: p = JOO 
J) H1: 11 * 300 
. X- Jl 
4)VIo = r. 
sl vn 
.;) Reject Ho if Ito I > laf:i.,n-1 wlterea = o.o.; and laf:i.,n-1 = 2.0.;6 for n = 2? 
6) X = j::i..5496.s = 198.786. n = 27 
325.496 - 300 
to = 198 786 I ffi 0.6665 
7) Because 0.666.; < 2.0.;6 we fa il to reject the null hypothesis. Tlbere is oot s-ufficient t'\idenoetooonrJu<teth:at t~.etrue mean body 
v;cight differs from 300 a t a= o.o.;. We ~.a\l! 2•0.2.5 < P·\'ahw < 2•04. 
Tr.at is. o.:; < P·\'Siu.e < 0.8 
b) We can reject the: null hypothesis if P-vsJue: <a. The P-vaJue: = -2•0.1554 = 0-5108. n.erefon. tlt.esmaUest lt'\l!l of signifieaooeat w~.ichwe 
can reject tlt.e null hypothesis isapprox.imateiyo,s1. 
c) 9.;% twosictedoonfidenoe intct\'aJ 
X - lo.02S,26 ( Jn) < J1 < X+ 10.025,26 [ .Jn) 
325.496 - 2.056(19~6) < J1 < 325.496 + 2.056[ 19~6) 
246.8409 < J1 < 404.1511 
We fa il to 11!ject the: null hypGthi'Sis bees use tP.e hypothesized \'SJueof 300 is included \\-ithin theconficteooe inteMI. 
9·59 a. 1) The paNimeter of interest is tf!.e true mean female bodytemper;ature.p. 
') H, :p=98.6 
3) H, :p•9Jl.6 
X- Jl 
•l lo = r 
sl -vn 
.;) Reject Ho if! to I > laji.n-1 w~.erea = o .o.; and laji.n-1 = ~.064 for n = ~ 
6) x = 98.264 , s = 0.4821, n = 25 
t = 98 264- 98 6 = - 3.48 
0 048211./2s 
7) Becau:se348 > :LOf>4 rfject the: null hypGthi'Sis. n.ere is sufficient evidence toooochr.de that t.Jt.e trlle mean femaJe bodytemperiltute 
is oot cqu:aJto98.6 )F' at a= o.o.;. 
P·\ 'Siu.e = 2:• 0 .001 = O.OOi 
Normal ProbJ~bility P lot for Q-3 1 
IH e .. u ... t, .. . Mil. <a 
.. 
.. 
" 
I I ; 
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Data 
b) Data appear to be oorm.aJiydistribnted. 
c) d = ~= 111- Jlo I= 198 - 986 1 = 1.24 
q q 0.4821 
Using tJ!.eOCcuM.ChanVII e) fora= o .o.;.d = 1 .~.aod n =~.obtain ~· o aod powerof1- o •1. 
d) d = ~ = I 11- J1o I= !982 - 9861 = 0.83 
q q 0.4821 
Using tlteOCcut\~.O.artVH e) for a= o.o.;.d = 0.8,3,and~c0.1 {Power= 0 .9), n = -2.0 
e) 95'% 1\\'"o-sidedoonfidenoe intet'\'31 
X - fo.oo5,24 ( .Fn) < J1 <X+ 10.025,24( .Fn) 
98.264 - 2 064( 0; 1) < Jl· < 98.264 + 2 064[ 0; 1) 
98 065 < J1 < 98.463 
Concllld:e tf!..at t~.e mean female body temperature differs from 98.6 boos~ t.J!.e,'Silleis oot ioclud.OO inside tlt.eoonfidence inte:n oal. 
9~1 a) t) Ttte paNimeter of interest is the true mean sodium oontent.p. 
:l) Ho : )! = ljO 
3) H1: 111: 130 
X - fl 
•l lo = r:: 
sl -vn 
.;) Reject Ho ifitQI > laj:!..o-1 where a= o.o.; andlaj:to-1 = ~L093 for n = :w 
6l x = 129.747 , s = 0.876 n = 20 
t = 129.747 - 130 = - 1.291 
0 0.8761../2o 
7) Ek!csuse 1.:191< :1.093 we fail to reject t~.e null hypothesis. There is not sufficient e~.idence that~ true mesnsodiu.m content is 
different from 130mgata = o .o.;.. 
From the rtable(rab!e V)th.cto \'Sine is ~,·eeo tlle\'Sine;of 0 .1 anci:0.2.) '~itf!. t9degrce.soffr«<iom. 
T}!.eref'ore. :!(O.l) < P-vahle < :!(0.2.;i)sod 0.2 < P-vsJne < o..;. 
b) Th.eassumptionofnonnalityappears to be reasonable. 
P.ct::.obii ty Plot of SodiU'I'I Contellt 
Nem4I.OBOO 
• 
-
·~' 
• ·~ f.' ,'(.I 
• 1·- -· -- - --- - - - - " » 
• 1- 1-· --- -- -- --- - ··- - ! -- -- --- - ~ 'M • 
·-
0110 
• 1- . --- -- -- --- - . ·--- -- --- -• 
• " 1·-
. 
-· 
. 
-- ----
, .. ,. 
- -- --~ • • . -- - ~~ . -l • . -- - . - ~ 0 1·- . -· -- -- - ... 
» 1·- . -· -- -- -.. . - ... c-
• 
. 
-- • --- - -- -
1·- • c-• . -· •• --- . - ... -
' 1·- . --- -- -- . -- - . -- -- -- --- -
• 
' 
"' "' 
lN 
"' '" "' "' sodt.rno:rtcnJ 
c) d = ~= l fl- flo l = l!30.5 -130 1 0.57! 
" " 0.876 Using th.eOC~.ChartVth)fora = o.o.;.d = 0,57,and n = 20, we obtain ~· O.:Jand tlti:powerofl- 0.:)0 = 0 .10. 
d) d = ~= lfl- flol = l 1 30 1- 1 30 1 = 0. 114 
(j (j 0.876 
Using theOCCIU'\oe, Chart vn e)(or a= o.o.:;. d = o .u .and ~· o.~ {Power = 0 .7.5), tltesamJ)!esiJ4do not o:teod to the point d = 0 .114 and 
~ = 0 .1j. W~canoonclll<ie th:a t n > 100. 
e) 9.5% two-sidedoon6dence intervsJ 
X - I 0.025,29 (.in) < f1 < X + 1o.!Yt5,29 [.in l 
129.747 -2.093(~)< /1 < 129.747 +2093[~6] 
129.337 < /1 < 130.157 
There is ooe>.idcnoe th:a t t.Jt.e mean differs from 130 because that \'SJil!e is inside fl!.eocmfidenceintenoal. n.e result is t~.esame as part (a). 
9~~ a) In order to use t statistics in ~.ypoth.esis testing, we ooed to assume that th;e undttl)ing distribl!tion is oormal. 
1) The patilmeter of interest is tlt.e true meanoxygenoonoentriltion.p. 
:l) Ho : Jl= 4 
3) Hl: JH4 
X - fl 
•l to= I r 
s -vn 
5) Reject Ho ifftol > rajz.n-1 where a= CJ.ou00to.oo.;.l9 = :!.86t (or n = ::!'0 
6) X =$.:!.6.;.s = :l.12J,n=:lO 
to= 3.265-4 = - 1.55 
2.1271../20 
7) Because- ::!.861 < - 1-55 fail to reject t~.e null hypothesis. There is insufficient eo.idcOCi' tooonclnde that the true meaooxygcndiffers 
from4atn=0.01. 
P.Vslue: 2•o.o.; < P·\'Sitte < 2•0.10 tJ!.crcfore0.10 < P·valne < 0.20 
b) From flt.e norms! probability plot. U..e nonnaJityas.sumption seems reasonable: 
• 
• 
• 
• 
-· 
c ~., 
•• ~ ~ 
• 
Probab ili ty Plot of ~Concentration 
...... 
•• r 
• I • • 
' 
• 
' • 
.. 
c) d = ~= 1./l - flol 13- 4I = 0.47 
17 " 2.127 
7.5 
Using tlbcOCcu.noe,CJ!.artV[( Ofot n= O.Ol ,d = 047.a.ndn = :!0, wegct jl.-0.70 aodpowerofl-0.70 = O.j:O. 
<l = d =~= IP- flo i = I2 5 - 4 I = 07l 
" (j 2.127 
Using tlbcOCcu.noe, CJ!.art V[( 0 fora= O.Ol,d = 0 .71,and tl• 0 .10 {Power = 0 .9() ), n = 40. 
T~.e9.;'%oonfidmoe ioten'8J is: 
X - fat2,1>-{Jn) <_it < X +to/2,n-{Jn) = 1.9 < p < 4.62 
8ecacse4 is \'lit~.intl.eoonfidence intervaJ. we fail to .rejcc:t the nuiJ hypotft.e.s.is. 
