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X p(3) =P(X= 3) =✓ 4 p(4) =P(X= 4) =✓ 18 p(5) =P(X= 5) =✓ 36. ✓ p ✓ X X ✓ 0 P5 i=3p(i)=1 ✓=3 E(X)= 5X i=3 ip(i) = 10/3 (X)=E(X2) E(X)2=7/18. FX(x)=P(Xx)= 8 >>>>>>< >>>>>>: 0 x<3 3 4 3x<4 11 12 4x<5 1 x 5 P(X>3) = 1 P(X3) = 1 3/4=1/4 P(X2) = 0 •Y ⇠B(30,3/4) •W ⇠B(30,1/4) P(Y= 30) =P(W= 0) = ⇣ P(X= 3) ⌘30 =( 3/4)30=0.00018 = 0.018% n= 30 30 T T= 30X j=1 Xj'N ⇣ 10·30/3,7·30/18 ⌘ P(T>105) =P 0 @ 30X j=1 Xj>105 1 A=1 P 0 @ 30X j=1 Xj105 1 A=1 P 0 @ 30X j=1 Xj105.5 1 A '1 P 0 @Z105.5 100q7·30 18 1 A=1 P Z5.5 r 3 35 ! =1 (1.61) = 1 0.9463 = 0.0537 Z⇠N(0,1) pI pN nI nN XI XN XI XN XI XN bpI bpN pI pN ↵ • • ↵ • • • • • • •XI ⇠B(nI,pI)'N(nIpI,nIpI(1 pI)) •XN ⇠B(nN,pN)'N(nNpN,nNpN(1 pN)) •bpI=XI/nI •bpN=XN/nN • H0:pIpN H1:pI>pN • ↵ RC↵= ( ˆpI>ˆpN+ s ˆp(1 ˆp) ✓1 nI +1 nN ◆ z↵ ) = 8 >>< >>: z0= ˆpI ˆpNr ˆp(1 ˆp) ⇣1 nI +1 nN ⌘>z↵ 9 >>= >>; , ˆp=XI+XN nI+nN . • nIbpI nI(1 bpI) nNbpN nN(1 bpN) • bpI=530 1000=0.53 bpN=576 1200=0.48. • nIbpI nI(1 bpI) nNbpN nN(1 bpN) XI XN • =P(Z>z0)=P(Z>2.34) = 0.0096. • 0.05> H0 pI>pN. • nIbpI nI(1 bpI) nNbpN nN(1 bpN) IC95%(pI pN)=ˆpI ˆpN±z0.025 s ˆpI(1 ˆpI) nI +ˆpN(1 ˆpN) nN =0.05±1.96·0.02⇡0.05±0.04 = (0.01,0.09). X Y X Y x30= 19.19, y30= 2607583, ●● ● ●●● ●● ● ●●● ●●●● ● ● ● ●● ● ●●● ● ●● ● ● 152025 0.0e+001.0e+072.0e+07 y vs. x x y ●● ● ● ● ● ●● ● ●●● ●●●●● ● ● ●● ● ●● ● ● ● ● ● ● 152025 −10123 Residui standardizzati vs. x x Residui standardizzati ●● ● ● ● ● ●● ● ●●● ●●●● ● ● ● ●● ● ●● ● ● ● ● ● ● −2−1012 −10123 Normal Q−Q Plot Theoretical Quantiles Sample Quantiles SW−pvalue = 7.325e−05 ● ● ● ● ● ● ●● ● ●●● ●●●● ● ● ● ●● ● ●● ● ● ● ● ● ● 2.42.62.83.03.2 10121416 log( y ) vs. log( x ) log(x) log(y) ● ●● ● ● ● ●● ● ●● ● ● ●●● ● ● ● ●● ● ● ● ● ● ●● ●● 2.42.62.83.03.2 −3−2−10123 Residui standardizzati vs. log( x ) log( x ) Residui…