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, 0*- w% AN0+ts (?)r-.rr=Jmarks) h=Yt+tTrt=@tK='@ X ecotogist wishes to investigate the levels of mercury pollution in four laKes. ne citches 4 fish from lake 1, 7 fish from lake 2, 5 fish from lake 3, and 5 fish from lake 4. The following table gives a summary of measurements (in parts per million) for each of four samples. Lake (treatment) Sample data 1 2 -l 4 tli 4 7 5 5 n. t' 1.,.. - T. ? ,'ll I ./ =l 52 69 7t 6t n. I z .i=1 2 1' . . 753 798 t248 9t2 Assume that concentr.tions of mercury in all fish in a lake constitute the population of values. Assume that all three populations are normal and have common variance. a. Construct the ANOVA table for the sample data. use the rest of page for the calculations required for ANovA table. b. Test at 5yo significance level to determine if the data provide sufficient evidence to indicate a difference in larr'-population means. lf required, use Tukey' method to determine which means differ' c. construct an individ ual g|%o confidence interva! for the difference between mean concentrations of mercury in the fish from lake # 2 and lake # 1. Write the solutions for questions b. and c. on the following page' OVA T Source of variation Degrees of freedom Sums of squares Mean squares F ratio Treatment (-l=3 60, t, gf LO, l6t 0. 16 q Error h* f. -- l1 60t. 1 rt 3S. Y 33 Tota I h-t=20 5 6 Z-. 9SZ Ffr =-- ?'5 Er+';g, "Y-s;,;:@ + a'a d7r r{h;,-: _--- ss; =rtd-+'z-3?rr E ry=@ ntJ- , ? ^- tlc, ssrr=E# #={.q*rF-q 9sE = sST- SSI-r = aqz,tg+ o

b. Test at 5% significance level to determine if the data provide sufficient evidence to indicate a difference in four population means. lf required, use Tukey' method to determine which means differ. a. State null and alternative hypotheses. (0.5 mark) F{o: F,= lrr = ft = 1, , I{,: 6+ ArAryI' *ru r"tun iE il1'tr1*t'* W W Yl4'trtt b. State the calculated value of test statistic and criticalvalue of test. (1 mark) .f*,. = 0. S59 f-.- !-o-o.sr3f l? =3.zt c. Decide whether null hypothesis is true and make the conclusions. lf required, use Tukey's procedure. (0.5 mark) Lrrrq l*. z ?, we da vuv+ W lfin. a+ sL = O'os- \t1,g ti,'"w[[iu.uN+ W wi)a'uu l4^/* ('u/+L a'&d''[4t'w PPru"'** ttt^'artq T,r4"" W te vw+ w,ywiva4. c. Construct an individ ual gil%confidence interval for the difference between.mean concentrationsofmercuryinthefishfromlake#1andIake#z.Lz,niin>".I- -f [,. = * = * = B t tr . =+" =+ = s-sb i F-[r.= 3.[Y, t_ to, zsllr MtE[*,*frJ = 2,uD @e4,s7- =l,ll0

":r fi.= [l.ht, ) Tr.=[st)6= t06t T.z.=-!9s)y=YY0, Il.=(F0)6=?a0 T,, =fl.nTz.*T3. : s 0o t yyo r 3 og :@ ) 5 SIr cnn ^$(. f,ut ,,i i" 1ry0 wu.atf,. @ ssrr- t#,- F_- -h*-#,H- F= #-#.y-w:- 5\t+09.8 t- 5qt806 - 9Y,?0t,8 = luo.za i',.=F=W=F2.3^ - Lake (treatment) sa m p re o ari'-------------- 1 2 a J tti 6 8 6 i. 51 55 50 s. I 5.4 4.9 5.2 Assume that concentrations of mercury in allfish in a lake constitute the population of values. Assume that all three populations are normal and have common variance. a. Construct the ANOVA table for the sample data. Use the rest of page for the calculations required for ANOVA table. b. Test at 5%o significance level to determine if the data provide sufficient evidence to indicate a difference in three population means. lf required, use Tukey' method to determine which means differ. c. Construct an individ ual95% confidence interval for the difference between mean concentrations of mercury in the fish from lake # 2 and lake # 1. Write the solutions for questions b. and c. on the following page. ANOVA TABLE Source of variation Degrees of freedom Sums of squares Mean squares F ratio Treatment lc-t = / t0 0, Lo F0, t0 4.{0 Error h-Kal+ o,{r. o} 26, 4z Total [1-] = l! s\s, ?\" SLSS-52.il2 + 6LST-sZ.3l2 = lo0,L q000 _ \Z . tJl = 15,605 +-Zyrlsotl o s sTr =Z,n;L[;-8, S 9l'r = 6 Ll t- SZ. 3)z + . SsE = Jt,.(n;-f) t=l "= S;,{1(s) * t{.gt(?) r ;.ZL(,s) =Vyg.O+ @