9~5 a) 1) Tlte par-ametero finteriSt is t.Jt.etrne mean tire life.11. 
:l) H() : _I!= 60000 
j ) Ht: ,l1 >60000 
4) !,) = :11 
.;) Reject 1--1() iff() > ta,0 •1 wJt.crea = o.o.; aod ro.o.;.1.; = 1.7.;:3 for n = 16 
6) It= 16 X= ()t),lj9.7 s = j 64_;.9<f 
= 601 39.7- 60000 = 0.15 
I<) 3645.941 Jf6 
7) Because 0.1.; < 1.7.53 fail to reject the nllll hypothesis. There is insufficient evidence tooonclcdetl"-.at t~true mean tire life is greater 
tltan 60.-000 kilo meters a t a= o .o.;. Tlte.P·vah:e > 0 40. 
b) d = .£. = I 11 - flo I= 161000-60000 I= 027 
(j " 3645.94 
Using tt.e.OCCI!n'e. CJun1 \i'll g)for a= -0.0.;. d = 0 .2J, and 13• 0 .1 {Poh'et = -0.9), o = 4. 
Yes., t.hesamplesit.eoft6 was sufficient. 
9~7 lnord.tr to cse a t statistic in hypotltais testing, we need to assume tJt.at theund;erlying distrib1:tio n is normal. 
1) Tlte par:ameter ofirrt:'!rest is tJt.e t:ru;e mesocurrent.p . 
2) fio: !l=jOO 
j) Ht : J1>300 
X - jl 
•J!o = I r 
s -vn 
.;) Reject Ho if to > fa.0•1 wh . .-:rea. = o .o.; aDd ro.o.;.9 = 1.8.:)3 for n = 10 
6) n=lO X =jl7.2S=lS .. i' 
t = 317.2 - 300 = 3.46 
0 15.7 1-./10 
7) Bec3ustj46 > 1,8J3 reject tJt.;e nl!ll hypothes.is. There is sufficient evid.enoe to indicate that tJt.;e true meancurrm is gr~ter thanjOO 
microamp.sat a= o.os. Tl:e0.00 2.,5 < P·\'alne < o .oo.; 
9-69 a ) I nord.et to use t statistics in hypotlt.esis testing. we nood to assume that th(! u.oderl)ing dis triblltion is oonnal. 
1) n.e panmeter of interest is the true mesndistance,p. 
l) Ho:p=~80 
3) H1: p >:.l80 
X- Jl 
•l to = s! Jii 
5) Reject Ho if to > ta.n-1 w~..erea = 0.05 and I<J.o5,1)9 = 1.6fJ04 for n = 100 
6) X = ~60,3s = 1341 n = 100 
= 260.3 - 280 =-14.69 
to 13.41/ .JiOo 
7) Be::ause -14.69 < 1.6604 fa iJ to n:ject the nniJ llypot~is. n.ere is imnfficicnt E'\i deooe to iodicate tha t the true meandistanoe is 
greater than :!.80 at a= 0.05. 
From Table V the to value inabsolntevalne is greater t.Jt.an tlt.e,'Sitteoorresponding to0.0005. 
Therefore. tlt.e P-valne > o.m s. 
b) From thenormaJ probabiJityplo t, tlt.enormaJitya.ssnmptions.-"'f!ms reasonable: 
P·ro bability Plot o f Distance for go lf balls 
Normal 
c) d = ~= I Jl- .uo l 1290- 280 1= 0.75 
17 17 13.41 
Using theOCCUT\<t,Chsrt vn g) for a= o.o5.d = o .n.andn = 100.obtain ~-o snd:po'"et = 1 
d) d = ~= I ll- .uo l = 1290 - 2801 = O 75 
17 17 13.41 
Using theOCCUT\<t, Cbsrt vn &.Hora= o .o5.d = o .n.and ~·O.l-0 {Po'"et = o.SO). n = 15 
9·7l a) a= 0 .01. n = 20. from TableVwefind tlt.efollowingcritical valnes6&andj.S,s8 
b) a= 0.05. n = 1.2. from Table Vwe fiod the follo,.,.;ng ctiticaJ \'ak-t.i 3.8~ a.nd ~1 .9~ 
c) a= 0 .10, n= 15. ftom TableVwefind the followingcritical \'8ltt-t.i6,S'J and ~.68 
? 
9-73 a) a= 0 .0!, n = ~0. from Tab.!eVwefiodXi-a,t!- 1 = 7.63 
2 4 "7 b) a=0.05,n =l.2.fromTableVwefiod X t- <U!- l = . )J 
? 
c) a=<UO.n = ls.fromTableVwefiodXi-a,.n- 1 = 7.79 
9·75 a) O.l<l-P<O,stlt.enO,s<P-vaJne<0:9 
b) O.l< l- P< O,s t1t+m0S <P·\ 'ahi.e < 0 .9 
c) 0 .99 < 1- P< 0.~5 then0.005 < P-valne < 0 .01 
2 9·n a ) tnorder to use t.Jt.ex sUttistic in hypotlw.s.is testing and oonfid.enoe intervaJ oonstnlction. we nero to assume that tft.e underlying 
distribution is oormal. 
1) Tbc parameter~ interest is the truestaod:arddf:'.iationofpertbnnancetimeo. How~w, t he answer can be found bypctform.inga 
llypotr..r.sis test ono . 
2 2 
l) Ho : o =<I.'JS 
2 2 3) H1 :o >O.J.'j 
' 2 (n - 1)s· 
4) Xo = ., 
17-
;l Reject fl<> ifXij > x;, _1 where o: = 0.05 and x5.o5,16 = 26.30. 
6) n = 17.s = 0 .0<) 
2 2 
2 = (n -1)s = 16(0.09) = 0 ?J Xo , ., ·--
17- .75• 
7) Becauseq_.2;) < 26..:)0 fa il to reject. f-lo. Tt.ere is insufficient f!\idmce tooonclude that~ tru.e \ 'Srianoeofperfonnanoe timeoontent 
exooodso.a5 ata=0.05. 
Be::auseX o = 0 .2$ the P-vahte> 0.995 
b) TJ:e9s%oOMi&doonfidmoe intervaJ gi~n below inchr.des tJ!.evah.'(:0.75· Tt.erd'ore. weare oot be able tooonclll.de tha t thestao&ard 
df:'.ia tion is gri'lltet U...ao o .7 5. 
2 16(.09) < 172 
26.3 -
0.07 <u 
• 9·79 a) r oor<h."'t to u.sethe.x statistic in hypothesis t e&ting andoonfideoceintm•aJ oonstn:ction. we oeed tosssume t~ll.'l t the. underlying 
distrib1:tion is norm.al. 
1) The parameter of ~tere;t is tl>.etrui!staodardde\iationoftitanillm percentage, a. Howeo.w, t.beanoh·er can be found by pcrtorming 
a hypotf!.es.iste.stono . 
• • :!) Ho :o = (0:1.;) 
• • j) Hl: o -'t (0 .2,;) 
)> -"(1'--1 ~'r:)s:_-4l xo = -
172 
.;l 
2 
Reject Ho ifX5 <x f-o.n,n-1 wh.er-ea = o.o:; and x5.99s,so = 32.36 or x5 > xl,2,n-t where a= o.os and 
Xo.oos,;o = 71.42 forn=.;1 
n = 51 s = 0 37 v 2 = (n -l)s2 (,) > • ...\ 0 2 
17 
= 50(037{ = 109.52 
(0.25) 
7) Because lOC),s.1 > 71<4.2. we reje:t Ho· The tru.estand3.rd C.e\'iationoftitanium percentage is significantly different from 0 .2,5 a t a= 
O.Ol. 
p.,oalne/1 < o.oo.;, tl".en p.,oalue < 0 .01 
b) 9.;9/tocmfid.elX'C intl!t\'31 foro: .2. 
Hrst find the confidence interval foro : 
For a= o.-o.s andn = .;l.n = 51, X.!n ,n-1 = t5.02S,SO = 71.42 and Xf-.:..n ,n-1 = x5.975,51) = 32.36 
S0{0.37i < 172 < 50(0.37)
2 
0_096 :;;o'l:;;o.2115 
7142 - 32.36 
TakiQg the square r.oot ofthe endpoints of this intervaJ w~obtaio.O.Jl < o < 046 
Bo::ause 0 .25 faJ is below the lO\'.·er confidence bou.ndwewou.Jd oonch:.deth.at tlt.e population standard deviation is oot f:<I)UI I too.~.;. 
9-81 a) tnordier to~ t~.ex 2 statistic-in h)'J)I)th.esis tc.<.ting andocmfictenc.cintervsl oonstruction. we oo.."d to assume that the underlyjng 
distriblttion is norma I. 
1) TJte pa.rameter ofinteN:St is t.bestaodarddc\iationof tire li fe,. a. HO\'le\'ef', t~.eanswer can be found byperfonning a hypothesis te&t 
ono . 
