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08 07 14

Study material for Statistica per L'ingegneria, shared by the Studwiz community and reviewed by moderators.

Statistica per L'ingegneriaSecond midterm

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X X λ n X1,...,X n λ = E[X] ˆλ = Xn. ˆλ σ = √ Var(X) ˆσ = √ ˆλ = √ Xn. ˆσ λ ˆλ n ˆλ − λ√ bλ n ≃ N (0, 1). γ λ X 23, 18, 20, 26, 20, 23, 12, 20, 16, 35. λ σ λ γ =0 .95 E[ˆλ] − λ =0 MSE(ˆλ)=V a r (ˆλ)= λ/n se(ˆλ)= √ Var(ˆλ)= √ λ/n ˆλ ≃ N ( λ, λ n ) E[ˆσ] − σ = E [√ ˆλ ] − λ ≃ √ E[ˆλ] − λ =0 MSE(ˆσ) ≃ Var(ˆσ) ≃ λ n ( 1 2 √ λ )2 = 1 4n se(ˆσ)= √ Var(ˆσ) ≃ √ 1 4n ˆλ ± √ ˆλ n z 1−γ 2 ˆλ =2 1.3 ˆσ =4 .6 se(ˆλ)=1 .5 se(ˆσ)=0 .2 21.3 ± 2.9 (18.4 , 24.2) p 120 120 α 120 78 120 102 p X2010 = = { 1, 0, ∼ B(p2010) X2014 = = { 1, 0, ∼ B(p2014) pk = k H0 : p2010 ≥ p2014 H1 : p2010 <p 2014 X2010,1,...,X 2010,120 B(p2010) X2014,1,...,X 2014,120 B(p2014) Rα : Z0 = ˆp2010 − ˆp2014√ ˆp ∗(1 − ˆp) ∗( 1 120 + 1 120 ) < −zα ˆpk = ∑120 j=1 Xk,j 120 , ˆp = ∑120 j=1 X2010,j + ∑120 j=1 X2014,j 240 . ˆp2010 =0 .65 ˆp2014 =0 .85 ˆp =0 .75 Z0 = −3.577708764 < −zα = −1.645, 0.05 p α Z0 = −3.577708764 = −zα zα =3 .577708764 1 − α = Φ(zα ) = Φ(3 .577708764) = 0 .9998266903 α =0 .0001733097 = 1 .7 · 10− 4 Y = y =2 1 3.6735 s2 y =1 1 5.6454. Y Y Y x Y = β0 + β1X + ϵ, ϵ ∼ N (0,σ 2). Y x 22◦ Y

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