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QUESTION 1.1) Obtain thelikelihoodfunctionfor thesampleobservationsY1,...,Yn
given X1,...,Xn if thenormal model is assumed tobeapplicable.2) Obtain themaximumlikelihood estimators foro and1.Solution:
1) Under normal modelYi hasNo 1Xi,2, withthecorrespondingdensity functiongiven by
fYiyi 1
2 2exp yio1Xi
2
22
Hencethelikelihoodfunctionfor thenormal error model, given thesampleobservationsY1,...,Yn, is:
Lo,1,2 i1
n1
22exp 1
22Yi o 1Xi2
2) In order to find theMLE weuse
lnLo,1,2 lni1
n1
22exp 1
22Yi o 1Xi2
n2 ln2 122 Yi o 1Xi2
olnLo,1,2 1
222Yi o 1Xi 1
1
2Yi no 1Xi
and1
lnLo,1,2 122
2Yi o 1Xi Xi
1
2XiYi oXi 1Xi2
Fromo
lnLo,1,2 01
lnLo,1,2 0
weget thefollowingequationsYi no 1Xi
XiYi oXi 1Xi2
Fromthefirst onewegeto Y
1X
Using it in thesecond wegetXiYi Y
1XXi
1Xi
2
hence
1 XiYi
Xi Yin
Xi2 Xi2
n
XiXYiY
XiX2
TheMLE foro and1 are
1
XiYi Xi Yi
n
Xi2 Xi2
n
ando Y
1X
thesameas estimatorsobtained usingleastsquares method.QUESTION 2.
Datafromastudy of therelationbetween thesizeof abid in million rands (X) and thecost to thefirmof preparing thebid in thousands rands (Y) for 12 recent bidsare
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presented in table below:
i 1 2 3 4 5 6 7 8 9 10 11 12
Xi 2.13 1.21 11.0 6.0 5.6 6.91 2.97 3.35 10.39 1.1 4.36 8.0
Yi 15.5 11.1 62.6 35.4 24.9 28.1 15.0 23.2 42.0 10 20 47.5
Thescatter plot strongly suggest that theerror varianceincreasewith X.Fit theweighted lest squares regressionlineusingweightswi 1
Xi2 .
Solution
Xi Yi wi 1/Xi2 wiXi wiYi wiXiYi wiXi
2
2.13 15.5 0.220415 0.469484 3.41643 7.276995 1
1.21 11.1 0.683013 0.826446 7.587449 9.173554 1
11 62.6 0.008264 0.090909 0.517355 5.690909 1
6 35.4 0.027778 0.166667 0.983333 5.9 1
5.6 24.9 0.031888 0.178571 0.794005 4.446429 1
6.91 28.1 0.020943 0.144718 0.588505 4.06657 1
2.97 15 0.113367 0.3367 1.700507 5.050505 1
3.35 23.2 0.089107 0.298507 2.067276 6.925373 1
10.39 42 0.009263 0.096246 0.389061 4.042348 1
1.1 10 0.826446 0.909091 8.264463 9.090909 1
4.36 20 0.052605 0.229358 1.0521 4.587156 1
8 47.5 0.015625 0.125 0.742188 5.9375 1
63.02 335.3 2.098715 3.871698 28.09667 72.18825 12 Totals
bi
wiXiYiwiXi wiYi wi
wiXi2 wiXi2
wi
72.18825 3.87169828.09667
2.098715
12 3.8716982
2.098715
4. 1906
bo wiYibiwiXi
wi
28.096674.19063.8716982.098715
5. 6568
HenceY 5.6568 4.1906X
QUESTION 3.Thefollowing datewereobtained in acertain study.
i 1 2 3 4 5 6 7 8 9 10 11 12
Xi 1 1 1 2 2 3 3 3 3 5 5 5
Yi 4.8 4.9 5.1 7.9 8.3 10.9 10.8 11.3 11.1 16.5 17.3 17.1
Summary calculational resultsare: Xi 34, Yi 126,Xi2 122Yi2 1554.66, XiYi 434.1) Fit alinear regression function2) Performan F test to determinewhether or not thereis lack of fit
of alinear regression function. Use 0.05.Solution
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1) Wehave
b1 XiYi
Xi Yin
Xi2 Xi2
n
434 34126
12
122 342
12
3
andbo Y b1X 1n Yi b1 1n Xi 112 126 3 112 34 2
ThereforeY 2 3 X2) F test for lack of fit.Wehavec 4 levels for X and 3 replicates for X 1 level,2 replicates for X 2 level, 4 replicates for X 3 leveland 3 replicates for X 5 level, and n 12.Hence
Yj 1nj
i1
nj
Yi,j - themean atj th level ofX.
Y 1 13 4.8 4.9 5.1 4. 9333 at level X 1Y 2
127.9 8.3 8. 1 at level X 2
Y 3 1410.9 10.8 11.3 11.1 11. 025 at level X 3
Y 4 1316.5 17.3 17.1 16. 967 at level X 5
SSPE j1
c
i1
nj
Yi,j Yj2 4.8 4.93332 4.9 4.93332 5.1 4.93332
7.9 8.12 8.3 8.12 10.9 11.0252 10.8 11.0252 11.3 11.0252 11.1 11.0252 16.5 16.9672 17.3 16.9672 17.1 16.9672 0. 62083
MSPE SSPEnc 0.62083
124 0.077604
SSE i1
n Yi Yi2 Yi2 boYi b1XiYi
1554.66 2 126 3 434 0. 66SSLF SSE SSPE 0.66 0.62083 0.03917MSLF SSLF
c2 0.03917
42 0.019585
Thehypothesis
Ho:EY o 1X
Ha:EY o 1X
Test statisticsF MSLF
MSPE
0.019585
0.077604
0. 25237Thedecision rule
If F F1 ;c 2,n c conclude Ho
If F F1 ;c 2,n c conclude HaF1 ;c 2;n c F0.95;2;8 4.46SinceF 0. 25237 F1 ;c 2;n c 4.46 weconcludeHo.
Thereis no lack of fit.QUESTION 4.1) Statethesimplenormal linear regressionmodel in matrix terms.2) Provethefollowingformula for SSE:
SSE Y
Y b
X
Y3) Provethat for
Yh Xh
b thevarianceis in matrix notation
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2Yh 2X h
X X1X hSolution1) Let
Y
Y1
Y2:
Yn
X
1 X1
1 X2: :
1 Xn
o1
1
2
:
n
then
n1
Y n2
X21
n1
where: is thevector of parametersX - matrix of knownconstants,namely,thevalues of theindependentvariable
is avector of independent normal randomvariableswithE 0
and2
2
I.2)We know thatSSE Yi2 boYi b1XiYiLet us noticethat
if Y
Y1
Y2
:
Yn
then Y Y1 Y2 .. . Yn
and
if X
1 X1
1 X2
: :
1 Xn
then X 1 1 ... 1
X1 X2 .. . Xn.
Hence
Y Y Y1 Y2 .. . Yn
Y1
Y2
:
Yn
Yi2 Yi2
and
X Y 1 1 ... 1
X1 X2 .. . Xn
Y1
Y2
:
Yn
YiXiYi
Using this with b bo
b1wehave
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Y Y bX Y Yi2 bo b1YiXiYi
Yi2 boYi b1XiYi SSEwhichcompletes theproof.
or2)Weknow thatA A, A B A B, and AB BA
andthenormal equation:X Xb X Y
hence
X Xb X Y 0
0
where b bo
b1
so
X Xb X Y bX X Y X 0
0
0 0
FromdefinitionSSE ee Y Xb Y Xb Y bX Y Xb
Y Y Y Xb bX Y bX Xb Y Y bX Y bX Xb Y Xb Y Y bX Y bX X Y Xb
Y Y bX Y 0 0 bob1
Y Y bX Y 0 Y Y bX Y
3)We know that:Let W bearandomvector obtained by premultiplyingtherandomvector
Y by aconstant matrix AW AY
Then*) 2W 2AY A2YA
Since
Yh X h
b using *) withA X h
weget
2Yh X h
2bX hUsingthefact that2b 2X X1
weget2
Yh X h
2X X1X h 2X h X X1X h
QUESTION 5.Thefitted values and residuals of aregression analysis aregiven below
t 1 2 3 4 5 6 7 8 9 10
Yt 21.96 4.15 7.36 22.11 10.98 22.06 47.35 47.05 73.40 69.79et -1.45 -0.26 -0.16 -0.20 0.32 0.63 0.24 0.55 -0.50 -0.65
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t 11 12 13 14 15 16 17 18 19 20Yt 83.83 87.09 75.64 76.15 69.08 32.24 47.30 52.29 78.03 77.78
et 0.06 -0.09 -0.24 -1.03 0.02 0.56 0.80 0.11 0.57 0.72
Assumethat thesimplelinear regressionmodel with therandomterms followingafirst-order autoregressiveprocessis appropriate.Conduct aformal test for positiveautocorrelation using 0.05.Solution
Thehypothesis:
Ho : 0
Ha : 0
TheDurbin-Watson test statistics:
D
t2
n
etet12
t1
n
et2
6.50256.7072
0. 96948
t1
n
et2
1.452 0.262 0.162 0.202 0.322 0.632
0.242 0.552 0.502 0.652 0.062 0.092 0.242 1.032 0.022 0.562 0.802 0.112
0.572 0.722 6. 7072
t2
n
et et12 0.26 1.452 0.16 0.262 0.20 0.162
0.32 0.202 0.63 0.322 0.24 0.632 0.55 0.242 0.50 0.552 0.65 0.502 0.06 0.652 0.09 0.062 0.24 0.092 1.03 0.242 0.02 1.032 0.56 0.022 0.80 0.562 0.11 0.802 0.57 0.112 0.72 0.572 6. 5025
Thedecision rule
If D dU conclude Ho
IfD dL concludeHa
IfdL D dU thetest is inconclusive
p 2 , dL 1.20 and dU 1.41SinceD 0.96948 dL 1.20 weconcludeHa, that theerror terms arepositively autocorrelated.
