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# INTRO STATISTICS THRY STA 4322

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This 22 page Class Notes was uploaded by Golden Bernhard on Friday September 18, 2015. The Class Notes belongs to STA 4322 at University of Florida taught by Ronald Randles in Fall. Since its upload, it has received 23 views. For similar materials see /class/206573/sta-4322-university-of-florida in Statistics at University of Florida.

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Chapter 11 Linear Models and Estimation by Least Sguares 111 112 113 115 116 Using the hint yo 30 3 y 12 lz 7 a slope 0 intercept 1 SSE 6 b The line with a negative slope should exhibit a better t c SSE decreases when the slope changes from 8 to 7 The line is pivoting around the point 0 1 and this is consistent with f 7 from part EX 111 d The best t is y 1000 0700x The summary statistics are 9 0 7 15 SW 76 S 10 Thus f 15 7 6x may The graph is above The summary statistics are 9 72 7 721 Scy 54243 S 54714 Thus f 072 099x When x 100 the best estimate ofy is f 072 099100 9972 The summary statistics are 9 45 7 433625 Sy 20335 S 42 Thus f 21575 4842x Since the slope is positive this suggests an increase in median prices over time Also the expected annual increase is 4842 a intercept 43362 SSE 1002839 b the data show an increasing trend so a line with a negative slope would not t well c Answers vary d Answers vary e 45 433625 139 The sum ofthe areas is the SSE a The relationship appears to be proportional to x2 b No c No it is the best linear model 231 Chapter 11 Linear Models and F timation bV Least Squares Instructor s Solutions Manual 118 The summary statistics are 9 15505 7 9448 Scy 1546459 S 2359929 Thus f 70712 0655x When x 12 the best estimate ofy is f 7712 065512 7148 119 a See part c b 3 71545 6517x pm By c The graph is above mm d Whenx 19 the best estimate ofy is f 71545 651719 108373 dSSE n n A 2 Jay 1110 d ZZ1y 1xx ZZ1xy 1x3 0 so 1 1 11 1 1111 Since ngly 134542 and 21le 53514 31 2514 1112 The summary statistics are 9 204 7 1294 Scy 7425571 S 18592 a The least squares line is f 17609 7 0229x Chapter 11 Linear Models and F timation bV Least Squares b The line provides a reasonable t 23 3 Instructor s Solutions Manual p1112x c When x 20 the best estimate ofy is j 17609 7 022920 13029 lbs 1113 The summary statistics are 7 6177 7 2705 Scy 7583004 S 19829 a The least squares line is f 452119 7 29402x 1391 g 398 1IE 1392 b The graph is above W m 1114 The summary statistics are 9 325 7 755 SW 727125 S 20625 a The least squares line is f 1182 7 1315x Chapter ll Linear Models and F timation bV Least Squares 23 4 Instructor s Solutions Manual I I I I n1 n2 n3 n4 n5 p11 14x The graph is above The line provides a reasonable t to the data Fquot 1115 a SSE 210 o x 2 210 r fmx aw 210 W izgm if 2 121y rx 210 W 458 2318 S 438 Since SSE SW 431 SW SSE lsxy SSE SW 2 Sm But S gt 0 and Sly2 2 0 So SW 2 SSE Fquot 1116 The summary statistics are 9 60 7 27 SW 71900 S 6000 a The least squares line is f 460 73166799 p11 Way 0 p11 16x b The graph is above Chapter ll Linear Models and E timation bV Least Squares 1117 1118 1119 1120 235 Instructor s Solutions Manual c Using the result in EX 1115a SSE Syy A31Sxy 792 7 73166771900 190327 So sz 19032710 19033 a With SW 10028388 and SW 20335 SSE 10028388 7 4 84220335 18286 So s2 182866 3048 b The tted line is 3 4335 242x The same answer for SSE and thus 32 is found a For EX 118 