• • 2) li():o =-4000 
a • 3) H1 :o <4000 
' 2 (n - l)s-
4l Xo = 172 
5) Reject Ho if x5 < xf-« r.- 1 wlt.erea =- 0.05 aoox5.9515 = 7.26 for n = !6 ~ 1 • • 
6) o = 16.s = tl64s.94) 
,2 = (n - l)s2 = 15(3645.94f = !? 46 Xo 172 40002 -· 
7) B«:ause 12.46 > 7;26 fail to reject f-lo. There is oot sufficient t".idenoe tooonclude the true standard deo.i.a tionoftire life is less titan 
4000 km at<2=- 0.1l5. 
P·\'alue =- P(x < t246)tbn5 d;:greesofftttdom. Thts.o..; < 1- P-va!tre-< 0 .9arxi0.1 < P-vaJne < o.s 
b) T'!f:95'%0DE:s.id.."'<loonfidence intervaJ bei0\\0" ioch:des t.be vshle4000. TJ!.erefore. we ate not able toocmc:h<.de f.!!-.at tlw\arisoce is less than 
4000 . 
173 < l 5(364 5.94)2 27464625 
- 7.26 
C7 < 5240 
• 9.S3 a) lnorder to use thex statistic in hypot~.csis tcsting.aOOoonfidenc:e intmal eonstruction, we need to assume tl>.at the und~rl)ing 
distribt:tioo is nonnal. 
) f • . • 1 The parameter of interest is tl':.e true.varisnceo sngaroontcnt. o . T!ti'amwer can be found byperforminga hypotlteiis test on a . 
• 2) Ho:o =- 18 
a 
3) H1: o '1':18 
2 
' (n - l)s 
4) x,; = 
172 
2 < 2 2 2 70 or2 > 2 . 5)? Reject l4J ifXo XJ- «12,n- l wft.erea =-0.05 andX0.975,9 = · Xo X«,1,n- i where a =O.os sod 
Xoms.9 = 19.02 for n = 10. 
(,) D =- lO,s =- 4.8 
' (n - l)s2 _ 9(4.8)2 xo = -'--,,:...__ 
17- IS 
JLS2 
7) Because n,s2 < 19.0~ fa il to rejo::t flo. Th.ere is insufficient t'\ideooe tooonc:lud~ that the true wrianoe:of sugar oontcnt is 
significantlyd!fferent from 18st a=- 0 .01. 
P-valwe: Tt.c:x. o is be1w00l 0 .10 ando.,so. T'herd'or~, o.~ < P-W~Ine < 1 
b) U.s.ing tlt..ech.art in thcAppendh:. wi t!;).= ~and n =- 10 wefiod jj = 04,5. 
c) Using tlt<charUnth<Appcndix, ~itn ). = ,m = 1.49 andp = 0 .10 , o = ;JO. 
9-85 a) A one-sidEd test booat:M! ~.be alternative hypothesis is p < 0.6 
b) Ttw:test is basOOontheoormal approxim.atioo. ltisapproptiate bcc:atsenp > 5 andn(1--p) > .;. 
X 287 
c) sample p = - =-= 0.574 
N 500 
20 
= X - np0 = 287 - 500(0.6) = - 1.1867 ~npo(l-p0 ) .Jsoo(0.6)(0.4) 
P-vaJt!f: = W{-1.1867) = o.nn 
Th.e%'% uppcroorJidence intm'8J is: 
p < p +zo~ 
,.,...,....,..,,...,..,.,_ 
•
[) < 0.574+ 1 65 0.574(0.426) 
500 
p < 0.6!05 
9-87 a) l ) n .e p.'l tameter of interest is the true fr-action of rejected parts. 
>) fl,j : p = 0.03 
3) Ht :p< 0 .03 
4) : citlt.er a pproach will yield thesameoonclusion. 
.;) Reject Ho if 1.() <- Zc where a= o .o.; and -t.a = -7.o.o.; = -li>S 
6) x=!On=;oo p= ;~O = 0.02 
Zo = x - 11p0 = 10 - 500(0.03) =-1.31 
Jnp0(1- p0 ) J soo(0 03)(0 97) 
7) 80\:ause -1.,31 > -1.6.5 fa.il to rqect the null hypotltes.is. Tllere is not enough e"'ideooe to conclude thatthe true frilctionof rejected 
parts is less t.han0 .03at a=-o.o.;. 
P-vaJc.e·= W{-1,31) = 0.095 
b) Tr.e upper one-sickd 9.;%confidence interval for tt£ fraction of rejected parts is: 
p <p - z r(l j ) 
0 ll 
r:-:-:-=-=:-
p < 0 02+ 1 65 0.02(0.98) 500 
p < 0.0303 
9-89 a) l) The psrameter ofinterestis the tntesuccess ra te 
~) Ho :p=0.78 
3) H1 : p >0.78 
4) ci t.r.er a pprosch \.,rill }ieldtlbesameronclu:sion. 
.5) Reject Ho ifr-4) > 7.a.Siooe ft,oe \'8JUtefor a.is not given. Weassu.mea = o.o.; and 1.a =- 'l().o.; =- 1.6.;. 
6) x=>B9n=3;0 p" = 289 !0< 0.83 
350 -
Zo = x - np0 = 289 - 350(0.78) = 2_06 
J np0 (1-p0 ) J350(0.78)(0.22) 
7) Becatse ~.06 > 1.6.; reject the oull hypot.hf:sisandoonclnd.e the tn:cst:OOtSS rate issigniticantlygreater than0 .78.at a= o.o.;. 
P·\'3h z.e = l-0:9SOj = O.Ol97 
b) The9,s% lower confidence inter\'al: 
p - z.~<p 
0 83 I 6' 0.83(0.17) < 
. - . ) 350 _ p 
0.7969 < p 
Ek!cattSC tt.c hypGtbes.iu:d \"Slt>i! is oot io tlt.i!confidmce intenaJ {0.78 < 0.79f:19), reject the null h}poth.es.is. 
9-91 a) 1) Th.e pal'ilmeter of interest is the trueperttntageof(ootball lbelmcts tr.at contain fla"'-s. p. 
>) fl,j :p =O.l 
J ) H1 :p >O.l 
x - npo 
4) :either a pprooclt "'ill )i eld tlt.esameoondusion. 
.5) Reject Ho ifz.o > 1.a where a= 0.01 and 7.a =ZI).oj = .i~ 
6) x=l6n='oo P = 21~0 = 0.08 
zo = r= .. P~P,l;o""7= ---r7o=c.oocsc==ooc.1oc0 """ = - 0.94 
p0 (1- p0 ) 0.10(1 - 0.10) 
n 200 
7) 80\:ause -0.94 < 2.3:3 fa il to reject tf!..e null hypothesis. Tlt.ere is ooteoou:ghe-.id.enre tooonc:hl:dethatthe proportion of football 
f!.elmets with !Ja \~'S exoeeds.10%. 
P·vt~J c.e·= 1 - q_.(- 0.9cf) = 0.8164 
b) The9!)~ IO\'Ietoonfidenoe interva l : 
j;(1- jJ) <p 
n 
.OS- 2.33 0.08(0.92) < 200 _ p 
0.035 <p 
Bo::ause theoonfi~ intmoaJ contains tlt.e ll)1)0thesiudva.lu-e{0.035 ~ p =O.l)we fa iJ to reject th-e nu.ll h)'J)Oth.esis. 
943 1) The pa101meter of interest is the true proportionofba«eries t l>.at fail before48 hours, p. 
2) Ht>: p ::0.()()2 
3) H1: p < 0 .00:! 
4) ; cither approach will )ield t~.esameoonclu.sion. 
.;) Reject Ho if'l(l < - 2a w~.ereo = 0 .01 and -Zo = -1.0.01 = - 2..')3 
6) x = 15 n=;ooo p· = ___!2.._ = 0 003 
5000 
0.003 - 0 002 = 1.58 
0.002(1 - 0.998) 
5000 
7) Because 1..;8 > - 2j3 fa.iJ to Ttjec:t lite null hypothesis. 'tf!.ere is oot sufficient e-.ideoce tooooclude that tft.e proportion of cell ph.one 
ba«eries th.st fai l is Jess than0 .2%at a= 0.01. 
9-95 a ) 1) The paNimeter of interest is the true proportion of enginec:ranksr.aft bearingsoc.f!.ibitingsu.tface roughness. 
2) Ho:p=O.lO 
3) H1 :p> O.lO 
4) 
.;) Reje::t Ho if Zo > 1.a where a= o.o.; anci1'0: = 7.o.o.; = 1J,.; 
f>) X=I00=8; p = 10 = 0.118 
85 
: f!itf!.er approach"'ill )icldtf!.esameoonclu.sion. 
