Probability and Statistics for Engineering and the Sciences (with CD-ROM and InfoTrac )

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(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. ( X 2 1 + X 2 2 ) / 2 = ( X 2 i ) = Var ( X i ) = 1 θ . Hence ˆ θ = 2 / ( X 2 1 + X 2 2 ) . (b) The likelihood func- tion is L ( θ ) = θ/ (2 π ) exp( - θ ( x 2 1 + x 2 2 ) / 2) . Get log- likelihood function ln L ( θ ) = const + ln θ - θ ( x 2 1 + x 2 2 ) / 2 . Differentiate with respect to θ to get the maxi- mum: d ln L ( θ ) = 1 θ - 1 2 ( x 2 1 + x 2 2 ) = 0 . Solving the equation, we get ˆ θ = 2 / ( x 2 1 + x 2 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 X i N ( μ, σ 2 ) . Since we do not know σ , we need t -statistic to construct a CI: t = x - μ ) / ( s/ n ) . After a lengthy algebra, CI will be ¯ x ± t α/ 2 , 9 s/ 10 . ¯ x = 21 . 6 , s = 15 . 6 , t α/ 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 = x - μ ) / ( σ/ n ) . After a lengthy algebra, CI will be ¯ x ± z α/ 2 3 / 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

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