QUESTION 6.Thefollowing datawereobtained in acertain experiment:
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i Xi,1 Xi,2 Yi
1 1 2 2.5
2 1 2 3
3 1 2 3.5
4 2 1 3
5 2 1 4
6 0 1 1
7 0 1 1.5
8 0 1 2
9 1 0 1.5
10 1 0 2
11 1 0 2.5Thedatasummary is given below in matrix form
X X
11 10 11
10 14 10
11 10 17
X X1
2354
527
16
527
1154
0
16
0 16
X Y
26.5
29
29.5
Y Y 72.25
Assumethat first-order regressionmodel with independentnormal errorsis appropriate.1) Find theestimated regressioncoefficients.2) Obtain an ANOVA table and useit totest whether thereis aregression
relation using 0.05.3) Estimate1 and2 jointly by theBonferroni procedureusing
80percentfamily confidencecoefficient.Solution
1) b
bo
b1
b2
X X1X Y
2354
527
16
527
1154
0
16
0 16
26.5
29
29.5
1.0
1. 0
0. 5
2) Y 1 1Y Yi 26.5 (weget it from X Y )
SSTO Y Y 1n Y11Y
72.25 111
26.5 26.5 8. 4091
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SSE Y Y bX Y 72.25 1 1 0.5
26.5
29
29.5
2.0
SSR bX Y 1n Y11Y
1 1 0.5
26.5
29
29.5
111
26.5 26.5 6. 4091
ANOVA table
sourceof variation SS df MS
regression SSR 6.4091 p 1 2 MSR SSRp1 3.2046
error SSE 2 n p 8 MSE SSEnp 28
0.25
total SSTO 8.4091 n 1 10Hypothesis:Ho : 1 2 0Ha : not both1 and2 areequal tozero
Test statisticsF MSR
MSE
3.20460.25
12. 818
Thedecision ruleIfF F1 ;p 1,n p, concludeHoIfF F1 ;p 1,n p, concludeHa
F1 ;p 1,n p F0.95,2,8 4.46SinceF 12.818 F0.95,2,8 4.46weconcludeHa (not both1 and2 areequal tozero), that means that thereis alinear associationbetween X and Y .3) Ifg parameters areto beestimated jointly (whereg p), theBonferroni confidencelimits withfamily confidencecoefficient 1 are:
bk BsbkB t1 /2g,n p
andweget s2b1, s2b2s2b MSEX X1
In our caseg 2, b1 1, b2 0.5
s2b MSEX X1 0.25
2354
527
16
527
1154
0
16
0 16
0. 10648 4. 6296 102 4. 1667 102
4. 6296 102 0.050926 0
4. 1667 102 0 0.041667
sos2b1 0.050926 and sb1 0.050926 0. 22567
s2b2 0.04166 and sb2 0.04166 0. 20411B t1 /2g,n p t1 0.2
22,8 t0.95,8 1.860
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Hencethelimits for b1 and b2 are0.58025,1.4197 and 0.12036,0.87964respectively, since1 1.860 0. 22567 0. 580251 1.860 0. 22567 1. 4197
0.5 1.860 0. 20411 0. 120360.5 1.860 0. 20411 0. 87964QUESTION 7.For acertain experiment thefirst-order regressionmodel with twoindependentvariables was used. Thecalculated diagonal elementsof thehat matrix are:
i 1 2 3 4 5 6 7 8
hi,i 0.237 0.237 0.237 0.237 0.137 0.137 0.137 0.137
i 9 10 11 12 13 14 15 16
hi,i 0.137 0.137 0.137 0.137 0.237 0.237 0.237 0.237
1) Describeuseof hat matrix for identifyingoutlyingX observations.2) Identify any outlyingX observations using thehat matrix method.Solution1) Thehat matrix H is given by:
H XX X1X
Thediagonal element hi,i in thehat matrix is called theleverageof the ith observation.Thus, alargeleveragevaluehi,i indicates that the ith observation is distantfromthecenter of theX observations. Themean leveragevalue
h hi,i
n pn
Hence, leveragevalues greater than2pn areconsidered by this ruletoindicateoutlying
observations with regard to theX values.2) In our casen 16, p 3 so thecritical value
2pn
616
0. 375
Sinceall leveragevalues in our casearelessthan 0.375 thereforethis methoddoes notidentified outlying observationsfor X.QUESTION 8.Provethat1) SSRX1,X2,X3 SSRX1 SSRX2,X3 X12) SSRX1 SSRX2 X1 SSRX2 SSRX1 X2Solution:1) Weknow that
SSRX2,X3 X1 SSEX1 SSEX1,X2,X3and
SSTO SSEX1 SSRX1 SSEX1,X2,X3 SSRX1,X2,X3HenceLHS SSRX1,X2,X3 SSTO SSEX1,X2,X3 SSEX1 SSRX2 SSEX1,X2,X3 SSRX1 SSEX1 SSEX1,X2,X3 SSRX1 SSRX2,X3 X1 RHSwhichcompletes theproof.2) Weknow that
SSRX2 X1 SSEX1 SSEX1,X2,SSRX1
X2 SSEX2
SSEX1,X2
andSSTO SSEX1 SSRX1 SSEX2 SSRX2
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HenceLHS SSRX1 SSRX2 X1 SSRX1 SSEX1 SSEX1,X2 SSTO SSEX1,X2 SSEX2 SSRX2 SSEX1,X2 SSRX2 SSRX1 X2 RHSwhichcompletes theproof.
QUESTION 9.Themeasurer2 is called thecoefficient of determination and is given by formula
r2 SSTOSSESSTO
SSR
SSTO 1 SSE
SSTO
Thesquareroot (with aplus or minus sign is attached to this measureaccording towhether theslopeof thefitted regressionlineis positiveor negative)
r r2 is called thecoefficient of correlation. Provethefollwing
r XiXYiY
XiX2YiY21/2Solution:Wearegoing to usethefollowingformulas:
SSR b1Xi XYi Y, b1 XiXYiY
XiX2 ,SSTO Yi Y2,
Let us noticethat b1 0 if Xi XYi Y 0and b1 0 if Xi XYi Y 0.
r2 SSRSSTO
b1XiXYiYYiY2
XiXYiYXiX2
XiXYiY
YiY2
XiXYiY2
XiX2YiY2Hence
r signb1 XiXYiY
XiX2YiY212
XiXYiYXiX2YiY2
12
QUESTION 10.Thefollowing datawereobtained in certain experiment:
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i Yi Xi,1 Xi,2
1 64 4 2
2 73 4 4
3 61 4 2
4 76 4 4
5 72 6 2
6 80 6 4
7 71 6 2
8 83 6 4
9 83 8 2
10 89 8 4
11 86 8 212 93 8 4
13 88 10 2
14 95 10 4
15 94 10 2
16 100 10 4
Thedatasummary is given below in matrix form
X X
16 112 48
112 864 336
48 336 160
X X1
9980
780
316
780 180 0
316
0 116
X Y
1308
9510
3994
Y Y 108896
Assumethat first-order regressionmodel with independentnormal errorsis appropriate.1) Find theestimated regressioncoefficients.2) Obtain an ANOVA table and useit totest whether thereis aregression
relation using
0.05.3) Estimate1 and2 jointly by theBonferroni procedureusing80 percentfamily confidencecoefficient.Solution
1) b
bo
b1
b2
X X1X Y
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9980
780
316
780
180
0
316
0 116
1308
9510
3994
75320
17740
358
37. 65
4. 425
4. 375
2) Y 1 1Y Yi 1308 (weget it from X Y )
SSTO Y Y 1n Y11Y
108896 116
1308 1308 1967
SSE Y Y bX Y 108896
37. 65
4. 425
4. 375
1308
9510
3994
94. 3
SSR bX Y 1n Y11Y
37. 654. 425
4. 375
13089510
3994
116
1308 1308 1872. 7
ANOVA table
sourceof variation SS df MS
regression SSR 1872.7 p 1 2 MSR SSRp1 1872.7
error SSE 94.3 n p 13 MSE SSEnp 94.313
7. 253846154
total SSTO 1967 n 1 15
Hypothesis:Ho : 1 2 0Ha : not both1 and2 areequal tozero
Test statisticsF MSR
MSE
1872.77.253846154
258. 1664899
Thedecision ruleIfF F1 ;p 1,n p, concludeHoIfF F1 ;p 1,n p, concludeHa
F1 ;p 1,n p F0.95,2,13 3.81SinceF 258. 1664899 F0.95,2,13 3.81
weconcludeHa (not both1 and2 areequal tozero), that means that thereis alinear associationbetween X and Y .3) Ifg parameters areto beestimated jointly (whereg p), theBonferroni confidencelimits withfamily confidencecoefficient 1 are:
bk BsbkB t1 /2g,n p
andweget s2b1, s2b2s2b MSEX X1
In our caseg 2, b1 4.425, b2 4.375
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s2b MSEX X1 7. 253846154
9980
780
316
780
180
0
316
0 116
8. 976634616 . 6347115385 1. 3600961540. 6347115385 0.09067307693 0
1. 360096154 0 0. 4533653846
sos2b1 0.09067307693 and sb1 0.09067307693 0. 3011197053s2b2 0. 4533653846 and sb2 0. 4533653846 0. 6733241304B t1 /2g,n p t1 0.20
22,13 t0.95,13 1.771
Hencethelimits for b1 and b2 are3. 8917,4. 9583 and3. 1825,5. 5675
respectively, since4.425 1.771 0. 3011197053 3. 89174.425 1.771 0. 3011197053 4. 95834.375 1.771 0. 6733241304 3. 18254.375 1.771 0. 6733241304 5. 5675QUESTION 11.
Thefollowing datawereobtained in certain experiment:
i Xi,1 Xi,2 Yi
1 1 2 2.5
2 1 2 3
3 1 2 3.5
4 2 1 3
5 2 1 4
6 0 1 1
7 0 1 1.5
8 0 1 2
9 1 0 1.5
10 1 0 2
11 1 0 2.5
Thedatasummary is given below in matrix form
X X
11 10 11
10 14 10
11 10 17
X X1
2354
527
16
527
1154
0
16
0 16
X Y
26.5
29
29.5
Y Y 72.25
Assumethat first-order regressionmodel with independentnormal errorsis appropriate.
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1) Find theestimated regressioncoefficients.2) Obtain an ANOVA table and useit totest whether thereis aregression
relation using 0.05.3) Estimate1 and2 jointly by theBonferroni procedureusing80 percentfamily confidencecoefficient.
Solution
1) b
bo
b1
b2
X X1X Y
2354
527
16
527
1154
0
16
0 16
26.5
29
29.5
1.0
1. 0
0. 5
2) Y 1 1Y Yi 26.5 (weget it from X Y )SSTO Y Y 1n Y
11Y 72.25 1
11 26.5 26.5 8. 4091
SSE Y Y bX Y 72.25 1 1 0.5
26.5
29
29.5
2.0
SSR bX Y 1n Y11Y
1 1 0.5
26.5
29
29.5
111
26.5 26.5 6. 4091 6.4091/2 3. 2046
ANOVA table
sourceof variation SS df MS
regression SSR 6.4091 p 1 2 MSR SSRp1 3.2046
error SSE 2 n p 9 MSE SSEnp 2
10 0.2
total SSTO 8.4091 n 1 10
Hypothesis:Ho : 1 2 0Ha : not both1 and2 areequal tozero
Test statisticsF MSR
MSE
3.20460.2
16. 023
Thedecision ruleIfF F1 ;p 1,n p, concludeHoIfF F1 ;p 1,n p, concludeHa
F1 ;p 1,n p F0.95,2,9 4.26SinceF 16.023 F0.95,2,13 4.26weconcludeHa (not both1 and2 areequal tozero), that means that thereis a
linear associationbetween X and Y .3) Ifg parameters areto beestimated jointly (whereg p), theBonferroni confidence
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limits withfamily confidencecoefficient 1 are:bk BsbkB t1 /2g,n p
andweget s2b1, s2b2s2b MSEX X1
In our caseg 2, b1 1, b2 0.5
s2b MSEX X1 3.2046
2354
527
16
527
1154
0
16
0 16
1. 3649 0. 59346 0. 53411
0. 59346 0. 65278 0
0. 53411 0 0. 53411
so
s2b1 0. 65278 and sb1 0. 65278 0. 80795s2b2 0. 53411 and sb2 0. 53411 0. 73083B t1 /2g,n p t1 0.2
22,9 t0.95,9 1.833
Hencethelimits for b1 and b2 are 0. 48097,2. 481 and 0. 83961,1. 8396respectively, since1 1.833 0. 80795 0. 480971 1.833 0. 80795 2. 4810.5 1.833 0. 73083 0. 839610.5 1.833 0. 73083 1. 8396
QUESTION 12.For acertain experiment thefirst-order regressionmodel with twoindependentvariables was used. Thecalculated diagonal elementsof thehat matrix are:
i 1 2 3 4 5 6 7 8 9 10
hi,i 0.91 0.194 0.131 0.268 0.149 0.141 0.429 0.067 0.135 0.165
i 11 12 13 14 15 16 17 18 19 20
hi,i 0.179 0.059 0.110 0.156 0.095 0.128 0.97 0.230 0.112 0.073
1) Describeuseof hat matrix for identifyingoutlyingX observations.2) Identify any outlyingX observations using thehat matrix method.
Solution1) Thehat matrix H is given by:
H XX X1X
Thediagonal element hi,i in thehat matrix is called theleverageof the ith observation.Thelargeleveragevaluehi,i indicates that theith observationis distant fromthecenter of theX observations. Themean leveragevalue
h hi,i
n pn
Hence, leveragevalues greater than2pn areconsidered by this ruletoindicateoutlying
observations with regard to theX values.2) In our casen 20, p 3 so thecritical value
2pn
620
0.36
Sinceleveragevalues corresponding totheresultsfrom1, 7, 17 trials aregreaterthan our critical valuethereforewecan classify 1,7, 17 as an outlyingobservation.
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QUESTION 13.Thefitted values and residuals of aregression analysis aregiven below
i 1 2 3 4 5 6 7Yi 2.92 2.33 2.25 1.58 2.08 3.51 3.34
ei 0.18 -0.03 0.75 0.32 0.42 0.19 0.06
i 8 9 10 11 12 13 14 15Yi 2.42 2.84 2.50 3.59 2.16 1.91 2.50 3.26
ei -0.42 0.06 -0.20 -0.39 -0.36 -0.51 -0.50 0.54
Assumethat thesimplelinear regressionmodel with therandomterms followingafirst-order autoregressiveprocessis appropriate.Conduct aformal test for positiveautocorrelation using 0.05.Solution
Thehypothesis:
Ho : 0
Ha : 0
TheDurbin-Watson test statistics:
D
t2
n
etet12
t1
n
et2
3.46042.21788
1. 560228687
t1
n
et2
0.182 0.032 0.752 0.322 0.422 0.192
0.062
0.422
0.062
0.202
0.392
0.362
0.512 0.502 0.542 2.21788
t2
n
et et12 0.03 0.182 0.75 0.032 0.32 0.752
0.42 0.322 0.19 0.422 0.06 0.192 0.42 0.062 0.06 0.422 0.20 0.062 0.39 0.202 0.36 0.392 0.51 0.362 0.50 0.512 0.54 0.502 3. 4604
Thedecision rule
If D dU conclude Ho
IfD dL concludeHa
IfdL D dU thetest is inconclusivep 2 , dL 1.08 and dU 1.36SinceD 1. 560228687 dU 1.36 weconcludeHo, that theerror terms arenot positively autocorrelated.QUESTION 14Provethefollowing statements:1) Thesumof theobserved values Y i equals thesumof thefitted values
Yi;
i1
n
Yi i1
n Yi
2) Theregressionlinealwaysgoes through thepoint(X,YSolution:
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1)This condition is implicit in thefirst normal equation
Yi nbo b1Xi bo b1Xi i1
n Yi.