10 - 85(0.10) = 0.54 
.J85(0.10)(0.90) 
7) Because 0..;4 < 1.6.; fail to rejo=t tf!.e nuJI hypotf!.esis. Tltere is not eooughf!'.idenoe tooonclu.M that tl«: true proportion of crankshaft 
bearings exhibiting surface rong,hness o:ceecis o .10. a t a = o .o.;. 
P·value = 1-~{0..;4) = 0 .29.; 
b) p = 0 .1.;,p0 = 0 .10.n = 8.;.and%of2 = 1.96 
/3= q,[ Po - P + z~n.JPo(1-Po) l n ] - ii![Po- p - Zan.JPo(1- Po) I n) 
.Jp!J - p) l n .Jp(1- p)ln 
= q,r0.10- 0.15 + 1.96.J0.10(1 - 0.10) / 85) _ 1>(01 0 - 0 15 - 1.96.)'-0 .,..,10"'(1-_-.,0....,._ 1""0)-;/8~51 .j0.15(1 - 01 5) 185 .j01 5(1-015) / 85 
= i!>(0.36) - i!>(- 2.94) = 0.6406 - 0 0016 = 0.639 
r-;:-.,- ' 
[
Zaf2.JPo(1-Po) - z~.Jp(1 -p)l-
c) n= 
P - Po 
= [1.96.J0.10(1 - 0 1 0~ - 1.28.J0.15(1- 0.15)]2 
O.b - 0.10 
= (IO s5i = 117.63 ~ 11s 
9~ E.xp0.."1ed trequencyis found by miog t.J!.e Poisson distribution 
e->_xx 
P(X = x) = where .A = [1(1)+ 2(11)+ . .. + 7(10)+8(9))/75 = 4.907 
x! 
~tim.ated mean= 4.907 
Value I 2 3 4 5 
Obs=ed 1 11 8 13 II 
Frequency 
ExpectedFrequency 2.7214 6.6770 10.9213 13.3977 13.1485 
Siooe the first category has an expected freqnax:y less th:anj. oombioe it withfl!.e next category: 
Value 1-2 3 4 5 
Ob!.m'~d Frequency 12 8 13 II 
Expecred Frequency 9.3984 10.9213 13.3977 13.1485 
Thedegreesoffrtedomar-e k- p-1= 7-1- 1 = 5 
.a) 1) Tlte, oariab!eof inter.-:.st is lf!.e form ohJ!.edistribntion for lite number offlav.-s. 
:l) Ho: n.e fonn oft~.edistribntion is Poisson 
j) H1: The form oftltedistribution is not Poisson 
4) Tr.e te:ststatisticis 
2 ~(0; - E;/ Xo = L- ,.,. 
f=l i!;i 
sl RejectH, ;fX~ > x 5.ol,S =15.09 ·rora=O.O> 
,2 -- (12 - 93984)2 (9 - 4.6237)2 -- 6 9)' )' 
6l Xo 9.3984 +"·+ 4.6237 · 
6 
12 
10.7533 
6 
12 
10.7533 
7 8 
10 9 
7.5381 4.6237 
7 8 
10 9 
7.5381 4.6237 
7) B«:ause6.955 < l5ll9 fail to rcj€\':1 H(). We are unable to tqect the null hypotltesis t!:;at t.hedistrib~tionofth.e number offlav.-s is 
Poisson. 
b) P-vaJne = 0.~7 (from Minitab) 
9-99 Use tf!.e bioomiaJ distribntion to get tlteocpo.."1ed frequencies "'ith tU mean= np = 6(0.~) = 1-5 
Value 0 I 2 3 4 
Observed 
Expected 
4 
8.8989 
2 1 
17.7979 
10 
14.8315 
13 
6.5918 
2 
16479 
Theo:pocted freql!.mc:yfor \'aJn.e4 is less tf!..anj. Combine tltisocll with \'aln.ej: 
Value 
Observed 
Expected 
Tf!.ed-egreesoffrcerdomare k - p- 1 = 4 -o- 1 = 3 
0 
4 
8.8989 
I 
21 
17.7979 
2 
10 
14.8315 
a) 1) TJ!.evarisbleofinterf'.St is the form oftlt~distri blltion for t~.e random variable X. 
:l) Ho: n.e(onn oftf!.edistribti.tionis bioomiaJ withn = 6and p = 0.'15 
j) H1: Th-e form of.Udistribu.tion is not binomiaJ t-ri tll n = 6and p = 0.2.5 
4)Th.e test statistic is 
2 ~(o, - E;l 
x(l =w 
i=J Ei 
sl Rejo::t H, ;rx~ > xt.os,J = 7.81 !ora= o.o; 
6) 2 (4-8.8989)2 (15- 82397)2 
Xo = 8.8989 + ... + 8.2397 = 10.39 
3-4 
IS 
8.2397 
7) Sinoe 10;,19 > 7.81 rfject Ho· Wecanoonchrde that tJ!.cdistribntion is not binomiaJ \\iilt n = 6and p = o.~ a t a= 0.05. 
b) P-vaJne = 0.01.55 (from Mini tab) 
9-101 Estimated mean= 49.6741 use Poissondistrib~tion"'ithA= 49.674 
AU expected frequencies are greater tlumj. 
Thedq:,rt"E$offroodom arek- p-l = 26- 1 -J = 24 
a) 1) Tlte,oariablt'of interest is th.e form oft~.edistribution for t!te number of cars passing through tf!.e intersection. 
2) Ho: The form ofthedistribn.tion is Poisson 
J) H1: Tr.c form of the distribution is not Poisson 
' 2 ~(0; -E,f 
4) Ttte teststatistic is:Xo = L.J E 
i= l . i 
sl Reject fi<J ;rx; > x~_05,24 = 36.42 for a = 0.05 
6) Estimated mean= 49.6741 
X~ = 769.57 
7) Becats.e 769-57 > > !)64~. reject Ho. Weeanoonclnde IJ!.a t tltedistribntion is oot Poisson a t a= 0 .05. 
b) P·t'Sin.e = o (found using Minitab) 
1 
> 
All 
9 -103 1 Tltevariab!eofintmst is hreakdov.<nS8mongs.hift. 
~ ~: Breakdown;areindepeod.entofs.hift. 
3 H1: Break<fao,.·nure oot iOO-epeodent of shift 
4 TJ:e test statistic is: 
s ThecriticaJ,'8JneisX~, = 12.592 : rora=o.o.s 
6 TJ:ecalculated test statistic is X; = 11.65 
? -.;. ' 7 Ekcsc.se XO 1 X0.05,6 fail torejoct f-lo. The data pto\ideins-ufficient evidmoetoclaim that macll.inebreakdownnnds"jft8tcd-epeodeot 
8ta=o.o.;;. 
P·\'alne = 0 .070 {using Minitab) 
9 ·10.5 1. Thevariab!eofinterest is statistics gradesandORg.rades. 
~. fio: Statistics gr-ades are independent of OR grades. 
3. H1: Statistics and OR grades are oot independent. 
4. The ti':St statistic is: 
.;. Tlbecritical valu.e is x: .. = 21.665 for a= 0.01. 
6. TlbecaJculatedteststatisticisX: = 25.55. 
v 2 > x2 
7. "' 0 -0.01,.9Th.cret0re. reject Ho and oooclndetJ!.at fl!.egradesare not iOOependent at a= 0 .01. 
P·\ 'altile = 0.002. 
9 ·107 1. Tf!.evuiableofintere.st is failurr.;of aoelectronicoomponmt. 
2. Ho: 1)-peoffailure is independent of mounting po5ition. 
3. H1: Type of failure is oot independent of mounting position. 
4. The test statistic is: 
r c (o - E)2 
2 "" lj u Xo = L.L-
i=l i =l Eq 
.;;. The critical valne is X;n, = 11.344 fora= 0 .01. 
6. TltecalculatedteststatisticisX; = 10.71. 
2 -l 2 
7. Be::ause '(o r XO.Ol,3 faiJ to reject Ho. n.ef!\idmceis ootst:.fficient to claim that ttoel)peoffailur-eis not irx$epeodentofthemounting 
position a t a= 0 .01. P·\'a lu.e = 0 .013. 
9 ·109 8) 1. Tr..evariab!eofinterestissn.ccesse.s. 
~ . ff(): Stl.<lOeSSOS are indq>rndent of sire of stooe. 
:). H1: stl<lOCSS('S8te not ind:.;:peodent of si1.eof stone. 
? 
2 r c (Ou-EuJ 
4. Theteststatisticis: Xo = LL g . 
i=l .i=l iJ 
.;;. Thecriticalvah:eisX~. = 3 .84 fora=o.o.s 
6. Ttt.ecaJculsted test statistic x: = 13.766 v.itltdetails below. 
2 2 
7. X 0 > XO.OS,!, reject Ho andoonclnde that the number of sucoess.es8nd tlt.estonesizear-e oot inckpendent. 
l > .... u 
55 25 •• 
66. 06 lJ. 94 BO. OO 
234 3G ,. 