Syy 11011686 and SW 1546459 SSE 11011686 7 65525281546459 8784701 So sz 87847018 1098 b Using the coding x x1 7c the tted line is f 9448 655x The same answer for 32 is found The summary statistics are 9 16 7 106 SW 1520 S 320 a The least squares line is f 300 475x p11 19y b The graph is above 0 s2 5025 The likelihood function is given by K 6x 2111 1 n 05045 KeXP ZH e 451x 2 so that 1 n 2 262 211y1 0 1x1 J or 1 39 39 ly the l g J with respect to Bo and Bl is identical to minimizing the positive quantity 21y1 50 31ny This is the leastisquares criterion so the estimators will be the same lnL O lan Note that maximizing the quot39 quot39 I39I I39I 23 6 Instructor s Solutions Manual 1121 1122 1123 1124 1125 1126 Chapter 11 Linear Models and F timation bV Least Squares Using the results of this section and Theorem 512 C0V 0 1 C0VY E1J C9 1 C0VYa 1 C0V 1f 1 0 J CV 1 Thus C0V 0 1 f62 Su Note that if Zilxl 0 7c 0 so COV o l 0 From EX 1120 let 0 62 so that the logilikelihood is 1 n lnL0 ln2n ln0 Zllyl 30 31x1 2 Thus 71 l n 29 WZI1y1 0 1x12 The MLE is 0 62 zlyl 50 51x1 2 but since Bo and Bl are unknown we can insert their MLEs from EX 1120 to obtain 62 1721y1 0 31x17 IVSSE lan From EX 113 it is found that Syy 0 a Since SSE 4 7 7676 4 32 43 1333 T0 testHo Bl 0 vsHa 51 0 1 t1 520 with 3 degrees offreedom Since tozs 3182 we can reject H0 b Since toos 5841 and t01 4541 01 ltpivalue lt 02 Using the Applet 2PT gt 520 200691 01382 c 76 i 31821333 76 i 367 or 7967 7233 To test H0 BI 0 vsHa 5175 0 SSE 6166766 and sz 513897 Then it 1 5775 with 12 degrees of freedom a From Table 5 POTl gt 3055 2005 01 gtpivalue b Using the Applet 2PT gt 5775 00008 c Reject H0 From EX 1119 to test H0 Bl 0 vs Ha B1 0 32 5025 and S 320 Then it 1 3791 with 8 degrees offreedom a From Table 5PlTl gt 3355 2005 01 gtpvalue b Using the Applet 2PT gt 3791 200265 0053 c Reject H0 d We cannot assume the linear trend continues ithe number of errors could level off at some point e A 95 CI for Bl 475 i 2306 502532 475 i 289 or 186 764 We are 95 con dent that the expected change in number of errors for an hour increase of 10st sleep is between 186 764 The summary statistics are 3 539 7 71 Scy 19894 S 168069 Syy 236 a The least squares line is j 072 0118x Chapter 11 Linear Models and F timation bV Least Squares 1127 1128 23 7 Instructor s Solutions Manual b SSE 236 7 11819894 125 so sz 013 A95 CI for Bl is 0118 i 2776 40005 0118 i 008 c When x 0 EY Bo 510 g S0 to test H0 Bo 0 vs Ha Bo i 0 the test statistic is l t 1 4587 with 4 degrees offreedom Since Loos 4604 and t01 3747 we know that 01 ltp7value lt 02 d Using the Applet 2PT gt 4587 200506 01012 e Reject H0 Assuming that the error terms are independent and normally distributed with 0 mean and constant variance oz x 10 6 a We know that Z has a standard normal distribution under H0 cl Furthermore V n 2S2 62 has a chi7square distribution with n 7 2 degrees of freedom Therefore by De nition 72 Z x 10 JV n 2 SJ has a t7distribution with n 7 2 degrees of freedom under Hg for 139 1 2 5quot Using the pivotal quantity expressed above the result follows from the material in Section 88 Restricting to 20 the likelihood function is 1 n 2 LQO2nn1ch eXp 214 y e It is not difficult to verify that the MLEs for Bo and 52 under the restricted space are 17 and fZLOI 172 respectively The MLEs have already been found for the 2 nZ SSE Syy unrestricted space so that the LRT simpli es to n A n A 2 x LQO 2110 y M9 2104 W So we reject if S k or