2) Theestimated regressionlineisY bo b1X
WehavetoshowthatY bo b1X.
Fromthefirst normal equationdiveded by n (both sides)Yi nbo b1Xi : n
weget1n Yi bo b1 1n Xi
and Y bo b1X whichcompletes theproof.QUESTION 15.
Thefollowing datewereobtained in acertain study.
i 1 2 3 4 5 6 7 8 9 10 11 12
Xi 1 1 1 2 2 3 3 3 3 5 5 5
Yi 4.8 4.9 5.1 7.9 8.3 10.9 10.8 11.3 11.1 16.5 17.3 17.1Summary calculational resultsare: Xi 34, Yi 126,Xi2 122Yi2 1554.66, XiYi 434.1) Fit alinear regression function2) Performan F test to determinewhether or not thereis lack of fit
of alinear regression function. Use 0.05.Solution1) Wehave
b1 XiYi
Xi Yin
Xi
2 Xi2
n
434 34126
12
122 342
12
3
andbo Y b1X 1n Yi b1 1n Xi 112 126 3
112
34 2
ThereforeY 2 3 X2) F test for lack of fit.Wehavec 4 levels for X and 3 replicates for X 1 level,2 replicates for X 2 level, 4 replicates for X 4 leveland 3 replicates for X 5 level, and n 12.Hence
Yj
1
nj i1
nj
Yi,j - themean atj th level ofX.Y 1
134.8 4.9 5.1 4. 9333 at level X 1
Y 2 127.9 8.3 8. 1 at level X 2
Y 3 1410.9 10.8 11.3 11.1 11. 025 at level X 4
Y 4 1316.5 17.3 17.1 16. 967 at level X 5
SSPE j1
c
i1
nj
Yi,j Yj2 4.8 4.93332 4.9 4.93332 5.1 4.93332
7.9 8.12 8.3 8.12 10.9 11.0252 10.8 11.0252 11.3 11.0252 11.1 11.0252 16.5 16.9672 17.3 16.9672 17.1 16.9672 0. 62083
MSPE SSPEnc 0.62083124 0.077604
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SSE i1
n
Yi Yi2 Yi2 boYi b1XiYi
1554.66 2 126 3 434 0. 66SSLF SSE SSPE 0.66 0.62083 0.03917MSLF SSLF
c2 0.03917
42 0.019585
ThehypothesisHo:EY o 1X
Ha:EY o 1X
Test statisticsF MSLF
MSPE
0.0195850.077604
0. 25237
Thedecision rule
If F F1 ;c 2,n c conclude Ho
If F F1 ;c 2,n c conclude HaF1
;c
2;n
c F0.95;2;8 4.46
SinceF 0. 25237 F1 ;c 2;n c 4.46 weconcludeHo.Thereis not lack of fit.QUESTION 16Consider thesimplelinear regressionmodel expressed in matrix terms.Provethefollowing formula for SSE:SSE Y Y bX YSolutionWeknow thatSSE Yi2 boYi b1XiYiLet us noticethat
if Y
Y1
Y2
:
Yn
then Y Y1 Y2 .. . Yn
and
if X
1 X1
1 X2
: :
1 Xn
then X 1 1 ... 1
X1 X2 .. . Xn.
Hence
Y Y Y1 Y2 .. . Yn
Y1
Y2
:
Yn
Yi2 Yi2
and
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X Y 1 1 ... 1
X1 X2 .. . Xn
Y1
Y2
:
Yn
YiXiYi
Using this with b bo
b1wehave
Y Y bX Y Yi2 bo b1YiXiYi
Yi2 boYi b1XiYi SSEwhichcompletes theproof.
QUESTION 17.
Provethefollowingtheorem:TheoremMSE is an unbiased estimator of2 for thesimple linearregressionmodel.SolutionProof: Weknow thatSSE Yi
Yi2 Yi bo b1Xi2
Yi Y b1X b1Xi2 Yi2 nY2 Xi2 nX2b12 2nXb1Y 2XiYib1ESSE EYi2 nY2 Xi2 nX2b12 2nXb1Y 2XiYib1
EYi2
nEY2
Xi2
nX2
Eb12
2nXEb1Y 2XiEYib1
EYi2 nEY2 Xi X2Eb12 2nXEb1Y 2XiEYib1EYi2 2 o 1Xi2 n2 no2 2no1X 12Xi2
nEY2 n 2
n o 1X2 2 no
2 2no1X n1
2X2
Xi X2Eb12 Xi X2 2
XiX2 1
2 2 1
2Xi X2
2 12Xi2 n12X2 Xi X2Eb12
Eb1Y 1n E
XiX
XiX2Yi Yj 1n E
ij
XiX
XiX2YiYj XiXXiX2
Yi2
1n ij
XiX
XiX2 EYi Yj 1n
XiX
XiX2 EYi2
1n
ij
XiX
XiX2o 1Xio 1Xj
1n
XiX
XiX22 o 1Xi2
1n
ij
XiX
XiX2o 1Xio 1Xj XiXXiX2
o 1Xi2
1n
XiX
XiX22 1n
i,j
XiX
XiX2o 1Xio 1Xj
1n
j
i
XiX
XiX2o 1Xio 1Xj
1n
j
o 1Xji
XiX
XiX2o
1n no 1Xj 1i
XiXXiXiX2 o 1X1iXiXXiXiX2 o 1X1
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2nXEb1Y 2nXo 1X1 2no1X 2n12X2 2nXEb1Y
EYib1 EXjX
XiX2YjYi
j,,ji
XjX
XiX2EYjYi
XiX
XiX2EYi
2
ji
XjX
XiX2o 1Xio 1Xj
XiX
XiX22 o 1Xi2
ji
XjX
XiX2o 1Xio 1Xj
XiX
XiX2o 1Xi2
XiX
XiX22
j
XjX
XiX2o 1Xio 1Xj
XiX
XiX22
o 1Xij
XjX
XiX2o 1Xj
XiX
XiX22
o 1Xi oj
XjX
XiX2 1
j
XjXXj
XiX2
XiX
XiX22
o 1Xi1 XiX
XiX22
2XiEYib1 2Xi o 1Xi1 XiXXiX22
2Xio 1Xi1 2 XiXXiXiX22
2o1Xi 212Xi2 22 2no1X 21
2Xi2 22 2no1X 212Xi2 22 2XiEYib1HenceESSE EYi2 nY2 Xi2 nX2b12 2nXb1Y 2XiYib1 EYi2 nEY2 Xi2 nX2Eb12 2nXEb1Y 2XiEYib1
EYi2
nEY2
Xi X2
Eb12
2nXEb1Y 2XiEYib1 n2 no
2 2no1X 1
2Xi2 2 no2 2no1X n12X2 2 1
2Xi2 n12X2 2no1X 2n12X2 2no1X 212Xi2 22 n 22
Weused thefollowingX
j
Xi X 0
j
XjXXj
XiX2
j
XjXXjXj
XiX
XiX2
j
XjXXjXXiX
XiX2
j X
jXXjX
XiX2 1
and Xi2 nX2 Xi X2 Xi2 nX2Finally
EMSE E SSEn2
ESSE
n2 n22
n2 2
QUESTION 18.Assumethat thenormal regressionmodel is applicable.For thefollowingdatagiven by:
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i 1 2 3 4 5 6
Xi 4 1 2 3 3 4
Yi 16 5 10 15 13 22
usingmatrix methodfind:1)Y Y2) X X3) X Y4) b
5)Testthe Ho : 1 0 versus Ha : 1 0usingANOVA.with 0.05
6) Covariance-variancematrix s2bSolution.
X
1 4
1 1
1 2
1 3
1 3
1 4
Y
16
5
10
15
13
22
1Y 1 1 1 1 1 1
16
5
10
1513
22
81
1)Y Y
16
5
10
15
1322
16
5
10
15
1322
1259
2) X X
1 4
1 1
1 2
1 3
1 3
1 4
1 4
1 1
1 2
1 3
1 3
1 4
6 17
17 55
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3) X Y
1 4
1 1
1 2
1 31 3
1 4
16
5
10
1513
22
81
261
4) X X1 6 17
17 55
1
A1 1detA
Aij
5541
1741
1741
641
b bo
b1 X X1X Y
5541
1741
1741
641
81
261
18
4118941
0. 439024. 6098
5) SSR bX Y 1nY11Y
0. 43902
4. 6098
81
261 1
6 812 145. 2
SSTO Y Y 1nY11Y 1259 1
6 812 331
2 165. 5
SSE Y Y bX Y 1259 1238. 7 20. 3
(ANOVA table):sourceof variation SS df MS
regression SSR 145.2 1 MSR SSR1
145.2
1 145.2
error SSE 20.3 4 MSE SSEn2
20.34
5. 075
total SSTO 165.5 5
Ho : 1 0Ha : 1 0
F MSRMSE
145.25.075
28. 611
Thedecision rule:IfF F1 ;1,n 2, concludeHo
IfF F1 ;1,n 2, concludeHaF1 ;1,n 2 F0.95,1,4 7.71SinceF 0. 39511 F1 ;1,n 2 F0.95,1,4 7.71weconcludeHo that means that thereis alinear associationbetween X and Y .
6) s2b SEX X1 5. 0755541
1741
1741
641
6. 8079 2. 1043
2. 1043 . 74268
QUESTION 19Provethefollowing statements:1) Thesumof residuals is zero:
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i1
n
ei 0
2) Thesumof theweighted residuals is zero when theresidual in theith trialis weighted by thelevel of theindependent variable in theith trial:
i1
n
Xiei 0
Solution:1) Let us recall thefirst normal equationYi nbo b1XiPuttingall terms on onesidewegetYi nbo b1Xi 0Hence
i1
n
ei i1
n
Yi Yi
i1
n
Yi bo b1Xi Yi nbo b1Xi 0
2) Let us recall thesecond normal equation
XiYi
boXi
b1Xi2
Puttingall terms on onesidewegetXiYi boXi b1Xi2 0Hence
i1
n
Xiei i1
n
XiYi bo b1Xi XiYi boXi b1Xi2 0
QUESTION 20.Theresults of acertain experiments areshown below
i 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 2
Xi 5.5 4.8 4.7 3.9 4.5 6.2 6.0 5.2 4.7 4.3 4.9 5.4 5.0 6.3 4.6 4.3 5.0 5.9 4.1 4.
Yi 3.1 2.3 3.0 1.9 2.5 3.7 3.4 2.6 2.8 1.6 2.0 2.9 2.3 3.2 1.8 1.4 2.0 3.8 2.2 1.Summary calculationresultsare:Xi 100.0,Yi 50.0,Xi2 509.12,Yi2 134.84,XiYi 257.66.a) Obtain theleastsquares estimates ofo and1, andstatetheestimated regression
function.b) Obtain thepoint estimatefor mean Y when X scoreis 5.0c) What is thepointestimateof changein themean responsewhen theX score
increasesby one.