212. 94 •1i. OG 270. C•O 
••• 61 350 
289. 00 6!.. 00 350. 00 
Cell \:!)!'\te:-,t.~ : Col~:':t 
£xpcct.cd C!m!'.t 
b) P-vall:e < o.oo.; 
9-m 8) 1. The parameter of inter-est is mcdiantitanit:.m oont:mt. 
2.. Ho : f1·=8s 
3. H1 :ji. t:8.; 
• 
4. The test statistic is t.beob.s.ernld number ofph.sdiffereocesor r = 7 for a= o.o.; . 
• 
.;. We reject H0 iftlte P-valneoorrespooeiing to J' = 7 is less than or equal to a= o .o.;; . 
• 
6. Using lf!.e binomisl distrib1:tion"'ith o = 20 aod p = os, p.,>aJn.e = lP(R !: 7 1 p = o.;) = 0 .1$1.5 
7. Conclusion: we fa il to reject H0 . There is not eoollg.hevidence to reject tt.e. manut8ctuw'sclaim thst tt.e. medinnoftt.e. titanium 
oonteotis8.;. 
b) J . Parameter ofintere.st is lf!.e median titanium content 
2.. H0 :fl.= 8.; 
3· Hl:Jl. t:8.; 
r+ - O.Sn 
4 . Teststatisticis Zo = O • r 
.) '\/ Jl 
.;. We reject Ho ifthe I~ I > Zo.02..5 = 1.¢fora = o .o.;. 
. - 7 - 0.5(20) 
6. Computatmn: .t0 = r.::;;:; = -1.34 .. 0.5-v20 
7. Condnsion: we fa il to r-eject H(J. Tt.ere is oot f!OOllgh: evidence tooonc:ludeth.at the median titanimn oontent differs from 8.;. Tt.e P· 
vsll:e = :i'P(Z < -1,34) = 0.180~. 
9·llj a ) 1. P<~Nmcter of interest is tt.e median ma.rguioe fat oontent 
2.. Ho : il· = 11.0 
3· H1:j1, •17.0 
4· 0=0.0.; 
. + 
.;. 'rf!.eteststatis tic is tlwob.sef'\'ed nu.mberofplu.sdiffereooesor r = ,3 . 
• (>. We rejtct Ho if tft.e P-vsh:.eoorresponding. to l' = 3 is less tJ!.anor eqll:31 ton= o .o.; . 
• 7. Using the binomial distributionv.itlt n = 6and p = o..;. tJt.e P·\'8lue = ~·F(R :.!:liP= o..;.n = 6) = 1. 
8. Conclc:sion: fail to reject Ho. Tft.ere is oot enough evidence: tooonclc.dethat tlte median fa t oontent differs from 17 .o. 
b ) 1. Parameter ofintcrest is tJ!.;e median marga rine fa tooment. 
2. Ho : fl.· =l7.0 
;3. H1 c p.. t: 17.0 
r+ - O.Sn 
4. 'r('Ststatistic is Zo = O.S.Jii 
.;. We reject Ho if tlte ~I > ~.02.5 = 1.9(>foro = o.o.;. 
- 3- 0.5(6) 
6. Comptltation: .t.o = O.S.[6 = 0 
7· Conclusion: fa il to.rejl'ct Ho . The P-valu.e = 1ll - <P<o )J = 1{1- o ,s) = 1. Tt.ere is not enough e\ideoce to conclude tJ!.at the modian fa l 
oontentdiffers from 17.0 . 
9·U5 a) 1) Tt:e paramcter ofioter('Stistt.e meanball diameter. 
2) Ho: JI<J = 0.165 
'3) ff.o: Jlo t: 0.16.5 
+ • 
4) w= min(w ,w ) 
. -
5) Reject Ho if\V < Wo.05,n=9 = ) for o= 0 .0 5. 
• (>) UsuaJiy z...~OISated:ropp.."'<i from tt.e ranking andtlt.e~mplesil.(' istOOli.oed. The sum oftlt.e posithoe ra nks is"' = (l + 4-5 + 4-5 ... 4-5 
... 4-5 ... Bs ... 8..;) = j6. Tt.esumofthencga ti\'E': Ninksis w = { 4..; ... 4-5) = 9. Tt.ercfore, '" = min{j6,9) = 9. 
Oiffercr.-ee xi - Si-;:r.ed 
ob:.er-1.1t.ior; 0. ~65 Rlt!\k 
1 
6 
9 
' 2 
' 5 
' 
• 
12 
10 
ll 
0 
0 
0 
O. COl 1 
- 0 . 001 -•LS 
0. 00:? ~ . 5 
0. 002 4. 5 
0 . 002' (, , 5 
0 . 001 <. 5 
- 0 . 002' - (, . 5 
0 . 003 0. 5 
0 . 003 • . 5 
. 
7) Conclusion: becau.w'~ = 9 is not less tl'..,n or equal to tJ!.;ecriticaJ \ '!l lt:e 
that t~.e mean baJI diameterisO.:l-65 a t tlt.e 0 .0 5 le\'el of significance. 
b z = rrr--n(n + l) / 4 = 36- 9(10) / 4 = 1 5993 
l 0 .Jn(n+l)(2n+ l) / 24 .J9(10Xl9)124 
. -
lt.b.05,rr=9 = ) , we fa iJ to reject t.lbe: mlll f!.ypot.lt.esis 
and71).0 :!:.5 = J .9().1k<:at~Se71'l = l-5993 < 71).0!5 = 1.¢ we fail to reject t~.;e null hypotJ!.esis th.at 1M meso ball diameter is.0.16s a t t~.;eo.o,s 
I~ of significance. Also, th-e P·\'Shtc = 1!1-~ < l-5993U = 0 .1098. 
9·ll7 1) The parameter of interest is tr.e mean d)ing timeoftlt.e primer. 
~) Ho: #() =l-5 
3) Ho: #(),. l..'j 
. 
4) w 
.5) Reject Ho if\V- < ub.o;, . .,.= l 7 = 41 fora= 0 .05 
• 6) Tltes.lflll of the positi\uank is'~ = (4 ... 4 ... 4 ... 4 ~ 4 ... 9-5 ... 9S .._ 13-5 ... 13-5 ... 13-5 + 13-5 .._ 16-5 ... 16-5) = 126. Tr.es-u.m oftJ!.;e n;:ga ti\-e 
rank.isw = (4 '"' 4 ~9.s- 9.S) = 17. 
Di f f~cr.ec xi -
Ob:-..cr-/~t. i~:'l 1.5 
l.S 
l.S 
1.5 
: . 6 
1.6 
L6 
u 
1.6 
'-' 
L6 
1.3 
1.1 
1.7 
u 
1.& 
: . e: 
l.e: 
,_. 
1.9 
: .9 
0 
0 
0 
0 . 1 
0 . 1 
G. l 
- C< . l 
0 . 1 
- O. i 
C. l 
- C< . :? 
0 . 2 
0 . 2 
- 0 . '1 
0 . 3 
0 . 3 
0 . 3 
0. 3 
••• 
0 . 4 
' 
' 
·1 
_, 
' _, 
' 
- 9 . 5 
9 . 5 
9 . 5 
- 9 . 5 
13. 5 
t3. 5 
13. 5 
u .s 
16. 5 
16. 5 
. . 
7) ConrJll.'Sion: Be::ausew = ~'7 is less thant~.ecritieaJ \ 'alne Wo.O?.n=l? = 41. we r-eject th.c null hypothesis that t~.e mean dying 
time of tlt.e primer o:ooeds t..;. · 
suoplemerua I E.xet.CM$ 
9·!19 a) Oq reesoffr.eedom = n- 1-= 16-1 = 15. 
bl SE Mean = ~ = 4~ = 1.1525 
-vN -v l6 
= 98.33 -100 =-1.4490 
lo 4.61 I .Jf6 
10 = - 1.4490 \\~th df= 15, so 2(0.05) < P-value < 2(0.1). That is, 0.1 < P-value < 0.2. 
9 •• , CI f th . - S - S ) "/o o e me.an 1s x - t0ms.1.s c < J1 < x + to.o25 15 r.: 
' '\I ll ' '\Ill 
98.33 -(2. 131)~ < p < 98.33 +(2.131)~ 
-v16 -v16 
95.874 < 11 < 100 786 
e) lkcause tlt-c P·\'alue >a= 0 .05 \,·efaiJ to reject tlt-cnuJI hypot.r.esisat theo .o.s le>.~l of significance. 
d) ro.o5.15 = 1.7.53. Be<:au.se to = -14490 < ro.o5,15 = 1.753 wefail to reject tile null }!.ypofl!.es.isattf!.eo.o.s level of significanoe. 
