100%(1)1 out of 1 people found this document helpful
This preview shows page 1 out of 1 page.
(b) Find the likelihood function and use it to obtain themaximum likelihood estimator ofθ. (5 points foreach question).Solution.(a) Match the second moments from the sam-ple and population.(X21+X22)/2 =(X2i) =Var(Xi) =1θ.Henceˆθ= 2/(X21+X22).(b) The likelihood func-tion isL(θ) =θ/(2π) exp(-θ(x21+x22)/2). Get log-likelihood functionlnL(θ) =const+ lnθ-θ(x21+x22)/2. Differentiate with respect toθto get the maxi-mum:dlnL(θ)dθ=1θ-12(x21+x22) = 0.Solving the equation, we getˆθ= 2/(x21+x22).3. Consider the sample of IQ of 10 dogs: 25, 21, 22, 17,29, 25, 16, 20, 19, 22. Suppose that these are from anormal population.(a) Compute a95%confidence interval of the meanIQ (5 points).(b) Somehow you figured out the variance of IQ to be9. Compute a95%confidence interval of the meanIQ. Explain why you have a smaller confidence in-terval (5 points for each question).Solution.(a) LetXi∼N(μ, σ2).Since we do notknowσ, we needt-statistic to construct a CI:t=(¯x-μ)/(s/√n). After a lengthy algebra, CI will be¯x±tα/2,9s/√10.¯x= 21.6, s=√15.6, tα/2,9= 2.26.After punching a calculator, CI is[18.8,24.4]. (b) Sinceσis known, we needZ-statistic to construc a CI:z=(¯x-μ)/(σ/√n). After a lengthy algebra, CI will be¯x±zα/23/√10. CI is[19.7,23.5]. In (a), two parame-ters are unknown while in (b), only one paramter is un-known. Hence your interval estimation should be more