Solution.
a) b1 XiYi
Xi
Yi
n
Xi2 Xi2
n
XiXYiY
XiX2
257.66 1005020
509.12 1002
20
0. 83991
bo 1n Yi b1Xi Y b1X 120 50 0. 83991 100 1. 6996
Y 1.6996 0. 83991 Xb)
Y 1.6996 0. 83991 5 2. 5
c) 0. 83991 ( b1)QUESTION 21.For thefollowingset of data:
Xi 30 20 60 80 40 50 60 30 70 60
Yi 73 50 128 170 87 108 135 69 148 132
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1) Obtain theestimated regressionfunction. 2) Interpret bo and b1.Solution
1)
Xi Yi XiYi Xi2
30 73 30 73 302
20 50 20 50 202
60 128 60 128 602
80 170 80 170 802
40 87 40 87 402
50 108 50 108 502
60 135 60 135 602
30 69 30 69 692
70 148 70 148 702
60 132 60 132 602
500 1100 61800 32261 Totals
b1 XiYi
Xi Yin
Xi2 Xi2
n
XiXYiYXiX2
61800 5001100
10
32261 5002
10
68007261
0. 93651
bo 1n Yi b1Xi Y b1X 110 1100 0. 93651 500 63. 175
Y 63. 175 0. 93651 X2) bo - sincewedont know theif thescopeof themodel cover X 0 wecan not
giveanyinterpretation ofbo.
b1 0.93651 - mean of Y increaseby 0.93651 when X increaseby 1.QUESTION 22.Provethefollowingformula for SSE:SSE Yi2 boYi b1XiYiSolution:By thedefinition
SSE i1
n
Yi Yi2
i1
n
Yi bo b1Xi2 i1
n
ei2
Hence
SSE YiYi2 Yi2 2Yi
Yi
Yi
2
Yi2
Yi
Yi Yi
Yi
Yi2
Yi2 Yibo b1Xi bo b1XiYi
Yi
Yi2 boYi b1XiYi boYi Yi b1XiYi
Yi
Yi2 boYi b1XiYiWeused thefollowingpropertiesYi
Yi 0,XiYi
Yi Xiei 0
QUESTION 231) Statethenormal error model2)Find thedistributionofYi under normal error model3)Show that bo as defined in (3.19) is an unbiased estimator ofo.
Solution1) Thenormal error model is as follows:
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Yi o 1Xi iwhere:
Yi - is thevalueof theresponsein theithtrialo and1 areparametersXi is theknown constant, namely,thevalueof theindependent
variable in theithtriali areindependent N0,2 i 1,2,...,n.
2) SinceYiis alinear transformationof anormally distributedrandomvariable ithereforeYi has normal distribution.EYi Eo 1Xi i o 1Xi Ei o 1XiVarYi Varo 1Xi i Vari 2
Henceunder normal modelYi hasNo 1Xi,23) Wehavetoshowthat Ebo oFrom2) abovewehaveYi hasNo 1Xi,2whereXi, o,1 - constants
b1XiXYiY
XiX2
Eb1 EXiXYiYXiX2
XiXEYiYXiX2
XiX1XiXXiX2
1XiXXiXXiX2
1
We usedEYi Y EYi 1n EYj o 1Xi 1n o 1Xjo 1Xi o 1 1n X1 X2 ....Xn 1Xi XSincebo Y b1XsoEbo EY b1X E 1n Yi XEb1
1n EYi X1 1n o 1Xi 1X
o 11n Xi 1X o
QUESTION 24Provethat1) SSRX1,X2,X3 SSRX1 SSRX2,X3 X12) SSRX1 SSRX2 X1 SSRX2 SSRX1 X2Solution:1) Weknow that
SSRX2,X3 X1
SSEX1 SSEX1,X2,X3andSSTO SSEX1 SSRX1 SSEX1,X2,X3 SSRX1,X2,X3
HenceLHS SSRX1,X2,X3 SSTO SSEX1,X2,X3 SSEX1 SSRX2 SSEX1,X2,X3 SSRX1 SSEX1 SSEX1,X2,X3 SSRX1 SSRX2,X3 X1 RHSwhichcompletes theproof.2) Weknow that
SSRX2 X1 SSEX1 SSEX1,X2,
SSRX1 X2 SSEX2 SSEX1,X2and
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SSTO SSEX1 SSRX1 SSEX2 SSRX2HenceLHS SSRX1 SSRX2 X1 SSRX1 SSEX1 SSEX1,X2 SSTO SSEX1,X2 SSEX2 SSRX2 SSEX1,X2 SSRX2 SSRX1 X2 RHS
whichcompletes theproof.QUESTION 25For thefollowingset of data:
Xi 30 20 60 80 40 50 60 30 70 60
Yi 73 50 128 170 87 108 135 69 148 132
1) Obtain theestimated regressionfunction.2) Interpret bo and b1.3) Find the95% confidenceinterval for:o4)Testthe Ho : 1 0 versus Ha : 1 0 using tand 0.055) Find the90% confidenceinterval for1 andinterpret it.
Solution:
Xi Yi XiYi Xi2
Yi bo b1Xi ei Yi
Yi ei
2
30 73 2190 900 70 3 9
20 50 1000 400 50 0 0
60 128 7680 3600 130 -2 4
80 170 13600 6400 170 0 0
40 87 3480 1600 90 -3 9
50 108 5400 2500 110 -2 4
60 135 8100 3600 130 5 25
30 69 2070 900 70 -1 1
70 148 10360 4900 150 -2 4
60 132 7920 3600 130 2 4
500 1100 61800 28400 1100 0 60 TOTALS
ThereforeX 1n Xi 50010 50Y 1n Yi 110010 110
To calculatebo and b1 weusethefollowing formulas
b1 XiYi
Xi Yin
Xi2 Xi2
n
61800 5001100
10
28400 5002
10
2.0
bo Y b1X 110 2 50 10SoYi 10 2 Xi2) Sincewedo not knowwhether thescopeof themodel includesX 0 weareunabletoprovideany particular meaning for bo.b1 2 this can beinterpret as follows: thechangein themean of theprobability distributionof Y is equal to 2 per unit increasein X.
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3) SSE Yi Yi2 ei2 60
MSE SSEn2
608
7.5
Then
s2bo MSE Xi2
n
XiX2
7.5 28400103400
6. 2647
0necan usealso
s2bo MSE1n
X2
XiX2 7.5 1
10
502
3400 6. 2647
so sbo s2bo 6.2647 2. 5029
The95% confidenceinterval foro isbo t1 /2;n 2sbo,bo t1 /2;n 2sbo 10 2.306 2.5029,10 2.306 2.5029 4. 2283,15. 772where t1 /2;n 2 t0.975;8 2.3064)Test for 1Ho;1 0
Ha;1 0s2b1
MSE
XiX2
MSE
Xi2 Xi2
n
7.5
28400 5002
10
7.5
3400 0.002206
sb1 0.002206 0.046968Thetest statistics
t b1
sb1
20.046968
42. 582
Thedecision rulein our caseisIf |t | t1 /2;n 2, concludeHoIf|t | t1 /2;n 2, concludeHaIn our case|t | 42.582 2.306 t0.975;8 t1 /2;n 1
so weconcludeHa (1 0), that means that thereis alinear associationbetween X and Y .5) Confidenceinterval foroFirst wecalculate
s2bo MSE Xi2
nXiX2 7.5 28400
103400 6. 2647
0necan usealso
s2bo MSE1n
X2
XiX2 7.5 1
10
502
3400 6. 2647
so sbo s2bo 6.2647 2. 5029
The90% confidenceinterval foro isbo t1 /2;n 2sbo,bo t1 /2;n 2sbo 10 1.860 2.5029,10 1.860 2.5029 5.34,14.66where t1 /2;n 2 t0.95;8 1.860QUESTION 26.
Provethefollowingformula for SSR:SSR b1
2Xi X2Solution:By definition:
SSR Yi Y2
Yi
2 2
YiY Y
2
Yi
2
2n 1n i1
n YiY nY2
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Yi
2 2nY2 nY2
Yi
2 nY2
bo b1Xi2 nY2 bo2 2bob1Xi b12Xi2 nY2 nY b1X2 2Y b1Xb1nX b1
2Xi2 nY2 nY2 2nb1X Y nb1
2X2 2nb1X Y 2nb12X2 b1
2Xi2 nY2
b12Xi2 nb12X2 b12Xi2
Xi2
n b12Xi X2We usedbo Y b1X ,
Yi bo b1Xi
and Y 1n i1
n
Yi 1n
i1
n Yi
QUESTION 27.Show that bo as defined by bo Y b1X is an unbiased estimator ofo.SolutionUnder normal modelYi hasNo 1Xi,2whereXi, o,1 - constants
b1 XiXYiY
XiX2
Eb1 EXiXYiYXiX2
XiXEYiYXiX2
XiX1XiXXiX2
1XiXXiXXiX2
1
EYi Y EYi 1n EYj o 1Xi 1n o 1Xjo 1Xi o 1 1n X1 X2 ....Xn 1Xi Xbo Y b1XsoEb
o EY
b
1X E 1
n Y
i
XEb1
1n EYi X1 1n o 1Xi 1X
o 11n Xi 1X o
QUESTION 28.In atest of thealternatives Ho : 1 0 versus Ha : 1 0, astudent concludedHo. Does this conclusion imply that thereis no linear association between X and Y ?Solution.
Thenull (Ho) hypothesis consists twocases1 0 and1 0. Only thesecond partsupports statement that thereis nolinear association between X and Y .
Thereforetheresult of thetest does not imply that thereis no linear associationbetween X and Y .
QUESTION 29.Thefollowing datewereobtained in acertain study.
i 1 2 3 4 5 6 7 8 9 10 11 12
Xi 1 1 1 2 2 2 2 4 4 4 5 5
Yi 6.2 5.8 6 9.7 9.8 10.3 10.2 17.8 17.9 18.3 21.9 22.1
Summary calculational resultsare: Xi 33, Yi 156,Xi2 117Yi2 2448.5, XiYi 534.1) Fit alinear regression function2) Performan F test to determinewhether or not thereis lack of fit
of alinear regression function. Use
0.05.Solution
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1) Wehave
b1 XiYi
Xi Yin
Xi2 Xi2
n
534 33156
12
117 332
12
4
andbo Y b1X 1n Yi b1 1n Xi 112 156 4 112 33 2
ThereforeY 2 4 X2) F test for lack of fit.Wehavec 4 levels for X and 3 replicates for X 1 level,4 replicates for X 2 level, 3 replicates for X 4 leveland 2 replicates for X 5 level, and n 12.Hence
Yj 1nj
i1
nj
Yi,j - themean atj th level ofX.
Y 1 13 6.2 5.8 6 6.0 at level X 1Y 2
149.7 9.8 10.3 10.2 10.0 at level X 2
Y 3 1317.8 17.9 18.3 18.0 at level X 4
Y 4 1221.9 22.1 22.0 at level X 5
SSPE j1
c
i1
nj
Yi,j Yj2 6.2 62 5.8 62 6 62
9.7 102 9.8 102 10.3 102 10.2 102 17.8 182 17.9 182 18.3 182 21.9 222 22.1 222 0.5
MSPE SSPEnc 0.5
124 0.0625
SSE i1
n Yi Yi2 Yi2 boYi b1XiYi
2448.5 2 156 4 534 0. 5SSLF SSE SSPE 0.5 0.5 0MSLF SSLF
c2 0
42 0
Thehypothesis
Ho:EY o 1X
Ha:EY o 1X
Test statisticsF MSLF
MSPE
0
0.0625
0Thedecision rule
If F F1 ;c 2,n c conclude Ho
If F F1 ;c 2,n c conclude HaF1 ;c 2;n c F0.95;2;8 4.46SinceF 0 F1 ;c 2;n c 4.46 weconcludeHo.QUESTION 30.
Theresults of acertain experiments areshown below
i 1 2 3 4 5 6 7 8 9 10
Xi 5.5 4.8 4.7 3.9 4.5 6.2 6.0 5.2 4.7 4.3Yi 3.1 2.3 3.0 1.9 2.5 3.7 3.4 2.6 2.8 1.6
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i 11 12 13 14 15 16 17 18 19 20
Xi 4.9 5.4 5.0 6.3 4.6 4.3 5.0 5.9 4.1 4.7
Yi 2.0 2.9 2.3 3.2 1.8 1.4 2.0 3.8 2.2 1.5
Summary calculational resultsare: Xi 100, Yi 50,Xi2 509.12,Yi2 134.84, XiYi 257.66.1) Obtain theestimated regressionfunction.2) Set up theANOVA table3) Conduct theF test of Ho : 1 0 versusHa : 1 0 using 0.05Solutions1) Wehave
b1 XiYi
Xi Yin
Xi2 Xi2
n
257.66 10050
20
509.12 1002
20
0. 83991
and
bo Y b1X 1n Yi b1 1n Xi 120 50 0. 83991 120 100 1. 6996ThereforeYi 1.6996 0. 83991 Xi2) ANOVA table
sourceof variation SS df MS
regression SSR 1 MSR SSR1
error SSE n 2 MSE SSEn2
total SSTO n 1
Where
SSR Yi Y2 b1 XiYi
XiYin
0. 83991 257.66 1005020
6. 4337
SSE i1
n
Yi Yi2 Yi2 boYi b1XiYi
134.84 1. 6996 50 0. 83991 257.66 3. 4088MSE SSE
n2 3.4088
18 0.18938
SSTO Yi Y2 Yi2 Yi2
n 134.84502
20 9. 84
Hence
sourceof variation SS df MS
regression SSR 6. 4337 1 MSR 6.4337
error SSE 3. 4088 18 MSE 0.18938
total SSTO 9. 84 19
3) F test: HypothesisHo : 1 0Ha : 1 0
Test statistics:F MSR
MSE
6.43370.18938
33. 972
Thedecision rule:IfF F1 ;1,n 2, concludeHo
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IfF F1 ;1,n 2, concludeHaIn our caseF1 ;1;n 2 F0.95;1;18 4.41SinceF 33. 972 F1 ;1;n 2 4.41weconcludeHa (1 0), that means that thereis alinear association
between X and Y .QUESTION 31.