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b. Test at 5% significance level to determine if the data provide sufficient evidence to indicate a difference in three population means. lf required, use Tukey' method to determine which means differ. ' -" State null and alternative hypotheses. (0.5 mark) F[: dr{- at W V+a,k uul '*kf,[+al WWyY State the calculated value of test statistic andtiiticalvalrie of test. (1 mark) f*= to.og,zrtL=3,50' Ff,: tlU= lh.=Vt ' *ltu"dr .fut, = 4, g0 Decide whether null hypothesis is true and make the conclusions. lf required, use Tukey's procedurs. iron '5 mark) t Jto.oqrr 2b.\rc(lrt) @_Jg: t*.,we Aav'olyw( ltro a**= o,oy d,vd w,t/),",rq il^rr* l4/'ptL r9 w d4[41^r/^/L &ful(/)^ la4puw@^/L wtA,t^^' Inr,&r,t' L Wdilrt,ls no4 wry.,rc/. c.Constructan,no,,',,ffinceinterva!forthedifferencebetweenmean concentrations of mercury in the fish from lake # 2 and lake # 1' tt- tl b L,ll _,1"g6 , ltr^ ! I *rhr*,) L6 ,tro (f +f) . l, - !, , ss-sl t z .,r

, $tnf -1n&r{ A[r|$17A , M ril/vi{1/^^k4 d-vian. 6)t, +3+2=(,r;;;"" --"T*'['=63," vt= 5+t{ +r=11' 0 \#order to study the differences in operation cost of different types of machines, an engineer examined the operation costs for three different types of machines. Random samples of operation costs (in $ per hour) were obtained for each type of machine. The data obtained by the engineer are represented in the table below. one observation is missed, not all sample sizes are the same. 1, 2 3 4 5 [, Type I machine L0.4 9.9 10.6 8.9 1,1,.0 50"8 Type ll machine 8.8 1,1,.6 L0.9 1,1,.4 Yz,a Type lll machine L1,.2 13.4 11,.9 13.3 13.1 62.5 Itb.Y h 9u= t0, ft ff2, = 10.6t k ,= lz'58 $$rl:1 77462 7,, '" SgT= 9SIr = 9$E= l4gh = IV'I9E = l6 r0?,1,' a 6. oll if ti-+= r3t 6:l v *t-t,= Er h; h L = [t. 3g g. Ir a. Construct ANOVA table. Leave 3 digits after decimal point for the values of SSTr, SSE, SST, MSTr, MSE, and 2 digits for the value of F ratio. Use the rest of page for the calculations required for ANOVA table. b. Test at 5% significance level to determine if differences exist in mean operation costs of 3 types of machines. lf required, use Tukey's method to determine which means differ. Use a = 0.05. c. Construct 90 % individual confidence interval for the difference in mean operation costs for Type I machine and Type 2 machine. Write the solutions for question b. - c. on the following two pages. ANOVA TABLE Source of variation Degrees of freedom Sums of squares Mean squares F ratio Treatment K-l:t t6,ozt f ,ou +,+\ Error h-tl- =11 ,1. e ,E 1,03r Total 1a-1 = l3 z+. t{09 lT+Y" 6L - lE6. Y I -. l4 = L+, rlo 3 5!-C*\z.lz r hr* 156.Y-t = lb, r,zt. #lr{ sgr - 5e. f(-l sE_ h -tu ssrr = L+.rurl-!'itliu = rt ,388 =. l.0et (0

