= 6degrees of freedom, if the test statistic value f = 4.53, then P-value =.

Transcription

1 Sectn In testng H : σ = σ versus Ha : σ > σ wth ν = 4 and ν = 6degrees f freedm, f the test statstc value f = 4.53, then P-value = The sample standard devatn f sdum cncentratn n whle bld (meq/l) fr m = 0 marne eels was fund t be s = 40.5, whereas the sample standard devatn f cncentratn fr n = 0 freshwater eels was s = 3.5. Assumng nrmalty f the tw cncentratn dstrbutns, test at level.0 t see whether the data suggests any dfference between cncentratn varances fr the tw types f eels. H : σ = σ wll be rejected n favr f Ha : σ σ f ether f F.05,9,9.8 r f f =.459. Snce f = (40.5) =.59, whch s nether.8 nr.459, H s nt rejected. The data des.8 (3.) nt suggest a dfference n the tw varances. 76. In a study f cpper defcency n cattle, the cpper values (ug Cu/00mL bld) were determned bth fr cattle grazng n an area knwn t have well-defned mlybdenum anmales (metal values n excess f the nrmal range f regnal varatn) and fr cattle grazng n a nnanmalus area, resultng n s =.5(m = 48) fr the anmalus cndtn and s = 9.45 (n = 45) fr the nnanmalus cndtn. Test fr the equalty versus nequalty f ppulatn varances at sgnfcance level.0 by usng the P-value apprach. H : σ = σ wll be rejected n favr f Ha : σ σ f ether f F.975,47,44.56 r f f F.05,47,44.8. Because f =., H s nt rejected. The data des nt suggest a dfference n the tw varances. Sectn An experment s cnducted t study the effectveness f three teachng methds n student perfrmance. In ths experment, the factr f nterest s, and there are dfferent levels f the factr. teachng methd, three 5. Sngle-factr ANOVA fcuses n a cmparsn f mre than tw ppulatn r treatment. means 6. In a ne-way ANOVA prblem nvlvng fur ppulatns r treatments, the null hypthess f nterest s H :. µ = µ = µ 3 = µ 4 7. In sngle-factr ANOVA, the s a measure f between samples varatn, and s dented by. mean square between grups, MSA 8. In sngle-factr ANOVA, the s a measure f wthn-samples varatn, and s dented by.

2 mean square fr errr, MSE r MSW 9. In ne-factr ANOVA, bth mean square fr treatments (MSA) and mean square fr errr (MSE) are unbased estmatrs fr estmatng the cmmn ppulatn varance σ when, but MSA tends t verestmate σ when. H s true, H s false 0. In sngle-factr ANOVA, SST SSA=. SSE. In ne-factr ANOVA, dented by s the part f ttal varatn that s unexplaned by the truth r falsty f H. sum square fr errr, SSE. In ne-factr ANOVA, dented by s the part f ttal varatn that can be explaned by pssble dfferences n the ppulatn means. sum square fr treatments, SSA 3. Let F =MSTr/MSE be the test statstc n a sngle-factr ANOVA prblem nvlvng fur ppulatns r treatments wth a randm sample f sx bservatns frm each ne. When H s true and the fur ppulatn r treatment dstrbutns are all nrmal wth the same varance σ, then F has an F dstrbutn wth degrees f freedm ν = and ν =. Wth f dentng the cmputed value f F, the rejectn regn fr level.05 test s. 3, 0, f In a sngle-factr ANOVA prblem nvlvng fve ppulatns r treatments, whch f the fllwng statements are true abut the alternatve hypthess? A. All fve ppulatn means are equal. B. All fve ppulatn means are dfferent. C. At least tw f the ppulatn mean are dfferent. D. At least three f the ppulatn mean are dfferent. E. At mst, tw f the ppulatn means are equal. C 4. Whch f the fllwng statements are true? A. In sme experments, dfferent samples cntan dfferent numbers f bservatns. Hwever, the cncepts and methds f sngle-factr ANOVA are mst easly develped fr the case f equal sample szes. B. The ppulatn r treatment dstrbutns n sngle-factr ANOVA are all assumed t be nrmally dstrbuted wth the same varance σ. C. In ne-way ANOVA, f ether the nrmalty assumptn r the assumptn f equal varances s judged mplausble, a methd f analyss ther than the usual F test must be emplyed. D. The test statstc fr sngle-factr ANOVA s F = MSA/MSE, where MSA s the mean square fr treatments, and MSE s the mean square fr errr. E. All f the abve statements are true. E 5. In sngle-factr ANOVA, MSA s the mean square fr treatments, and MSE s the mean square fr errr. Whch f the fllwng statements are nt true?