Thefollowing datewereobtained in thestudy of solution concentration.
i 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Xi 9 9 9 7 7 7 5 5 5 3 3 3 1 1 1
Yi 0.07 0.09 0.08 0.16 0.17 0.21 0.49 0.58 0.53 1.22 1.15 1.07 2.84 2.57 3.1
Summary calculational resultsare: Xi 75, Yi 14.33,Xi2 495Yi2 29.2117, XiYi 32.77.1) Fit alinear regression function
2) Performan F test to determinewhether or not thereis lack of fitof alinear regression function. Use 0.05.Solution.1) Wehave
b1 XiYi
Xi Yin
Xi2 Xi2
n
32.77 7514.33
15
495 752
15
0. 324
andbo Y b1X 1n Yi b1 1n Xi 115 14.33 0. 324
115
75 2. 5753
ThereforeYi 2. 5753
0. 324 Xi
2) F test for lack of fit.Wehavec 5 levels for X and 3 replicates for eachlevel (henceeachnj 3and n 15.Hence
Yj 1nj
i1
nj
Yi,j - themean atj th level ofX.
Y 1 133.1 2.57 2.84 2. 8367 at level X 1
Y 2 1
3Y1.07 1.15 1.22 1. 1467 at level X 3
Y 3 130.53 0.58 0.49 0. 53333at level X 5
Y 4 1
3
0.21 0.17 0.16 0.18 at level X 7
Y 4 13 0.08 0.09 0.07 0.08 at level X 9
SSPE j1
c
i1
nj
Yi,j Yj2 2.84 2.83672 2.57 2.83672
3.1 2.83672 1.22 1.14672 1.15 1.14672 1.07 1.14672 0.49 0.533332 0.58 0.53332 0.53 0.53332 0.16 0.182 0.17 0.182
0.21 0.182 0.07 0.082 0.09 0.082 0.08 0.082 0.1574MSPE SSPEnc
0.1574155 0.01574
SSE i1
n
Yi Yi2 Yi2 boYi b1XiYi
29.2117 2. 57530 14.33 0. 324 32.77 2. 9251SSLF SSE SSPE 2. 9251 0.1574 2. 7677
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MSLF SSLFc2
2.767752 0. 92257
Thehypothesis
Ho:EY o 1X
Ha:EY o 1X
Test statisticsF MSLF
MSPE
0.922570.01574
58. 613
Thedecision rule
If F F1 ;c 2,n c conclude Ho
If F F1 ;c 2,n c conclude HaF1 ;c 2;n c F0.95;3;10 3.71SinceF 58.613 F1 ;c 2;n c 3.71 weconcludeHa.QUESTION 31.A largediscount departmentstorechain advertises ontelevision(X1),on theradio (X2), and in newspapers (X3). A sample of 12 of its stores inacertain areashowed thefollowingadvertising expenditures andrevenuesduringagiven month. ( All figures arein thousands of rands) (Table 1.)1) Find theestimated regressioncoefficients.2)Testwhether thereis aregressionrelation using 0.01.3) Estimate1,2 and3 jointly by theBonferroni procedureusing99 percentfamily confidencecoefficient.4) Obtain an interval estimateofEYh when Xh,1 11, Xh,2 6 and Xh,3 2.Usea90 percent level of confidence.5) Obtain an ANOVA table and useit totest whether thereis aregressionrelation using 0.01.
6) Obtain theresiduals.7) Calculatethecoefficient of multipledeterminationR2.8) Obtain thesimultaneous interval estimates for two levels ofX :
i 1 2
X1 11 15
X2 7 9
X3 2 3
using 90 percent level of confidence.
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Table1
i Revenues (Yi) Xi,1 Xi,2 Xi,3
1 84 13 5 2
2 84 13 7 13 80 8 6 3
4 50 9 5 3
5 20 9 3 1
6 68 13 5 1
7 34 12 7 2
8 30 10 3 2
9 54 8 5 2
10 40 10 5 3
11 57 5 6 2
12 46 5 7 2
Thedatasummary is given below in matrix form
X X
12 115 64 24
115 1191 610 222
64 610 362 129
24 222 129 54
X Y
647
6393
3600
1292
Y Y 39973
X X1
5182112801
208912801
332612801
21664267
208912801
16412801
1612801
724267
332612801
1612801
62612801
8312801
21664267
724267
8312801
730638403
Solution
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1) X
1 13 5 2
1 13 7 1
1 8 6 3
1 9 5 3
1 9 3 1
1 13 5 1
1 12 7 2
1 10 3 2
1 8 5 2
1 10 5 3
1 5 6 2
1 5 7 2
Y
84
84
80
50
20
68
34
30
54
40
57
46
X X
1 13 5 2
1 13 7 1
1 8 6 3
1 9 5 3
1 9 3 1
1 13 5 1
1 12 7 2
1 10 3 2
1 8 5 2
1 10 5 3
1 5 6 2
1 5 7 2
1 13 5 2
1 13 7 1
1 8 6 3
1 9 5 3
1 9 3 1
1 13 5 1
1 12 7 2
1 10 3 2
1 8 5 2
1 10 5 3
1 5 6 2
1 5 7 2
12 115 64 24
115 1191 610 222
64 610 362 129
24 222 129 54
det
12 115 64 24
115 1191 610 222
64 610 362 12924 222 129 54
115209
X X1 1detA
Ai,j
12 115 64 24
115 1191 610 222
64 610 362 129
24 222 129 54
1
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5182112801
208912801
332612801
21664267
208912801
16412801
1612801
724267
332612801
1612801
62612801
8312801
21664267
724267
8312801
730638403
X Y
1 13 5 2
1 13 7 1
1 8 6 3
1 9 5 3
1 9 3 1
1 13 5 1
1 12 7 2
1 10 3 2
1 8 5 2
1 10 5 3
1 5 6 2
1 5 7 2
84
84
80
50
20
68
34
30
54
40
57
46
647
6393
3600
1292
b X X1X Y
5182112801
208912801
332612801
21664267
208912801
16412801
1612801
724267
332612801 1612801 62612801 8312801 2166
426772
4267 83
128017306
38403
647
6393
3600
1292
11518753
1973753
5690753
42942259
5) ANOVA
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Y Y
84
84
80
5020
68
34
30
54
40
57
46
84
84
80
5020
68
34
30
54
40
57
46
39973 Y 1
84
84
80
5020
68
34
30
54
40
57
46
1
1
1
11
1
1
1
1
1
1
1
647
bX Y
11518753
1973753
5690753
42942259
647
6393
3600
1292
82483577
2259 36513.
SSTO Y Y 1n Y11Y 39973 1
126472 61067
12 5088 11
12 5088. 9
SSE Y Y bX Y 39973 36513 3460.0SSR bX Y 1n Y
11Y 36513 112
6472 1954712
1628. 9
TheANOVA table
Sourceof variation SS df MS
Regression SSR 1628. 9 p 1 3 MSR SSRp1 542. 97
Error SSE 3460.0 n p 8 MSE SSEnp 432. 5
Total SSTO 5088. 9 n 1 11
Test of regression relationHypothesis: Ho : 1 2 3 0
Ha : not all k 0
Weusethetest statisticsF MSRMSE
542.97432.5
1. 2554
Thedecision ruleis
IfF F1 ,p 1,n p, concludeHo
IfF F1 ,p 1,n p, concludeHaAssumingthat 0.01 fromtable wegetF1 ,p 1,n p F0.99;3;8 7.59SinceF 1.2554 F1 ,p 1,n p F0.99;3;8 7.59weconcludeHo, Thereforethereis notalinear regressionrelation.
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3) s2b MSEX X1 432. 5
5182112801
208912801
332612801
21664267
208912801
16412801
1612801
724267
332612801
1612801
62612801
8312801
21664267
724267
8312801
730638403
1750. 8 70. 58 112. 37 219. 54
70. 58 5. 541 . 54058 7. 2979
112. 37 . 54058 21. 15 2. 8043
219. 54 7. 2979 2. 8043 82. 281
Hences2b1 5.541 sb1 5.541 2. 3539s2b2 21.15 sb2 21.15 4. 5989s2b3 82.281 sb3 82.281 9. 0709
Ifg parameters aretobeestimated jointly (whereg p), theconfidencelimits withfamily confidencecoefficient 1 are:
bk Bsbkwhere
B t1 /2g,n p
Nextfor g 3 fromthetableB t1 0.01/2 3;8 t0.9833;8 t0.99,8 2.8961 0.01/2 3 0. 99833 0.99So for 1 1973
753
2.896 2. 3539, 1973753
2.896 2. 3539 4. 1967,9. 4371for 2 5690
753 2.896 4. 5989, 5690
753 2.896 4. 5989 5. 762,20. 875
for 3 4294
2259 2.896 9. 0709, 4294
2259 2.896 9. 0709 24. 368,28. 17
4)
X h
1
11
6
2
Thepoint estimateof mean forY is
Yh Xh
b 1 11 6 2
11518753
1973753
5690753
42942259
141563
2259 62. 666
Theestimated varianceb is:s2
Yh X h
s2bX h
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1 11 6 2
1750. 8 70. 58 112. 37 219. 54
70. 58 5. 541 . 54058 7. 2979
112. 37 . 54058 21. 15 2. 8043
219. 54 7. 2979
2. 8043 82. 281
1
11
6
2
57.
586andsYh 57. 586 7. 5885
t1 /2;n p t1 0.01/2,8 t0.995;8 3.53Yh t1 /2;n ps
Yh
Hence62. 666 3.53 7. 5885,62. 666 3.53 7. 5885 35. 879,89. 4536) Fitted values
Y Xb
1 13 5 2
1 13 7 1
1 8 6 3
1 9 5 3
1 9 3 1
1 13 5 1
1 12 7 2
1 10 3 2
1 8 5 21 10 5 3
1 5 6 2
1 5 7 2
11518753
1973753
5690753
42942259
1363312259
1661772259
42700753
38983753
742212259
1320372259
1645522259
844342259
1067362259
13652251
1060492259
1231192259
60. 35
73. 562
56. 707
51. 77
32. 856
58. 449
72. 843
37. 377
47. 24954. 39
46. 945
54. 502
residuals
e Y Y
84
84
80
50
20
68
34
30
54
40
5746
60. 35
73. 562
56. 707
51. 77
32. 856
58. 449
72. 843
37. 377
47. 249
54. 39
46. 94554. 502
23. 65
10. 438
23. 293
1. 77
12. 856
9. 551
38. 843
7. 377
6. 751
14. 39
10. 0558. 502
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7) Coefficient of multipledeterminationR2 SSR
SSTO
1628.95088.9
0. 32009
8) simultaneous interval estimates for twolevelsaofX
A B
Xh,1 11 15Xh,2 7 9
Xh,3 2 3
In this caseg 2. To determinewhichsimultaneous prediction intervals arebesthere,we
shall find S and B assumingtheconfidencecoefficient0.90.S2 gF1 ;g;n p 2F0.99;2;8 2 8.65 17. 3
soS 17. 3 4. 1593
and
1 0.01/4 . 9975B t1 /2g;n p t0.9975;8 t0.995 3.250
Hence, theBonferroni limitsaremoreefficient here.(They giveshorter intervals)For explanatory variables level A wehave
X A
1
11
7
2
Thepoint estimateof mean forY is
YA X A
b 1 11 7 2
11518753
1973753
5690753
42942259
158633
2259 70. 223
ands2
YA X A
s2bX A
1 11 7 2
1750. 8 70. 58 112. 37 219. 54
70. 58 5. 541 . 54058 7. 2979
112. 37 . 54058 21. 15 2. 8043
219. 54 7. 2979 2. 8043 82. 281
1
11
7
2
108.
