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Solution Manual for An Introduction to Mathematical Statistics and Its Applications, 6th Edition

Solution Manual for An Introduction to Mathematical Statistics and Its Applications, 6th Edition gives you all the tools you need to solve your textbook problems effectively.

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SOLUTIONSMANUALANINTRODUCTION TOMATHEMATICALSTATISTICSANDITSAPPLICATIONSSIXTHEDITIONRichard LarsenVanderbilt UniversityMorris MarxUniversity of West Florida

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iiiContentsChapter 2:Probability ..............................................................................................................12.2Samples Spaces and the Algebra of Sets.............................................................................12.3The Probability Function.....................................................................................................62.4Conditional Probability .......................................................................................................72.5Independence.....................................................................................................................132.6Combinatorics ...................................................................................................................172.7Combinatorial Probability .................................................................................................24Chapter 3:Random Variables ................................................................................................273.2Binomial and Hypergeometric Probabilities .....................................................................273.3Discrete Random Variables...............................................................................................363.4Continuous Random Variables..........................................................................................413.5Expected Values ................................................................................................................443.6The Variance .....................................................................................................................523.7Joint Densities ...................................................................................................................573.8Transforming and Combining Random Variables.............................................................693.9Further Properties of the Mean and Variance....................................................................733.10Order Statistics ..................................................................................................................793.11Conditional Densities ........................................................................................................823.12Moment-Generating Functions..........................................................................................88Chapter 4:Special Distributions ............................................................................................934.2The Poisson Distribution ...................................................................................................934.3The Normal Distribution ...................................................................................................984.4The Geometric Distribution.............................................................................................1054.5The Negative Binomial Distribution ...............................................................................1074.6The Gamma Distribution.................................................................................................109Chapter 5:Estimation...........................................................................................................1115.2Estimating Parameters: The Method of Maximum Likelihoodand Method of Moments .................................................................................................1115.3Interval Estimation ..........................................................................................................1185.4Properties of Estimators ..................................................................................................1235.5Minimum-Variance Estimators: The Cramér-Rao Lower Bound ...................................1285.6Sufficient Estimators .......................................................................................................1305.7Consistency .....................................................................................................................1335.8Bayesian Estimation ........................................................................................................135

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ivChapter 6:Hypothesis Testing .............................................................................................1376.2The Decision Rule ...........................................................................................................1376.3Testing Binomial Data................................................................................................................................................................1386.4Type I and Type II Errors................................................................................................1396.5A Notion of Optimality: The Generalized Likelihood Ratio...........................................144Chapter 7:Inferences Based on the Normal Distribution ....................................................1477.3Deriving the Distribution of/YSn....................................................................................1477.4Drawing Inferences About.........................................................................................1507.5Drawing Inferences About2........................................................................................156Chapter 8:Types of Data: A Brief Overview ......................................................................1618.2Classifying Data ..............................................................................................................161Chapter 9:Two-Sample Inference .......................................................................................1639.2Testing0:XYH.......................................................................................................1639.3Testing220:XYH—TheFTest .................................................................................1669.4Binomial Data: Testing0:XYHpp.............................................................................1689.5Confidence Intervals for the Two-Sample Problem ........................................................170Chapter 10:Goodness-of-Fit Tests ......................................................................................17310.2The Multinomial Distribution .........................................................................................17310.3Goodness-of-Fit Tests: All Parameters Known...............................................................17510.4Goodness-of-Fit Tests: Parameters Unknown.................................................................17810.5Contingency Tables.........................................................................................................185Chapter 11:Regression ........................................................................................................18911.2The Method of Least Squares..........................................................................................18911.3The Linear Model............................................................................................................19911.4Covariance and Correlation.............................................................................................20411.5The Bivariate Normal Distribution..................................................................................208Chapter 12:The Analysis of Variance .................................................................................21112.2TheFtest.........................................................................................................................21112.3Multiple Comparisons: Tukey’s Method.........................................................................214

