231HW11-S10 - HWH (I <2 GMhmmfl (b) .; ... : ) Jim: 7;...

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Unformatted text preview: HWH (I <2 GMhmmfl (b) .; ... : ) Jim: 7; (If-’f) : es 1% : MCL~W, a": z “(J kfi » M JMP mffut. I’) \3 K WNW/m 13. 1C!) M50 J66 JMP mséut. (c) 39*mm%(+flfi% 30 = ~ M20452, @, .3 H'fi 55/ 5/; HWH’N <4 HTML/Elm [33/ ) I JEV/ MP ouQ‘H/fl - <5 (Pm/Mm 15.36) '9; M" Y: me! + ‘6 + + Kid” '5‘ . J h 1 1,5 lrb V '3 5" Po + (’2. v5 + P: J1, "f (" V Jae J??? 'mjm/fi . I {6 (WM 13.37)’ (w 447.50»: —o.z’soo+ 0.060X5Of‘ mom 2 4.9 w Mm t/u mméer of Mo'ww (17;) r; W yak/b 042 air?ny Miagfi/fi r r i n ’4‘ ‘ M ~ a .u . 57",» [1% £de Mg U W W Wrwi 9%» 0L 0’25?” M 0349’” M156} I 4 g I.“ \ ( ,1- " . {tiff/10m m «312mm, aim/€3,953 W/ U 0900 ’me’ ” "' , w . ) MM, , 'bjflm (/Lc (/fiiam WWW ()3) [3 KW my” me Marya J ' .. , ,r , WW {n 1le vhf/J" 1&0.er m (j) afiomczzim[ mm ‘ W WW MW?” (W. (L V DMZ—MT int/WE 1% la.) {I 0.960 Wm I {E} IN] Z Max} 3 W M Mom ($9, (152) 5.5” PUSH : Mil—i <flz) 12/)(352‘2) :: 0.785] ‘ 0.5 <7 HroM'm 5.35) (m mmfl'jc: 125+175x40+ 0.0?50XH00 -. own/Momma : H15 Kb) fi'mc, dz mm {ye Wm «mi—30 is ; (2'5 -% 7.7M30 w? 0.0%50 ~ 0,959:szka : 5337.5 ~ 0175;;a M {/u chi/ye m Wm MW’M’z Wit/1 02h, find/MR yr 1 (A [Mi to * M75 . Emmi, M3: mm {m tan/,er @140 6'3 ; 125 + mm + 0,5?252ovxc; -» owoww; : 455» 0.26%, In m r/zflfyt {11 mm his Wehran M m mmm 1 m {M L3 * 0265, 2% <8 ( Pro/Mm 15.42) Jet JMP ougmt. (M 3192: 0.45212!" , it: 5/ 42 3*- tags/H ~ 5%: : 3 i [Lynx 05212; : [Wzé/ $057M Homh jammy, irififfiCa/HUE‘, f/WCI , M Uta/{hm i/(U/f mm (1 J 9“ p l h f (MW [71/ tarp/firmiwa ‘r/fiwm on aria, m mime, 2M, 55 JWJKKU/Hfi‘ Mam 1‘ 0742!; mi M74. ,— (C) Mm I) 3 L300) Ilz/ In.me — 199.5556 + 0.21 x I300 + 3X7 2' 91+. 4444 A 7570 U (for WW .- 5Lir-Isbo/7 "11 “:95, 5 ' 5? I Wfl44 iZ-WX 0-355 ' _—_ L 75.5606, 95.30ng (ll) A 9570 PI for 4M 1300,? W526 MM? 1. : WWW, 97.117432] MHWVN @Y‘mm i 9,44% 3: 244M /20+ 03553 (9 ( Pro/Vim fEL JW (5.47) outfut . H!” 4/11- 20,25,40, 45 3 2245.0? Jr MM 33: ‘1' 7:64 EAT 13.30 J65 £5155 l4 : 2245.0? + 'wzxzo + m M + wax 40 "" 5W 2 2505.03 ' ‘ A 75% c1 jar I 4.7-2ogzé,40,45 (5 ; taggers x 5? :— 150507 i WNW /\ %Y‘20,2\5/1}l01 i. I [1459122 , 1520.458} MUM/LIE: (A1 WWW/[J L‘J KMDHQ/Jgf NUMW , we MOM/M coming/1; J , 0 (Mt M2 mm WW con/tent: AM im wme wtfmazefl. (m A 957., P1 Jew r/a Wm met if a {/MW’: Wm {£5163le .: /\ fiY'20/251'40r45 i— . .1 505.00) i 2.0m { 77! New? 757/755] , by I to.025,25 X (51+ 8’}; I: WE: Ja/gf/i flaw #{fi HW 11 Problem 3: (a) Bivariate Fit of Y By X -————- Linear Flt Linear Fit Y = 72.958547 + 0.0410338*X Summary of Fit RSquare RSquare Adj Root Mean Square Error Mean of Response Observations (or Sum Wgts) Analysis of Variance 0.79994 0.788826 0.665331 78.74 20 Mean Square Source DF Sum of Squares ModeI 1 31 .860014 Error 18 7.967986 C. Total ' 19 39.828000 Parameter Estimates Term Estimate Std Error intercept 72.958547 0.697528 0.004837 X 0. 