7/11/23,w1:42wA SolutionswManualwforwStatisticalwInferen
M ce
Solutionsw Manualw for
Statisticalw Inference,w Secondw Edition
GeorgewCasellaw Rogerw L.w Berger
Universityw ofw Florid Northw Carolinaw Statew UniversitywDam
a arisw Santana
Universityw ofw Florida
about:blank 1/196
,7/11/23,w1:42wA SolutionswManualwforwStatisticalwInferen
M ce
0-2 Solutions w Manualw for w StatisticalwInference
“WhenwIwhearwyou wgivewyourwreasons,”wIwremarked, w“thewthingwalwayswappearswtowm
ewtowbewsowridiculouslywsimplewthatwIwcouldweasilywdowitwmyself,wthoughwatweachwsuc
cessivewinstancewofwyourwreasoningw Iw amw baffledw untilw youw explainw yourw process.”
Dr.w Watsonw tow Sherlockw Holmes
Aw Scandalw inw Bohemia
0.1 Description
Thiswsolutionswmanualwcontainswsolutionswforwallwoddwnumberedwproblemswpluswawlargewnu
mberwofwsolutionswforwevenwnumberedwproblems.wOfwthew624wexerciseswinwStatisticalwInfere
nce,wSecondwEdition,wthiswmanualwgiveswsolutionswforw484w(78%)wofwthem.wTherewiswanwobt
usewpatternwaswtowwhichwsolutionswwerewincludedwinwthiswmanual.wWewassembledwallwofwth
ewsolutionswthatwwewhadwfromwthewfirstwedition,wandwfilledwinwsowthatwallwodd-
numberedwproblemswwerewdone.wInwthewpassagewfromwthewfirstwtowthewsecondw edition,w pro
blemsw werew shuffledw withw now attentionw paidw tow numberingw (hencew now attentionwpaidw to
w minimizew thew new w effort),w butw ratherw wew triedw tow putw thew problemsw inw logicalw order.
Awmajorwchangewfromwthewfirstweditionwiswthewusewofwthewcomputer,wbothwsymbolicallywt
hroughwMathematicatmwandwnumericallywusingwR.wSomewsolutionswarewgivenwaswcodewinweit
herwofwthesewlan-
wguages. wMathematicatmwcanwbewpurchasedwfrom wWolfram wResearch, wandwRwiswawfreewdow
nloadwfromwhttp://www.r-project.org/.
Herew isw aw detailedw listingw ofwthew solutionswincluded.
Chapterwww Numberw ofw Exerciseswww Numberw ofw Solutions Missing
1 55 51 26,w30,w36,w42
2 40 37 34,w38,w40
3 50 42 4,w6,w10,w20,w30,w32,w34,w36
4 65 52 8,w14,w22,w28,w36,w40
48,w50,w52,w56,w58,w60,w62
5 69 46 2,w4,w12,w14,w26,w28
allw evenw problems w fromw 36 w−w
68
6 43 35 8,w16,w26,w28,w34,w36,w38,w42
7 66 52 4,w14,w16,w28,w30,w32,w34,
36,w42,w54,w58,w60,w62,w64
8 58 51 36,w40,w46,w48,w52,w56,w58
9 58 41 2,w8,w10,w20,w22,w24,w26,w28,w30
32,w38,w40,w42,w44,w50,w54
10 48 26 ,w56
allw evenw problems w except w 4w an
dw 32
11 41 35 4,w20,w22,w24,w26,w40
12 31 16 allw even w problems
0.2 Acknowledgement
Manywpeoplewcontributedwtowthewassemblywofwthiswsolutionswmanual.wWewagainwthankw
allwofwthosewwhowcontributedwsolutionswtowthewfirstweditionw–
wmanywproblemswhavewcarriedwoverwintowthewsecondwedition.wMoreover,wthroughoutwthew
yearswawnumberwofwpeoplewhavewbeenwinwconstantwtouchwwithwus,wcontributingwtowbothwt
hewpresentationswandwsolutions.wWewapologizewinwadvancewforwthosewwewforgetwtowmention,
wandwwewespeciallywthankwJaywBeder, wYongwSungwJoo, wMichaelwPerlman, wRobwStrawderma
n,wandwTomw Wehrly.w Thankwyouw allw forwyourw help.