3 A. MSE s a measure f between-samples varatn. B. MSE s a measure f wthn-samples varatn. C. MSA s a measure f between-samples varatn. D. The value f MSA s affected by the status f H (true r false). E. All f the abve statements are true A 7. In a sngle-factr ANOVA prblem nvlvng fur ppulatns r treatments, the fur sample standard devatns are 5.6, 30.4, 8.7, and Then, the mean square fr errr s 56 A. 9.3 B. 7. C D E. Nne f the abve answers are crrect. C 9. In ne-way ANOVA, whch f the fllwng statements are true? A. SST s a measure f the ttal varatn n the data. B. SSE measures varatn that wuld be present wthn treatments even f H were true, and s thus the part f ttal varatn that s unexplaned by the truth r falsty f H. C. SSA s the amunt f varatn between treatments that can be explaned by pssble dfferences n the ppulatn r treatments means. D. If explaned varatn s large relatve t unexplaned varatn, then H s rejected n favr f H. a E. All f the abve statements are true. E 4. In an experment t cmpare the tensle strengths f I = 5 dfferent types f cpper wre, J = 4 samples f each type were used. The between-samples and wthn-samples estmates f σ were cmputed as MSA = and MSE = 394, respectvely. Use the F test at level.05 t test H : µ = µ = µ = µ = µ versus H : at least tw µ 's are unequal a H wll be rejected f f F = 3.06 (snce I = 4, and I (J ) = (5)(3) = 5). The cmputed value f F s.05,4, f = =.85. Snce.85 s nt 3.06, H s nt rejected. The data des nt ndcate a dfference n the mean 394. tensle strengths f the dfferent types f cpper wres. 43. The lumen utput was determned fr each f I = 3 dfferent brands f 60-watt sft-whte lght bulbs, wth J = 8 bulbs f each brand tested. The sums f squares were cmputed as SSE = 475 and SSA = 590. State the hyptheses f nterest (ncludng wrd defntns f parameters), and use the F test f ANOVA ( α =.05) t decde whether there are any dfferences n true average lumen utputs amng the three brands fr ths type f bulb by btanng as much nfrmatn as pssble abut the P-value. (nte: MSTr = MSA) Wth µ = true average lumen utput fr brand bulbs, we wsh t test H: = = 3 H a µ µ µ versus : at least tw µ 's are unequal MSTr = ˆ σb = = 95, MSE = ˆ σw = = 5, s

4 MSTr 95 f = = =.3 Fr fndng the p-value, we need degrees f freedm (I ) = and MSE 5 I (J ) =. Snce f =.3 < F.0,, =.57, then the p-value >.0. Snce.0 s nt <.05, we cannt reject H. There are n dfferences n the average lumen utputs amng the three brands f bulbs. 44. In an experment t nvestgate the perfrmance f fur dfferent brands f spark plugs ntended fr use n a 5-cc tw-strke mtrcycle, fve plugs f each brand were tested and the number f mles (at a cnstant speed) untl falure was bserved. The partal ANOVA table fr the data s gven belw. Fll n the mssng entres, state the relevant hyptheses, and carry ut a test by btanng as much nfrmatn as yu can abut the P-value. Surce df SS MS f Brand Errr 4,700 Ttal 30,00 Surce df SS MS f Treatments 3 75,000 5, Errr 6 35,00 4,700 Ttal 9 30,00 The hyptheses are H : µ = µ = µ 3 = µ 4 vs. Ha : at least tw µ 's are unequal..70 < F.0,3,6 =.46, s p-value >.0, and we fal t reject H. 45. Sx samples f each f fur types f cereal gran grwn n a certan regn were analyzed t determne thamn cntent, resultng n the fllwng data (ug/g): Wheat Barley Maze Oats Des ths data suggest that at least tw f the grans dffer wth respect t true average thamn cntent? Use a level α =.05 based n the P-value. Surce df SS MS f Treatments Errr Ttal = F < < 4.94 = F,.0 < p value <.05 and H s rejected at level.05. Snce.05,3,0.0,3,0 Sectn 0.

5 36. Cnsder calculatng a 95% cnfdence nterval fr a ppulatn mean µ based n a sample frm a ppulatn, and then a 95% cnfdence nterval fr a ppulatn prprtn p based n anther sample selected ndependently frm the same ppulatn. Whch f the fllwng statements are true? A. Prr t btanng data, the prbablty that the frst nterval wll nclude µ s.95. B. Prr t btanng data, the prbablty that the secnd nterval wll nclude p s.95. C. The prbablty that bth ntervals wll nclude the values f the respectve parameters s abut.90. D. All f the abve statements are crrect. D 37. If three 90% cnfdence ntervals fr a ppulatn prprtn p are calculated based n three ndependent samples selected randmly frm the ppulatn, then the smultaneus cnfdence level wll be A. abut 73% B. exactly 90% C. exactly 8% D. exactly 70% E. Nne f the abve answers are crrect. A 38. The assumptns f sngle-factr ANOVA can be descrbed succnctly by means f the mdel equatn X j = µ + εj where ε represents a randm devatn frm the ppulatn r true treatment mean µ j. Whch f the fllwng statements are true? A. The ε j's are assumed t be ndependent. B. The ε j's are nrmally dstrbuted randm varables. C. E( ε j ) = 0 fr every and j. D. V( εj ) = σ fr every and j. E. All f the abve statements are true. E 39. Whch f the fllwng statements are nt true? A. ANOVA can be used t test H : µ = µ versus Ha : µ µ. B. ANOVA cannt be used t test H : µ = µ versus Ha : µ µ. C. ANOVA can be used t test H : µ = µ = µ 3 versus Ha : at least tw f the µ ' sare dfferent. D. The tw-sample t test can be used t test H : µ = µ versus Ha : µ µ. E. All f the abve statements are true. B 40. Whch f the fllwng statements are nt true? A. The tw-sample t test s mre flexble than the F test when the number f treatments r ppulatns s. B. The tw-sample t test s vald wthut the assumptn that the tw ppulatn varances are equal. C. The tw-sample t test can be used t test Ha : µ > µ r H : a µ < µ as well as H a : µ µ. D. The F test can be used t test Ha : µ > µ r Ha : µ < µ as well as H a : µ µ. E. When the number f treatments r ppulatns s at least 3, there s n general test prcedure knwn t have gd prpertes wthut assumng equal ppulatns varances. D 58