041 0338 81 .8600 0.4427 t Ratio 104.60 8.48 F Ratio 71.9731 Prob > F <.OOO1* Prob>|t| <.0001* <.0001* 1.5 _. Ill 9 05 u a“ 32 ' u 1' I. g I " 2 a E DC'O.5 II B an 4.5 ""—1‘ ‘1‘“ 1‘— 1’" 1 50 . 100 150 200 X This plot indicates that the variance of e is reasonably constant, there are no outliers, and the e are normally distributed. A straight~line regression function is a reasonable choice for a model“; (b) We E113; Tdalcs Row; 5315,0015 Analyze Graph Tools View Whdi Heb 01.111111111111110” 11100111111 11 i ~1_\ 2‘ E — x . ' , “ii-.243 4" . I + U a U' mun-mmwgnm..1.mm...1.n....m....mm.m,.....1....m....m.unu.1.1a.;.s..;;..;i...;.;....m.1 1“ , . ineswimsv r 31111111511101: ‘1 411331! ‘ x v mammal (BJ) F 117111.111 5x): (e_|‘i “.31 LJ} - » . 77.9 78.100071”! 4120007114 140.895 1892182515 473107954 0,04376222 75.8 76.0000532i -fl.1800032 140.895 1092102515 413000827 0.51441051 81.5 81.22274035 0.2772517; 140.095 1892111295 1147533820 1157840759 78.8 78.002821?! 0.7971707 1:10.595 189213295 1131111275 1104742170 ' A. -1 I»; .1 .-.-.: 70.2 78.915374: 145374 140.805 18921.8295 ~1.1507835_-_2._Bll771001_ 70‘._3_ -JB.D7_545_§_1 l 4511§L 14D.89§ _18921,0285 0.310994011 0.01341500 77.5 77.56253 _ 41.052536 140.895 189218295 410007707 0.033143% 77 77.890006L‘. 00908001 140.395 189213295 4.3810205 “9.040393% 00.1.:7957319071 05250001 11011113 111921112551113215011~ 11510505511 ufwu 1302 8029918841 43.080481“ 140.1395 1092182515 4116005133 0.62157516 - ouiiimns (11/1) ‘1‘ _ 79.3 _ 030055071 110.005 113321.0qu 051701015 0.641850%» ' ‘ a“ - 1?; 70 70.111127013‘1 0.050720% 110.095 1992111295 0.0908691) 0.5177270: ‘ "1F,edmwy¢ 1 70,7 7130101031: 005911109 110.005 1992111295 1.10033930 0350171012 419051.,"me i1: 711.2 7917105901 00710590 1101105 109310295 4.5027910 001016019 Ameané} :1 79.5 79.0755150l 0.52130330 110.095 1092111295 0.90300905 0.51796107 09nd? 125 1131 71100775112111.01223177 110005 1092111295 11010110552 0.51358071 .11 gldflnjuklulswfi If 1921.0 1 01.5 01.11005051030393055 1411.005 _10921.11295 055570230 0.501159an "figmmfiflwm ‘ 1212.5 77 7039552151 13055215 1101195 109210295 4.157270 0541100900 3, 159.5 79 79507531171-050753117’ 110095 1119210295 0.7907735 0.51102an 1 5;; 110,7 7011 7751109053: 10011011119 110.095 1092111295 111020510 011315101 _ mm; H A _M__ _ ' _ ‘ mums" "'20 1 v1 8019:1120 , 20 1511011111111] 0 . 1 Hidden ll" 1 Labelied 0 i 1 1 E i r 1 i 1 (c) Bivariate Fit of Std Residuis (e__i*) By X o .3 OCH-JOIN (e_i*) .9 :x .12 w a) D: "a u U) This plot looks very much the same as the one in part a. Add: Distributions Std Residuls -2.5 -2 ~15 —1 ~05 0 0.5 1 1.5 2 Normal(0.0135,1.01842) Quantiies 100.0% maximum 1.7403 99.5% 1 .7403 97.5% 1 .7403 90.0% 1.22502 75.0% quartile 0.78088 50.0% median 0.05484 25.0% quartile -0.6708 10.0% —1 .4916 2.5% 2.1573 0.5% 62.1573 0.0% minimum ‘ -2.1573 Moments Mean 0.0135007 Std Dev 1.0184208 Std Err Mean 0.2277257 Upper 95% Mean 0.4901361 Lower 95% Mean 0463135 N 20 Fitted Normal Parameter Estimates Type Parameter Estimate Lower 95% Upper 95% Location 0 0.0135007 —O.463135 0.4901361 Dispersion 0 1.0184203 0.7744989 1.4874758 -2|og(Likelihood) = 56.4876503191666 Diagnostic Plot 0.95 53.0 COCO Normal Probability O 01 —2.5 —2 —1.5 -1 —0.5 O 0.5 1 1.5 2 Std Residuls The normal probability plot shows that the straight line fits all the standard residuals. Therefore, the standard residuals seem to have the normal distribution: Therefore;the'assum'ption of~ normality of the error term seems to hold. Problem 14: Bivariate Fit of Y By X 100 150 200 250 300 350 400 450 500 550 X ———————Llnear F—‘ll Linear Fit Y = 21272093 ~ 0.2355118*X Summary of Fit RSquare 0.696714 RSquare Adj 0.67144 Root Mean Square Error 24.56456 Mean of Response 137.3571 Observations (or Sum Wgts) 14 Lack Of Fit Source DF Sum of Squares Mean Square F Ratio Lack Of Fit 2 2880.0132 1440.01 3.3020 Pure Error 10 4361.0000 436.10 Prob > F Total Error 12 7241 .01 32 0.07392 Max RSq 0.8173 Anaiysis of Variance Source DF Sum of Squares Mean Square F Ratio Model 1 16634.201 16634.2 27.5666 Error 12 7241.013 603.4 Protr> F C. Total 13 23875.214 0.0002* Parameter Estimates Term Estimate Std Error t Ratio' Probsitr' Intercept 212.72093 15.78406 13.48 <.0001* X —0.23551 2 0.044856 —5.25 0.0002* (3) Since p-value 0.0792 is bigger than 0.05, we do not reject Ho. This formal test procedure does not suggest that a linear model is inappropriate. (b) The scatter plot clearly reveals a curved pattern which suggests that a nonlinear model would be more reasonable and provide a better fit than a linear model. Problem 19: (a) Bivariate Fit of Lifetime By Temp. (D E E 3000 .3 No, there is definite curvature in the plot. 0)) Bivariate Fit of In(Liietime) By 1/Temp. ln(Lifetime) .m s» >1 90 UlUiCDOImeU‘iLO 4‘5 —I 1—“! "r—“I "I— r—1 ‘1‘“ 0004100043 0.0045 0.0047 0.0049 0.0051 1fl’emp. ——-—- Linear Fit Linear Fit In(Lifetime) = —10.20451 + 3735.4511*1/Temp. Summary of Fit RSquare 0.954303 RSquare Adj 0.951447 Root Mean Square Error 0.295367 Mean of Response 6.869139 Observations (or Sum Wgts) 18 Lack Of Fit Source DF Sum of Squares Mean Square F Ratio Lack Of Fit 1 0.0290733 0.029073 0.3191 Pure Error 15 1.3667952 0.091120 Prob > F Total Error 16 1.3958684 0.5805 Max RSq 0.9553 Analysis of Variance Source ‘ DF Sum of Squares Mean Square F Ratio Model 1 29.150072 29.1501 334.1297 Error 16 1.395868 0.0872 Prob > F " C. Total 17 30.545941 <.0001* Parameter Estimates Term Estimate Std Error t Ratio Prob>lt| intercept ~10.20451 0.936638 —1 0.89 <.0001* 1/Temp. 3735.4511 204.3551 18.28 <.0001* 1 temp appearance (and in fact R2 = .9543 for the straight line fit). Let x’ = and y, = ln(lifetime) . Plotting