uww\ s!.tnl4oyu c*vl4 rrtl- tuul *anvw W dll (W 4 W4ltt4dta b. ' il\WT"iti"to determine if the data provide sufficient evidence to Test at 5% y$ni in operation costs of 3 types of machines. State null and alternative hypoth (1 mark) r{,: *the* e\t- vyuwieu w,r+ $ d'4[t+rr+/^rf fu fua1 olfuS-S' State the calculated value of test statistic and critical value of test. (1 mark) f*.= {e,asrzrn = 3.98' .foor.= +r ?t{ Decide whether nun hypothesis is true Jt'm;.[]l"r.,rsions. lf required, use Tukey's procedure. Draw the diagram for corresponding population mean and underline the means that do not differ significantly. (3 marks) W-"fu y @ vNn-c is w; il^*'*n Mek@r.FT,r,,-,w%% T*. = {*,r, vr-,e ry,f r. ,6, r. (f, I (a- ,Ta <T trr lr ri >li >l L< r, E le.g 8 - [0, 16 =2.Y2, L, 2 lz,gt - 10:68 z,t.g0 Ir =,0.68- 10. 15= 0,S h ,{ I * [i Ir F 3. ,2t 6,.Conftru.t 6O % individual confidence interval for the difference in mean operation costs for Type : 0, 6tL 0,5L- {.T96 ( o.sl - 4" L\ t,+q6 r0,6gY) Z9 o.1r t f, - tn r. = [0. [6 t )=16s6 i+;, {EsT-' ind icate a I

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' $''M. - hrn^r All o VA @trmarks) P"^")*^^^fV'l $L'dL MY,'. Four groups of tennis players, each of which contains five tennis players, have been randomly selected from a racquet facility club in a city. Using the same tennis ball, each of five players from a group hits a serve using one of five racquet models (model A, model B, model C, model D, and model E)' The players from the same group are randomly assigned different racquet models' The speed of each serve is measured with a machine for all players in four groups, and the result is recorded. a. What type of design was employed in this experiment? (1 mark) b. ldentify the response variable for this experiment. (1 mark) r^qlr* *^rdel. d. What are the levels of a factor? (1 mark) c. ldentify the factor that should influence the response? (1 mark) uedll I f4 od'LL e. What general name can you give to each group of tennis players? (1 mark) Y^fl^t @

H.o-,,1.,;:ffiI:l",lff#o,"W{*,W,,rW,W"mexperiment with 6 blocks is shown below. The missing values in ANOVA table are identified by asterisks. @ complete the ANOVA table by placing the missing vatue below corresponding asterisk. @ ,o the data provide sufficient evidence to indicate the differences among the treatment means? Test an appropriate hypothesis at t% significance level. O't there sufficient sample evidence to indicate the differences between the block means? Test an appropriate hypothesis at L% significance levet. Was the blocking usefut design strategy for this experiment? @ tt the difference between sample means for the treatment 3 and for the treatment 2 is lfor u -u 2.40 , find individ ual99% fid l4SIr = SL ,= g.03 =) K-l SSE= gSf - S9T'r -SgB; g,TS [f ) = ZT.0q. Ll.gg - ?{,,S = 33 t{l SSj^r = t3s. 0 - e Hotli.= $*^. * tt.oI | $," = $o.rr, r f ru.. fr. =) Ho is * i5rlrWuqt*ry wr dPuat t, c)'rrililf'L Pl/tart' r15 =@' Jrtwlca. alr d.= 0. O f , rlrlo lnOlVlOUal 93Yo COnItOence tntefva or p_ -lt lwD )VA TABLE J. 2. Source of variation Degred of freedom Sums of squares Mean squares F ratio Treatment ,Fl k-[ =6) * I+.0r. 9.03 q.or Block 6! 6-r 71.5 + lu,g a.6g' Error (}(r-'l|bil * 33.q I z,23 Total 01, 135.0 @t"vrto I Fr-r,grt-r)(G-r) =Ea,,y= ffi hu*+twvrlr v\At ,lnl o\^{- di&Wk o+u$" o