6 4. In a sngle-factr ANOVA prblem nvlvng 4 ppulatns, the sample szes are 7,5,6, and 6. If SST = 65.7 and SSA = 3.49, then the test statstc value f s A B..09 C D E A 50. The fllwng data refers t yeld f tmates (kg/plt) fr fur dfferent levels f salnty: salnty level here refers t electrcal cnductvty (EC), where the chsen levels were EC =.6, 3.8, 6.0, and 0. nmhs/cm: Use the F test at level α =.05 t test fr any dfferences n true average yeld due t the dfferent salnty levels. Surce df SS MS f Treatments Errr Ttal Snce 7. F.05,3,4 = 3.34, 0 4 the fur dfferent levels f salnty. H : µ =... = µ s rejected at level.05. There s a dfference n yeld f tmates fr 5. The fllwng partal ANOVA table s taken frm a study n whch the abltes f three dfferent grups t dentfy a perceptual ncngruty were assessed and cmpared. All ndvduals n the experment had been hsptalzed t underg psychatrc treatment. There were ndvduals n the depressve grup, 3 ndvduals n the functnal ther grup, and ndvduals n the bran-damaged grup. Cmplete the ANOVA table and carry ut the F test at level α =.0. Surce df SS MS f Grups 75 Errr Ttal 44 Surce df SS MS f Grups Errr Ttal F 4.94, reject H : µ = µ = µ at level.0. Snce.0,,7 3

7 Sectn. 7. The parameters fr the fxed effects mdel wth nteractn are α = µ. µ, β j = µ. j µ, and γ j = µ j ( µ + α + β j ). Thus the mdel s µ = µ + α + β + γ. The α ' sare called the fr factr A, whereas the β ' s are the j j j fr factr B. The γ ' s are referred t as the parameters. man effects, man effects, nteractn j j 4. Whch f the fllwng statements are nt true? A. The mdel specfed by X = α + β + ε and µ = α + β ( =, KK, I and j =, KK, J) s called an addtve mdel. B. The mdel X j j j j j I = µ + α + β + ε where α = 0, β = 0, and the ε ' s j j j j j = j= J are assumed ndependent, nrmally dstrbuted wth mean 0 and cmmn varance σ s an addtve mdel n whch the parameters are unquely determned. C. In tw-way ANOVA, when the mdel s addtve, addtvty means that the dfference n mean respnses fr tw levels f ne f the factrs s the same fr all levels f the ther factr. D. All f the abve statements are true. E. Nne f the abve statements are true. D 6. In a tw-factr experment where factr A cnssts f 4 levels, factr B cnssts f 3 levels, and there s nly ne bservatn n each f the treatments, whch f the fllwng statements are nt true? 60 A. SST has degrees f freedm B. SSA has 3 degrees f freedm C. SSB has degrees f freedm D. SSE has 6 degrees f freedm E. Nne f the abve statements are crrect. A 30. In the fxed effects mdel wth nteractn, assume that there are 5 levels f factr A, 4 levels f factr B, and 3 bservatns (replcatns) fr each f the 0 cmbnatns f levels f the tw factrs. Then the number f degrees f freedm f the nteractn sum f squares (SSAB) s A. 60 B. 0 C. 5 D. E. 59 D 35. The fllwng equatn SST = SSA + SSB +SSAB +SSE apples t whch ANOVA mdel? A. One-factr ANOVA B. Tw-factr ANOVA wth nteractn C. Three-factr ANOVA D. Randmzed blck desgn E. All f the abve B