equivalently if Syy zkizVl 2k SSE Using the result from 1115 SSE A S A S A2 2 1 7 1 1 y 1 1S 1 T SSE SSE n 2S n 2 31 1 1s large 1n magmtude where 011 11011 S 0 So we see that 7 is small whenever T This is the usual t7test statistic so the result has been proven 23 8 Instructor s Solutions Manual 1129 1130 1131 Chapter ll Linear Models and F timation bV Least Squares Let A51 and 71 be the leastisquares estimators for the linear models Y BO 51x g and Wi Y0 10 s as de ned in the problem Then we have that E 1 7 91 51711 V1131 11 51 1 SI Where SE 2161 52 A51 7 11 follows a normal distribution so that under H0 Bl 7 1 0 so that 131 11 6 L L Z is standard normal Sc Let V SSEY SSEW Edal 462 Z1VK 4122 Then We2 has a chiisquare distribution with n m 7 4 degrees of freedom By De nition 72 we can build a random variable with a tidistribution under H 0 Z 3 1 T 1 1 whereSSSE SSE nm74 1 Vnm 4 31 Y W H0 is rejected in favor ofHa for large values oflTl a For the rst experiment the computed test statistic for H 0 Bl 0 vs Ha Bl i 0 is t1 1550202 767 with 29 degrees offreedom For the second experiment the computed test statistic is t 1900193 984 with 9 degrees of freedom Both of these values reject the null hypothesis at or 05 so we can conclude that the slopes are signi cantly different from 0 b Using the result from Ex 1129 S 204 186 3 1 11 4 1026 We can extract the values of S and SCC from the given values of V631 SSE n 2 20429 V 1 02022 so similarly SCC 554825 So to test equality for the slope parameters the computed test statistic is 172397 0 1155 1901 1024Wm with 38 degrees of freedom Since 025 Z 2025 196 we fail to reject H0 we cannot conclude that the slopes are different m 125 Here R is used to t the regression model gt x lt c191 382 gt y lt c095 174 gt sumary1myx 573 256 762 95 348 114 131 150 429 500 580 170 651 722 Call 1mf0rmu1a y x Chapter 11 Linear Models and F timation bV Least Squares 1132 1133 1134 1135 23 9 Instructor s Solutions Manual Residuals in 1Q Median 3Q Max 1333e02 4278e03 2314e05 8056e03 9811e03 Coefficients Estimate Std Error t value Prgtltl Intercept 1875602 6129e03 0 00183 x 4215603 5771605 73040 23397e11 Signif codes 0 3939 0001 3939 001 3939 005 3939 01 39 39 1 Residual standard error 0008376 on 7 degrees of freedom Multiple R Squared 09987 Adjusted R squared 09985 F statistic 5335 on 1 and 7 DP p value 2372e 11 From the output the tted model is j 01875 004215x To test H0 Bl 0 against Ha BI i 0 note that the pevalue is quite small indicating a very signi cant test statistic Thus H 0 is rejected and we can conclude that peak current increases as nickel concentrations increase note that this is a oneesided alternative so the pevalue is actually 2 37e 11 divided by 2 a From EX 115 31 48417 and S 42 From EX 1115 s2 30476 so to testHo Bl 0 vs Ha Bl gt 0 the required test statistic is I 1797 with 6 degrees offreedom Since t01 3143 H0 is rejected there is evidence of an increase b The 99 CI for Bl is 484 i 100 or 384 584 Using the coded x s from 1118 655 and sz 1097 Since SM xiz 655 10 97 2350 2388 freedom Since 025 2306 we can conclude that there is evidence ofa linear relationship 23602388 the computed test statistic is l t l 962 with 8 degrees of a Since toos 3355 we have thatgkvalue lt 2005 01 b Using the Applet 2PTlt 961 200001 00002 With a0 1 and a1 x the result follows since 1amp2le x2 2xrc 2 Z1xf m2xquot2 2xquotrc7c2 2 S S is x Tc2 6 1x f2 62 S n S 0 0 V6 x xx 240 Instructor s Solutions Manual 1136 1137 1138 1139 