And,w aswwew saidwthewfirstwtimewaround,w althoughwwew havewbenefitedw greatlyw fromw thewassistanc
ewand
about:blank 2/196
,7/11/23,w1:42wA SolutionswManualwforwStatisticalwInferen
M ce
ACKNOWLEDGEMENT 0-3
commentswofwotherswinwthewassemblywofwthiswmanual, wwewarewresponsiblewforwitswultimatew
correctness.wTowthiswend,wwewhavewtriedwourwbestwbut,waswawwisewmanwoncewsaid, w“Youwp
ayswyourwmoneywandwyouwtakesw yourw chances.”
GeorgewCasella
wRogerwL.wBer
gerwDamariswSa
ntana
December, w 2001
about:blank 3/196
, 7/11/23,w1:42wA SolutionswManualwforwStatisticalwInferen
M ce
Chapterw 1
ProbabilitywTheory
“Ifw anyw littlew problemw comesw yourw way,w Iw shallw bew happy,w ifw Iw can,w tow givew youw aw
hintw orw twow aswtow itsw solution.”
SherlockwHolmes
Thew Adventurew ofw thew Threew Students
1.1 a.w Eachw samplew pointw describesw thew resultw ofw thew tossw (Hw orw T)w forw eachw ofw thew f
ourw tosses.w So,wforw examplew THTTw denotesw Tw onw 1st,w Hw onw 2nd,w Tw onw 3rdw and
w Tw onw 4th.w Therew arew 24w =w 16wsuchw samplew points.
b. Thew numberw ofw damagedw leavesw isw aw nonnegativew integer.w Sow wew mightw usew Sw =w{0,w1,w
2,w.w.w.}.
c. Wewmightwobservewfractionswofwanwhour.wSowwewmightwusewSw=w{tw:wtw≥w0},wthatw
is,wthewhalfwinfinitew intervalw [0,w∞).
d. Supposewwewweighwthewratswinwounces.wThewweightwmustwbewgreaterwthanwzerowsowwewmig
htwuse
Sw =w (0,w∞).w Ifw wew knoww now 10-day-
oldw ratw weighsw morew thanw 100w oz.,w wew couldw usew Sw =w (0,w100].
e. Ifw nw isw thew numberw ofw itemsw inw thew shipment,w thenw Sw =w {0/n,w1/n,w.w.w.w,w1}.
1.2 Forweachwofwthesewequalities,wyouwmustwshowwcontainmentwinwbothwdirections.
a. xw ∈wA\Bw ⇔wxw ∈wAw andw xw ∈ /w Bw ⇔wxw ∈wAw andw xw ∈ /w Aw∩wBw ⇔wxw ∈wA\(Aw∩wB).w
Also,w xw ∈wAw and
xw ∈
/w Bw ⇔wxw ∈wAw andw xw ∈wB cw ⇔wxw ∈wAw∩wB c .
b. Supposew xw ∈w B.w Thenw eitherw xw ∈w Aw orw xw ∈w Ac.w Ifw xw ∈w A,w thenw xw ∈w B
w ∩wA,w and,w hencewxw ∈w(Bw∩wA)w∪w(Bw∩wAc).w Thusw Bw ⊂w(Bw∩wA)w∪w(Bw∩wAc).
w Noww supposew xw ∈w(Bw∩wA)w∪w(Bw∩wAc).wThenw eitherw xw ∈w (Bw ∩wA)w orw xw ∈w
(Bw ∩wAc).w Ifw xw ∈w (Bw ∩wA),w thenw xw ∈w B.w Ifw xw ∈w (Bw ∩wAc),wthenw xw ∈wB.w T
husw (Bw∩wA)w∪w(Bw ∩wAc)w ⊂wB.w Sincew thew containmentw goesw bothw ways,w wew hav
ewBw=w(Bw∩wA)w∪w(Bw∩wAc).w(Note,wawmorewstraightforwardwargumentwforwthiswpa
rtwsimplywuseswthew Distributivew Laww tow statew thatw (Bw ∩wA)w∪w(Bw ∩wAc)w =w Bw ∩
w(Aw∪wAc)w =w Bw ∩wSw =w B.)
c. Similarw tow partw a).
d. Fromw partw b).