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v12.4Testing Subhypotheses with Constrasts ..........................................................................21612.5Data Transformations ......................................................................................................218Appendix 12.A.2The Distribution of/(1)/()SSTRkSSEnkWhen1HIs True ............................................218Chapter 13:Randomized Block Designs .............................................................................22113.2TheFTest for a Randomized Block Design ...................................................................22113.3The PairedtTest..............................................................................................................224Chapter 14:Nonparametric Statistics...................................................................................22914.2The Sign Test ..................................................................................................................22914.3Wilcoxon Tests................................................................................................................23214.4The Kruskal-Wallis Test .................................................................................................23614.5The Friedman Test...........................................................................................................23914.6Testing for Randomness..................................................................................................241

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1Chapter 2: ProbabilitySection 2.2: Sample Spaces and the Algebra of Sets2.2.1S=( , , ), ( , ,), ( ,, ), (, , ), ( ,,), (, ,), (,, ), (,,)ss ss s fs fsfs ss fff s fff sfffA=( ,, ), (, , )sfsf s s;B=(,,)fff2.2.2Let (x, y, z) denote a redx, a bluey, and a greenz.ThenA=(2, 2,1), (2,1, 2), (1, 2, 2), (1,1,3), (1,3,1), (3,1,1)2.2.3(1,3,4), (1,3,5), (1,3,6), (2,3,4), (2,3,5), (2,3,6)2.2.4There are 16 ways to get an ace and a 7, 16 ways to get a 2 and a 6, 16 ways to get a 3 and a5, and 6 ways to get two 4’s, giving 54 total.2.2.5The outcome sought is (4, 4). It is “harder” to obtain than the set {(5, 3), (3, 5), (6, 2), (2, 6)}of other outcomes making a total of 8.2.2.6The setNof five card hands in hearts that are not flushes are calledstraight flushes. These arefive cards whose denominations are consecutive. Each one is characterized by the lowestvalue in the hand. The choices for the lowest value are A, 2, 3, …, 10. (Notice that an ace canbe high or low). Thus,Nhas 10 elements.2.2.7P= {right triangles with sides (5,a,b):a2+b2= 25}2.2.8A= {SSBBBB,SBSBBB,SBBSBB,SBBBSB,BSSBBB,BSBSBB,BSBBSB,BBSSBB,BBSBSB,BBBSSB}2.2.9(a)S= {(0, 0, 0, 0) (0, 0, 0, 1), (0, 0, 1, 0), (0, 0, 1, 1), (0, 1, 0, 0), (0, 1, 0, 1), (0, 1, 1, 0),(0, 1, 1, 1), (1, 0, 0, 0), (1, 0, 0, 1), (1, 0, 1, 0), (1, 0, 1, 1, ), (1, 1, 0, 0), (1, 1, 0, 1),(1, 1, 1, 0), (1, 1, 1, 1, )}(b)A= {(0, 0, 1, 1), (0, 1, 0, 1), (0, 1, 1, 0), (1, 0, 0, 1), (1, 0, 1, 0), (1, 1, 0, 0, )}(c) 1 +k2.2.10(a)S= {(1, 1), (1, 2), (1, 4), (2, 1), (2, 2), (2, 4), (4, 1), (4, 2), (4, 4)}(b) {2, 3, 4, 5, 6, 8}2.2.11Letp1andp2denote the two perpetrators andi1,i2, andi3, the three in the lineup who areinnocent.ThenS=11121321222312121323(,), (,), (,), (,), (,), (,), (,), ( ,), ( ,), (,)pipipipipipippi ii iii.The eventAcontains every outcome inSexcept (p1,p2).2.2.12The quadratic equation will have complex roots—that is, the eventAwill occur—ifb24ac< 0.2.2.13In order for the shooter to win with a point of 9, one of the following (countably infinite)sequences of sums must be rolled: (9,9), (9, no 7 or no 9,9), (9, no 7 or no 9, no 7 or no 9,9),…