1140 Chanter ll Linear Models and F timation bV Least Squares 1 1 Th1s1s m1n1m1zed when x x2 0 so x x From EX 1113 and 1124 when x 5 3 452119 7 294025 30511 so that VG is estimated to be 40298 Thus a 90 Cl forEY is 30511 i 178244029 30511 i 35773 From EX 118 and 1118 when x 12 3 715 so that VG is estimated to be 12 155042 2359929 715 r 2477 or 467 963 10971 1154 Thus a 95 CI forEY is 715 i 23064115 Referto EX 113 and 1123 where 32 1333 3 15 7 6x SM 10 and 7c 0 When 95 0 the 90 CI forEY is 15 1 2353 0r112 188 When 95 72 the 90 CI for EY is 27 1 2353 or 203 337 o When 95 2 the 90 CI forEY is 3 1 2353 or 737 97 Pllav On the graph note the interval lengths Refer to EX 1116 When xii 65 3 25395 and a 95 CI forEY is 1 65 602 25395 i 2228 19033 or 25395 i 2875 12 6000 Refer to EX 1114 When xii 3 3 7878 and with SSE 0155 S 20625 and 2 x325the90CIforEYis78781186 1 w or 76 81 10 20625 Chapter ll Linear Models and F timation bV Least Squares 1141 1142 1143 1144 1145 1146 1147 1148 24 l Instructor s Solutions Manual so that a Using 00 17 01 and 01 as estimators we have y 17 0101ux n 7 1fH b Calculate VlayII77 itZI lf H2 g cze From EX 114 32 71057 and SM 54714 so that y 7219974 72 7408 and the variance ofthis estimate is calculated to be 71057LL0 711 The twoistandard deviation error bound is 2 L711 169 Similar to EX 1135 the variance is minimized when x 9 Refer to EX 115 and 1117 Whenx 9 year 1980 3 6515 and the 95 PI is 6515i2447 l30511 i 6515 i 542 or 5973 7057 For the year 1981 x 10 So 3 6999 and the 95 PI is 6999 i 24471 30511 10125 6999 i 580 For the year 1982 x 11 So 3 7483 and the 95 PI is 7483i 2447305l1 l 7483 i 624 Notice how the intervals get wider the further the prediction is from the mean For the year 1988 this is far beyond the limits of experimentation So the linear relationship may not hold note that the intervals for 1980 1981 and 1982 are also outside of the limits so caveat emptor From EX 118 and 1118 also see 1137 when x 12 j 715 so thatthe 95 PI is 1 12 155042 715 i 2306 10971 715 i 803 or786 1518 10 2359929 From 1116 and 1139 when x 65 3 25395 so that the 95 PI is given by 2 25395 i 2228 190331LM 25395 i 10136 12 6000 From EX 1114 when x 6 3 3933 so thatthe 95 PI is given by 1 6 3252 3933 i 2306 00194 1 3933 i 12 or 27 51 10 20625 The summary statistics are S 3805 Scy 25560 and SW 192636 Thus r 944 To test H0 p 0 vs Ha p gt 0 t 80923 with 8 degrees offreedom From Table 7 we nd that pivalue lt 005 242 Instructor s Solutions Manual 1149 1150 1151 1152 1154 1155 1156 Chapter ll Linear Models and F timation bV Least Squares a r2 behaves inversely t0 SSE since r2 1 7 SSESW b The best model has r2 817 so r 90388 since the slope is positive r is as well a r2 increases as the fit improves b For the best model r2 982 and so r 99096 c The scatterplot in this example exhibits a smaller error variance about the line The summary statistics are S 2359929 Sy 1546459 and SW 11011686 Thus r 9593 T0 testHo p 0 vs Ha p i 0 M 9608 with 8 degrees offreedom From Table 7 we see that pivalue lt 2005 01 so we can reject the null hypothesis that the correlation is 0 a Since the slope 0fthe line is negative r 1r2 V6 7781 b This is given by r2 so 61 c To test H0 p 0 vs Ha p lt 0 t W 433 w1th 12 degrees offreedom Since 7 t05 71782 we can reject H0 and conclude that plant density decreases with increasing altitude a This is given by r2 82612 68244 or 68244 b Same answer as part a 82613 39 39 c To test H0 p 0 vs Ha p gt 0 t m 4146 w1th 8 