@ Vl4/Lrrtl>. T$r+ v H"futvutc; lLn I (rc-r)(6-r) = Fr, rtr = f4ig hsE &. = 6.66 t f*.= F.r, r f,r r, =@ fl t f *', Hs rs Y ak*=o.o( lhr^c ,t *buo^4 e,,^r+t" *ia,J,ru t1*t fiu \t*,rtc ratA,wt uW' yr.*'[cata+ 'ilt-[b-r^r* $u-6l*k vtrt?A,t^^ W# k'd" \'Y'aa W dl'"r % f fi'us e Uu*^ , Tr,- [r, = z,i- *o.ooIr li = Z.9IT ,[G tT L,$Y+[o.Bb2) _ 0. l1 2,!o b 2. tl0 @

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6. (3+4+3+2=12marks) K=3,6 =t{ t'h, k* = 12. three methods of soil An experiment was conducted to determine the effects preparation on the first - year growth of slash pine seedlings. Four locations (provincial forest lands) were selected, and each location was divided into three plots. Since it was felt that soil fertility within a location was more homogeneous than between locations, a randomized block design was employed using the locations as blocks. The methods of soil preparation were A (no preparation), B (light fertilization), and C (burning). Each soil preparation was randomly applied to a plot within each location. On each plot, the same number of seedlings was planted and the average first - year growth of the seedlings was recorded on each plot for each method of soil preparation. The data obtained in the experiment are represented in the table below. Soil ------\ Location P repa ratio"----------- 1 ) J 4 l-'i, A 1t 13 t6 10 s0 B 15 t7 20 t2 6q C 10 15 13 10 tl 6 36 vt 14g 37 I bz. t<b It was found that I I i:l j:l Construct ANOVA table. Leave 3 digits after decimal point for the values of SSTr, SSB, SSE, SST, MSTr, MSB, and MSE. Leave 2 digits after decimal point for the values of F - ratio. Do the data provide evidence to indicate a difference in the mean growths for the three methods of soil preparation? Test an appropriate hypothesis at5% significance level. lf required, use Tukey's method to determine which means differ. Use a = 0.05. ls there evidence to indicate a difference in mean rates of growth for the four locations? d. Use a 95% individual confidence interval to estimate the difference in mean growth for methods A and B. 0^, - Yft N 0 VA, kar,"*w,i*c,A - (r,b.& d,.e+a^ y, 2,298 = of ..2 v.. - ,1J a. b. 11

fl , (. i,r',stlrrci i\NO\/,A rablc. I I ,"'U"e ) 0rJ_:3 tK-r)10-r =$ -l= ll 5ST'= 5=3r 0='{' s <,r1'f}ecdonr S.r'rrs ,,l i.lullrus i \1.,,,r lc-t = / 3$o4o I tq lo , o,t 6 t, 66+ 26, SSi 4,991 €,f,t'IT# -- = ZLIE - F bt*,rt<- = WL = zo ' 916 = Wb i-. egs Y = ZzSs- zts+.@ E LLLi - Lt8+ =@ Isb =,* + * : ! [lo.u !sr+,rBrr3zz\ -zt8{; = zzv,.66+_ zr&r =@ ssE_ = sgr-ssrr-sss= {1 r-ss*6(,6(+=@ hgf.: S,3{-r = 33- = k-r a ll'llt ll,t , oo0 Soirlce oi veLriiLtiou ij ;,iiiii.'- "l'rtitintc n I la, pg l.:. rro r ['o ta i