8 47. The number f mles useful tread wear (n 000 s) was determned fr tres f fve dfferent makes f subcmpact car (factr A, wth I = 5) n cmbnatn wth each f fur dfferent brands f radal tres (factr B, wth J = 4), resultng n IJ = 0 bservatns. The values SSA = 30, SSB = 45, and SSE = 60 were then cmputed. Assume that an addtve mdel s apprprate. a. Test H : α = α = α3 = α4 = α5 = 0(n dfferences n true average tre lfetme due t makes f cars) versus Ha : at least ne α 0usng a level.05 test. b. H : β = β = β3 = β4 = 0(n dfferences n true average tre lfetme due t brands f tres) versus H : at least ne β 0 usng a level.05 test. a j a. MSA = = 7.50, MSE = = 5.0, f A = =.50. Snce.50 s nt F.05,4, = 3.6, dn t reject H B. There s n dfference n true average tre lfetme due t dfferent makes f cars b. MSB = = 5.0, f B = = 3.0. Snce 3.0 s nt F.05,3, = 3.49, dn t reject H B There s n dfference n true average tre lfetme due t dfferent brands f tres. 48. In an experment t see whether the amunt f cverage f lght-blue nterr pant depends ether n the brand f pant r n the brand f rller used, galln f each f fur brands f pant was appled usng each f three brands f rller, resultng n the fllwng data (number f square feet cvered). Rller Brand Pant Brand a. Cnstruct the ANOVA table. b. State and test hyptheses apprprate fr decdng whether pant has any effect n cverage. Use α.05. c. Repeat part (b) fr brand f rller. d. Use Tukey s methd t dentfy sgnfcant dfferences amng brands. Is there ne brand that seems clearly preferable t the thers? a Surce df SS MS f f.05 A B Errr Ttal 38.5 b. Snce , reject H0A : α = α = α3 = α4 = 0 : The amunt f cverage depends n the pant brand. c. Snce.80 s nt 5.4, d nt reject HA : β = β = β 3 = 0. The amunt f cverage des nt depend n the rller brand. d. Because H B was nt rejected. Tukey s methd s used nly t dentfy dfferences n levels. f factr A (brands f pant). Q.05,4,6 = 4.90, w= : 4 3 x g :

9 50. The strength f cncrete used n cmmercal cnstructn tends t vary frm ne batch t anther. Cnsequently, small test cylnders f cncrete sampled frm a batch are cured fr perds up t abut 8 days n temperature- and msture-cntrlled envrnments befre strength measurements are made. Cncrete s then bught and sld n the bass f strength test cylnders. The accmpanyng data resulted frm an experment carred ut t cmpare three dfferent curng methds wth respect t cmpressve strength (MPa). Analyze ths data. Batch Methd A Methd B Methd C Surce df SS MS f Methd Batch Errr Ttal F = 6.0 < 8.69 < F = 0.39, s.00 < p value<.0, whch s sgnfcant. At least tw f the curng.0,,8.00,,8 methds prduce dfferng average cmpressve strengths. (Wth p-value <.00, there are dfferences between batches as well.) Q.05,3,8 = 3.6; w= (3.6).34 =.3 0 Methd A Methd B Methd C Methds B and C prduce strengths that are nt sgnfcantly dfferent, but Methd A prduces strengths that are dfferent (less) than thse f bth B and C. 6

10 Sectn 4.. In a tw-way cntngency table, f the secnd rw ttal s 5, the thrd clumn ttal s 60, and the ttal number f bservatns s 375, then the estmated expected cunt n cell (, 3) s A tw-way cntngency table has 3 rws and 5 clumns. Then, the number f degrees f freedm asscated wth the ch-squared test fr hmgenety s A tw-way cntngency table has r rws and c clumns. Then, the number f degrees f freedm asscated wth the ch-squared test fr ndependence s. (r-)(c-) 5. The number f degrees f freedm fr a tw-way cntngency table wth I rws and J clumns s A. I J I B. ( ) J C. I ( J ) D. ( I ) ( J ) E. I + J D 6. In a tw-way cntngency table wth 3 rws and 5 clumns, assume that the secnd rw ttal s 0 and the furth clumn ttal s 50, and the ttal number f bservatns s 600. Then, the estmated expected cunt n cell (, 4) s A. 50 B. 40 C. 30 D. 0 E. 0 E 9. The number f degrees f freedm n testng fr ndependence when usng a cntngency table wth 6 rws and 4 clumns s: A. 4 B. 0 C. 5 D. 0 E. C 3. A statstcs department at a state unversty mantans a tutrng servce fr students n ts ntrductry servce curses. The servce has been staffed wth the expectatn that 40% f ts students wuld be frm the busness statstcs curse, 30% frm engneerng statstcs, 0% frm the statstcs curse fr scal scence students, and the ther 0% frm the curse fr agrculture students. A randm sample f n=0 students revealed 50, 40, 8, and frm the fur curses. Des ths data suggest that the percentages n whch staffng was based are nt crrect? State and test the relevant hyptheses usng α =.05.