y' vs. x, gives a plot which has a pronounced linear HW 11 Problem 31: (a) Bivariate Fit of Y By X --—- F‘olynomial Fit Degree=2 Polynomial Fit Degree=2 Y = 30.580587 + 0.6234489*X ~ 1.7154788*(X-3.14286)’\2 Summary of Fit RSquare 0.947253 RSquare Adj 0.92088 Root Mean Square Error 1.428435 Mean of Response 27.42857 Observations (or Sum Wgts) 7 Lack Of Fit Source DF Sum of Squares Lack Of Fit 1 8.3500371 Pure Error 3 4.8116667 Total Error 4 8.1617038 Analysis of Variance Mean Square 3.35004 1.60389 Source DF Sum of Squares Mean Square Model 2 14657258 73.2863 1 Error" 4 8.16170 2.0404 G. Total 6 15473429 Parameter Estimates Term Estimate Std Error t Ratio intercept 30.580587 1 .221005 25.05 X 0.6234489 0.315254 1 .98 (X—3.14286)/\2 —1 .715479 0.203636 —8.42 F Ratio 2.0887 Prob > F 0.2441 Max RSq 0.9689 F Ratio 35.9.1172... Prob > F 0.0028* Prob>ltl <.0001* 0.1191 0.0011" Regression function is: Y = 30.580587 + O.6234489*X - l.7154788*(X—8.14286)"2 (C) s2 = 2.0404 R2 = 0.947253 Yes. The quadratic model thus explains 94.7253% of the variation in the observed y’s, which suggests that the model fits the data quite well. Add: E3}?39%‘4'}j3éi‘fiii‘431fifi" £5352”? 3W 52% “"l 1W” l‘l‘ l‘ ‘1 l ’ ,filflmflfl ’ l N “N l 1H‘ W“ i H atltfittii File Edi Table: lbw; (ck DOE Amiga: Graph Tools View 07rd)“ Heb . ‘ ’ a H: . . , t z . pun-n. «.mmmmmmmuuumwmwwww AWtM...tWM.. _ Y x54. sunlenuzednesluY anewmmmn‘! mum smut-:qu Lowevos‘aluuw‘l Umm 9525!)“th - _ W I - 23.32093”: 43269374 1 0.307501 ’ 20.876507?) 25.077007? __' 105508645 23.00U8104I 3 24.5 23,220837-H 1‘17315258 ‘ 1 H0397”! 100750076 _ 25.077057? _ 195558645 ‘ 20.095810” 2 20 295005507} 4.5068597 t -1.29|34 216105101 315632003 251557332 34,0}79952 __ ____ __ 4 __ 30,9 3mm”; 419140312 tG_ 01521435 29,§555305 * magmas” 2 9371e§__ 36.32934_3_q_ 4 32 3111149312! ultussaunz t§_ _. 041513843 29.655302 33.91253” __v__ 2 .307an aaazgagq 4 3376 ‘313140322' 1.7059500! 15 _ 1.4003555! 2033555306 31577253” 271907185 363293439 , p 21‘ “"23 203173739! -n,3z13m___._ _ 3g _ _ 4.29m m 16.4:99793 ‘ gtgggggsa ____1_q.]§uuan‘ “magmas v__,_~___ x L _ mi- _ ______ irculumneoofl) M — _ t _ w _ _ __ _ ____ _~_ 37;! h '''' ' ' ’ V ' r t "— JY , , I ,d PmdlctedYé} "‘ | " " .d Reslduats V ‘ 41x34: _ ‘ ,. , A stunqeultzedneslm — -—— .—-A - T J— —- a — ‘ ‘ —~1—' ‘ -JJ " ’ .J LUNEIE5% many . .. .1! UpnerES‘s, Menn‘l ‘ i " ’ ALowerflfi%lndivY _ _ __ _ l ; ___ _n_ .41 Unpurgfi'll "HWY _ t l n 1 ' ‘ t l l "7 sexacmn o } actudud a “ " L ~— —- _. Hlddun ' I) - Labelled 0 t - — f 4’ .-———~—-— - T1 . - .