K=3,4=Y I lUr,rrru+*) b Do the data;rrovide evidence to irrdicate a difference in the mearr growths for the three nrethods of soil preparation? Test an appropriate hypothesis at 59t significance level. lf required, use Tukey's method to deterrnine which nre.tns difier. Usc rt = {1.0,:. ho i t't^&-(/ 19 rnn e;kk-"r^^" Ltih^/{*a W u^,e.a,14 y* fi^* $ur \u^r*til,+ ? s-0,(0 y\wn"4+w H{ i vutxtrL YW f,e* a*+ Urr+ &-"* o,t,k-tf,5 d, =.0,0 f \n,l4 r4.^4{il4c I F* -= 5 --; , !* j os.. 0,or l4^-a-+ <* hAL {+,t &*t u*t *l- ^ v'1 aq. ws?tr*"t is *8P* tK-r)t{o-l) 3 FzrO J,a'>fo'+ wLWy (: llv,t{ tl Wua* *--fe 4ptd*ta,tt % tdr" \^^r,U/wd* % s,oat- ytp,Aarkpu b,- f,= l6-12.s=3,s>L fr_t-, ftz [,,-F,= lz,S-]2,0:()'5(fcr ==*e tr.. r.tt- I^,, *www6 b k^& *qt+".a B ii Wpqt4:Ao{*^@ il wfr.n^ ww w lw v,,w rw*un*i A aAL*c. m-,*^r" t */tal $* F{4,e- t^Ailvirva,cll & a-v,a- L et r<- -tv++ "]*,e ,"r{&$d-*.,{- W "tl)r't'L o{'ukr' -- @ I^/\tAl - ay*a<t Vtt de(f+<"^} $*",r" vt+** . q': :+ i .5,,i ry =@ f,,= *= @ il 11 f:::+ i .5,, 1e=@ {,,=f = @ x--,,_= Y=@ l vv\Lpr^t>i@ /L-;\'_ Tl,=12.0r f',=le,.5r t'2,=l8,olL:4*,or rr-r \ft-r) \tr- fo.offi ttr--@

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(: r"r,t") C,ls there evidence to indicate a clifference in nrean rates of growth for thc four locations? @W"^-l!/1) Test an appropriate hyprrthesis at 5% significarrce level. ls rt reasonable to use tlre locatiorrs as l:iock:;? W di&P+{n+ bcc*w ttt : l.{* h^&&(^- qc*^/tr, #*"{ U* ol L % L&e-1<-o45 ir raq^,a"rdi crultt^dd4{4{rtrd* &ro..r& a'14u5 . TCI't !to-fue4-16 ; Fg-, r'(v-r) {4-r) ; Fs,s = M,B HEE [, l- l- [, } ,n" a 957o confiderlce interval to estimate thb diffe,.cnce in rr.le;r'r growth for nrethoLls A and B' (Z "4 = [0, 0r[r, = lf,,t - f"/", [K-l)C0-l) f9,o?f,,6 {! I,. 3. t l. tl rlt [ 0 ,g+L) a @- + 2.I1?h,97tI @ z,g8 , 14z' ^t, 3,S- 2.3s .lnr, < 3., 3.5 t r(. |cr = 5-o,o,lI Tl^, i* il''cllrnt ww(t wiil'wtz e( {4* lnoq/ttryW wL WL n/t't&lu X,trurn' MWww-M ?, Ll .E89 ) @

@ 13*, +2+3+2+2= rymarks, MN0-vA **o d,rrxr^b4. t ntfiaifi of jewelry stores conducted an exferimenfto investigate the effect of locition and price on the demand for its diamonds. Six small-town stores were selected for the study, as well as six stores located in large suburban malls. Two stores in each of these locations were assigned to each of three item percentage markups. The percentage change (gain or loss) for each store was recorded at the end of one month. The data are shown in the table below. -----.\ Markup (Factor B) --\-:.:___--: Locatio n (facto r A)------.- 2 3 T;, Smalltowns 12,6 Jir, = 16 f,',.,= I -1,9 I,z, = E l,z, = 4 -8,-22 Trr, = -30 ['rr,= -tS -h Suburban Malls 't 6, 20 Ti,.= 36 Tr,' = 16 l_0, 5 Tr,z.: lf, f,zr.= T'5 - 2,5 'lzsr= 3 lrg, =4's ,1 sl 23 _z+ @ following information is available ill in ANOVA table on the next pae 23 2 t- =1,E, P-,!iy =1620 -, I,t. \r,., T,r. . For SS and MSTeave thr, ee dieits after decimal For F ratios [r.. = -'1. Te'= 5tt T;, ,=.90 The Fill in ANOVA table on the next page leave two digits after decimal point. digits after decimal point. b. Test the hypothesis at the 5% significance level to determine if factors A and B interact. Show that there is insufficient sample evidence to conclude that there is interaction between the factors A and B. c. Test at the 5% significance level to deternnine if differences exist between mean percentage gain (or !oss) for two different locations (differences between factor A means). d. Test at the 5% significance level to determine if differences exist among mean percentage gain (or Ioss) across different levels of markup (differences among factor B means). If there are differences among factor B means, use Tukey's procedure to determine which means differ. e. Find an individual 95% confidence interval for the difference in the mean change in sales for stores i" Vnf6rrnk,,t" hdll1 ,ersus l4.c+6 i" ftnno(l |ei JtSif the stores are using price markupt. f. Construct the interaction plot for given set of data. Place the levels of factor $on horizontal axis. Write the solutions to questions a - f on the following 3 pages 15 t