Aw∪wBw =w Aw ∪w[(Bw ∩wA)w∪w(Bw ∩wAc)]w =w Aw∪w(Bw ∩wA)w ∪wAw ∪w(Bw ∩wAc)w =w Aw ∪w[
Aw∪w(Bw ∩wAc)]w =
Aw∪w(Bw ∩wAc).
1.3 a.w xw ∈w Aw∪wBw ⇔w xw ∈w Aw orw xw ∈w Bw ⇔w xw ∈w Bw ∪wA
xw ∈w Aw ∩wBw ⇔w xw ∈w Aw andw xw ∈w Bw ⇔w xw ∈w Bw ∩wA.
b.w xw ∈w Aw ∪w(Bw ∪wC)w ⇔w xw ∈w Aw orw xw ∈w Bw ∪wCw ⇔w xw ∈w Aw ∪wBw orw xw ∈w C
w ⇔w xw ∈w (Aw ∪wB)w ∪wC.w(Itw canw similarlyw bew shownw thatw Aw∪w(Bw ∪wC)w =
w (Aw∪wC)w∪wB.)
xw ∈w Aw ∩w(Bw ∩wC)w ⇔w xw ∈w Aw andw xw ∈ w Bw andw xw ∈w Cw ⇔w xw ∈w (Aw ∩wB)w ∩wC.
c.ww xw ∈w(Aw∪wB)cw ⇔wwwxw ∈
/w Aw orw xw ∈
/w Bw ⇔wwxw ∈wAcwwandw xw ∈wB cw ⇔wwxw ∈wAcw ∩w
Bc
xw ∈w(Aw∩wB)cw ⇔wwxw ∈
/w Aw∩wBw ⇔wwxw ∈ /w Aw andw xw ∈
/w Bw ⇔wwxw ∈wAcwworw xw∈wB cw
⇔wwxw ∈wAcw ∪wB c .
1.4 a.w “Aw orw Bw orw both”w isw A∪B.w Fromw Theoremw 1.2.9bw wew havew Pw(A∪B)w =w Pw(A)+Pw(B)−P
w(A∩B).
about:blank 4/196
M ce
Solutionsw Manualw for
Statisticalw Inference,w Secondw Edition
GeorgewCasellaw Rogerw L.w Berger
Universityw ofw Florid Northw Carolinaw Statew UniversitywDam
a arisw Santana
Universityw ofw Florida
about:blank 1/196
,7/11/23,w1:42wA SolutionswManualwforwStatisticalwInferen
M ce
0-2 Solutions w Manualw for w StatisticalwInference
“WhenwIwhearwyou wgivewyourwreasons,”wIwremarked, w“thewthingwalwayswappearswtowm
ewtowbewsowridiculouslywsimplewthatwIwcouldweasilywdowitwmyself,wthoughwatweachwsuc
cessivewinstancewofwyourwreasoningw Iw amw baffledw untilw youw explainw yourw process.”
Dr.w Watsonw tow Sherlockw Holmes
Aw Scandalw inw Bohemia
0.1 Description
Thiswsolutionswmanualwcontainswsolutionswforwallwoddwnumberedwproblemswpluswawlargewnu
mberwofwsolutionswforwevenwnumberedwproblems.wOfwthew624wexerciseswinwStatisticalwInfere
nce,wSecondwEdition,wthiswmanualwgiveswsolutionswforw484w(78%)wofwthem.wTherewiswanwobt
usewpatternwaswtowwhichwsolutionswwerewincludedwinwthiswmanual.wWewassembledwallwofwth
ewsolutionswthatwwewhadwfromwthewfirstwedition,wandwfilledwinwsowthatwallwodd-
numberedwproblemswwerewdone.wInwthewpassagewfromwthewfirstwtowthewsecondw edition,w pro
blemsw werew shuffledw withw now attentionw paidw tow numberingw (hencew now attentionwpaidw to
w minimizew thew new w effort),w butw ratherw wew triedw tow putw thew problemsw inw logicalw order.
