Mathematics and statistics stat3502a final

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Essentials of Business Analytics
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Chapter 5 / Exercise 23
Essentials of Business Analytics
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Mathematics and StatisticsSTAT3502A Final ExaminationAugust 2014910. To test the ability of auto mechanics to identify simple engine problem, an automobile witha single such problem was taken in turn to 72 different car repair facilities. Only 42 of the 72mechanics who worked on the car correctly identified the problem.a.[2] Does this strongly indicate that the true proportion of mechanics who could identifythis problem is less than 0.75?(α= 0.05)b.[2] Calculate an upper prediction bound for population proportion using a confidence levelof 95%.c.[1] Does the result of part (b) confirm part (a)?11. A sample of 12 radon detectors of a certain type was selected, and each was exposed to 100pCi/L of radon. The resulting readings were as follows104.389.689.995.695.290.098.8103.798.3106.4102.091.1a.[4] Does this data suggest that the population mean reading under these conditions differsfrom 100? State and test the appropriate hypotheses usingα= 0.05b.[3] Construct a 99% confidence interval for the population mean reading.xi= 1164.9,x2i= 113493.25
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Essentials of Business Analytics
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Chapter 5 / Exercise 23
Essentials of Business Analytics
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Mathematics and StatisticsSTAT3502A Final ExaminationAugust 201410Formulae Sheets2=1n1ni=1(xi¯x)2=Pni=1x2i(Pni=1xi)2nn1E(X) =μ=xxp(x) =integraltextxf(x)dx,σ2=x(xμ)2p(x) =xx2p(x)μ2=integraltextxx2f(x)dxμ2IfA1, A2,· · ·, Akare mutually exclusive and exhaustive events. Then for an event B,P(B) =P(B|A1)P(A1) +· · ·+P(B|Ak)P(Ak) =ksummationdisplayi=1P(B|Ai)P(Ai)P(Aj|B) =P(AjB)P(B)=P(B|Aj)P(Aj)ki=1P(B|Ai)P(Ai)j= 1,· · ·, kfX(x) =yP(x, y) =integraltextyf(x, y)dy,fX|Y(x|y) =fX,Y(x,y)fY(y)E(XY) =integraltextxintegraltextyxyf(x, y)dydx,Cov(X, Y) =E(XY)μXμY,ρX,Y=Cov(X,Y)σXσYIfE(X) =μandV(X) =σ2,thenE(¯X) =μ;V(¯X) =σ2/n;Z=XE(X)V(X)V(a1X1+a2X2+· · ·+anXn) =ni=1nj=1aiajCov(Xi, Xj)DistributionsBinomial:P(X=x) =n!x!(nx)!px(1p)nxμ=np,σ2=np(1p)HypergeometricP(X=x) =(Mx)(N-Mn-x)(Nn),μ=nMN,σ2=nparenleftBigMNparenrightBigparenleftBig1MNparenrightBigparenleftBigNnN1parenrightBigPoissonP(X=x) =eλ λxx!,E(X) =λ= Var(X)Gammaf(x) =1βαΓ(α)xα1ex/β,xgreaterorequalslant0E(X) =αβ,V(X) =αβ2Exponentialf(x) =λeλxxgreaterorequalslant0E(X) = 1/λ, V(X) = 12Uniformf(x) =1baalessorequalslantxlessorequalslantbE(X) = (a+b)/2, V(X) = (a+b)2/12
Mathematics and StatisticsSTAT3502A Final ExaminationAugust 201411Important Test StatisticsConfidence IntervalXμ0σ/nparenleftBigX±zα/2σnparenrightBigXμ0s/nparenleftBigX±tn1,α/2snparenrightBigX1X2D0rσ21n1+σ22n2parenleftbiggX1X2±zα/2radicalBigσ21n1+σ22n2parenrightbiggX1X2

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