-?_ _3, L_ L t -, r*ro ,,tz2,$=3, fzz ;E/rttln=t'uzn a. Fill in missing values ANOVA table. For SS and MS Ieave 4 digits after decimal point. For F ratios leave two digits after decimal point. (2 marks) ANOVA table Source of variation Degrees of freedom Sum of squares Mean square F ratio Location q-l = 1 lgo, 37?g l,to, ?353 T,g? Markup 0-l z E35.166t tt lT. sg 3Y [r " g T Interaction [a-r){6-r) = Z 8t, t66+ \2, Essv ,1 "21 . Error "6tr-r) = 6 7-ll , 000 0 3S,166A Total olt-r = ll l4u , bb6+' I l= 3, V,= z { zo g, 9? 33 z F' I"t = dhr lbzo :. / z .t0 IZ -5!-' t2- = tk*)'.ril - .t- *lgE, 6 6 6+ = { Lsu'r z3z+ [-z?17) - # SgIr = = lv0g 99rta = - S, Uo6 I,," = L dbr 2 ,y',rY3.5 (,r'+ 6t+(- 30) L + Z6'* tf?+g)- #'= ^ SSAB = lzoo. 606I - 2&o ,3?)i - I35, t66+-- gSE = 991^ gSIr = lytt c 666*_, ZoO,656+ =@ 16 rCI} 5 TL ,G 1,g0,?133 )

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b. Test the hypothesis at the 5% significance level to determine if factors A and B interact. Show that there is insufficient sample evidence to conclude that there is interaction between the factors A and B. State null and alternative hypotheses and test statistic. (0.5 mark) f{r: llr,t,lr ir Ig rntV,n<4rioq !urL-u^ W ^ auf' b I{,: tlnrrc ir v)prwrfuiou Url<Nq,ra W A dld,b {= 0.0f Test statistic: hltg = F 2,6 ntE State the calculated value of test statistic and critical value of test. (1 mark) f*,. = 4. zl f*. = &n' rs, ?, o = f,tY Decide whether null hypothesis is true and make the conclusions. (0.5 mark) hv,,u fo,tt, < !,tr, wc AA ry W ut * = o'of I1^r^r is ,n*r,iiu'; W dt,dt w h ea^,r)il)e l4A'* l'l'anev v^lt^a*1*-W; tdfr* A^',,.Ab ' c. Test at the 5% significance i6vel to determine if differences exist between mean percentage gain (or loss) for two different locations' State null and alternative hypotheses and test statistic' (0'5 mark) FIo: wdW-WY-ryf,wunr* 11, -[lffi ; i a,,-@ari Wn- i,'Ara,'1^ Test statistic: [4SA = Ft,a jaE ' --r e State the calculri& ,rtre of test statistic and critical value of test' (1 mark) f*,.= 1.9 + L: te,osrtrb = r.q7. Decide whether null hypoihesis is tiue and make the conclElo1r::q'S mark) br"* f *,s ., f*. wa */g* t4o of &'= 0'.0t t ..tt .tt.. W ai u^k W "w n'd,n nrrrr* h c&d,brAL 44^r)'+ tJh;P is t^ t"r^^ PW Y;n'[bYJ fi' Pun n

