STA3/-Fall 8 Mdterm Test October, 8 Last Name: Frst Name: Studet Number: Erolled (Crcle oe) STA3 STA INSTRUCTIONS Tme allowed: hour 45 mutes Ads allowed: A o-programmable calculator A table of values from the t dstrbuto s o the last page (page 7). Total pots: 5 Some useful formulae r = ( )( Y Y) ( ) ( Y Y) ( )( Y Y) s b = = r Y ( ) s b = Y b Var( b ) σ = + ( ) ( ) SSTO = Y Y Var( b ) = σ ( ) Cov( b, b ) = σ ( ) SSE = ( Y Yˆ ) SSR = b ( ) = ( Yˆ Y ) ( ) h ˆ ˆ σ { Yh} = Var( Yh) = σ + ( ) ( ) h ˆ σ { pred} = Var( Yh Yh) = σ + + ( )
) A chemst s terested determg the weght loss pouds (Y) of a partcular compoud as a fucto of the amout of tme the compoud s exposed to the ar. She collected data o the weght loss assocated wth = settgs of the depedet varable, exposure tme hours (). Some useful summary statstcs obtaed from the SAS outputs (from proc uvarate ad proc corr) from ths study are also gve below. The SAS output for the regresso o Y o gave SSE = 8.368. Whe aswerg the questos below, you may assume that the data satsfes the ecessary assumptos. The SAS System The UNIVARIATE Procedure Varable: Momets N Sum Weghts Mea 5.6666667 Sum Observatos 6 Std Devato.673446 Varace.6666 Skewess -.47643 Kurtoss -.8787469 Ucorrected SS 338 Corrected SS 7.6666667 Coeff Varato 4.584735 Std Error Mea.3658399 The SAS System The UNIVARIATE Procedure Varable: Y Momets N Sum Weghts Mea 5.5833333 Sum Observatos 66. Std Devato.786357 Varace.95744 Skewess -.569374 Kurtoss.777 Ucorrected SS 396.57 Corrected SS 3.469667 Coeff Varato 3.964 Std Error Mea.495963 Pearso Correlato Coeffcets, N = Prob > r uder H: Rho= Y..659.98 Y.659..98
[5 pots] a) Calculate the least squares predcto equato for the model Y = β + β +ε wth ε s satsfyg the usual assumptos. Show your workgs clearly. b s Y = r =.659*(.786357/.673446) =.893395335 (3 pots) s b = Y b = 5.5833333 -.893395335*5.6666667 =.89457495 ( pots) Here s a SAS out to compare these calculatos The REG Procedure Model: MODEL Depedet Varable: Y Aalyss of Varace Sum of Mea Source DF Squares Square F Value Pr > F Model 4.77 4.77 7.68.98 Error 8.3684.83684 Corrected Total 3.4697 Root MSE.3553 R-Square.4343 Depedet Mea 5.5833 Adj R-Sq.3777 Coeff Var 4.6455 Parameter Estmates Parameter Stadard Varable DF Estmate Error t Value Pr > t Itercept.8945.73.5.634.8934.345.77.98 [3 pots] b) Calculate ad terpret the value of the coeffcet of determato for ths model. Show your workgs clearly. R-sq =.659^ =.4348 [ pots] Ths model wth exposure tme explas 43.4% of the varato the wt loss. [ pot) [5 pots] c) Coduct a F-test to decde whether or ot there s a lear assocato betwee ad Y. State the ull ad alteratve hypotheses. Use a level of sgfcace of.5. Show your workg clearly. 3
H : β = ( pot) H : β ( pot) a SSR =SST SSE = 3.469667 8.368= 4.667 Note SST = Corrected SS SST ca also be foud from SSE = (-) 3.469667 (from the sas output above) s Y Test statstc F = MSR?MSE = 4.667/(8.368/(-))= 4.667/(8.368/(-)) = 7.677899 ( pots) Table F(,,.95) = 4.96 F calculates > 4.96 ad so we reject the ull hypothess. I.e. a lear regresso relato exsts. ( pot) [4 pots] c) Calculate a 95% cofdece terval for β. Show your workg clearly. s= sqrt(mse) = sqrt(8.368/(-)) =.35585948 ( pot) b ± ts + = ( ) = 5.6666667.89457495 ±.8.35585948 +.6666 pot for table readg.e..8 ad pots for substtutg values correctly) The aswer maybe left ths form. Note : ( ) (=7.6666667 ) s also the corrected SS for the varable the SAS = uvarate output above. [3 pots] d) Calculate a 95% cofdece terval for the mea weght loss whe = 5 hours. Show your workg clearly. ; (5-5.6666667) (.89457495+.893395335 5) ±.8.35585948 +.6666 4