47and MSE 432. 5Hence
s2YAnew MSE s2YA 432. 5 108. 47 540. 97
andsYAnew 540. 97 23. 259
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X B
1
15
9
3
Thepoint estimateof mean forY is
YB X B
b 1 15 9 3
11518753
1973753
5690753
42942259
24527
251 97. 717
ands2
YB X B
s2bX B
1 15 9 3
1750. 8 70. 58 112. 37 219. 5470. 58 5. 541 . 54058 7. 2979
112. 37 . 54058 21. 15 2. 8043
219. 54 7. 2979 2. 8043 82. 281
115
9
3
645.
24Hence
s2YBnew MSE s2YB 432. 5 645. 24 1077. 7
andsYBnew 1077. 7 32. 828
Wefound beforethatB
3.250. Thesimultaneous Bonferroni predictionintervals with confidencecoefficient 0.90 are:Yh BsYhnew so70. 223 3.250 23. 259 YAnew 70. 223 3.250 23. 25997. 717 3.250 32. 828 YBnew 97. 717 3.250 32. 828or5. 3688 YAnew 145. 818. 974 YBnew 204. 41
QUESTION 32Assumethat thenormal regressionmodel is applicable.
For thefollowingdatagiven by:i 1 2 3 4 5
Xi 8 4 0 -4 -8
Yi 7.8 9 10.2 11 11.7
usingmatrix methodfind:1)Y Y2) X X3) X Y4) b5)Testthe Ho : 1 0 versus Ha : 1
0usingANOVA.
with 0.056) covariance-variancematrix s2b
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Solution
X
1 8
1 4
1 0
1 4
1 8
Y
7.8
9
10.2
11
11.7
1) Y Y 7.8 9 10.2 11 11.7
7.8
9
10.2
11
11.7
503. 77
2) X X 1 1 1 1 1
8 4 0 4 8
1 81 4
1 0
1 4
1 8
5 0
0 160
3) X Y 1 1 1 1 1
8 4 0 4 8
7.8
9
10.2
11
11.7
49. 7
39. 2
4) X X1 5 0
0 160
1
1
detAAij
1
800
160 0
0 5
15
0
0 1160
b
X
X
1
X
Y
15
0
0 1160
49. 7
39. 2
9. 94
0. 245
5) ANOVA tableSSTO Y Y 1n Y
11Y
7.8 9 10.2 11 11.7
7.8
9
10.2
11
11.7
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15 7.8 9 10.2 11 11.7
1
1
1
1
1
1 1 1 1 1
7.8
9
10.2
11
11.7
9. 752
SSE Y Y bX Y 503. 779. 94
0. 245
49. 7
39. 2 0.15
SSR SSTO SSE 9. 752 0.15 9. 602
(ANOVA table):
sourceof variation SS df MS
regression SSR 9.602 1 MSR SSR1 9.6021 9.602error SSE 0.15 3 MSE SSE
n2 0.15
3 0.05
total SSTO 9. 752 4
Ho : 1 0Ha : 1 0
F MSRMSE
9.6020.05
192. 04
Thedecision rule:IfF F1 ;1,n 2, concludeHo
IfF
F1 ;1,n 2, concludeHaF1 ;1,n 2 F0.95,1,3 10.13SinceF 192. 04 F1 ;1,n 2 F0.95,1,3 10.13weconcludeHa that means that thereis alinear associationbetween X and Y .
6) s2b MSEX X1 0.0515
0
0 1160
0.01 0
0 0.0003125
QUESTION 33.ProvethatSSE Y Y bX YSolution:
Weknow thatA A, A B A B, and AB BA
Alsothenormal equation:X Xb X Y
hence
X Xb X Y 0
0
where b bo
b1
so
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X Xb X Y bX X Y X 0
0
0 0
FromdefinitionSSE ee Y Xb Y Xb Y bX Y Xb
Y
Y Y
Xb b
X
Y b
X
Xb Y Y bX Y bX Xb Y Xb Y Y bX Y bX X Y Xb
Y Y bX Y 0 0bo
b1
Y Y bX Y 0 Y Y bX YQUESTION 34.Let
Yh Xh
bProvethat:
ThevarianceofYh, is in matrix notation:
2Yh 2X h X X1X h
Proof: Weknow thatLet W bearandomvector obtained by premultiplyingtherandomvector
Y by aconstant matrix A*) W AY
Then2W 2AY A2YA
SinceYh X h
b using*) with A X h weget
2Yh X h
2bX h
Usingthefact that2b 2X X1
weget2
Yh X h
2X X1X h 2X h X X1X h
QUESTION 35Theresults of acertain experiments areshown below
i 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 2
Xi 5.5 4.8 4.7 3.9 4.5 6.2 6.0 5.2 4.7 4.3 4.9 5.4 5.0 6.3 4.6 4.3 5.0 5.9 4.1 4.
Yi 3.1 2.3 3.0 1.9 2.5 3.7 3.4 2.6 2.8 1.6 2.0 2.9 2.3 3.2 1.8 1.4 2.0 3.8 2.2 1.
Summary calculationresultsare:Xi 100.0,Yi 50.0,Xi2 509.12,Yi2 134.84,XiYi 257.66.a) Obtain theleastsquares estimates ofo and1, andstatetheestimated regression
function.b) Obtain thepoint estimatefor mean Y when X scoreis 5.0c) What is thepointestimateof changein themean responsewhen theX score
increasesby one.
Solution.
a) b1 XiYi
Xi Yin
Xi2
Xi
2
n
XiXYiY
XiX2
257.66 10050
20
509.12 1002
20
0. 83991
bo 1n Yi b1Xi Y b1X 120 50 0. 83991 100 1. 6996
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Y 1.6996 0. 83991 X
b)Y 1.6996 0. 83991 5 2. 5
c) 0. 83991 ( b1)QUESTION 36.For thefollowingset of data:
Xi 30 20 60 80 40 50 60 30 70 60
Yi 73 50 128 170 87 108 135 69 148 132
1) Obtain theestimated regressionfunction. 2) Interpret bo and b1.Solution
1)
Xi Yi XiYi Xi2
30 73 30 73 302
20 50 20 50 202
60 128 60 128 602
80 170 80 170 802
40 87 40 87 402
50 108 50 108 502
60 135 60 135 602
30 69 30 69 692
70 148 70 148 702
60 132 60 132 602
500 1100 61800 32261 Totals
b1 XiYi
Xi Yin
Xi2 Xi2
n
XiXYiYXiX2
61800 5001100
10
32261 5002
10
68007261
0. 93651
bo 1n Yi b1Xi Y b1X 110 1100 0. 93651 500 63. 175
Y 63. 175 0. 93651 X2) bo - sincewedont know theif thescopeof themodel cover X 0 wecan not
giveanyinterpretation ofbo.b1 0.93651 - mean of Y increaseby 0.93651 when X increaseby 1.
QUESTION 37Theresults of acertain experiments areshown below
i 1 2 3 4 5 6 7 8 9 10
Xi 1 0 2 0 3 1 0 1 2 0
Yi 16 9 17 12 22 13 8 15 19 11
1) Obtain theestimated regressionfunction. 2) Plottheestimated regressionfunctionand thedata. 3) Interpret bo and b1. 4) Find the95% confidenceinterval for:o,1,and interpret them. 5)Testthe Ho : 1 0 versus Ha : 1 0 using t and ANOVA.using 0.05 6) Find 95% confidenceintervals for mean of responsevariablecorresponding tothelevel of theexplanatory equal to3.7)Find 90% predictionlimitsfor new observationof theresponsevariable
corresponding tothelevel of theexplanatory equal to3.8) Obtain theresiduals ei. 9) Estimate2 and. 10)Computeei2
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Solution1)
X1 Yi XiYi Xi2 Yi
2Yi ei ei
2
1 16 16 1 256 14.2 1. 8 3. 24
0 9 0 0 81 10.2 1.2 1.442 17 34 4 289 18.2 1.2 1.44
0 12 0 0 144 10.2 1.8 3.24
3 22 66 9 484 22.2 0.2 0.04
1 13 13 1 169 14.2 1.2 1.44
0 8 0 0 64 10.2 2.2 4.84
1 15 15 1 225 14.2 0. 8 0. 64
2 19 38 4 361 18.2 0. 8 0. 64
0 11 0 0 121 10.2 0.8 0.64
10 142 182 20 2194 0 17. 6 totals
b1 XiYi
Xi Yin
Xi2 Xi2
n
182 10142
10
20 102
10
4
bo 1n Yi b1Xi 110 142 4 10 10. 2
Y 10.2 4 X2)
Regression
X vs. Y (Casewise MD deletion)
Y =10.200 +4.0000 * X
Correlation: r =.94916
X
Y
6
8
10
12
14
16
18
20
22
24
-0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5
3) bo 10.2 - since0 is included in scopefor X (wehaveobservations)this represent theestimated mean valuefor Y at X 0.
b1 4 - themean of Y increaseby 4 when X increaseby 1.4)
SSE i
1
n
Yi Yi2
i
1
n
Yi bo b1Xi2 i
1
n
ei2
17.6
MSE SSEn2
17.6102 2. 2
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s2bo MSE Xi2
nXiX2 MSE 1n
X2
XiX2
MSE 1n X
2
Xi2 Xi2n
2.2 110
1
20 102
10
0. 44
sbo 0.44 0. 66332confidenceinterval for o isbo t1 /2;n 2sbo95% hence 0.5 and t1 /2;n 2 t0.975,8 2.306and 95% confidenceinterval foro is(10.2 2.306 0.66323,10.2 2.306 0.66323 8.6706,11.729
Interpretation- thereis 95% chancethat themean of Y corresponding tothevalue0 ofexplanatory variableX is in8.6706,11.729.
Confidenceinterval for1.
s2b1 MSE
XiX2
MSE
Xi2 Xi2
n
2.2
20 102
10
0. 22
sb1 0.22 0. 46904andb1 t1 /2;n 2sb1
wheret1 /2;n 2 t0.975,8 2.306
and 95% confidenceinterval for1 is4 2.306 0.46904,4 2.306 0.46904 2. 9184,5. 0816Interpretation- thereis 95% chancethat the1isin2. 9184,5. 0816.Sincethhis interval does notcontain 0 thereforethereis 95% chancethatther is alinear association between Y and X.5) Using t
Ho : 1 0Ha : 1 0
teststatisticst b1
sb1
40.46904
8. 5281
The decision rule(at thelevel of significance) isif|t | t1 /2;n 2, concludeHoif|t | t1 /2;n 2, concludeHa
wheret1 /2;n 2 t0.975,8 2.306Since |t | 8. 5281 t1 /2;n 2 t0.975,8 2.306 weconclude Ha (1 0), that means that thereis alinear associationbetween X and Y .ANOVA
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sourceof variation SS df MS
regression SSR Yi Y2 1 MSR SSR1
error SSE Yi Yi2 n 2 MSE SSEn2
total SSTO Yi Y2
n 1
SSR Yi Y2 b1 XiYi
XiYin 4 182
1014210
160.0
SSE i1
n
Yi Yi2
i1
n
Yi bo b1Xi2 i1
n
ei2
17.6
MSE SSEn2
17.6102 2. 2
SSTO Yi2 Yi2
n 21941422
10 177. 6
sourceof variation SS df MS
regression SSR 160 1 MSR 160error SSE 17.6 8 MSE 2.2
total SSTO 177.6 9
Ho : 1 0Ha : 1 0
Test statistics:F MSR
MSE
1602.2
72. 727
Thedecision rule:IfF F1 ;1,n 2, concludeHoIfF F1
;1,n
2, concludeH
awhereF1 ;1,n 2 F0.95,8 5.32
SinceF 72. 727 F1 ;1,n 2 F0.95,8 5.32weconcludeHa (1 0), that means that thereis alinear associationbetween X and Y .6) 5)A 1 confidenceinterval for EYh is given by:
Yh t1 /2;n 2sYh
whereYh bo b1Xh 10.2 4 3 22. 2,
s2Yh MSE
1n
XhX2
XiX2
2.2 110
3 10
102
20102
10
1. 1
and sYh 1.1. 1. 0488
t1 /2;n 2 t0.975,8 2.306Hencethe95% confidenceinterval for mean of theresponsevariablecorresponding tothelevel of explanatory variable equal to3 is22.2 2.306 1. 0488,22.2 2.306 1. 0488 19. 781,24. 6197)90% confidenceinterval for new observation of theresponsevariablecorrespondingtothelevel of explanatory variable equal to3 is given by
EYh z1 /2In casewhen theparameters areunknown1- predictionlimits are
Yh t1 /2;n 2sYhnew
where
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s2Yhnew s2Yh MSE MSE1 1n
XhX2
XiX2
2.2 1 110
3 10
32
20 102
10
2. 4444
sYhnew 2.4444 1. 5635
and t1 /2;n 2 t0.95,8 1.860Hencetheconfidenceinterval is22.2 1.860 1. 5635,22.2 1.860 1. 5635 19. 292,25.108QUESTION 38.Datafromastudyof computer-assisted learning by 12 students, showingthetotalnumber of responses in completingalesson (X) and thecost of computer time(Y, in cents), follow
i 1 2 3 4 5 6 7 8 9 10 11 12
Xi 16 14 22 10 14 17 10 13 19 12 18 11
Yi 77 70 85 50 62 70 52 63 88 57 81 54
Thescatter plot strongly suggest that theerror varianceincreasewith X.Fit theweighted lest squares regressionlineusingweightswi 1
Xi2 .