11 Usng the number fr busness, fr engneerng, 3 fr scal scence, and 4 fr agrculture, let p = the true prprtn f all clents frm dscplne. If the Statstcs department s expectatns are crrect, the relevant null hypthess s H0 : p =.35, p =.30, p3 =.0, p4 =.5, versus H a : The Statstcs department s expectatns are nt crrect. Wth d.f = k-=4-=3, we reject H f 0 χ χ.05,3 = Cell 3 4 n Observed ( ) Expected ( ) np X term Snce all the expected cunts are at least 5, the ch-squared test can be used. The value f the test statstc s k ( n np) ( bserved exp ected) = χ = = = = np exp ected allcells H. (alternatvely, p-value = ( ) Snce X = s nt 7.85, we fal t reject P χ.57 whch s >.0, and snce 0 the p-value s nt <.05, we reject H ). Thus we have n evdence t suggest that the statstcs department s 0 expectatns are ncrrect. 4. A study reprts n research nt the effect f dfferent njectn treatments n the frequences f audgenc sezures. Treatment N respnse Wld runnng Clnc sezure Tnc sezure Thenylalanne Slvent Sham Unhandled Des the data suggest that the true percentages n the dfferent respnse categres depend n the nature f the njectn treatment? State and test the apprprate hyptheses usng α =.005. Wth pj dentng the prbablty f a type j respnse when treatment s appled, H : pj = pj = p3j = p fr j =,, 4j 3, 4 wll be rejected at level.005 f χ χ.005,9 = 3,587. The cmputed value f the test statstc E ˆj χ s χ = = 7.4 Snce χ = , s reject at level.005 H 4. Each ndvdual n a randm sample f hgh schl and cllege students was crss-classfed wth respect t bth pltcal vews and marjuana usage, resultng n the data dsplayed n the accmpanyng tw-way table. Des the data supprt the hypthess that pltcal vews and marjuana usage level are ndependent wthn the ppulatn? Test the apprprate hyptheses usng level f sgnfcance Usage Level Pltcal Never Rarely Frequently

12 Vews Lberal Cnservatve Other H : pltcal vews and marjuana usage level are ndependent wthn the ppulatn. 0 H : pltcal vews and marjuana usage level are nt ndependent wthn the ppulatn. a H wll be rejected f 0 X X.0,4 = 3.77 The cmputed value f the test statstcs E ˆj χ s χ = = Snce X = , the ndependence hypthess s rejected n favr f the cnclusn that pltcal vews and level f marjuana usage are dependent (related).

13 Sectn The s a measure f hw strngly related tw varables x and y are n a sample. sample crrelatn ceffcent 36. Gven n pars f bservatns ( xy ),( xy ),...,( xn, y n), f large x s are pared wth large y s and small x s are pared wth small y s, then a relatnshp between the varables s mpled. Smlarly, t s natural t speak f x and y havng a relatnshp f large x s are pared wth small y s and small x s are pared wth large y s. pstve, negatve 37. The value f the sample crrelatn ceffcent r s always between and. -, The sample crrelatn ceffcent r equals f and nly f all ( x, y ) pars le n a straght lne wth slpe. pstve 39. The sample crrelatn ceffcent r equals - f and nly f all ( x, y) pars le n a straght lne wth slpe. negatve 40. If the sample crrelatn ceffcent r equals -.80, then the value f the ceffcent f determnatns s When H : ρ = 0 s true, the test statstc 0 T = R n / R has a t dstrbutn wth degrees f Freedm, where n s the sample sze. n- 84. A data set cnssts f 5 pars f bservatns ( x, y),( x, y),...( x5, y 5). If each x s f each y s replaced by 4 y, then the sample crrelatn ceffcent r replaced by 3x and A. ncreases by 3/5 B. ncreases by 4/5 C. remans unchanged D. decreases by 3/5 E. decreases by 4/5 C 85. A data set cnssts f 0 pars f bservatns ( x, y),( x, y),...( x0, y 0). If each x s x and f each y s replaced by y, then the sample crrelatn ceffcent r replaced by A. decreases by.05 B. decreases by.0 C. ncreases by.05 D. ncreases by.0 E. remans unchanged 66

14 E 03. The Turbne Ol Oxdatn Test (TOST) and the Rtatng Bmb Oxdatn Test (RBOT) are tw dfferent prcedures fr evaluatng the xdatn stablty f steam turbne ls. The accmpanyng bservatns n x = TOST tme (hr) and y = RBOT tme (mn) fr l specmens have been reprted: TOST RBOT TOST RBOT a. Calculate and nterpret the value f the sample crrelatn ceffcent. b. Hw wuld the value f r be affected f we had let x = RBOT tme and y = TOST tme? c. Hw wuld the value f r be affected f RBOT tme were expressed n hurs? d. Nrmal prbablty plts ndcate that Bth TOST and ROBT tme appear t have cme frm nrmally dstrbuted ppulatns. Carry ut a test f hyptheses t decde whether RBOT tme and TOST tme are lnearly related. a. Summary values: x= 44,65, x = 70,355,45, y= 3,860, y =, 84, 450, xy = 4, 755, 500, n =. Usng these values we calculate S = 4, 480,57.9, S = 4,86.67, and S = 404, S xx yy xy Sxy r = =.933. S S xx yy b. The value f r des nt depend n whch f the tw varables s labeled as the x varable. Thus, had we let x = RBOT tme and y = TOST tme, the value f r wuld have remaned the same. c. The value f r des n depend n the unt f measure fr ether varable. Thus, had we expressed RBOT tme n hurs nstead f mnutes, the value f r wuld have remaned the same. r n d. H : ρ = 0 vs Ha : ρ 0. t = ; r Reject H shuld be rejected. The mdel s useful Hydrgen cntent s cnjectured t be an mprtant factr n prsty f alumnum ally castngs. The accmpanyng data n x = cntent and y = gas prsty fr ne partcular measurement technque have been reprted: x y x y MINITAB gves the fllwng utput n respnse t a CORRELATION cmmand: Crrelatn f Hydrgen and Prsty = a. Test at level.05 t see whether the ppulatn crrelatn ceffcent dffers frm 0. b. If a smple lnear regressn analyss had been carred ut, what percentage f bserved varatn n prsty culd be attrbuted t the mdel relatnshp? a. H : ρ = 0 vs Ha : ρ 0, Reject H f; Reject at level.05f ether r n (.449) t t.05, =.79 r t.79. t = = =.74. Fal t reject H / the data r (.449) des nt suggest that the ppulatn crrelatn ceffcent dffers frm 0.

