Stat312 Midterm I Solutions: Probability and Statistics, Exams of Mathematical Statistics

Download Stat312 Midterm I Solutions: Probability and Statistics and more Exams Mathematical Statistics in PDF only on Docsity! Stat312: Midterm I Solution Moo K. Chung, Yulin Zhang mchung@stat.wisc.edu, yulin@stat.wisc.edu March 5, 2003 1. Let X1, · · · , Xn be a random sample from Bernoulli distribution with parameter p. (a) What is (S2/p2)? S2 is the sample variance. Ex- plain your results (10 points). (b) Find an unbiased estimator of p2. Explain your results (5 points). Solution. (a) The sample variance is an unbiased esti- mator of the population variance. Hence (S2/p2) = (S2 )/p2 = Var(Xi)/p 2. The variance for a Bernoulli random variable can be esily computed as Var(Xi) = p(1−p). So (S2/p2) = (1−p)/p. (b) We know S2 and X̄ will be unbiased estimators of population variance p(1 − p) and mean p respecively. So E(S2) −E(X̄) = p(1 − p) − p = −p2. Hence, X̄ − S2 is an unbiased estimator of p2. 2. Let X1, X2 be a random sample from N(0, 1/θ). Note that the sample size is 2 and the density function for Xi is f(xi) = √ θ √ 2π exp(−θx2i /2). (a) Obtain an estimator of θ using the method of mo- ments. Explain your results (5 points). (b) Find the likelihood function and use it to obtain the maximum likelihood estimator of θ. (5 points for each question). Solution. (a) Match the second moments from the sam- ple and population. (X2 1 + X2 2 )/2 = (X2i ) = Var(Xi) = 1 θ . Hence θ̂ = 2/(X2 1 + X2 2 ). (b) The likelihood func- tion is L(θ) = θ/(2π) exp(−θ(x2 1 + x2 2 )/2). Get log- likelihood function ln L(θ) = const + ln θ − θ(x2 1 + x2 2 )/2. Differentiate with respect to θ to get the maxi- mum: d ln L(θ) dθ = 1 θ − 1 2 (x2 1 + x2 2 ) = 0. Solving the equation, we get θ̂ = 2/(x2 1 + x2 2 ). 3. Consider the sample of IQ of 10 dogs: 25, 21, 22, 17, 29, 25, 16, 20, 19, 22. Suppose that these are from a normal population. (a) Compute a 95% confidence interval of the mean IQ (5 points). (b) Somehow you figured out the variance of IQ to be 9. Compute a 95% confidence interval of the mean IQ. Explain why you have a smaller confidence in- terval (5 points for each question). Solution. (a) Let Xi ∼ N(µ, σ2). Since we do not know σ, we need t-statistic to construct a CI: t = (x̄ − µ)/(s/ √ n). After a lengthy algebra, CI will be x̄ ± tα/2,9s/ √ 10. x̄ = 21.6, s = √ 15.6, tα/2,9 = 2.26. After punching a calculator, CI is [18.8, 24.4]. (b) Since σ is known, we need Z-statistic to construc a CI: z = (x̄ − µ)/(σ/ √ n). After a lengthy algebra, CI will be x̄ ± 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 accurate in (b). 4. 47 out of 102 doctors knew the generic name for the drug methadone. (a) Compute a 98% confidence interval for the propo- tion of all doctors who know the generic name for methadone (10 points). (b) Compute a 98% confidence interval for the propo- tion of all doctors who do not know the generic name for methadone (5 points). >qnorm(0.02) [1] -2.053749 >qnorm(0.01) [1] -2.326348 Solution. (a) Since the sample size is large, CI compu- ation will be based on Z = (p̂ − p)/ √ p(1 − p)/n ∼ N(0, 1). This requires solving a quadratic equation so we will approxmate this by substituting p in the denom- inator to p̂. Then CI becomes p̂ ± zα/2 √ p̂(1 − p̂)/n. p̂ = 47/102. So CI is [0.35, 0.58]. (b) The propotion of doctors who do know know is q = 1 − p. Hence CI would be [1 − 0.58, 1− 0.35] = [0.42, 0.65]. 1