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2Chapter 2: Probability2.2.14Let (x,y) denote the strategy of puttingxwhite chips andyred chips in the first urn (whichresults in 10xwhite chips and 10yred chips being in the second urn). ThenS=( ,) :0,1,...,10,0,1,...,10, and 119x yxyxy. Intuitively, the optimal strategiesare (1, 0) and (9, 10).2.2.15LetAkbe the set of chips put in the urn at 1/2kminute until midnight.For example,A1= {11, 12, 13, 14, 15, 16, 17, 18, 19, 20}. Then the set of chips in the urn at midnight is1({1})kkAk .2.2.16move arrow on first figure raise B by 12.2.17Ifx2+ 2x8, then (x+ 4)(x2)0 andA= {x:4x2}. Similarly, ifx2+x6, then(x+ 3)(x2)0 andB= {x:3x2). Therefore,AB= {x:3x2} andAB= {x:4x2}.2.2.18ABC= {x:x= 2, 3, 4}2.2.19The system fails if either the first pair fails or the second pair fails (or both pairs fail). Foreither pair to fail, though, both of its components must fail. Therefore,A= (A11A21)(A12A22).2.2.20(a)(b)(c) empty set(d)2.2.21402.2.22(a) {E1,E2}(b) {S1,S2,T1,T2}(c) {A,I}2.2.23(a) Ifsis a member ofA(BC) thensbelongs toAor toBC. If it is a member ofAorofBC, then it belongs toABand toAC.Thus, it is a member of (AB)(AC).Conversely, choosesin (AB)(AC). If it belongs toA, then it belongs toA(BC). If it does not belong toA, then it must be a member ofBC.In that case it also is a member ofA(BC).

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Section 2.2: Sample Spaces and the Algebra of Sets3(b) Ifsis a member ofA(BC) thensbelongs toAand toBC. If it is a member ofB,then it belongs toABand, hence, (AB)(AC). Similarly, if it belongs toC, it isa member of (AB)(AC). Conversely, choosesin (AB)(AC). Then itbelongs toA. If it is a member ofABthen it belongs toA(BC). Similarly, if itbelongs toAC, then it must be a member ofA(BC).2.2.24LetB=A1A2…Ak. Then12...CCCkAAA= (A1A2…Ak)C=BC. Then theexpression is simplyBBC=S.2.2.25(a) Letsbe a member ofA(BC). Thensbelongs to eitherAorBC(or both). Ifsbelongs toA, it necessarily belongs to (AB)C. Ifsbelongs toBC, it belongs toBorCor both, so it must belong to (AB)C. Now, supposesbelongs to(AB)C. Then it belongs to eitherABorCor both. If it belongs toC, it mustbelong toA(BC). If it belongs toAB, it must belong to eitherAorBor both, soit must belong toA(BC).(b) Supposesbelongs toA(BC), so it is a member ofAand alsoBC. Then it isamember ofAand ofBandC. That makes it a member of (AB)C. Conversely, ifsis a member of (AB)C, a similar argument shows it belongs toA(BC).2.2.26(a)ACBCCC(b)ABC(c)ABCCC(d) (ABCCC)(ACBCC)(ACBCC)(e) (ABCC)(ABCC)(ACBC)2.2.27Ais a subset ofB.2.2.28(a) {0}{x: 5x10}(b) {x: 3x< 5}(c) {x: 0 <x7}(d) {x: 0 <x< 3}(e) {x: 3x10}(f){x: 7 <x10}2.2.29(a)BandC(b)Bis a subset ofA.2.2.30(a)A1A2A3(b)A1A2A3The second protocol would be better if speed of approval matters. For very important issues,the first protocol is superior.2.2.31LetAandBdenote the students who saw the movie the first time and the second time,respectively. ThenN(A) = 850,N(B) = 690, and[() ]CNAB= 4700(implying thatN(AB) = 1300). Therefore,N(AB) = number who saw movie twice= 850 + 6901300 = 240.

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4Chapter 2: Probability2.2.32(a)(b)2.2.33(a)(b)2.2.34(a)A(BC)(AB)C(b)A(BC)(AB)C2.2.35AandBare subsets ofAB.

