Exam Let X be the random variables representing the marks achieved by students in an econometric theory paper an let the statistical model b )0={(G0=如(=) 6∈6 (b)x=(X1, X2,, Xn, n=40 is random sample from p The hypothesis to be tested is H0:6=60(e.X~N(60,64),Oo={60} against H1:≠60(ie.X~N(p,64),p≠60),O1=[0.,100-{60} or the sample realization t takes a value 'around60 then we will be inclined, Common sense suggests that if some 'good'estimator of 8, say Xn=(1/n)isai accept Ho. Let us formalise this argument The accept region takes the form:60-ε≤Xn≤60+e,E>0,or C0={x:|Xn-60≤e} and C1=a: Xn-602e, is the rejection region Formally,rifr∈Cl( reject Ho)andb∈eo( Ho is true)- type I error;ifc∈Co (accept Ho) and 0 E O1(Ho is false)-type II error. The hypothesis to be tested is formally stated as follows Ho:6∈60,eos. Against the null hypothesis Ho we postulate the alternative Hi which takes theExample: Let X be the random variables representing the marks achieved by students in an econometric theory paper an let the statistical model be: (a) Φ = n f(x; θ) = 1 8 √ 2π exp h − 1 2 ￾ x−θ 8
2 io , θ ∈ Θ ≡ [0, 100]; (b) x = (X1, X2, ..., Xn) 0 , n=40 is random sample from Φ. The hypothesis to be tested is H0 : θ = 60 (i.e. X ∼ N(60, 64)), Θ0 = {60} against H1 : θ 6= 60 (i.e. X ∼ N(µ, 64), µ 6= 60), Θ1 = [0, 100] − {60}. Common sense suggests that if some ’good’ estimator of θ, say X¯ n = (1/n) Pn i=1 xi for the sample realization x takes a value ’around’ 60 then we will be inclined to accept H0. Let us formalise this argument: The accept region takes the form: 60 − ε ≤ X¯ n ≤ 60 + ε, ε > 0, or C0 = {x : |X¯ n − 60| ≤ ε} and C1 = {x : |X¯ n − 60| ≥ ε}, is the rejection region. Formally, if x ∈ C1 (reject H0) and θ ∈ Θ0 (H0 is true)–type I error; if x ∈ C0 (accept H0) and θ ∈ Θ1 (H0 is false)–type II error. The hypothesis to be tested is formally stated as follows: H0 : θ ∈ Θ0, Θ0 ⊆ Θ. Against the null hypothesis H0 we postulate the alternative H1 which takes the form: H1 : θ ∈ Θ1 ≡ Θ − Θ0. 2