15 b. (.449).0 = s 0 percent f the bserved varatn n gas prsty can be accunted r by varatn n hydrgen cntent. Sectn 3. k 9. The regressn ceffcent β n the multple regressn mdel Y = β0 + βx+ βx + LL + βk x + ε s nterpreted as the expected change n asscated wth a -unt ncrease n,whle are held fxed. Y, x, ( x, x3, x4, KK, x k ) 0. A dchtmus varable, ne wth just tw pssble categres, can be ncrprated nt a regressn mdel va a r varable x whse pssble values 0 and ndcate whch categry s relevant fr any partcular bservatns. dummy, ndcatr 3. If a data set n at least fve predctrs s avalable, regressns nvlvng all pssble subsets f the predctrs nvlve at least dfferent mdels 3 5. If SSE s the errr sum f squares cmputed frm a mdel wth k predctrs and n bservatns, then the mean k squared errr fr the mdel s MSE = /. SSE n-k- k k 6. When the numbers f predctrs s t large t allw fr an explct r mplct examnatn f all pssble subsets, several alternatve selectn prcedures generally wll dentfy gd mdels. The smplest such prcedure s the, knwn as BE methd. backward elmnatn 7. In many multple regressn data sets, the predctrs x, x3, x4, KK, x are hghly nterdependent. When the sample x k values can be predcted very well frm the ther predctr values, fr at least ne predctr, the data s sad t exhbt. multcllnearty 8. Whch f the fllwng statements are true? A. One way t study the ft f a mdel s t supermpse a graph f the best-ft functn n the scatter plt f the data. B. An effectve apprach t assessment f mdel adequacy s t cmpute the ftted r predcted values y ˆ and the resduals e ˆ = y y, then plt varus functns f these cmputed quanttes, and examne the plts ether t cnfrm ur chce f mdel r fr ndcatns that the mdel s nt apprprate. C. Multple regressn analyss nvlves buldng mdels fr relatng the dependent varable y t tw r mre ndependent varables. D. All f the abve statements are true. E. Nne f the abve statements are true. D 68

16 3. A multple regressn mdel has A. One ndependent varable. B. Tw dependent varables C. Tw r mre dependent varables. D. Tw r mre ndependent varables. E. One ndependent varable and ne ndependent varable. D 33. In multple regressn mdels, the errr term ε s assumed t have: F. a mean f. G. a standard devatn f. H. a varance f 0. I. negatve values. J. nrmal dstrbutn. E 38. The ceffcent f multple determnatn R s A. SSE/SST B. SST/SSE C. -SSE/SST D. -SST/SSE E. ( SSE + SST ) / C 5. A frst-rder n-nteractn mdel has the frm Yˆ = 5+ 3x+ x. As x ncreases by -unt, whle hldng x fxed, then y wll be expected t A. ncrease by 0 B. ncrease by 5 C. ncrease by 3 D. decrease by 3 E. decrease by 6 C