~.. , g l y n. l- t . humantutuna:1tatI:at;uathtttttttttttttu Fit Y by X Group Bivariate Fit of Studentized Resid Y By Predicted Y Studentized 26 28 30 32 Predicted Y Studentized Distributions Studentized Resid Y Normai(-0.1262,1.10398) Quantiles 100.0% maximum 1 .49037 99.5% 1 .49037 97.5% 1 .49037 90.0% 1 .49037 75.0% quartile 1.10397 50.0% median -O.3076 25.0% quartile -1.2813 10.0% -1.2813 2.5% -1.2813 0.5% -1 .2813 0.0% minimum —1 .2813 Moments Mean -0.126209 Std Dev 1.1039753 Std Err Mean 0.4172634 Upper 95% Mean 0.8947981 Lower 95% Mean —1 .147216 N 7 Fitted Normal Parameter Estimates Type Parameter Estimate Lower 95% Upper. 95% Location 11 ~0.126209 ~1.147216 0.8947981 Dispersion 0 1.1039753 0.7113944 2.4310265 -2log(Likelihood) 7. 20.249598463603293 Diagnostic Plot Normal Probability Studentized Resid Y It shows that the standardized residuals are randomly scattered along the horizontal zero line. The Normal Probability Plot of standardized residuals shows that the linear line could fit the standardized residuals. Therefore, the assumptions of both constant variance and normality of the error term hold. The end. Problem 35: Bivariate Fit of Y By X Transformed Fit Log Transformed Fit Log Log(Y) = 2.0397251 + 0.1798991*X ~ 0.002242*X/\2 Summary of Fit RSquare 0.998072 RSquare Adj 0.996144 Root Mean Square Error 0.044807 Mean of Response 4.628789 Observations (or Sum Wgts) 5 Analysis of Variance Source DF Sum of Squares Model 2 2.0787835 Error 2 0.0040153 C. Total 4 2.0827988 Parameter Estimates Term Estimate Std Error intercept 2.0397251 0.178103 X 0.1798991 0.019369 X02 0.002242 0.000479 Fit Measured on Original Scale Sum of Squared Error 85.312728 Root Mean Square Error 5.3326894 RSquare 0.9963245 , Sum of Residuals 0.4108717 Mean Square 1 .03939 0.00201 t Ratio 1 1 .45 9.29 -4.68 F Ratio 517.7169 Prob~> F- 00019" Prob>|tl 0.0075* 0.01 14* 00427“ Problem 42: Response Y Whole Model Actual by Predicted Plot 75 80 85 90 95 100 105 110 Y Predicted P<.0001 RSq=O.99 RMSE=1.0585 Summary of Fit RSquare 0.990692 RSquare Adj 0.98759 Root Mean Square Error 1.058475 Mean of Response 94.44444 Observations (or Sum Wgts) 9 Analysis of Variance Source DF Sum of Squares Mean Square F Ratio Model 2 715.50000 357.750 319.3140 Error 6 6.72222 1.120 Prob > F C. Total 8 72222222 <.0001* Parameter Estimates Term Estimate Std Error 1 Ratio Prob>|t| Intercept —199.5556 11.64056 -17.14 <.0001* X1 0.21 0.008642 24.30 <.0001* X2 3 0.432121 6.94 00004" Effect Tests Source Nparm DF Sum of Squares F Ratio Prob..>.-,E X1 1 1 661 .50000 590.4298 <.0001* X2 1 1 54.00000 48.1983 0.0004" Residual by Predicted Plot "a? 3 E (n (D 0: >.. .75 80 85 90 95 100105110 YPredicted Prediction Expression -199.55555555556+0.21*X1+3*X2 X1 Leverage Plot 110 YLeverage Residuals _x _\ 00 to (D O O U! O 01 O U”! 