9 ·l2l a) Th.eou.ll l<ypothes.isisp= 12\l!l'SllSJl > 1:1 
x = 12.4737 , S = 3.6266, and N = 19 
. -
12
·
4737
-
12
- 0 '694 ' hdf-19 1- 18 10 - - . ) \\~t - - - . 3.6266 / ..Jl9 
The P-valudaJis between twovaJtles0.:1.57 (a= 04)a.nd0.688{a = 0:2.5). Thl.s, 0.~ < P·value < 04. Bee.<~ use the P·value >a= 0.05 '"e fail to 
rej~ th;e rmll h)poth.esisat t~.e 0.05 le>.'el of signifieaooe. 
- s - s 
b) 95%two-sidOOCl oftltemean is X - lo.O'-.S,JS Jji < P < X + fo.OlS,lS .Jn 
12.4737 - (2101) 3~6 < ,u < 12.4737 + (2.101) 3~6 
19 19 
10.7257 < 11·< 142217 
a) u = 25 {3 = ifl(z001 + 
85 'J!s] = <1>(2.33 - 0.3 1) = <1>(2.02} = 0.9783 
. 16 / 25 
n = 100 8 = <P[z0 01 + 
85~)= ifi(2.33 -0.63) = <P(L70) = 0.9554 
. 16 / 100 
u = 400 8 = <P[z0 01 + ~~~~) = .P(2.33 - 125) = if/(1.08) = 0.8599 
n = 2500 /3 = <P(z0 01 + 168J~]= <P(233 -3.1 3) = <P(-0.80) = 0211 9 
b) u = 25 z0 = 1~61~ = 0.31 ?-value: 1- <1>(0.31) = 1- 0.6217 = 0.3783 
~-85 -
u = 100 z0 = = = 0.63 P-value: 1- •'P(0.63) = 1- 0.73)7 = 0.2643 16/ -v lOO 
86- 85 
n = 400 z0 = 161
.../400 1.25 ? -value 1- <1>(1.25) =1-0.8944 = 0.1056 
- 86 -85 
n = 2' 00 z0 = 16
J.J
2500 
= 3.13 ?-value: 1- if/(3.13) = 1- 0.9991 = 0.0009 
The data would ix:statisticaJJysignifieant when n = 2500 at a= 0 .01 
9-125 
n I est statistic P-value conclusion 
a. 50 0.095-0.10 
Zo = ,jO. IO(J-0.10) / 50 - 0.12 
0.4522 Fail to reject H0 
b. 100 0.095 -0.10 
= - 0.15 0.4404 Fail to rej•ct H~ z -
o - ,/0.10(1 0.10) / 100 
c. 500 0.095 - 0.10 
--0.37 0.3557 Fai1to reject Ho =o-
,/0.10(1 - 0.10) 1500 
d. 1000 0.095 - 0.10 
Zo = ,/0.10(1-0.10) / 1000 - 0.53 
0.2981 Fail to reject Ho 
e) Th• P-va1ue decre.asts as the sampl• siu increases. 
a 
9·127 o = 14,6 = U>5- ::l:OO = 5· 2 .:: 0 .02..j,Z(),02..5 = 1.9(>. 
a) n= 20 ,8 = <1>(1.96 -~~= 1>(0.362) = 06406 
b) n = 50 ,8 = <1>(1.96 - S:)= <P(- 0 565) = 1- .P(0565) = 1- 0 7123 = 0 2877 
c) n= 100: .8 = <~>[1.96 - sJIOO] = 1>(-1611) = 1-1>(1 611) = 1- 0.9463 = 0.0537 
14 
d) TJ!.e prob.!lbilityof a 1)'))1! I {error increases with an irx:rease in tf!.estsndard deviation. 
9·129 a) a=o.o.; 
n = 100 .8 = <~>( z0.05 + v'o~:o~~~~oo) = <1>(1.65 - 2 O) = <P(- 035) = 03632 
Power = 1- {3 = 1-0.3632 = 0.6368 
n= ISO ,8 = <P(zo.o; + v' 0~5-06 ] = <1>(1.65 - 2.45) = 1>(- 0.8) = 0.2119 0.)(0.5) / 100 
Power = 1- !3= 1-0.2119 = 0.7881 
n= 300 .8= <P(z0.05 + v'O.~~O~~/~OO ) = <1>(1.65 - 3.46) = 1>(- 1.81) = 003515 
~wer = 1 - p = 1 - 0 :0$.5.l5 = o .¢48.; 
b) a =O.Ol 
n= 100 ,8 = <P[z001+ 0.5 - 0.6 ] = 1>(2.33 - 2.0) = 1>(0.33) = 0.6293 
. v'0.5(0.5)1100 . 
Power = 1- {3 = 1-0.6293 = 0.3707 
n= !50 ,S = <P(z001 + v' 0.5 - 0.
6 ) = <I>(2.33 - 2.45) = <I>(- 0.12) = 0.4522 
. 0.5(0.5) 1100 
Power = 1- (3 = 1-0.4522= 0.5478 
n= 300 ,8 = <P(z0.0 1 + v' ~.5-0.6 ) = <1?(2.33 - 3.46) = <1>(- 113) = 0.1292 0.) (0.5) / 300 
Power = 1- /3 = 1-0.1292= 0.8702 
Ottreasing the vaJueof a decresses tlt.e po,..-er of thete.st for the different samplcsi7.e.s. 
c) a = o.os 
n = I 00 ,8 = <P(zo.o; + v' 0·5-:_ 0 8 J = <1>(1.65 - 6.0) = <I>( - 4.35) ~ 0.0 0.5(0.)) / 100 
Power = 1- /3 = 1-o~ I 
n.e true value of p i'.asa large effect on tJt.;e power. The greater is t.J!.edifferenoeof p from Po t~.e larger is tlt.e power oft.J!.e test. 
d) n = (Za12 v'Po(J - Po) - zpv'P(l - P) J
2 
P - Po 
= (2.58v'o 5(1 - o.so) -J.~5v'o 6(1- o.6))2 = (4_82)2 = 23 _2 ~24 0.6- 0 ) 
n=(Zanv'PofJ - Po) - zpv'PO - P) J2 
P - Po 
= (2.5&v'0.5(1-0.50) - 1_6Sv'O 8(1 - 0.8) ]
2 
= (2_1f = 4.4! ~ 5 0.8 - 0.) 
The true value of p i'.asa large effect on tJt.;esamp!es.ize. Tl>.egreater is tlt.edlstaooeof ''from J'JO tJ!.esm.sJ!er is tft.es.amp!esiu that is 
reqt~i.rOO. 
9.131 a) d = ~= J ~t- l'o i = J73-75 j 2 
u u I 
Using ilw!OCcundora = o.o;;.d = :!..andn = 10. ~-o.o and po\'o·erofl-o.o •1. 
d = ~= l~t- 11ol = l 72-75 j 3 
u u I 
Using. th.eOCcurvefora = o.o.;.d = j .and n = 10. ~-o.o andpowerofJ-0.0 •1. 
b) d = ~= l~t- l'ol = l 73 -75 1 = 2 
u u I . 
Using th.eOCCIZ.n.:,Chart Vll (!)fora= o.o.:;.d = 2,and!l•O.l (Power= 0 .9).n .;. 
11. + I 5+ 1 The.refore, n= 
2 
= 2 = 3 
d = ~= l~t- !lol = l 72-75 j 3 
u u I 
Using tlt.eOCa:.noe.O.artVU e) fora= o.o:;.d = 3.and ~•0.1 (Power= 0 .9),n• = 3 . 
• 11 + I 3+1 The.refore., n = 
2 
= 2 = 2 
c) o=~ 
d = ~= l~t- !lol = 1 73 - 75 ) =1 
(j (j 2 
Using theOCCIZ.n-eforo = o.os.d = 1.andn = lO,Il•<UO andpowerofl- 0.10 · -0 .91). 
d = ~= l~t- !lol = 1 72-75 1 ' 1.) 
(j (j 2 
Using t~.eOC<:t!.noe:fora = o.os.d= 1,s.and n = JO, ~•0.04 and powerof1- 0 .04 • 0.96. 
d = ~= l~t- !loi = I 73 - 7S J =I 
(j (j 2 
Using t~.eOCcurve.ChartVll e)fOra= 0.05.d = 1,and ~·O.l {Power= 0 .9).n• = 10 . 
• 11 + I 10+ 1 , Therefore., n = = = ) . 5 n "' 6 
2 2 
d)~= I 11·- llo I= 172- 751 = 1.5 
" (j 2 Using tfteOCCUf'\l!, CJt.art VH e) for a= o.o.s.d = j.and ll • 0.1 {Po\\' et = 0 .9).n• = 7 . 
• 
Therefore, n = n : 1 7 ~ 1 = 4 
Increasing tf!.estsodarddeviationdec:reases tf!.e power oftf!.e test and inc:resses tl".esample size r~uired to obtain a certain power. 
