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Unformatted text preview: ISYE6414
Summer 2010
Exam I Dr. Kobi Abayomi June 10, 2010 You must show all work to receive full credit. kQK] >0 1 A Climatologist investigates the effect of Solar Radiation, Wind, and Temperature on Ozone +i levels. She takes n = 153 daily observations from May 1st until September 30th. The
multiple regression output is: Fill in the missing values above. Write the estimated linear model. A +5 {.2 494.34 +0.0;759L Xu , 3.33 X;; +1447“ X23 2 For the model Y1 : [30 + [31XZ —— 61; with Q ~ N(O, 02) the following statistics are computed: 9
i=1 9 9
255?: 324 = 2y?
1'21 i=1 9 9
i=0: i +5 !) Compute the least squares estimates [30,51 J.)Compute the SSE, SSR, and SST. Compute the R2 value. Compute the observed F
statistic and perform the appropriate hypothesis test. 'HO +§ ?) Compute the prediction error for a single response a; = x0 for this model. Compute a
prediction interval for a single response when :1: = 5 2(X;~T<‘)C ~~9’) 2 20996...) 443 01": :(Xr—Yﬁ 20(7): : 33.54 2 00!“ J Ln = 0
A) SST = g: C‘ﬂwﬂ >317‘ SSR = éQQ'Wft 5f§xf=w
SSE—‘~/‘H.0$'
,Q‘: i: = earl*5“
H" A” V" H" ‘3’” Fag?“ x = 7} 5.7567
,. Reject H.
3) ‘66 laxo i 1;.oatm—a. ‘Spgi
5;"4=MSE‘(’+%+(3%9 H§E=%= 95713: 30.3.? >0 3 + ‘ In a regression of Number of New Accounts (Y) on Size of Minimum Deposit (X) the
following ANOVA table is computed: Source l SS l df l MS Regression g] 40.4,. 1 $1 40 4
Error 14,7416 9 I637. 74
‘l‘ ’5' Lack of Fit 13,593.13 4 35 98. 4( Pure Error [(48 5 2.3.7. 4
Total 19,882 10 Test the hypothesis 51— — 0 vs. the two sided alternative; test the hypothesis E(Y ) = ﬁ0+ﬂ1X
vs. the two sided alternative What a1e your conclusions? / FL: PIT—‘0 VS Ha: Fl 4:0 s ._
+7 F= 11%;: 3.883 < 5.757%?) "5” 51¢th Ho
2.11. : eéy)= p. +p.x vs. In .~ ECY) + F» +sz +7 91.].
F: ,HSLF :______35q, 2/ .‘3 , (1/5 ‘9’
‘ MSPE £27,4 4 >51: ) 7
‘. l’éjto'b Ho
6 4 This data is collected X y
1 13.59141
2 36.94528
3 100.42768
4 272.99075
5 742.06580 Here is a scatter plot with a linear ﬁt o
o
to
o
o
> <1"
0
o
N
o
1 2 3 4 5
X I) Suggestan appropriate transformation and compute the new regression of Y on X using the
transformed values. Perform these tests on your model: :1) I)H0251=0andH)HO:Y=ﬁ0+ﬁ1X+e 3) Interpret the coefﬁcients of your model in terms of the original, untransformed data. >0 )This page intentionally left blank
I
+3 An «Ppmpr‘ld’a ‘lnns‘ﬁrn‘mlim : K’s ex/JX)
b; _ :(X ——x',~~)c:j a)
+1; ? Zéx~ ——~x ”T
Y" ~/~7'7¢0 IO ‘ + 5')" ~£ =S’ b.=/.7047./o no ’) I) H. . ISM—0 vs. H,:p,:¢=o
+15 F: 55R >> F.4t(l’3) I lecmxe QEQO M$ PE gr V7] ”a J7me 4!“ Model ‘H‘ﬁr JA'ILA Mt]? SSLFAD, S‘féko
(If you eff/Am “r 7‘» have ’ (10“ qef 'ﬁul” CNJI‘h) +2 .9) 77‘? an?! (LIIMae Ty» €xP(X) make: 75 c/unje 5— ,‘q'fNWS/W?! 5 Show work
Assume Q ~ N(0, 02) I) . For the model K: = 50 + ﬁlXi + 67; with , 21:1 6 z 0.
.1). For the above model7 when R2 : 0, yi : g). 3). For the above model, when the 65 are independent, they are also independent if 1 / X
is substituted for X. 47) What is an appropriate transformation for this model Y; : exp{ﬁo +ﬁ1X1Y1 +ﬁ2Xi,2+eZ}
i X Extra Credit: For the model K : e$p{ﬁo + 31X“ + ﬂQXiag + 61}, what is the expected
Change in the response variable for a unit change in X1? +1; I) M7h283=2(‘1;'b.—b.><z)‘ a}: =23(7;—b.~hx;)=0 ~<D Se == zLj;—b.—b.><:) =0 (2'6?) +§ A) o=p‘=_§_55_lfi ,‘IJ‘eto $51230. .ssR=:()?,~—7)“=o mam: 5‘: =31“ ﬁr all a.
CmrI neceuartlj ‘j;==‘? ﬁrm‘5 +5” 4) '{Ji=9n‘31=P°*P:Xn *P>Xf=+57 Erlm) E[ Yi/X,=1,H — {($11} = czf (p,+/2,(2,+I) + PM;
~o‘f (AdPOO +P3Xz)
’3 (9/3"!) ‘fo(/5p+ﬂ.x/ +I3+X2.) +5 ~’— (y’all) wrap» 4—13.)“ Mb) 10 ...
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