Solution.
Xi Yi wi 1/Xi2 wiXi wiYi wiXiYi wiXi
2
16 77 0.003906 0.0625 0.300781 4.8125 1
14 70 0.005102 0.071429 0.357143 5 1
22 85 0.002066 0.045455 0.17562 3.863636 1
10 50 0.01 0.1 0.5 5 1
14 62 0.005102 0.071429 0.316327 4.428571 1
17 70 0.00346 0.058824 0.242215 4.117647 1
10 52 0.01 0.1 0.52 5.2 1
13 63 0.005917 0.076923 0.372781 4.846154 1
19 88 0.00277 0.052632 0.243767 4.631579 1
12 57 0.006944 0.083333 0.395833 4.75 1
18 81 0.003086 0.055556 0.25 4.5 1
11 54 0.008264 0.090909 0.446281 4.909091 1
176 809 0.066619 0.868988 4.120748 56.05918 12 Totals
b1
wiXiYiwiXi wiYi wi
wiXi2 wiXi2
wi
56.05918 0.8689884.120748
0.066619
12 0.8689882
0.066619
3. 4711
bo wiYibiwiXi
wi
4.1207483.47110.8689880.066619
16. 578
HenceY 16. 578 3. 4711XQUESTION 39.Consider normal simple regressionmodel expressed in matrix terms.
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Provethat:1) SSTO Y Y 1n Y
11Y2) SSE Y Y bX YSolution1)We know that
SSTO Yi2 nY2 Yi2 Yi2
n
Wealso known thatY Y Yi2
Let 1
1
1
:
1
n 1
Using this wehave
1n Y11Y 1n Y1 Y2 .. . Yn
11
:
1
1 1 . .. 1
Y1
Y2
:
Yn
1n Y1 Y2 .. .YnY1 Y2 .. .Yn Yi 2
n
HenceSSTO Y Y 1n Y
11Y2) Weknow thatA A, A B A B, and AB BA
Alsothenormal equation:X Xb X Y
hence
X Xb X Y 0
0
where b bo
b1
so
X Xb X Y bX X Y X 00
0 0
FromdefinitionSSE ee Y Xb Y Xb Y bX Y Xb
Y Y Y Xb bX Y bX Xb Y Y bX Y bX Xb Y Xb Y Y bX Y bX X Y Xb
Y Y bX Y 0 0bo
b1
Y
Y b
X
Y 0 Y
Y b
X
YQUESTION 40.
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Theresults of acertain experiments areshown below
i 1 2 3 4 5 6 7 8 9 10
Xi 1 0 2 0 3 1 0 1 2 0
Yi 16 9 17 12 22 13 8 15 19 11
Summary calculational resultsare: Xi 10, Yi 142,Xi2 20Yi2 2194, XiYi 182.1) Obtain theestimated regressionfunction.2) Find the95% confidenceinterval for:13) Test the Ho : 1 0 versus Ha : 1 0 using tand 0.05Solution1) To calculatebo and b1 weusethefollowingformulas
b1 XiYi
Xi Yin
Xi2 Xi2
n
182 10142
10
20 102
10
4
bo Y b1X 1n Yi b1 1n Xi 110 142 4 110 10 10.2SoY 10.2 4 X2)In our case
SSE i1
n
Yi Yi2 Yi2 boYi b1XiYi
2194 10.2 142 4 182 17. 6MSE SSE
n2 17.6
8 2. 2
s2b1 MSE
XiX2
MSE
Xi2 Xi2
n
2.2
20 102
10
0. 22
sb1 0.22 0. 46904The95% confidenceinterval for 1 isb1 t1 /2;n 2sb1,b1 t1 /2;n 2sb1 4 2.306 0. 46904,4 2.306 0. 46904 2. 9184,5. 0816wheret1 /2;n 2 t0.975;8 2.306.3)Test for 1Ho;1 0Ha;1 0
Thetest statisticst
b1sb1
4
0.46904 8. 5281
Thedecision rulein our caseisIf |t | t1 /2;n 2, concludeHoIf|t | t1 /2;n 2, concludeHaIn our case|t | 8. 5281 2.306 t0.975;8 t1 /2;n 2so weconcludeHa (1 0), that means that thereis alinear associationbetween X and Y .QUESTION 41.
Thefollowing datawereobtained in acertain experiment:
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i Xi,1 Xi,2 Yi
1 1 2 2.5
2 1 2 3
3 1 2 3.5
4 2 1 3
5 2 1 4
6 0 1 1
7 0 1 1.5
8 0 1 2
9 1 0 1.5
10 1 0 2
11 1 0 2.5Thedatasummary is given below in matrix form
X X
11 10 11
10 14 10
11 10 17
X X1
2354
527
16
527
1154
0
16
0 16
X Y
26.5
29
29.5
Y Y 72.25
Assumethat first-order regressionmodel with independentnormal errorsis appropriate.1) Find theestimated regressioncoefficients.2) Obtain an ANOVA table and useit totest whether thereis aregression
relation using 0.05.3) Estimate1 and2 jointly by theBonferroni procedureusing
80percentfamily confidencecoefficient.4) Test thelack of fit for theEY o 1X1 2X2 using 0.05Solution
1) b
bo
b1
b2
X X1X Y
2354
527
16
527
1154
0
16
0 16
26.5
29
29.5
1.0
1. 0
0. 5
2) Y 1 1Y Yi 26.5 (weget it from X Y )SSTO Y Y 1n Y
11Y 72.25 1
11 26.5 26.5 8. 4091
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SSE Y Y bX Y 72.25 1 1 0.5
26.5
29
29.5
2.0
SSR bX Y 1n Y11Y
1 1 0.5
26.5
29
29.5
111
26.5 26.5 6. 4091
ANOVA table
sourceof variation SS df MS
regression SSR 6.4091 p 1 2 MSR SSRp1 3.2046
error SSE 2 n p 8 MSE SSEnp 28
0.25
total SSTO 8.4091 n 1 10Hypothesis:Ho : 1 2 0Ha : not both1 and2 areequal tozero
Test statisticsF MSR
MSE
3.20460.25
12. 818
Thedecision ruleIfF F1 ;p 1,n p, concludeHoIfF F1 ;p 1,n p, concludeHa
F1 ;p 1,n p F0.95,2,8 4.46SinceF 12.818 F0.95,2,8 4.46weconcludeHa (not both1 and2 areequal tozero), that means that thereis alinear associationbetween X and Y .3) Ifg parameters areto beestimated jointly (whereg p), theBonferroni confidencelimits withfamily confidencecoefficient 1 are:
bk BsbkB t1 /2g,n p
andweget s2b1, s2b2s2b MSEX X1
In our caseg 2, b1 1, b2 0.5
s2b MSEX X1 0.25
2354
527
16
527
1154
0
16
0 16
0. 10648 4. 6296 102 4. 1667 102
4. 6296 102 0.050926 0
4. 1667 102 0 0.041667
sos2b1 0.050926 and sb1 0.050926 0. 22567
s2b2 0.04166 and sb2 0.04166 0. 20411B t1 /2g,n p t1 0.2
22,9 t0.95,8 1.860
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Hencethelimits for b1 and b2 are0.58025,1.4197 and 0.12036,0.87964respectively, since1 1.860 0. 22567 0. 580251 1.860 0. 22567 1. 4197
0.5 1.860 0. 20411 0. 120360.5 1.860 0. 20411 0. 879644) First level
i Xi,1 Xi,2 Yj,1
1 1 2 2.5
2 1 2 3
3 1 2 3.5
n1 3, Y 1 3,
Squaredeviationat this level2.5 32 3 32 3.5 32 0. 5Secondlevel
i Xi,1 Xi,2 Yj,2
4 2 1 3
5 2 1 4
n2 2,Y 2 3.5
Squaredeviationat this level3 3.52 4 3.52 0. 5Third level
i Xi,1 Xi,2 Yj,3
6 0 1 1
7 0 1 1.5
8 0 1 2
n3 3,Y 3 1.5
Squaredeviationat this level1 1.52 1.5 1.52 2 1.52 0. 5Fourth level
i Xi,1 Xi,2 Yj,4
9 1 0 1.5
10 1 0 2
11 1 0 2.5
n4 3,Y 4 2
Squaredeviationat this level1.5 22 2 22 2.5 22 0. 5c 4,n 11,p 3SSPE Yj,i Y i2 0.5 0.5 0.5 0.5 2MSPE SSPEnc
2114 0. 286
SSE 2.0SSLF SSE SSPE 0MSLF SSLFcp 0
Test statisticsF MSLF
MSPE 0
Thehypothesis
Ho:EY o 1X1 2X2
Ha:EY o1X1
2X2
Thedecision rule
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If F F1 ;c p,n c conclude Ho
If F F1 ;c p,n c conclude HaF1 ;c p,n c F0.95,1,7 5.59SinceF F1 ;c p,n c weconcludeHo
Thereforetherein no lack of fit.QUESTION 42.For acertain experiment thefirst-order regressionmodel with twoindependentvariables was used. Thecalculated diagonal elementsof thehat matrix are:
i 1 2 3 4 5 6 7 8
hi,i 0.237 0.237 0.237 0.237 0.137 0.137 0.137 0.137
i 9 10 11 12 13 14 15 16
hi,i 0.137 0.137 0.137 0.137 0.237 0.237 0.237 0.237
1) Describeuseof hat matrix for identifyingoutlyingX observations.
2) Identify any outlyingX observations using thehat matrix method.Solution1) Thehat matrix H is given by:
H XX X1X
Thediagonal element hi,i in thehat matrix is called theleverageof the ith observation.Thus, alargeleveragevaluehi,i indicates that the ith observation is distantfromthecenter of theX observations. Themean leveragevalue
h hi,i
n pn
Hence, leveragevalues greater than2pn areconsidered by this ruletoindicateoutlying
observations with regard to theX values.
2) In our casen 16, p 3 so thecritical value2pn 616 0. 375
Sinceall leveragevalues in our casearelessthan 0.375 thereforethis methoddoes notidentified outlying observationsfor X.QUESTION 43.Consider thefollowing functions of therandomvariables
Y 1,Y 2 and Y 3 :W1 Y1 Y2 Y3W2 Y1 Y2W3 Y1 Y2 Y3
whereY1,Y2,Y3 areindependentidentically distributedrandomvariableswithN0,1 distribution.