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Section 2.2: Sample Spaces and the Algebra of Sets52.2.36(a)()CCAB=ACB(b)()CCBABAB(c)()CCAABAB2.2.37LetAbe the set of those with MCAT scores27 andBbe the set of those with GPAs3.5.We are given thatN(A) = 1000,N(B) = 400, andN(AB) = 300.Then()CCN AB=[() ]CNAB= 1200N(AB) = 1200[(N(A) +N(B)N(AB)]= 1200[(1000 + 400300] = 100. The requested proportion is 100/1200.2.2.38N(ABC) =N(A) +N(B) +N(C)N(AB)N(AC)N(BC) +N(ABC)2.2.39LetAbe the set of those saying “yes” to the first question andBbe the set of those saying“yes” to the second question. We are given thatN(A) = 600,N(B) = 400, andN(ACB) = 300. ThenN(AB) =N(B)()CN AB= 400300 = 100.()CN AB=N(A)N(AB) = 600100 = 500.2.2.40[() ]CNAB= 120N(AB) = 120[N(CAB) +N(ACB) +N(AB)]= 120[50 + 15 + 2] = 53

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6Chapter 2: ProbabilitySection 2.3: The Probability Function2.3.1LetLandVdenote the sets of programs with offensive language and too much violence,respectively. ThenP(L) = 0.42,P(V) = 0.27, andP(LV) = 0.10.Therefore,P(program complies) =P((LV)C) = 1[P(L) +P(V)P(LV)] = 0.41.2.3.2P(AorBbut not both) =P(AB)P(AB) =P(A) +P(B)P(AB)P(AB)= 0.4 + 0.50.10.1 = 0.72.3.3(a) 1P(AB)(b)P(B)P(AB)2.3.4P(AB) = P(A) + P(B)P(AB) = 0.3; P(A)P(AB) = 0.1. Therefore, P(B) = 0.2.2.3.5No. P(A1A2A3) = P(at least one “6” appears) = 1P(no 6’s appear) =351162.The Ai’s are not mutually exclusive, soP(A1A2A3)P(A1) +P(A2) +P(A3).2.3.6P(AorBbut not both) = 0.5 – 0.2 = 0.32.3.7By inspection,B= (BA1)(BA2)…(BAn).2.3.8(a)(b)(b)

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Section 2.4: Conditional Probability72.3.9P(odd man out) = 1P(no odd man out) = 1P(HHHorTTT) = 123842.3.10A= {2, 4, 6, …, 24};B= {3, 6, 9, …, 24);AB= {6, 12, 18, 24}.Therefore,P(AB) =P(A) +P(B)P(AB) =12841624242424.2.3.11LetA: State wins Saturday andB: State wins next Saturday. ThenP(A) = 0.10,P(B) = 0.30,andP(lose both) = 0.65 = 1P(AB), which implies thatP(AB) = 0.35. Therefore,P(AB) = 0.10 + 0.300.35 = 0.05, soP(State wins exactly once) =P(AB)P(AB)= 0.350.05 = 0.30.2.3.12SinceA1andA2are mutually exclusive and cover the entire sample space,p1+p2= 1.But 3p1p2=12, sop2=58.2.3.13LetF: female is hired andT: minority is hired. ThenP(F) = 0.60,P(T) = 0.30, andP(FCTC) = 0.25 = 1P(FT). SinceP(FT) = 0.75,P(FT)= 0.60 + 0.300.75 = 0.15.2.3.14The smallest value ofP[(ABC)C] occurs whenP(ABC) is as large as possible.This, in turn, occurs whenA,B, andCare mutually disjoint. The largest value forP(ABC) isP(A) +P(B) +P(C) = 0.2 + 0.1 + 0.3 = 0.6. Thus, the smallest value forP[(ABC)C] is 0.4.2.3.15(a)XCY= {(H,T,T,H), (T,H,H,T)}, soP(XCY) = 2/16(b)XYC= {(H,T,T,T), (T,T,T,H), (T,H,H,H), (H,H,H,T)} soP(XYC) = 4/162.3.16A= {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}ABC= {(1, 5), (3, 3), (5, 1)}, soP(ABC) = 3/36 = 1/12.2.3.17AB, (AB)(AC),A,AB,S2.3.18LetAbe the event of getting arrested for the first scam;B, for the second. We are givenP(A) = 1/10,P(B)= 1/30, andP(AB) = 0.0025. Her chances of not getting arrested areP[(AB)C] = 1P(AB) = 1[P(A) +P(B)P(AB)] = 1[1/10 + 1/300.0025]= 0.869Section 2.4: Conditional Probability2.4.1P(sum = 10|sum exceeds 8) =(sum10 and sum exceeds 8)(sum exceeds 8)PP=(sum10)3363(sum9,10,11, or 12)43633623613610P/P////.