Test at the 5% significance level to determine if the differences exist among mean percentage gain (or loss) across different levels of markup. lf the differences exist across different levels of markup, use Tukey's procedure to determine which means differ. Write down the sample means in ascending order. Draw the diagram for corresponding population mean and underline the means that do not differ significantly. State null and alternative hypotheses and test statistic. (0.5 mark) Fr,tM_!W-u4'":wY-- I{,: lfuris {tc eryuultr fu al 0'!4*t t,ru ?o"' { (l'w 't*oat'5 Test statistic: ' t- ztb = State the calcula W 14sE ted value of test statistic and critical value of test. (1 mark) -fou, L.= fs.o;rz,o =9, l1- f,.r;.H= +:=+=-6.Ts - !l, = 13"50 - 5'U = +.+S<Irr' , - g',3n a Er,?F * 0-6*fl z la. S0 < l?n = 41. 8+ Decide whether null hypothesis is true and make the conclusions. Use Tukey's procedure and dram the diagram if required. (1.5 mark) b'rs, Ir. $,, T,, l- -[ 6; +S , n,r, = e, ?f, I i'. 1, = 13.s0 .?. E .;3,t0 - t- 6.15\ 3 20,\S > l;n.' 18 fl*^. ;, -Aer^fu ? iy, W,y^,is w M&y,rr^^+prr@:+ oAt\L Y' % {4^t"W ' - #= !!r ="=Fl = i.i,: r3.F0 I rc", = Ir,a, albr-1, W = o-" of -Z.Z -l I - t_

,!,i,-td-*f,-{*t", i=Lr h s { j=j',P=3 €. Find an individull 95% confidence inte rvaifor the cjifference in the mean change in sales for stores in Wrtufio,t Vlrrllb Versus those lavv*alL FUUr lf th.r si ,'.'1rc are using price markup !. . Ilr. ='f.5, 1^,0, = * l; i T-.. - !-or, z 1uS* t-vt] " @ f,.r. - lo.s [b 'f Xir, f o. ozr, 6 ,j aU* lr" 2,qq+,1 A13s.r*rj ? Yt,* > , - fr. oz{,t lzusg ,[i; rq+ s rl b I '?" fr I ;.{ 1r s i, L{ J' {l interaction plot foi. 1;iv,:r.se i'li ciai,r. iriace the levels of factor B on horizontal axis. Mean response Ttl ='g -f f't ,lh * fiir,- [,, Jb"t- L. + tt.r f. Construct the f I I ,rl ?!. c {rS lw** 't o{ {u*+era -t5 !'n r'= ti"' V^ e g \rr,- ,1 ,T,*. = ". tt E tI fbr*.: +.t S*rs. * +,5 ,&**r *J $a**+r $ *ai E{, ; } tl. 4 brfo u) &ul'm"B, tlrr, : t{ @

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Iwo -wu AN0YA @,omarks) @ruruA"ounaddl'YY An experiment was conducted to stirdy starting salaries of assistant professors. Five male assistant professors and five female assistant professors at each of three types of faculties (Faculty of Arts, Faculty of Sciences, and Faculty of Engineerirrg) r,rrere polied and their initial starting salaries were recorded under conditions of anonymity. a. What are two factors involved in this experiment? {L rnark) b, What are the levels of each factor? (:1 r'nark] c. How many treatments are there it'i tht o'.irt,,iittrtiri? iCI.5 mark) 0-xB=6 [q=L,&=8, *,0=6 d. What are experimental rrnits in thi:; e r,i'e i-iitiei:i? i*,; mark) e. what is the response variable in this L,"ljr.iir)rt)iri, ,*.5 mark) f. How many observations were obtained in this experiment? (9:5_:1.:rk) -