[ pots] e) Calculate a 95% predcto terval for the weght loss whe = 5 hours. Show your workg clearly. (5-5.6666667) (.89457495+.893395335 5).8.35585948 ± + +.6666 [3 pots] f) Calculate the boudary values of the 95% cofdece bad (Workg- Hotellg) for the regresso le whe = 5 hours. Show your workg clearly. sol (5-5.6666667) (.89457495+.893395335 5) ± W.35585948 +.6666 Ths questo s smlar to.3 d p 9 ( pot) W = sqrt(*f(, -,.95)) = sqrt( *F(, -,.95)) = sqrt(*4.) =.8635643 ( pot for the correct F value ad pot for the correct substtuto.) [5 pots] g) The researcher beleves that the average weght loss per hour (.e. the slope of the regresso le of Y o ) s more tha.8 pouds. Coduct a test to decde whether the data ths sample provdes evdece favour of ths belef. State the ull ad alteratve hypotheses. Use α =.5. Show your workg clearly. : H : β =.8 ( pot) (Note H : β.8 s also correct) H : β >.8 ( pot) a b.8.893395335.8 t = = =.93395335/((.35585948)/sqrt(7.666 s/ S.35585948 / 7.6666667 66))=.8964867 ( pots) ot sgfcat at.5 (or eve ay usual value of alpha) ( pot) 5
[4 pots] h) We ow covert all the weght losses the data set above to klograms (usg the coverso: poud =.45 klograms) ad call ths varable (.e. weght loss klograms) Y. Calculate the least squares predcto equato for the model Y = β + β + ε. (Note: the varable s stll the same) sy.45sy b = r = r =.45*.893395335=.4798 ( pots) s s b = Y b =.45 *.89457495 =.4658433 ( pots) The ext questo cossts of three multple-choce questos (-), each carryg pots. ) The followg SAS output was obtaed from a study of the relatoshp betwee the salary ( thousads of dollars) ad legth of servce () years based o a radom sample of 5 employees from a large frm. The REG Procedure Root MSE 3.44985 R-Square.5384 Depedet Mea 9.88 Adj R-Sq.583 Parameter Estmates Parameter Stadard Varable DF Estmate Error t Value Pr > t Itercept.3368.85.98 <..9534.7673 5.8 <. ) If the average legth of servce of the 5 employees ths sample s 9.7 years, what s the average salary of the 5 employees? Choose your aswer from the optos below. As B A) t wll be less tha $ B) t wll be greater tha $ but less tha $3 C) t wll be greater tha $3 but less tha $4 D) t wll be greater tha $4 E) t caot be determed from the formato gve. 6
a= y bx substtutg for a, b ad x, ad solvg for y we get.3+.95*9.7 = 9.938 Here s the MINITAB output to compare aswers. Descrptve Statstcs: Legth, Salary Varable N N* Mea SE Mea StDev Mmum Q Meda Q3 Legth 5 9.7.797 3.985 3. 6.5 9..5 Salary 5 9.9.994 4.97.353 6.47 8.446 3.7 ) Calculate the value of SSE for the above regresso model. Choose your aswer from the optos below. As C A) t wll be less tha B) t wll be greater tha but less tha C) t wll be greater tha but less tha 3 D) t wll be greater tha 3 E) t caot be determed from the formato gve. 3.44985^ =.9465 ANS*3 = 73.7336955 Here s the MINITAb output to compare aswers Regresso Aalyss: Y versus The regresso equato s Y =.3 +.95 Predctor Coef SE Coef T P Costat.33.85.98..953.767 5.8. S = 3.44985 R-Sq = 53.8% R-Sq(adj) = 5.8% Aalyss of Varace Source DF SS MS F P Regresso 39.6 39.6 6.8. Resdual Error 3 73.73.9 Total 4 59.99 7
) Calculate the value of the correlato betwee the salary ad the legth of servce. Choose your aswer from the optos below. A) t wll be less tha.5 B) t wll be greater tha.5 but less tha.6 C) t wll be greater tha.6 but less tha.7 D) t wll be greater tha.7 E) t caot be determed from the formato gve. As D sqrt(.538) =.733484838 3) Cosder the smple lear regresso model: Y = β + β + ε wth the usual assumptos (.e. E( ε ) = for all, V ( ε ) = σ for all, Cov ( ε, ε j) = wheever, j. The ormalty of ε s s ot requres for the results below.). Let b ad b be the least squares estmators of β ad β respectvely. [3 pots] a) Prove that Cov( b, b ) = = σ ( ) =, where =. Cov( b, b ) = Cov[ k Y + k Y ] j j = = where k j ( j ) = ad S = kkcovy (, Y) = kk σ = σ kk = σ k k = = = = = because k =, Cov( Y, Y ) =, j ad = j = k k = k. = = σ ( ) = = S ( ). 8
[3 pots] b) Prove that EMSR [ ] = σ + β ( ), where MSE s the mea square error. : ( ) = MSR = b = ] = = E[ MSR] = E[ b ( ) ] = ( ) E[ b ( ) Var( b) ( E[ b] ) = = + = ( ) σ + = β ( ) = = σ + β ( ) = [4 pots] c) Prove that Var[ e ] = σ = ( ) ( ), where e ˆ = Y Y. Var[ e ] [ ˆ] [ ] [ ˆ = Var Y Y = Var Y + Var Y] Cov[ Y, Yˆ ] Cov[ Y, Yˆ ] = Cov[ Y, b + b ] = Cov[ Y, k Y + k Y ] j j j j j= j= k j = kj. where Cov[ Y, Yˆ ] = Cov[ Y, b + b ] = Cov[ Y, k Y + k Y ] j j j j j= j= [ ] = kcovy (, Y) + kcovy (, Y) = σ k + k = σ + = σ + = σ + ad so ( ) k k ( ) k ( ) = k j ( j ) = ad S 9
Var[ e ] = Var[ Y Yˆ] = Var[ Y ] + Var[ Yˆ] Cov[ Y, Yˆ] ( ) ( ) = σ + σ + σ + ( ) ( ) = = σ = ( ) = ( )