17 Sectn. 0. The vertcal devatns y yˆ ˆ ˆ, y y, KK, yn y frm the estmated regressn lne are referred t as the n. resduals. When the estmated regressn lne s btaned va the prncple f least squares, the sum f the resduals y ˆ y ( =, 3,.., n) shuld n thery be. zer 3. In smple lnear regressn analyss, the, dented by, can be nterpreted as a measure f hw much varablty n y left unexplaned by the mdel - that s, hw much cannt be attrbuted t a lnear relatnshp. errr sum f squares, SSE 4. In smple lnear regressn analyss, a quanttatve measure f the ttal amunt f varatn n bserved y values s gven by the, dented by. ttal sum f squares, SST 5. If SSE = 36 and SST = 500, then the prprtn f ttal varatn that can be explaned by the smple lnear regressn mdel s In smple lnear regressn analyss, SST s the ttal sum f squares, SSE s the errr sum f squares, and SSR s the regressn sum f squares. The ceffcent f determnatn r s gven by r = / SST r r = ( / SST). SSR, SSE. A 00( - α ) % cnfdence nterval fr the slpe β f the true regressn lne s ˆβ ± s ˆ β. t α /, n. Gven that ˆ β =.5, =., and n = 5, the 95% cnfdence nterval fr the slpe β f the true regressn lne s ˆ β (, ) , The null hypthess H0 : β = 0can be tested aganst Ha : β 0by cnstructng an ANOVA table, and rejectng H at 0 α level f sgnfcance f the test statstc value f, where n s the sample sze. 70 F α,, n 5. In smple lnear regressn mdel Y = β0 + β x+ ε, whch f the fllwng statements are nt requred assumptns abut the randm errr term ε? A. The expected value f ε s zer. B. The varance f ε s the same fr all values f the ndependent varable x. C. The errr term s nrmally dstrbuted. D. The values f the errr term are ndependent f ne anther. E. All f the abve are requred assumptns abut ε.

18 E 53. A prcedure used t estmate the regressn parameters β and β and t fnd the least squares lne whch prvdes, the best apprxmatn fr the relatnshp between the explanatry varable x and the respnse varable Y s knwn as the A. least squares methd B. best squares methd C. regressn analyss methd D. ceffcent f determnatn methd E. predctn analyss methd A 54. The prncple f least squares results n values f ˆ β ˆ 0 and β that mnmzes the sum f squared devatns between F. the bserved values f the explanatry varable x and the estmated values ˆx G. the bserved values f the respnse varable y and the estmated values ŷ H. the bserved values f the explanatry varable x and the respnse varable y I. the bserved values f the explanatry varable x and the respnse values ŷ J. the estmated values f the explanatry varable x and the bserved values f the respnse varable y B 63. If the errr sum f squares s and the ttal sum f squares s 400, then the prprtn f bserved y varatn explaned by the smple lnear regressn mdel s A B C D E. Nne f the abve answers are crrect. C 64. Whch f the fllwng statements are nt crrect? A. The ceffcent f determnatn, dented by r, s nterpreted as the prprtn f bserved y varatn that cannt be explaned by the smple lnear regressn mdel. B. The hgher the value f the ceffcent f determnatn, the mre successful s the smple lnear regressn mdel n explanng y varatn. C. If the ceffcent f determnatn s small, an analyst wll usually want t search fr an alternatve mdel (ether a nnlnear mdel r a multple regressn mdel that nvlves mre than a sngle ndependent varable). D. The ceffcent f determnatn can be calculated as the rat f the regressn sum f squares (SSR) t the ttal sum f squares. E. All f the abve statements are crrect. A 65. The quantty ε n the smple lnear regressn mdel Y = β0 + βx+ ε s a randm varable, assumed t be nrmally dstrbuted wth E( ε) = 0 and V( ε) = σ. The estmated standard devatn ˆ σ s gven by A. SSE / (n ) B. SSE /( n ) C. [ SSE /( n )]

19 D. SSE / n E. SSE /( n ) B 66. In smple lnear regressn analyss, f the resdual sum f squares s zer, then the ceffcent f determnatn r must be A. - B. 0 C. between - and zer D. E. between - and D 87. The accmpanyng bservatns n x = hydrgen cncentratn (ppm) usng a gas chrmatgraphy methd and y = cncentratn usng a new sensr methd were btaned n a recent study x y x y Cnstruct a scatter plt. Des there appear t be a very strng relatnshp between the tw types f cncentratn measurements? D the tw methds appear t be measurng rughly the same quantty? Explan yur reasnng. A scatter plt f the data appears belw. The pnts fall very clse t a straght lne wth an ntercept f apprxmately 0 and a slpe f abut. Ths suggests that the tw methds are prducng substantally the same cncentratn measurements y x 7

20 90. Suppse that n a certan chemcal prcess the reactn tme y (hur) s related t the temperature (F) n the chamber n whch the reactn takes place accrdng t the smple lnear regressn mdel wth equatn y = x and σ =.075. a. What s the expected change n reactn tme fr a F ncrease n temperature? Fr a 0 F ncrease n temperature? b. What s the expected reactn tme when temperature s 00 F? When temperature s 50 F? c. Suppse fve bservatns are made ndependently n reactn tme, each ne fr a temperature f 50 F. What s the prbablty that all fve tmes are between.4 and.6 hurs? d. What s the prbablty that tw ndependently bserved reactn tmes fr temperatures apart are such that the tme at the hgher temperature exceeds the tme at the lwer temperature? a. β = expected change fr a ne degree ncrease = -.0, and 0 β =. s the expected change fr a 0 degree ncrease. b. µ Y = (00) = 3, and µ Y = 50 c. The prbablty that the frst bservatn s between.4 and.6 s P(.4 Y.6) = P Z = P(.33 Z.33) =.864. The prbablty that any partcular ne f the ther fur bservatns s 5 between.4 and.6 s als.864, s the prbablty that all fve are between.4 and.6 s (.864) =.367. d. Let Y and Y dente the tmes at the hgher and lwer temperatures, respectvely. Then Y Y has expected value ( x+ ) ( x) =.0. The standard devatn f Y + = Thus Y s (.075) (.075) (.0) PY ( Y > 0) = P z> = PZ ( >.09) = The accmpanyng data n x = current densty (ma/cm ) and y = rate f depstn ( µ m/mn) appeared n a recent study. D yu agree wth the clam by the artcle s authr that a lnear relatnshp was btaned frm the tn-lead rate f depstn as a functn f current densty? Explan yur reasnng. x y Fr ths data, n= x = y = x = y = 4, 00, 5.37,.000, 9.350, (00) (5.37) xy = 333. Sxx =,000 = 000, Syy = =.40875, 4 4 (00)(5.37) and ˆ Sxy 64.5 Sxy = 333 = β = = =.035 and 4 S 000 ˆ β 0 = (.035) = SSE = S ˆ β S =.4085 (.035)(64.5) = yy xy SSE r = = =.97. Ths s a very hgh value f SST.4085 strng lnear relatnshp between the tw varables. xx r, whch cnfrms the authrs clam that there s a