1240 1260 1280 1300 1320 1340 1360 X1 Leverage, P<.0001 CO O X2 Leverage Plot 110 Y Leverage Residuals (O 01 6.0 6.5 7.0 7.5 8.0 X2 Leverage, P:0.00,04 (a) To test H 0 : ,61 = ’62 = 0 vs. H a I at least one fll. :5 O, the test statistic is f =MSR/MSE = 319.3140 (from output). The associated p—valtie is <.0001, so at any reasonable level of significance, H0 should be rejected. There does appear to be a useful linear relationship between temperature difference and at least one of the two predictors. ' Problem 47: Response Energy Content Whole Model Actual by Predicted Plot R E. 0 < w C (D 2. C O 0 9001000 1200 1400 Energy Content Predicted 1600 P<.0001 RSq=0.96 RMSE=31.48 Summary of Fit RSquare RSquare Adj Root Mean Square Error Mean of Response Observations (or Sum Wgts) Analysis of Variance 0.964079 0.958331 31 .48037 1 281 .267 30 Mean Square 166234 Source DF Sum of Squares Model 4 664934.53 Error 25 24775.34 C. Total 29 689709.87 Parameter Estimates Term Estimate Std Error Intercept 2245.0932 177.8922 Plastics 28.922384 2.823488 Paper 7.6428559 2.313652 Garbage 4.2969127 1.916106 Water -37.35594 1 .834172 ’1 Ratio 12.62 10.24 3.30 2.24 20.37 991 F Ratio 167.7410 Prob > F <.OOO1* Prob>|t| <.0001* <.0001* 00029" 0.0340* <.0001* Residual by Predicted Plot Residual a. C m 4.. C: o O >. 9 a) C LLl 9001000 1200 1400 1600 Energy Content Predicted Prediction Expression 2245.09316498552 +28.9223837892672* Plastics + 7.64285589353425* Pa per +4.29691270419331* Garbage + 67355935465749" Water Plastics Leverage Plot Energy Content Leverage Residuals 12 14 16 18 20 22 24 26 Plastics Leverage, P<.0001 Paper Leverage Plot Energy Content Leverage Residuals 20 25 Paper Leverage, P=0.0029 Garbage Leverage Plot 4.. c a: u C o O >~. E’ a) C UJ Leverage Residuals 4O 45 50 Garbage Leverage, P=0.034O Water Leverage Plot Energy Content Leverage Residuals 45 50 55 Water Leverage, P<.0001 (a) For a 1% increase in the percentage plastics, we would expect a 28.9 kcal/kg increase in energy content: A‘l'so‘, fora“ 1% increase in the moisture, we would expect a 37.4 kcal/kg decrease in energy content. Both of these assume we have accounted for the linear effects of the other three variables. (13) The appropriate hypotheses are H O 2 A = flz = [33 = ’34 = 0 vs. H a : at least onefl; i 0-. The-value-of-the- F test statistic is 167.7410, with a corresponding p~value that is extremely small (<.0001). So, we reject Ho and conclude that at least one of the four predictors is useful in predicting energy content, using a linear model. (C ) l H 0 : ‘63 = 0 vs. H a 2 H3 i 0 . The value of the t test statistic is t = 2.24, with a corresponding p—value of 0.0340, which is less than the significance level of .05. So we can reject H0 and conclude that percentage garbage provides useful information about energy consumption, given that the other three predictors remain in the model. ...
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This note was uploaded on 02/11/2012 for the course STAT 231 taught by Professor Staff during the Fall '08 term at Iowa State.

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231HW11-S10 - HWH (I <2 GMhmmfl (b) .; ... : ) Jim: 7;...

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