9·133 Assume the dahl folio,~· a oonnaJ distribli.tion. 
1) Tlbe paramcterofinterest is the staod:ard de\iationoft~.e.ooncentra tion. o. 
• • 2) Ho:o =4 
• • j) H1 :o <4 
4) T~.e teststa tisticis: 
2 (n - 1)s2 
xo = 2 (j 
5) Since oogh>en,-slueof a.lpha,so oocMtical \'Slue is gh>en. We,'liJJ calculate the P·\ 'Siue. 
6) s=0.004andn=lO 
• 
2 9(0 004)2 
Xo = (4)2 
P·\'aJue=/lx <0.00009): P-valtM•O. 
0.000009 
7) Conclusion: T~.e P·\'aluc isapproximatelyo. Therefore we reject the null hypothesis and ooncludethst t~.estandard deviation of theoonoentf"iltion is less than4 grams per li ter. 
9·13.5 Create a table for the number of ~rrors in a string of 1000 bits{valu.e)and theobsen'l'd number of times tl:-e number appeared. One 
possible table is: 
Value 0 I 2 3 4 5 
Obs 3 7 4 5 I 0 
Tlw!,alu.eof p mnst be estimated. tet t~.eestimate hedeootod by P samp!e 
S3mplem03n= 0(3)+ 1(7)+ 2{4)+3(5)+ 4{1)+ 5(0) = 1.7 
20 
• . . = sample mean =_!2_= 00017 P.a•v>" n I 000 . 
Value 
Observed 
Exgected 
0 
3 
3.64839 
Value 
Observed 
Exgected 
I 
7 
6.21282 
0 
3 
3.64839 
2 
4 
5.28460 
I 
7 
6.21282 
3 
5 
2.99371 
2 
4 
5.28460 
4 
I 
1.27067 
6 
4.69541 
5 
0 
0.43103 
Thed.egreesoffreedomare k - p -1 = 4- 1 -1 = ~ 
a) 1) Thevuiableofinterest is t ile (onn oft.f!..edistrihntion for tft.e number of errors ina string ohooo bits. 
~) Ho: TJ:e form oft~.edistri bu.tionis binomial 
j) H1: Th~fonnofthedistributionis oot binomial 
4) Tlteteststatisti<:is: 
2 ~(O; -E;)2 
Xo = L..t 
i= l E~ 
sl Rei"" H, ;1x6 > x6.o5,2 = 5.99 for cr. = 0.05 
2 (3- 3.64839)2 (6-4.69541)2 
f>l Xo = 3.64839 + .. . + 4.69541 0.88971 
7) Because 0.88971 < 949 fail to rejOC't Ho. Weare unable to reject the nu.JJ llypot~is that thedistrib.:tionofthe number of ~rrors is 
binomiaJ at a= o.os. 
b) p., oaJue = 0.6409(fou.nd using Minitab) 
9·137 a) In order to use t statistics in hypothesis testing, we oeOO to,usu.me that tt.e u.oderl)ing distribtr.tionis normal 
1) Tlte parameter of interest is the true meanoonoentrationof stl.'Spendedsolids. }1. 
i) Ho :p =so 
j) H, : ~ <SO 
4) Becausen >> jO wecanu.se tknonnal distribtr.tion 
.5) Reject Ho ift.o < - 1.6.5 for a= o.os 
f>l X= 59.87 s = 12.50 n = 6o 
X - /1 
Z() = 
s t.Jii 
= 59.87- 50 = 6.12 
Zo 12.501../60 
7) Because 6.12- > -1.65 fs iJ to reject t.ft.e: null ~.ypot.lt.esis. Th.ere is insufficient evicteoce to indicate th:at the truemeanooncentrationof 
suspended solids is less IJ!.anso ppm a t a= o.o.;. 
b) p.,-a)ue = W{6.l~)., 1 
c) Wecandivid~ the real line under a standard normal distribtr.tion into eight intm'als with tqU:aJ probability. T'J.A::Sf! int:!rvsJsaro: IO.O.,l-i). 
10.,12. 0.67S). IOJ)75. l .lS). Il.tS. e)and their n<:g,Sti\~toounterparts. 'fl!..e probability for each interval is p = 1/ 8 = .l~S so tlt.eo:pe<:tcdoell 
freqtl.eJX:ies: ar~ E.= np = (()() ){0 .1~.5) = 7.5. Tlte tableofnmgesand their oorrespooding fri!Q.uencies isoompletedas follows. 
r r.t.erv.ll Ob::.. F:cqt:er.cy. !:xp. F:cqt:er.cy. 
)( (.. ,5. 50 9 ' . 5 
45. 50 < )( <. 51.43 5 7. 5 
51.1,3 < )( <. 55.~1 7 7. 5 
55. 13'1 < )( < 59. 87 11 7. 5 
59. 87 < )( (.. 63. 87 
' 
' . 5 
63. 87 < )( " 6€. 31 9 7. 5 
68. 31 < )( <. 71. . 2·1 
' 
7. 5 
)( ), 14 . 14 6 7. 5 
The test statistic i.s.: 
2 = (9 - 7 5i +(S - 75 i + · .. (8 - 75i + (6- 7.5f = 5_06 
,\ . 7 ' 7' + 7 ' 75 ) ) ,) . 
aodwewou.Jd reject if this va !ti.~txooods x2 0.0$,5 = 11.07 .. Becatseitdocs oot. we fa il to reject the ~.ypot.J!..esis thatthedsta are normally 
distributed. 
9·1J9 a) [oord-er to use t shltis tics in t.ypotf!..f':.Sis testing, we need to assume that tr.e uOOetlying distribntion is normal. 
1) n .e parameter of interest is tlt.e true meanooefflcient ofre.stitlltion,p. 
~) H.o :p=0.6.:).5 
3) H1: p: > 0 .63.5 
4) Since o >;}O we can use tlu:nonnal di.st ribntion 
X- f.l 
Zo = -s -, .Jii--;n= 
.;) Reject Ho ifzo > 7-awlter-ezo.os = ~..)3 fora= 0.0 1 
Gl x = 0.624 s = 0.0131 n = 40 
0.624-0.635 -31 zo = =-) . 
0.01311 ,J40 
7) Because -5.31 < ~.33 fa il to reject th-e nt<ll ~.)poth.esis. There is insufficient e-.idmce toooorJt>d.e thst fl!.e true meanoodficient of 
restitlltion i.s gr-eater t~n 0.63..5 a t a= O.Ol. 
b) P-vaJtre:cl>{5jl) • 1 
c) J f tlt.e lower ix:mnd of tlt(' Cl w-a.s a bo\"t tfte \'a lt>e o .63.5 tr.en weoould ooncl udf: t~.-a t the mean ooef:ficient of restitlltion was grester th.an 
0.63.;. 
9·141 a) 1) Tlte parameter ofinterest i.s t.f!.e true mean s-uga r oonoentn'ltioo.p. 
~) fio: p= ll..'j 
3) H1: Jl * ll.S 
. X - Jl 
•l 'o = r 
sl "n 
S) Reject Ho ifltol > laj:tn -l wltere~'2.n- 1 = 2.093 (or a= o .o.; 
(,) X = ll47,S = 0:01~ n = ::!0 
_ 11.47 -liS - 6 10 to - = --. 0.022/ '1 20 
7) Eklcal!Si!6.10 > ::!.09:} reject t.f!.e nt!ll hypothesis. There is .sufficient evid.cnoe that t.J!.e true mean sugar concentration is different from 
n..;atn=o.o.;. 
From Table V U..e to \'Slue in absolntcvaJne is greater t.h.sn t.lw:\'8lueoorresponding too.ooos ,.,.;th 19degreeiof fr.troom. TJ!.erd'ore 
::~•o.ooos = 0 .001 > p.,'8Juc. 
bl d = ~= 1 11- Jlo i = J1 14- 115 1 = 454 
u u 0.022 
Using tlbe OCcut\l:, Chart vn e)for n = o.o.;,d = 4,s4,a nd n = :w we fi nd P•O and Power • 1. 
c) d = ~= I f.l - 11o I= 11145- 1151 = 2 27 
u u 0.022 . 
Usipg the OCcunl:, Clt.art VI ( d for a= o.o.s. d = 1.27 .and 1 -IJ > 0.9(P < 0 .1), we fi nd tha t oshould bea t least.:;. 
d) 95'% l\l.'o-sidE:dconfideoce intm'SI 
X - fo.025,l9( .;,) < f.l < x +toms,l9 [ .;,) 
11 47 -2093[~) < f.1 < 11.47 +2 093[~] 
1146 < f.1 <11.48 
Weconclu.d;e that t~.e mean snga r ooncentra tionoontent is ooteq-u:aJ to 11-5 OOcause th:ah'a lue is not ins.id.etheoonfide:ooe interval. 
a ) Th.e normality plot below indicate;: that t.J!.e normality assumption is reasonable. 