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8Chapter 2: Probability2.4.2P(A|B) +P(B|A) = 0.75 =()()10()5()()()4P ABP ABP ABP ABP BP A, which impliesthatP(AB) = 0.1.2.4.3IfP(A|B) =()()()P ABP AP B, thenP(AB) <P(A)P(B).It follows thatP(B|A) =()()()()()P ABP AP BP AP A=P(B).2.4.4P(E|AB) =(())()()()0.40.13()()()0.44P EABP EP ABP ABP ABP ABP AB.2.4.5The answer would remain the same. Distinguishing only three family types does not makethem equally likely; (girl, boy) families will occur twice as often as either (boy, boy) or (girl,girl) families.2.4.6P(AB) = 0.8 andP(AB)P(AB) = 0.6, soP(AB) = 0.2.Also,P(A|B) = 0.6 =()()P ABP B, soP(B) = 0.210.63andP(A) = 0.8 + 0.21233.2.4.7LetRibe the event that a red chip is selected on theith draw,i= 1, 2.ThenP(both are red) =P(R1R2) =P(R2|R1)P(R1) = 313428.2.4.8P(A|B) =()()()()()()()P ABP AP BP ABabP ABP BP Bb.ButP(AB)1, soP(A|B)1abb.2.4.9LetiWbe the event that a white chip is selected on theith draw,i= 1,2 .ThenP(W2|W1) =121()()P WWP W. If both chips in the urn are white,P(W1) = 1;if one is white and one is black,P(W1) = 12 .Since each chip distribution is equally likely,P(W1) = 111132224.Similarly,P(W1W2) = 111152428, soP(W2|W1) = 5 / 853 / 46.2.4.10P[(AB)| (AB)C] =[()() ]()0[() ][() ]CCCPABABPPABPAB2.4.11(a)P(ACBC) = 1P(AB) = 1[P(A) +P(B)P(AB)] =1[0.65 + 0.550.25] = 0.05

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Section 2.4: Conditional Probability9(b)P[(ACB)(ABC)] =P(ACB) +P(ABC) =[P(A)P(AB)] + [P(B)P(AB)] = [0.650.25] + [0.550.25] = 0.70(c)P(AB) = 0.95(d)P[(AB)C] = 1P(AB) = 10.25 = 0.75(e)P{[(ACB)(ABC)]|AB} =[()()]()CCPABABP AB= 0.70/0.95 = 70/95(f)P(AB)|AB) =P(AB)/P(AB) = 0.25/0.95 = 25/95(g)P(B|AC) =P(ACB)/P(AC) ] = [P(B)P(AB)]/[1P(A)] = [0.550.25]/[10.65]= 30/352.4.12P(No. of heads2| No. of heads2)=P(No. of heads2 and No. of heads2)/P(No. of heads2)=P(No. of heads = 2)/P(No. of heads2) = (3/8)/(7/8) = 3/72.4.13P(first die4|sum = 8) =P(first die4 and sum = 8)/P(sum = 8)=P({(4, 4), (5, 3), (6, 2)}/P({(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}) = 3/52.4.14There are 4 ways to choose three aces (count which one is left out). There are 48 ways tochoose the card that is not an ace, so there are 448 = 192 sets of cards where exactly threeare aces. That gives 193 sets where there are at least three aces. The conditional probability is(1/270,725)/(193/270,725) = 1/193.2.4.15First note thatP(AB) = 1P[(AB)C] = 10.2 = 0.8.ThenP(B) =P(AB)P(ABC)P(AB) = 0.80.30.1 = 0.5.FinallyP(A|B) =P(AB)/P(B) = 0.1/0.5 = 1/52.4.16P(A|B) = 0.5 impliesP(AB) = 0.5P(B).P(B|A) = 0.4 impliesP(AB) = (0.4)P(A).Thus, 0.5P(B) = 0.4P(A) orP(B) = 0.8P(A).Then, 0.9 =P(A) +P(B) =P(A) + 0.8P(A) orP(A) = 0.9/1.8 = 0.5.2.4.17P[(AB)C] =P[(AB)C] +P(ABC) +P(ACB) = 0.2 + 0.1 + 0.3 = 0.6P(AB|(AB)C) =P[(ABC)(ACB)]/P((AB)C) = [0.1 + 0.3]/0.6 = 2/32.4.18P(sum8|at least one die shows 5)=P(sum8 and at least one die shows 5)/P(at least one die shows 5)=P({(5, 3), (5, 4), (5, 6), (3, 5), (4, 5), (6, 5), (5, 5)})/(11/36) = 7/112.4.19P(Outandout wins|Australian Doll and Dusty Stake don’t win)=P(Outandout wins and Australian Doll and Dusty Stake don’t win)/P(Australian Doll andDusty Stake don’t win) = 0.20/0.55 = 20/552.4.20Suppose the guard will randomly choose to name Bob or Charley if they are the two to gofree. Then the probability the guard will name Bob, for example, isP(Andy, Bob) + (1/2)P(Bob, Charley) = 1/3 + (1/2)(1/3) = 1/2.The probability Andy will go free given the guard names Bob isP(Andy, Bob)/P(Guardnames Bob) = (1/3)/(1/2) = 2/3. A similar argument holds for the guard naming Charley.Andy’s concern is not justified.