' 't,i r 9.(1+1+1+1=4marksl Answer the following questions. Ytxr do not need to show calculations for futl mark. You can use free space below for rough work. a. A researcher wants to determine whether there are differences among the mean increases in heart rates after specific amount of exercises for four age groups. He selects random sam each of r-- :ases of heart rates after specific amount of exercises for each su four age groups and records the increases of heart ra - participatinffi the study. He uses ANOVA to test for a difference in heart rates due to the age. The critical value for the test is the value of test statistic which follows F - distribution with v, = 3 degrees of freedom in the numerator and ,, = Brb degrees of freedom in the denominator. -- n obiained in-tEEEEEiliffidnt is i7 * = l. l K-l= 3 [ =' lo (,\J = 1o It-t= t{ + K=f, h-K = 12. =+ h = l(+tL= 5+lz = [T c. Suppose that certain random experiment is performed using randomized block design with b = 3 blocks. The number of treatments k = 4. The values of response variable obtained in the experiment are placed in the table with 4 rows and 3 columns The number of observations in each treatment total is The number of observations in each block total is Y Total number of observations obtained in the experiment is d. ln a two - factor experiment, factor A is at 4 levels, factor p is at 5 levels, and there are 3 observations for each combination of factor A and factor B levels. The number of treatments involved in this experiment is equal to ttx 5' = zD Total number of observations in this experiment is equal to IL h-r( = 3b b. Assume that in a completely randomized design for one - way ANOVA we test the hypotheses that all treatment means are equal against the hypothesis that at least two treatment means are different using fi observations obtained in the experiment. lf test statistic for this test follows F - distribution with;!:]egleelof freedom in the numerator and l2degrees of freedom in the denominator, then total numbe, of-oIffiffiii Lo x 3: g0 21

ln a completely randomized design, 9 experimental units were assigned to the first treatment, 11 units were assigned to the second treatment, and 10 units were assigned to the third treatment A partial ANOVA table for this experiment is shown below: Nsfr/rtsa =.F + ilgrr Source of variation Degrees of freedom Sums of squares Mean squares F ratio Treatments {< k-l= e {< lz"oo {< 6, 0D 1.50 Error h-t(=L+ ,r loB.00 4 Total {< h-l = f.9 {. YLO, 00 - FiuinthemissingvatuesintheANovAtabte. SSE = I'tSE(n-u) = Y (t+1 = l0b Note: missing values are identified by the asterisks; Place the missing values below the asterisks' you can use free space below for rough work. You do not need to show calculations for full mark. "' - t . 10.(3 marks) 11. (3 marks) h=hrt[z*hr= S;f+10=30 1' k=3. t , oh Lr- r) : f o, oes 1a\_: = lltF. Fl ot=tl (=3, h=30 =)r= #= *=, @ ln 2x3 factorial experiment with factor A at 2 levels a levels, total number of observations of response variable was n = 30. Calculated value lf the mean of all observations at level 1 of factor A and the mean of all observations at interval for corresponding difference Lr j * z.o6t_{4ia+) 1 ,95 0.r52+ fit- l^. < \[ri4G- , Gr -/ llrE = fffo=* = 0. r+b z.g + z.o6Y Io,tgz+) 2 .62. 22 was equal to 4.2.

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Ct,=3rf =3r r-=Z- 12.(3+2=5marks) The table below gives experimental data for 3x3 factorial experiment, with 2 replications for each treatment. s offactor B Levels of factor A 1 2 J 1 5,7 i I [rr.: 6 9'7 flv' = I 4' 6 7t''' t 2 8'7 ]'o, --T'I t2, t3 f,4. =45 7,to F, =A,, J 14' rl f,ar. = lz.f, 8' e \n'= I't ),2, ts l*; I tr,s a. Construct the interaction plot for given set of data. Place the levels of factor B on horizontal axis ,r,l*otl$WA , bn, t 1Ptuh b. lf SSAB =71.45 and SSE =ZL.OO, dothe data provide sufficient evidenceto indicate interaction between [=3 A =E- f = L factors A and B? Test appropriate hypothesis ata = 0.05. [4, \ vrp i*,ra,,*iun Ais^, ry\. Pt't'stt tr ur^d g il i tt*--,, rv,,4-u*k;n;u[;; W kd''"Lhr *b i-r,, glahyk, i Frq"-r)t6-i ,aubcr-rJ = Fl1 t r' ffi $*o = $!eafi^-,t*-,1 z +t"t,s/q - +, 'f ,1,! - tr,yg. \J ssE /ai,(r-r) Ttr T' ffi7: Jr" = 1ilr,r,r = ?.bt + W= ff^ :i t4,o ioW*:4as i^t* I, Srultic*U* @*,We'nd""rll tW+ lut-*it lo,t*w-qr,rv €tr+ztot* K ur-bL b

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