Probability Plot of Sugar Concentration 
,.,.mal 
wTI,-o:-ol-o,-,,-.1---,!-,l-,,-,1~~ 
" L _ _l _____ L_j_L__L __ _l ___ - _t ___ , - _L 
I I I I I I I I . I I I 
<JO t--·-t---~---.-·--·---.----.-·-·1'-·--t--··· ..t--~----~ 
I I ; I I I I I ! I I I 
.. j---i---t---1---i-+-++--~- - -T--+-~--~ 7.) ..._ _ _._ _____ .,._ __ .. __ • ___ .. ____ ~--· - _.._ ___ .._ _ _., ___ ~ 
5 60 +---+---+--~--i--~---11----' ·---1---~--+--~---~ 
1::! so t·-~ -r--,---t--- -· - -· -t--t·--r--t-r 11. 40 .--·~---.----r--·--.---,--- -·-·T-·-7----r·-·r·-.----r 
J' -~--~---~---~--~- -- - --r---·+-·--i----t---~----}----~ 
0 0 + I I I I t I I I 1 " r··-T-- ___ T ___ ,_ · · ---r··-r·- -··r--T-r··--r 
10 :.---.!---..:..---.!._ - '1---i---~----!---~----~---:.--~----~ 
' ' ' . ' ' ' ' ' ' ' ' ' 
, 1---j- t -- ,_ ___ l -j- +---t·--r--+-r ·~-- ,. 
, ! I : I . I ! I . I . ! ! 
11.4) t1.42 11.44 11.46 11.48 u .so 1!.52 
Sugar COn¢enl r &tion 
9·143 a ) 1) The par-ameter ofintcrest is t.Jt..e tme mean peroent protein.11. 
2) fi<> :p=80 
:)) H1 :p>80 
X- 11 
•J to = 51 ,fii 
.:;) Reject. Ho ifft, > fa.n _ 1 wf!.ere f<J .O,S.lS = 1.753 foro= 0.05 
6) X = 80.6Ss = 7.38 n = 16 
l = 80.68- 80 = 0.37 
0 738 / .j!6 
7) BecsuseO.J7 < 1.7.53 faiJ to reject t ile null }!,ypotr.esis. Tf!..ere is not sufficient E!\idenoe to indicate that the true mean percent protein 
is greater than80 s to = o.o.:;. 
b) From th-e normal probability plot. t~.enonnalityassnmptionseems reasonable: 
Probability Plot of percent protein 
,.,. 
: .LI _ L __ L_L __ i._ i 
,. -~ I -+-·-j---f.--- . J -- r--r 
.. ·r· -- c·- -----1> · ··r -r-··r 
~ » r- . . -· . . --.-- .... --r- .a.---~ 
i. oo .,__. --r----~-- !L.---+----->---• 
... +-- __ ,______ --1----+----->---1 ! .. ~--r--t-·-t-··· 1 ----r--·t··--t-··t 
"' · - ,- --,- ---.--·r-r-1 
"' f--! 1 -·· ----;- ---r----r ·-r--r 
' ' ' . ' ' ' . ' 
: F--· t-:~r:=r ==F=r=f==f 
' 
' ' 
l l j 
.. 
9ercont protoln 
100 
c) From TahleV,0 .2.; < P·\'SJtl<! < 04 . 
• 9·145 a ) In order to uset ltex statistic in l;ypot~.-es.is tosti~ andoonfid'I'!DOI'! intervaJ oonstruction, we need to assume tlta t tlte t:.nderlying 
distribution is normal. 
• 1) Tile p;~r-ameter of interest is tJt,e true\'8riatttoft.hc: ratio between tlt-encmbersof s~mmrtricaJ and total synaps.es,o . 
• :!) Ho: 0 = ().0 :!. 
• 3) H1 : o * 0 .0:! 
) 2 (n -1)s2 4 Xo = ., 
u · 
; J Rcjoctfi<> ;rxZ < x L n,n-1 where a = 0.05 and x5.97s,;o = 16.79 or x5 > X~12;1-1 ·wh<rea=o.o;aoo 
2 
XO.<rlS,JO = 46.98 forn=j>. 
6) n = .}1.s = 0.1!)8 
yij = (n - ~)s2 = 30{0198)2 = 58.81 
(j 0.02 
7) Be::allSe.sS.Sl > 46.91) reject Hi). T~.e true \'atianoeofth.e r<~ tio hm··ecn tioe numbers of S)mtnttricaJ and total synapses is 
significantly differ-ent from 0.<>2 at a= o.o.:;. 
b) P-valne/ :! < 0.005 so tf!.at P-value < 0 .01. 
" 9·l47 a) 1) Tlte parameter ofintcrest is t~.e\'Srial'lCtoffa ttyacid messurements.o ~. 
' 2) ~:o =1.0 
• 3) H1 :o 't.l.O 
? 
2 (n -1)s· 
xo= 2 (T 
4) Th.ete.st statistieis: 
.sl Reject Ho if X~ < X~s9s,s = 0.4 1 or rejoo Ho if x5 > X~.oo;.,s = 16.75 fora= o.o1 andn = 6. 
6) n = 6,s = 0-319 
2 - 5(0319)2 - 0 '09 Xo - 2 - .J 1 
P·\'S!ne: o .oo.s < P-vah:ef :!. < 0 .01 so th.at 0 .01 < P·\ oalue < 0 .02 
7) lkeause 0-509 > 041 fa il to reject t lbc null ~.ypothe.sisat a= 0.01. Tltereis i ~ufficient eo.idence to conclude t!tat t lbcvariancediffers 
from 1.0 . 
• b) 1) the psr~meter of interest is tf!.f:varianoeoffatty acid meast:rements.o {oo\,. n =51) 
> 
:!) ~ :o = J .{) 
• 
.3) H1 :o * 1.0 
? 
4) Tltete;tstatis ticis: xt = (n - !)s· 
u · 
' ? 
.sl Reject f-1() irxO <x0.99s,so 2 2 =~·?.9;>or reje::t.H,o i f Xo > Xo.oos,so =7949fora=0.01aodn=.:;l. 
6) n = SJ.s = 0.319 
? 
2 50(0.319)• - 09 
Xo = 2 = ' · 1 
P·\'Shti':/~ < 0 .005 so that P·\'Shte-< 0 .01 
7) Because.s.09 < ~7 .9C) rtject t.Jt.e nuiJ h)pothes.is. TJ•.ere is sufficient E!\ideoce toocmch.'-de tf!..a t thevarianocis ootequaJ to 1.0 s t o = 
().01. 
c) Th.essm_plesizech:ange; theoonclllSion that isdrav.-n. With a small samplesir.ewe fa il to reject tf!.e null h)'potlte:s.is. Ho'•"E!\'el' ,a larger 
sample sizeaJIO\'IS us toooncl~ t~.e nu.JI Itypothesis is fa lse. 
~149 a) Ho: p =!fa 
H1 P~ !10 
Ill> r.el = t:and JlZ < - 1.a _ t:l ={a-t:). Tltcrefore Jv. > Tf:OT Z < - 7.a _ t:l ={a-t:) • t: =a 
b) J> = P(- ~_ , < Z < z. l !to + 5) 
9-151 1) Th.e parnmeterofioterESt is the tru.e mesn number of open circuits. A 
2) Ho : A=2 
3) Hl:). >:!. 
4) ~'ince n > .30 \~·ecem t.se t.r..e normal distribution 
X - >. 
Z<) = --,..J ).= ,=11 
.5) RejEX:t Ho if'T.Q > 7.a where to.o.; = 1.f'.5 for a= o.o.; 
6) X= 1038/,soo = 2.M6n=.;oo 
2.076 - 2 
Z<) = .../21500 1.202 
7) Because 1.202 < 1.6.; fa il to rejEX:t tl'te nu.JI ltypott.es.is. Tf!..ere is insufficient t'\idm:e to indicate thst tl'te tru.e mesn number of open circuits 
is grester th.an :lata= 0.01 
9·l.5.J 1) TJ!.e parameter of interest is the parameter of ano:ponentiaJ distribction,A 
') '"" :1.=/,o 
3) H1 :).*i.o 
4) tESt statistic 
v 2 > v 2 2 < > 
.5) Rejectfi()ifAO l\ al2.2n or Xo X1--a l2.2nfora=0.05 
n 
6) Compute 2.x E: xi aoo plug. into 
i =! 
2 i-1 Xo =----i~~-
" 2>.L X; 
i=l 
7) OrawConclnsion; 
Tlt.eooe-sided hypotheses below can also be te.sted"'it~th.ed.eri\Ui: tESt statistic as follows: 
1) Ho :),=AQ H1 :A>)-.o 
, ? 
Reject Ho if xo > x;~2n 
2) Ho:A=AoH1:A<i.o 
'2 2 
RejEX:t H0 ifXo < Xa,2n 
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