21 A scatter plt, alng wth the least squares lne, f x = ranfall vlume (m ) and y = runff vlume (m ) fr a partcular lcatn were gven. The accmpanyng values were read frm the plt. x y x y a. Des a scatter plt f the data supprt the use f the smple lnear regressn mdel? b. Calculate pnt estmates f the slpe and ntercept f the ppulatn regressn lne. c. Calculate a pnt estmate f the true average runff vlume when ranfall vlume s 50. d. Calculate a pnt estmate f the standard devatn σ. e. What prprtn f the bserved varatn n runff vlume can be attrbuted t the smple lnear regressn relatnshp between runff and ranfall? a. Runff vlume Ranfall vlume Yes, the scatterplt shws a strng lnear relatnshp between ranfall vlume and runff vlume, thus t supprts the use f the smple lnear regressn mdel. (798) b. x = 53.00, y = 4.867, S xx = = 0, 586.4, 5 (643) (798)(643) Syy = 4, 999 = 4, 435.7, and Sxy = 5, 3 = 7, ˆ Sxy 7, 04.4 β ˆ = = =.8697 and β0 = (.8697)53. =.78. Sxx 0, c. µ y 50 = (50) = d. SSE = S ˆ yy βsxy = 4,435.7 (.8697)(7,34.4) =

22 e. SSE s = ˆ σ = = = 5.4. n 3 SSE r = = = S 97.53% f the bserved varatn n runff vlume SST 4, can be attrbuted t the smple lnear regressn relatnshp between runff and ranfall. 93. The accmpanyng data was read frm a graph that appeared n a recent study. The ndependent varable s SO depstn rate (mg/m / day) and the dependent varable s steel weght lss (g/m ). x y a. Cnstruct a scatter plt. Des the smple lnear regressn mdel appear t be reasnable n ths stuatn? b. Calculate the equatn f the estmated regressn lne. c. What percentage f bserved varatn n steel weght lss can be attrbuted t the mdel relatnshp n cmbnatn wth varatn n depstn rate? d. Because the largest x value n the sample greatly exceeds the thers, ths bservatn may have been very nfluental n determnng the equatn f the estmated lne. Delete ths bservatn and recalculate the equatn. Des the new equatn appear t dffer substantally frm the rgnal ne (yu mght cnsder predcted values)? a y x Accrdng t the scatter plt f the data, a smple lnear regressn mdel des appear t be plausble. b. The regressn equatn s y = x c. The desred value s the ceffcent f determnatn, r = 99.0%. d. The new equatn s y* = x*. Ths new equatn appears t dffer sgnfcantly. If we were t predct a value f y fr x = 50, the value wuld be 567.9, where usng the rgnal data, the predcted value fr x = 50 wuld be An nvestgatn f the relatnshp between traffc flw x (000 s f cars per 4 hurs) and lead cntent y bark n trees near the hghway ( µ g/ gdry wt) yelded the data n the accmpanyng table. x y

23 The summary statstcs are:, n=, x = 98.3, y = 7034 x = , y = 5,390,38, xy = 49,354.4 In addtn, the least squares estmates are gven by: ˆ β ˆ 0 =.8459, and β = Carry ut the mdel utlty test usng the ANOVA apprach fr the traffc flw/lead-cntent data f Example.6. Verfy that t gves a result equvalent t that f the t test. SSE = 5,390,38 (-.8459)(7034)-( )(49,354.4)=76,49.54, and SST = 5,390,38 - (7034) / = 89, Surce df SS MS F Regressn 85, , Errr 9 76, Ttal 0 89, Snce n α s specfed, let s use α =.0. Then F.0,,9 = 0.56 < 96.0, s H : β = 0 s rejected and the mdel s judged useful. Nw, s= = 9.9, ( x x) = (98.3) / = 63.8, Then, t = = t = = = f 9.9 / , and (9.80)