degrees of freedom S1nce rm 2896 we can reject H0 and conclude that heights and weights are positively correlated for the football players d pivalue PT gt 4146 00161 a The MOM estimators for 6 and 6 were given in Ex 972 b By substituting the MOM estimators the MOM estimator for p is identical to r the MLE Since 01 SW SW and r 011 s Syy we have that the usual titest statistic is A51 VSxx 1vn 2VSxxSyy 1vn 2r n 2 ISS 1 s 4313 1 filsy syy T l rzi Here r 8 a For n 5 t 2309 with 3 degrees offreedom Since t05 2353 fail to reject H0 b For n 12 t 42164 with 10 degrees offreedom Here 1 05 1812 reject H0 For part a pvalue PT gt 2309 05209 For part b pvalue 00089 Different conclusions note the Vn 2 term in the numerator 0f the test statistic The larger sample size in part b caused the computed test statistic to be more extreme Also the degrees of freedom were larger was Chapter 11 Linear Models and F timation bV Least Squares 1157 1158 1159 1160 1161 1162 243 Instructor s Solutions Manual a The sample correlation r determines the sign b Both r and 7 determine the magnitude of M For the test H0 p 0 vs Ha p gt 0 we reject ift g 2 t05 292 The smallest value of r that would lead to arej ection of H 0 is the solution to the equation 2 92 2 r 7V1 r Numerically this is found to be r 9000 For the testHo p 0 vs Ha p lt 0 we reject ift 31E S 405 71734 The largest value of r that would lead to a rejection of H 0 is the solution to the equation r 39iEAVI rz 18 Numerically this is found to be r 73783 Recall the approximate normal distribution of In given on page 606 Therefore for sample correlations V1 and r2 each being calculated from independent samples of size 711 and m respectively and drawn from bivariate normal populations with correlations coefficients p1 and p2 respectively we have that 1rl 1r2 1 1 1 2 Z 2 aria aria 94 91 is approximately standard normal for large m and 712 Thus to testHo p1 p2 vs Ha pl 75 p2 with r1 9593 m 10 r 85 n2 20 the computed test statistic is Z 1n1 1n z 1 52 L L 39 39 J7 J5 Since the rejection region is all values lzl gt 196 for at 05 we fail to reject H0 Refer to Example 1110 and the results given there The 90 PI is 979 i 2132045 1 979 i 104 or 875 1083 Sxy Using the calculations from Example 1111 we have V m 9904 The proportion of variation described is r2 99042 9809 a Observe that lnEY lnoto 7 11x Thus the logarithm of the expected value of Y is linearly related to x So we can use the linear model Wi 50 519 t 8i where w lny Bo lnoto and Bl 7 11 In the above note that we are assuming an additive error term that is in effect after the transformation Using the method of least squares the summary statistics are 244 Instructor s Solutions Manual 1164 1165 Chapter ll Linear Models and F timation bV Least Squares 7c 55 2x2 385 W 35505 Sm 77825 S 825 and SW 008448 Thus 11 70095 10 3603 and 611 470095 0095 do exp3603 3670 Therefore the prediction equation is j 3670639 0095 b To nd a CI for 010 we rst must nd a CI for Bo and then merely transform the endpoints of the interval First we calculate the SSE using SSE SW 7amp1 SW 008448 7 700957782481 0010265 and so sz 00102658 0001283 Using the methods given in Section 115 the 90 CI for Bo is 36027 i 186100012831 335 or 35883 36171 So the 90 CI for 010 is given by 858 e361 7 3617 3723 This is similar to EX 1163 Note that lnEY 410x and ln7lnEY lnoto allnx So we would expect that ln7lny to be linear in 111x De ne w ln7lny t lnxi Bo 11100 Bl 011 So we now have the familiar linear