Awmajorwchangewfromwthewfirstweditionwiswthewusewofwthewcomputer,wbothwsymbolicallywt
hroughwMathematicatmwandwnumericallywusingwR.wSomewsolutionswarewgivenwaswcodewinweit
herwofwthesewlan-
wguages. wMathematicatmwcanwbewpurchasedwfrom wWolfram wResearch, wandwRwiswawfreewdow
nloadwfromwhttp://www.r-project.org/.
Herew isw aw detailedw listingw ofwthew solutionswincluded.
Chapterwww Numberw ofw Exerciseswww Numberw ofw Solutions Missing
1 55 51 26,w30,w36,w42
2 40 37 34,w38,w40
3 50 42 4,w6,w10,w20,w30,w32,w34,w36
4 65 52 8,w14,w22,w28,w36,w40
48,w50,w52,w56,w58,w60,w62
5 69 46 2,w4,w12,w14,w26,w28
allw evenw problems w fromw 36 w−w
68
6 43 35 8,w16,w26,w28,w34,w36,w38,w42
7 66 52 4,w14,w16,w28,w30,w32,w34,
36,w42,w54,w58,w60,w62,w64
8 58 51 36,w40,w46,w48,w52,w56,w58
9 58 41 2,w8,w10,w20,w22,w24,w26,w28,w30
32,w38,w40,w42,w44,w50,w54
10 48 26 ,w56
allw evenw problems w except w 4w an
dw 32
11 41 35 4,w20,w22,w24,w26,w40
12 31 16 allw even w problems
0.2 Acknowledgement
Manywpeoplewcontributedwtowthewassemblywofwthiswsolutionswmanual.wWewagainwthankw
allwofwthosewwhowcontributedwsolutionswtowthewfirstweditionw–
wmanywproblemswhavewcarriedwoverwintowthewsecondwedition.wMoreover,wthroughoutwthew
yearswawnumberwofwpeoplewhavewbeenwinwconstantwtouchwwithwus,wcontributingwtowbothwt
hewpresentationswandwsolutions.wWewapologizewinwadvancewforwthosewwewforgetwtowmention,
wandwwewespeciallywthankwJaywBeder, wYongwSungwJoo, wMichaelwPerlman, wRobwStrawderma
n,wandwTomw Wehrly.w Thankwyouw allw forwyourw help.
And,w aswwew saidwthewfirstwtimewaround,w althoughwwew havewbenefitedw greatlyw fromw thewassistanc
ewand
about:blank 2/196
,7/11/23,w1:42wA SolutionswManualwforwStatisticalwInferen
M ce
ACKNOWLEDGEMENT 0-3
commentswofwotherswinwthewassemblywofwthiswmanual, wwewarewresponsiblewforwitswultimatew
correctness.wTowthiswend,wwewhavewtriedwourwbestwbut,waswawwisewmanwoncewsaid, w“Youwp
ayswyourwmoneywandwyouwtakesw yourw chances.”
GeorgewCasella
wRogerwL.wBer
gerwDamariswSa
ntana
December, w 2001
about:blank 3/196
, 7/11/23,w1:42wA SolutionswManualwforwStatisticalwInferen
M ce
Chapterw 1
ProbabilitywTheory
“Ifw anyw littlew problemw comesw yourw way,w Iw shallw bew happy,w ifw Iw can,w tow givew youw aw
hintw orw twow aswtow itsw solution.”
SherlockwHolmes
Thew Adventurew ofw thew Threew Students
1.1 a.w Eachw samplew pointw describesw thew resultw ofw thew tossw (Hw orw T)w forw eachw ofw thew f
ourw tosses.w So,wforw examplew THTTw denotesw Tw onw 1st,w Hw onw 2nd,w Tw onw 3rdw and
w Tw onw 4th.w Therew arew 24w =w 16wsuchw samplew points.
b. Thew numberw ofw damagedw leavesw isw aw nonnegativew integer.w Sow wew mightw usew Sw =w{0,w1,w
2,w.w.w.}.
c. Wewmightwobservewfractionswofwanwhour.wSowwewmightwusewSw=w{tw:wtw≥w0},wthatw
is,wthewhalfwinfinitew intervalw [0,w∞).
d. Supposewwewweighwthewratswinwounces.wThewweightwmustwbewgreaterwthanwzerowsowwewmig
htwuse
Sw =w (0,w∞).w Ifw wew knoww now 10-day-
oldw ratw weighsw morew thanw 100w oz.,w wew couldw usew Sw =w (0,w100].
e. Ifw nw isw thew numberw ofw itemsw inw thew shipment,w thenw Sw =w {0/n,w1/n,w.w.w.w,w1}.