1) Stateabovein matrix notation
2) Find theexpectationof therandomvector W
W1
W2
W3
3) Find thevariance-covariancematrix ofW.Solution
1) W
W1
W2
W3
1 1 1
1 1 0
1 1 1
Y1
Y2
Y3
AY
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2) EW
EW1
EW2
EW3
EY1 EY2 EY3
EY1 EY2
EY1 EY2 EY3
0
0
0
3) Let us notice
2Y
2Y1 Y1,Y2 Y1,Y3
Y2,Y1 2Y2 Y2,Y3
Y3,Y1 Y3,Y2 2Y3
1 0 0
0 1 0
0 0 1
SinceW AY then
2W 2AY A2YA
1 1 1
1 1 0
1 1 1
2Y
1 1 1
1 1 1
1 0 1
1 1 11 1 0
1 1 1
1 0 00 1 0
0 0 1
1 1 11 1 1
1 0 1
3 0 10 2 2
1 2 3
QUESTION 44.Consider themultiple regressionmodel:
Yi 1Xi,1 2Xi,2 iwherei areindependentnormally distributed randomerrorswithN0,2.Obtain themaximumlikelihood estimators for 1 and 2.Solution
Thedistribution of Yi is given byfyi,1Xi,1 2Xi,2,2 1
2 2exp yi1Xi,12Xi,2
2
22
Thelikelihood function
L i1
n1
2 2exp yi1Xi,12Xi,2
2
22
lnL nln 12 2
yi1Xi,12Xi,22
22
lnL1
yi1Xi,12Xi,2Xi,122
lnL2
yi1Xi,12Xi,2Xi,222
Hence lnL1 0 lnL2
0
gives us thefollowing equations
yi1Xi,12Xi,2Xi,122
0
yi1Xi,12Xi,2Xi,222
0
Hence1Xi,12 2Xi,1Xi,2 YiXi,11Xi,1Xi,2 2Xi,22 YiXi,2Solving weget
1 YiXi,12 Xi,1Xi,2 Xi,12
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YiXi,12 Xi,1Xi,2 Xi,12
Xi,1Xi,2 2Xi,22 YiXi,2
2 Xi,22 Xi,12 Xi,1Xi,2
2
Xi,12 YiXi,2
YiXi,1 Xi,1Xi,2 Xi,12
2 Xi,22
Xi,12
Xi,1Xi,22
Xi,12 YiXi,2 Xi,1
2
YiXi,1 Xi,1Xi,2 Xi,12Hence
2 YiXi,2 Xi,12 YiXi,1 Xi,1Xi,2 Xi,22 Xi,12 Xi,1Xi,2
2
and
1
YiXi,1YiXi,2 Xi,12 YiXi,1 Xi,1Xi,2 Xi,22 Xi,12 Xi,1Xi,2
2 Xi,1Xi,2
Xi,12
QUESTION 45.
1) Obtain thelikelihoodfunctionfor thesampleobservationsY1,...,Yngiven X1,...,Xn if thenormal model is assumed tobeapplicable.
2) Obtain themaximumlikelihood estimators foro and1.Solution1) Under normal modelYi hasNo 1Xi,2, withthecorrespondingdensity functiongiven by
fYiyi 1
2 2exp yio1Xi
2
22
Hencethelikelihoodfunctionfor thenormal error model, given the
sampleobservationsY1,...,Yn, is:Lo,1,2
i1
n1
22exp 1
22Yi o 1Xi2
2) In order to find theMLE weuse
lnLo,1,2 lni1
n1
22exp 1
22Yi o 1Xi2
n2
ln2 122
Yi o 1Xi2o
lnLo,1,2 122
2Yi o 1Xi 1
1
2Yi no 1Xi
and1 lnLo,1,
2
1
22 2Yi o 1Xi Xi
1
2XiYi oXi 1Xi2
Fromo
lnLo,1,2 01
lnLo,1,2 0
weget thefollowingequations
Yi no 1XiXiYi oXi 1Xi2
Fromthefirst oneweget
o Y 1XUsing it in thesecond weget
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XiYi Y 1XXi
1Xi
2
hence
1
XiYi Xi Yi
n
Xi
2 Xi2
n
XiXYiYXiX2
TheMLE foro and1 are
1
XiYi Xi Yi
n
Xi2 Xi2
n
ando Y
1X
thesameas estimatorsobtained usingleastsquares method.QUESTION 46.Datafromastudy of therelationbetween thesizeof abid in million rands (X) and thecost to thefirmof preparing thebid in thousands rands (Y) for 12 recent bidsare
presentedin tablebelow:
i 1 2 3 4 5 6 7 8 9 10 11 12
Xi 2.13 1.21 11.0 6.0 5.6 6.91 2.97 3.35 10.39 1.1 4.36 8.0
Yi 15.5 11.1 62.6 35.4 24.9 28.1 15.0 23.2 42.0 10 20 47.5
Thescatter plot strongly suggest that theerror varianceincreasewith X.Fit theweighted lest squares regressionlineusingweightswi 1
Xi2 .
Solution
Xi
Yi
wi
1/X
i
2 wiX
iw
iY
iw
iX
iY
iw
iX
i
2
2.13 15.5 0.220415 0.469484 3.41643 7.276995 1
1.21 11.1 0.683013 0.826446 7.587449 9.173554 1
11 62.6 0.008264 0.090909 0.517355 5.690909 1
6 35.4 0.027778 0.166667 0.983333 5.9 1
5.6 24.9 0.031888 0.178571 0.794005 4.446429 1
6.91 28.1 0.020943 0.144718 0.588505 4.06657 1
2.97 15 0.113367 0.3367 1.700507 5.050505 1
3.35 23.2 0.089107 0.298507 2.067276 6.925373 110.39 42 0.009263 0.096246 0.389061 4.042348 1
1.1 10 0.826446 0.909091 8.264463 9.090909 1
4.36 20 0.052605 0.229358 1.0521 4.587156 1
8 47.5 0.015625 0.125 0.742188 5.9375 1
63.02 335.3 2.098715 3.871698 28.09667 72.18825 12 Totals
bi
wiXiYiwiXi wiYi wi
wiXi2 wiXi2
wi
72.18825 3.87169828.09667
2.098715
123.8716982
2.098715
4. 1906
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bo wiYibiwiXi
wi
28.096674.19063.8716982.098715
5. 6568
HenceY 5.6568 4.1906XQUESTION 47.
Thefollowing datewereobtained in acertain study.
i 1 2 3 4 5 6 7 8 9 10 11 12
Xi 1 1 1 2 2 3 3 3 3 5 5 5
Yi 4.8 4.9 5.1 7.9 8.3 10.9 10.8 11.3 11.1 16.5 17.3 17.1
Summary calculational resultsare: Xi 34, Yi 126,Xi2 122Yi2 1554.66, XiYi 434.1) Fit alinear regression function2) Performan F test to determinewhether or not thereis lack of fit
of alinear regression function. Use 0.05.
Solution1) Wehave
b1 XiYi
Xi Yin
Xi2 Xi2
n
434 34126
12
122 342
12
3
andbo Y b1X 1n Yi b1 1n Xi 112 126 3
112
34 2
ThereforeY 2 3 X2) F test for lack of fit.
Wehavec 4 levels for X and 3 replicates for X 1 level,2 replicates for X 2 level, 4 replicates for X 3 leveland 3 replicates for X 5 level, and n 12.Hence
Yj 1nj
i1
nj
Yi,j - themean atj th level ofX.
Y 1 134.8 4.9 5.1 4. 9333 at level X 1
Y 2 127.9 8.3 8. 1 at level X 2
Y 3 1410.9 10.8 11.3 11.1 11. 025 at level X 3
Y 4 1316.5 17.3 17.1 16. 967 at level X 5
SSPE j1
c
i1
nj
Yi,j Yj2 4.8 4.93332 4.9 4.93332 5.1 4.93332
7.9 8.12 8.3 8.12 10.9 11.0252 10.8 11.0252 11.3 11.0252 11.1 11.0252 16.5 16.9672 17.3 16.9672 17.1 16.9672 0. 62083
MSPE SSPEnc 0.62083
124 0.077604
SSE i1
n
Yi Yi2 Yi2 boYi b1XiYi
1554.66 2 126 3 434 0. 66SSLF SSE SSPE 0.66 0.62083 0.03917MSLF SSLF
c2
0.03917
42 0.019585
Thehypothesis
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Ho:EY o 1X
Ha:EY o 1X
Test statisticsF MSLF
MSPE
0.0195850.077604
0. 25237
Thedecision ruleIf F F1 ;c 2,n c conclude Ho
If F F1 ;c 2,n c conclude HaF1 ;c 2;n c F0.95;2;8 4.46SinceF 0. 25237 F1 ;c 2;n c 4.46 weconcludeHo.
Thereis no lack of fit.QUESTION 48.1) Statethesimplenormal linear regressionmodel in matrix terms.2) Provethefollowingformula for SSE:SSE Y Y bX Y3) Provethat forYh Xh b thevarianceis in matrix notation2
Yh 2X h
X X1X hSolution1) Let
Y
Y1
Y2
:
Yn
X
1 X1
1 X2
: :
1 Xn
o
1
1
2
:
n
then
n1
Y n2
X21
n1
where: is thevector of parametersX - matrix of knownconstants,namely,thevalues of theindependentvariable
is avector of independent normal randomvariableswithE 0and2 2I.
2)We know thatSSE Yi2 boYi b1XiYiLet us noticethat
if Y
Y1Y2
:
Yn
then Y Y1 Y2 .. . Yn
and
if X
1 X1
1 X2
: :
1 Xn
then X 1 1 ... 1
X1 X2 .. . Xn.
Hence
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Y Y Y1 Y2 .. . Yn
Y1
Y2
:
Yn
Yi2 Yi2
and
X Y 1 1 ... 1
X1 X2 .. . Xn
Y1
Y2
:
Yn
YiXiYi
Using this with b bo
b1wehave
Y Y bX Y Yi2 bo b1 Yi
XiYi
Yi2 boYi b1XiYi SSEwhichcompletes theproof.or2)Weknow thatA A, A B A B, and AB BA
andthenormal equation:X Xb X Y
hence
X Xb X Y 0
0
where b bo
b1
so
X Xb X Y bX X Y X 0
0
0 0
FromdefinitionSSE ee Y Xb Y Xb Y bX Y Xb
Y Y Y Xb bX Y bX Xb Y Y bX Y bX Xb Y Xb Y Y bX Y bX X Y Xb
Y Y bX Y 0 0bo
b1
Y Y bX Y 0 Y Y bX Y
3)We know that:Let W bearandomvector obtained by premultiplyingtherandomvector
Y by aconstant matrix AW AY
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Then*) 2W 2AY A2YA
SinceYh X h
b using *) withA X h weget
2Yh X h
2bX hUsingthefact that
2b 2X X1weget2
Yh X h
2X X1X h 2X h X X1X h
QUESTION 49.Thefitted values and residuals of aregression analysis aregiven below
t 1 2 3 4 5 6 7 8 9 10Yt 21.96 4.15 7.36 22.11 10.98 22.06 47.35 47.05 73.40 69.79
et -1.45 -0.26 -0.16 -0.20 0.32 0.63 0.24 0.55 -0.50 -0.65
t 11 12 13 14 15 16 17 18 19 20
Yt 83.83 87.09 75.64 76.15 69.08 32.24 47.30 52.29 78.03 77.78
et 0.06 -0.09 -0.24 -1.03 0.02 0.56 0.80 0.11 0.57 0.72
Assumethat thesimplelinear regressionmodel with therandomterms followingafirst-order autoregressiveprocessis appropriate.Conduct aformal test for positiveautocorrelation using 0.05.Solution
Thehypothesis:
Ho : 0
Ha : 0
TheDurbin-Watson test statistics:
D
t2
n
etet12
t1
n
et2
6.50256.7072
0. 96948
t1
n
et2
1.452 0.262 0.162 0.202 0.322 0.632
0.242 0.552 0.502 0.652 0.062 0.092 0.242 1.032 0.022 0.562 0.802 0.112
0.572 0.722 6. 7072
t2
net et12 0.26 1.452 0.16 0.262 0.20 0.162
0.32 0.202 0.63 0.322 0.24 0.632 0.55 0.242 0.50 0.552 0.65 0.502 0.06 0.652 0.09 0.062 0.24 0.092 1.03 0.242 0.02 1.032 0.56 0.022 0.80 0.562 0.11 0.802 0.57 0.112 0.72 0.572 6. 5025
Thedecision rule
If D dU conclude Ho
IfD dL concludeHa
IfdL
D
dU thetest is inconclusivep 2 , dL 1.20 and dU 1.41
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SinceD 0.96948 dL 1.20 weconcludeHa, that theerror terms arepositively autocorrelated.QUESTION 50.
Thefollowing datawereobtained in acertain experiment:
i Xi,1 Xi,2 Yi
1 1 2 2.5
2 1 2 3
3 1 2 3.5
4 2 1 3
5 2 1 4
6 0 1 1
7 0 1 1.5
8 0 1 29 1 0 1.5
10 1 0 2
11 1 0 2.5
Thedatasummary is given below in matrix form
X X
11 10 11
10 14 10
11 10 17
X X1
2354
527
16
527
1154
0
16
0 16
X Y
26.5
29
29.5
Y Y 72.25
Assumethat first-order regressionmodel with independentnormal errorsis appropriate.1) Find theestimated regressioncoefficients.2) Obtain an ANOVA table and useit totest whether thereis aregression
relation using 0.05.3) Estimate1 and2 jointly by theBonferroni procedureusing
80percentfamily confidencecoefficient.
Solution
1) b
bo
b1
b2
X X1X Y
2354
527
16
527
1154
0
16
0 16
26.5
29
29.5
1.0
1. 0
0. 5
2) Y
1 1
Y Yi 26.5 (weget it from X
Y )
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