model WiBOBll i again we are assuming an additive error term that is in effect after the transformation The methods of least squares can be used to estimate the parameters The summary statistics are t 7112805 W 714616 SM 36828 and Squot 151548 S0 31 724142 30 12617andthus 611 724142 and do eXp12617 35315 This tted model is j eXp 35315x24142 lfy is related to t according to y 1 7 64339 then 7ln1 7y Br Thus let wi 7ln1 7y and we have the linear model w Bti 8 again assuming an additive error term This is the n07intercept model described in A n I w EX 1110 and the least squares estimator for B is given to be 5 Now using 11 1 A 2 SSE quot W W s1m1lar methods from Sect10n 114 note that V c 2 and 2 2114 2 6 6 11 1 is chi7square with n 7 1 degrees of freedom So by De nition 72 the quantity T n 2 a S I where S SSEn 7 1 has a t7distribution with n 7 1 degrees of freedom A 1001 7 01 CI for B is Chapter 11 Linear Models and F timation bV Least Squares 1166 1167 1168 245 Instructor s Solutions Manual and ram is the upperi 0L2 critical value from the tdistribution with n 7 1 degrees of freedom Using the matrix notation from this section 1 2 3 1 1 2 75 5 0 X1 0 Y1 X39Y XX 6 010 1 1 1 1 2 5 75 5 5 X39Y XX 15 5 15 1 A 21 1 so that 6 Notthat the 3 2 Y1 1 5 LANHO 3 The student should verify that X X 391 slope is the same as in Ex 1166 but the yiintercept is different Since XX is not a diagonal matrix as in Ex 1166 computing the inverse is a bit more tedious 1 3 9 1 1 2 4 0 1 11 0 1 7 0 28 X10 0 Y 1 X39Y 4 XX 0 28 0 1 1 1 1 8 28 0 196 1 2 4 0 1 3 9 0 The student should verify either using Appendix I or a computer 246 Instructor s Solutions Manual 1169 1170 Chapter 11 Linear Models and F timation bV Least Squares 3333 0 04762 714285 X39X391 0 035714 0 so that 142857 and the tted 04762 0 011905 142859 model is 7 714285 142857x142859x2 p11 68y n The graphed curve is above For this problem R will be used gt x lt c 7 5 3 1 1 3 5 7 gt y lt c185226272312330 49494350 a Linear model gt 1myx Call 1mformu1a y x Coefficients Intercept x 32725 1812 lt j327251812x b Quadratic model gt 1m yxIxA2H Call 1mformu1a y x 1XA2 Coefficients Intercept x 1XA2 355625 18119 01351 935562518119x 1351x2 A 719805 a The student should verify that Y 39Y 105817 X 39Y and 106155 991392 S0 SSE 105817 7105760155 56845 and sz 568458 7105625 Chapter 11 Linear Models and F timation bV Least Squares 1171 1172 1173 247 Instructor s Solutions Manual b Using the coding as speci ed the data are xj 42 404345 725 40 736 169 713 95 y19 14 7 29451094023860 70 721 10 0 The student should verify that X Y X X and 54243 0 54714 721 A So SSE 105817 7105760155 56845 same answer as part a 991392 Note that the vector a is composed ofk 0 s and one 1 Thus E Ea39 a39EW 3W 15 VB1 may a39Ea a3962X39X 1a 62a39XX 1a 61162 Following EX 1169 more detail with the R output is given by gt summarylmyx1x02 Call lmformula y x 1XA2 Residuals 8 2242 0525 1711 2415 4239 5118 8156 6625 Coefficients Estimate Std Error t value PrgtltH Intercept 355625 31224 11390 913e05 X 18119 04481 4044 000988 Ix 2 01351 01120 1206 028167 Signif codes 0 3939 0001 3939 001 3939 005 3939 01 39 39 1 Residual standard error Multiple R Squared 07808 F statistic 8904 on 2 and 5 DE 5808 on 5 degrees of freedom Adjusted R squared 06931 p value 00225 a To test H0 B 0 vs Ha B i 0 the computed test statistic is t 71206 andpevalue 28167 Thus H0 would not be rejected no quadratic effect b From the output it is presented that 1VB2 1120 So with 