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10Chapter 2: Probability2.4.21P(BBRWW) =P(B)P(B|B)P(R|BB)P(W|BBR)P(W|BBRW) =356515 14 13 12 11=0.0050P(2, 6, 4, 9, 13) =11111115 14 13 12 11360,360.2.4.22LetKibe the event that theith key tried opens the door,i= 1, 2, …,n. ThenP(door opensfirst time with 3rd key) =123121312()()()()CCCCCCCP KKKP KP KKP KKK=121112nnnnnn.2.4.23(a) The complementary event is that the team loses three or four games. Assume the gamesare independent. The probability of the event is(0.6)(0.50)(0.6)(0.3) + (0.4)(0.50)(0.6)(0.3) + (0.4)(0.50)(0.4)(0.3) + (0.4)(0.50)(0.6)(0.7)+ (0.4)(0.50)(0.6)(0.3) = 0.234.The probability of a bowl appearance is 1 – 0.234 = 0.766.(b) LetkA= probability team wins exactlykgames,k= 3,4.Then4344344434343434[()][()]()()(|)()()()P AAAPAAAPP AP AAAP AAP AAP AA But this does not equal4()P A.(c) Yes, the two events are independent.2.4.24(1/2)(1/2)(1/2)(2/3)(3/4) = 1/162.4.25LetAibe the event “Bearing came from supplieri”,i= 1, 2, 3. LetBbe the event “Bearing intoy manufacturer’s inventory is defective.”ThenP(A1) = 0.5,P(A2) = 0.3,P(A3) = 0.2 andP(B|A1) = 0.02,P(B|A2) = 0.03,P(B|A3) = 0.04Combining these probabilities according to Theorem 2.4.1 givesP(B) = (0.02)(0.5) + (0.03)(0.3) + (0.04)(0.2)= 0.027meaning that the manufacturer can expect 2.7% of her ball-bearing stock to be defective.2.4.26LetBbe the event that the face (or sum of faces) equals 6. LetA1be the event that a Headcomes up andA2, the event that a Tail comes up. ThenP(B) =P(B|A1)P(A1) +P(B|A2)P(A2)= 115162362= 0.15.2.4.27LetBbe the event that the countries go to war. LetAbe the event that terrorism increases.ThenP(B) =P(B|A)P(A) +P(B|AC)P(AC) = (0.65)(0.30) + (0.05)(0.70) = 0.23.2.4.28LetBbe the event that a donation is received; letA1,A2, andA3denote the events that the callis placed to Belle Meade, Oak Hill, and Antioch, respectively.ThenP(B) =31100010002000()()(0.60)(0.55)(0.35)0.46400040004000iiiP B A P A.2.4.29LetBdenote the event that the person interviewed answers truthfully, and letAbe the eventthat the person interviewed is a man.ThenP(B) =P(B|A)P(A) +P(B|AC)P(AC) = (0.78)(0.47) + (0.63)(0.53) = 0.70.
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Statistics - Introduction to Statistics

Quick Guides

Statistics - Graphic Displays

Quick Guides

Statistics - Cummulative Reviews

Quick Guides

Statistics - Common Mistakes and Tables

Quick Guides

Statistics - Bivariate Relationships