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1、Satis (I)ialExmJa. 1, 解答：d d a ， cc dd cMtipeie uesios(30)1. ich o te fllowig ne ohe samplng dstrbuton of the samemn?a.Themean o th amlng sriution is equal to hpoplaineab. Tesanrd devtion f the splingdisributio is equ to te sandard eition ofth oulain diied he square ootof nc.he shape of thesamlin ds
2、tribution sapproxiaely norma fthe sml ize isuffcielylargl ofhe aoe areprprties f tesamplin diiuton of the ampl a.2. hy s e Cta Liherem soimranto the tud sampling dsrbutions？. It allows us to disgar the sze of h sal lectwhe thpopultio is prximaely norml.b.I llosus o drear teshapeof the sampingdistrib
3、ution when the sze o pouliis lare.c. lows us to disrega the siz f the pultione ar ampingfm.d. Italls t disrgardthe shae of pouaion when is larg. Asampetat doesno rovide goorepreentation o thpopultonrohch it ascollecte is referd toas a（n) _ apl. iasd b Empirica .ttistic d. Ieenal4upposa 5% nfiden int
4、erfor turns outto e (1,0, 2,100). Baed o the inter，doyou bievetheaveraesequa to ,20?. Yes, d I a 95% sur of it.bYes, ad I m10% sur of it. N, n m95% uref it. 1，20 i possible value fo he en.5.Intheconsuction ocfide interals, f all oter antties arencanged,a increasei th sample sze wl lt _ intervala. Nr
5、oer b.ider . Lsssignifcnt iase6. h sampesze formul are usu fr _. Calcatinga vaue oft teststtistic wen ttinb. Calculating avlue o a large samle tt o n.c. Caculatinthe sapleize toetime parametetwithia peified bnd.d ompn fet onfidnceintervls ctruted si dfferentsaple ze.7.Supose we wish to tet: 4 vsHa:
6、4. ht wil resul i we cocld ht th men s grtrthn 4when ittrevalue i real 5?a. Wehave made a Tpe II r.b We haemadeaType I errr. W hve mde correct decisond. oe of teabvearcorct.8 Wich he flwing statemns is Notrueabu the lvel of sifianc in a es o hpothsis?a. h arger the levl ofsinifince, moreiel yo are o
7、 ejecthenull hpothess. . The levl ofsigniicncs he mxu rkwere iligtaept nmakng a yp errorc. hlevl of igniance aso clld taphalvel.d Teeve of significne another or p-vaue. Whichpar of the ohpothessrocdure temes the vlueoftep-ve?. he altrtive hypohesisb.T et statiticc. Thsaling distribution of te e sati
8、tic Al fabe10.eae creted a 5% confide interval or wth e resut (10,5).Wht cocsionwil make i test H0: = 16 v Ha： 16 at= 5？a. RejecH in faor f Ha.b. ActH0 infavoof H0.c Fil to reject H0.dW annot elwat decision wil bewihtheiotion ive.Esayeios. A local akepoted h federal governmenthat its 5,246 saings ac
9、ots ave a mea baae of $，000 and a stndard deviaon o $40. Governmet auditors have askedto radly smp 64 of he banacounts o asess th elabilitf the mean bance epd byte bank. If h banksformatin is corect, fin theobabiity tat the sape men bane ould b lstan$,072. (%)答：N=54(有限母體隨機樣本並非完全獨立變異數要做修正）n=64N(1000,
10、 )P(1072）= () =P(Z45)099 Let X a Ybe idependetrandom vabls ith(X)3, E()=, Var(X)=Var()=2,fin thevlue f wich akM( - Y）Y2 be an unedestiat f 2.（1%)答:須使M（X2 - 2)+Y2之盼望值等於s2 Q E(X)=， E(Y)5 Vr（X)V(Y)=s2 E M(X2 - Y)2= M(X)+ （1-M)E（Y）其中,Vr(X)=E(2)-E(X)2 s2=E（2)-9 E（X2)= s+a()=E(Y2)-()2 s=(Y)-25 E(Y2)= s2+2
11、5E M(X2 -Y)+Y2 = ME(2)+(1-M）E（Y2) = s2 (Q不偏估計元) Ms+ (1-M) s2+2 = s2 s -16M +25 = s2 16M= 25 M = 25/163A na ppultohas unnownaiace,2. If (148, 8.5) s a 95 onfidence intealor2se onrnm sample f size 1，find te vaueof . （1%)答：s2=(14, 8.5) 148s2.5 推想s2信賴區間之計算措施 c2統計量 c2(-)Qn=2, a=005 信賴區間之建立: =8.86=32.436或 =1.482.92=32.4164.he brach manage of utlt f lge,natonwid oostor chinants t udy carcteistics custoers of er store,wch locted near te campus f alarge stae nivesityn ptcuar, h deids to focs n to vriabe:he ount f mese