5 degrees of freedom t05 3365 so a 90 for B is 71351 i 33651120 71351 i 3769 or 7512 2418 Note that this interval contains 0 agreeing with part a If the minimum value is to occur at x0 1 then this implies B1 2B 0 To test this claim let a 0 1 2 for the hypothesis H0 B1 2B 0 vs Ha B1 2B i 0 From 24 8 Instructor s Solutions Manual 1174 1175 1176 Chapter 11 Linear Models and F timation bV Least Squares EX 1168 we have that 31 2amp2 7 142861 s2 7 14285 and we calculate a39XX 1a 7 083334 So the computed value ofthe test statistic is l t l 131 with 4 degrees offreedom Since t05 2776 H0 is not rejected a Each transformation is de ned for each factor by subtracting the midpoint the mean and dividing by one7ha1f the range b Using the matrix de nitions of X and Y we have that 338 16 0 0 0 0 21125 502 0 16 0 0 0 31375 X39Y 194 XX 0 0 16 0 0 sothat B 12125 26 0 0 0 16 0 1625 204 0 0 0 0 16 1275 The tted model is j 21125 31375x1 12125x2 1625x3 1275x4 c First note that SSE Y39Y 7 B39X39Y 744652 7 73477075 988125 so that sz 98812516 7 5 898 Further tests ofHo B 0 vs H0 B7 0 fori 1 2 3 4 are 4 based on the statistic t1 5 and H0 is rejected ifltil gt toos 3106 The s H 898 four computed test statistics are t1 7419 t 7162 t3 722 and t4 7170 Thus only the rst hypothesis involving the rst temperature factor is signi cant With the four given factor levels we have a 1 71 1 71 1 and so a39X39X391a 516 The estimate of the mean of Y at this setting is j 21125 31375 12125 1625 1275 219375 and the 90 con dence interval based on 11 degrees of freedom is 219375 i1796M516 219375 i 301 or 1893 2495 First we calculate sz SSEn 7 k71 11070111 100637 a To testHo B 0 vs H0 B lt 0 we use the t7test with c 81104 t 73222 1110063700081 With 11 degrees of freedom 7t05 71796 so we reject H0 there is suf cient evidence that B lt 0 5quot Similar to EX 1175 With the three given levels we have a 1 914 65 6 and so a39X39X 1a 9276617 The estimate ofthe mean on at this setting is j 3883 0092914 926511566 399812 and the 95 Cl based on 11 degrees offreedom is Chapter ll Linear Models and F timation bV Least Squares 1177 1178 1179 1180 1181 1182 1183 249 Instructor s Solutions Manual 399812 i 22014100637 9276617 399812 i 212664 Following EX 1176 the 95 PI is 399812 i 2201410063749376617 399812 i 213807 From EX 1169 the tted model is j 355625 18119x 1351x2 For the year 2004 x 9 and the predicted sales is 3 355625 181199 713592 409346 With we have a 1 9 81 and so a39X39X 1a 194643 The 98 PI for Lexus sales in 2004 is then 409346 i 3365580811 19464 409346 i 335475 Forthe given levels j 219375 a39X39X 1a 3135 and sz 898 The 90 PI based on 11 degrees offreedom is 219375 i1796 89813135 219375 i 617 or 1577 2811 Following EX 1131 SW 3748 and SSE Syy 31Sxy 3748 7 004215888 000508 Therefore the Fitest is given by F W 515757 with 1 numerator and 7 denominator degrees of freedom Clearly pvalue lt 005 so reject H 0 From De nition 72 let Z N Nor0 1 and W xi and let Z and Wbe independent Then T has the tidistribution with v degrees of freedom But since Z2 N x12 Z IW v by De nition 73 F Ta has aFidistribution with 1 numerator and v denominator degrees offreedom Now specific to this problem note that ifk 1 SSER Syy So the reduced modelFitest simpli es to SW Syy 1Sxy T2 SSEC n 2 sz S 39 F a To test H0 Bl B 53 0 vs Ha at least one B 0 the Fistatistic is F 2 1096546 1107013 32653 1 10701 1 1 with 3 numerator and 11 denominator degrees of freedom From Table 7 we see that pivalue lt 005 so there