1.2 Forweachwofwthesewequalities,wyouwmustwshowwcontainmentwinwbothwdirections.
a. xw ∈wA\Bw ⇔wxw ∈wAw andw xw ∈ /w Bw ⇔wxw ∈wAw andw xw ∈ /w Aw∩wBw ⇔wxw ∈wA\(Aw∩wB).w
Also,w xw ∈wAw and
xw ∈
/w Bw ⇔wxw ∈wAw andw xw ∈wB cw ⇔wxw ∈wAw∩wB c .
b. Supposew xw ∈w B.w Thenw eitherw xw ∈w Aw orw xw ∈w Ac.w Ifw xw ∈w A,w thenw xw ∈w B
w ∩wA,w and,w hencewxw ∈w(Bw∩wA)w∪w(Bw∩wAc).w Thusw Bw ⊂w(Bw∩wA)w∪w(Bw∩wAc).
w Noww supposew xw ∈w(Bw∩wA)w∪w(Bw∩wAc).wThenw eitherw xw ∈w (Bw ∩wA)w orw xw ∈w
(Bw ∩wAc).w Ifw xw ∈w (Bw ∩wA),w thenw xw ∈w B.w Ifw xw ∈w (Bw ∩wAc),wthenw xw ∈wB.w T
husw (Bw∩wA)w∪w(Bw ∩wAc)w ⊂wB.w Sincew thew containmentw goesw bothw ways,w wew hav
ewBw=w(Bw∩wA)w∪w(Bw∩wAc).w(Note,wawmorewstraightforwardwargumentwforwthiswpa
rtwsimplywuseswthew Distributivew Laww tow statew thatw (Bw ∩wA)w∪w(Bw ∩wAc)w =w Bw ∩
w(Aw∪wAc)w =w Bw ∩wSw =w B.)
c. Similarw tow partw a).
d. Fromw partw b).
Aw∪wBw =w Aw ∪w[(Bw ∩wA)w∪w(Bw ∩wAc)]w =w Aw∪w(Bw ∩wA)w ∪wAw ∪w(Bw ∩wAc)w =w Aw ∪w[
Aw∪w(Bw ∩wAc)]w =
Aw∪w(Bw ∩wAc).
1.3 a.w xw ∈w Aw∪wBw ⇔w xw ∈w Aw orw xw ∈w Bw ⇔w xw ∈w Bw ∪wA
xw ∈w Aw ∩wBw ⇔w xw ∈w Aw andw xw ∈w Bw ⇔w xw ∈w Bw ∩wA.
b.w xw ∈w Aw ∪w(Bw ∪wC)w ⇔w xw ∈w Aw orw xw ∈w Bw ∪wCw ⇔w xw ∈w Aw ∪wBw orw xw ∈w C
w ⇔w xw ∈w (Aw ∪wB)w ∪wC.w(Itw canw similarlyw bew shownw thatw Aw∪w(Bw ∪wC)w =
w (Aw∪wC)w∪wB.)
xw ∈w Aw ∩w(Bw ∩wC)w ⇔w xw ∈w Aw andw xw ∈ w Bw andw xw ∈w Cw ⇔w xw ∈w (Aw ∩wB)w ∩wC.
c.ww xw ∈w(Aw∪wB)cw ⇔wwwxw ∈
/w Aw orw xw ∈
/w Bw ⇔wwxw ∈wAcwwandw xw ∈wB cw ⇔wwxw ∈wAcw ∩w
Bc
xw ∈w(Aw∩wB)cw ⇔wwxw ∈
/w Aw∩wBw ⇔wwxw ∈ /w Aw andw xw ∈
/w Bw ⇔wwxw ∈wAcwworw xw∈wB cw
⇔wwxw ∈wAcw ∪wB c .
1.4 a.w “Aw orw Bw orw both”w isw A∪B.w Fromw Theoremw 1.2.9bw wew havew Pw(A∪B)w =w Pw(A)+Pw(B)−P
w(A∩B).
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