is evidence that at least one predictor variable contributes 110701 1096546 variation in percent yield Y is explained by the model b The coef cient of determination is R2 1 899 so 899 ofthe a To test H0 B B3 0 vs Ha at least one 5175 0 the reduced modelFitest is 25 0 Instructor s Solutions Manual 1184 1185 1186 1187 1188 Chapter 11 Linear Models and F timation bV Least Squares F 547007 1107012 11070111 with 2 numerator and 11 denominator degrees of freedom Since F05 398 we can reject H0 21677 SSER 1107012 b We must find the value of SSER such that 11070111 398 The solution is SSER 190808 a The result follows from n k1in k11 SSESW k 1 R2 k L SSESW b The form isFTa SW SSEk SSEn k 1 Here n 15 k 4 4 1 942 10 denominator degrees of freedom From Table 7 it is clear that pivalue lt 005 so we can safely conclude that at least one of the variables contributes to predicting the selling price a Using the result from EX 1184 F 39942 j 40603 with 4 numerator and b Since R2 1 SSES W SSE 16382217942 9501676 To test H0 B 53 B4 0 vs Ha at least one B 75 0 the reducedimodel Fitest is 1553 950163 950167610 with 3 numerator and 10 denominator degrees of freedom Since F 05 371 we fail to reject H 0 and conclude that these variables should be dropped from the model 2115 a The Fistatistic using the result in EX 1184 is F 45 with 4 numerator and 2 denominator degrees of freedom Since F1 924 we fail to reject H0 b Since k is large with respect to n this makes the computed Fistatistic small c The Fistatistic using the result in EX 1184 is F 2353 with 3 numerator and 40 denominator degrees of freedom Since F 1 223 we can reject H 0 d Since k is small with respect to n this makes the computed Fistatistic large a False there are 15 degrees offreedom for SSE b False the t R2 cannot improve when independent variables are removed c True Chapter 11 Linear Models and F timation bV Least Squares 1189 1190 1191 1192 1193 25 l Instructor s Solutions Manual d False not necessarily since the degrees offreedom associated with each SSE is different e True 139 False Model 111 is not a reduction of Model I note the xlxz term a True b False not necessarily since Model 111 is not a reduction of Model I note the xlxz term c False for the same reason in part b Refer to EX 1169 and 1172 a We have that SSER 2177112 and SSEC 168636 ForHo B 0 vs Ha B i 0 2177112 168636 1686365 5 denominator degrees offreedom With F05 661 we fail to rejectHo the reduced model Fitest is F 1455 with 1 numerator and b Referring to the R output given in EX 1172 the Fistatistic is F 8904 and the p7 value for the test is 0225 This leads to a rejection at the or 05 level The hypothesis of interest is H0 Bl B4 0 vs Ha at least one B i 0 139 1 4 From EX 1174 we have SSEC 988125 To find SSER we fit the linear regression model with just an and x3 so that 338 116 0 0 X39Y 194 X39X391 0 116 0 26 0 0 116 and so SSER 744652 7 7164195 282325 The reducedimodelFitest is F 282325 9881252 1021 98812511 with 2 numerator and 11 denominator degrees of freedom Thus since F 05 398 we can reject H 0 can conclude that either T1 or T2 or both affect the yield To test H0 B3 B4 B5 0 vs Ha at least one 5145 0 the reducedimodel Fitest is 465134 1521773 15217718 with 3 numerator and 18 denominator degrees of freedom Since F 005 592 we have that pivalue lt 005 1234 Refer to Example 1119 For the reduced model sz 3266238 4083 Then 111 0 0 XX 1 0 217 0 a391 171 0 0 217

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