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

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The log-likelihood function is then log L ( c, σ ) = const. - n log σ + 1 2 σ 2 n X i =1 ( y i - cx i ) 2 . (c) Letting log L/∂c = 0 , we get n i =1 x i ( y i - cx i ) = 0 so we get the same answer. (d) ˆ c = n i =1 x i Y i / n i =1 x 2 i =c. (e) ˆ c = n i =1 x i Y i / n i =1 x 2 i . Var ˆ c = n i =1 x 2 i Var Y i / [ n i =1 x 2 i ] 2 = σ 2 / n i =1 x 2 i . (f) The sample correlation is given by r = S xy / p S xx S yy , where S xy = n i =1 ( x i - ¯ x )( y i - ¯ y ) . Let ( z i , w i ) = ( x i - ¯ x, y i - ¯ y ) . Note that ¯ z = ¯ w = 0 . So the sample correlation for ( z i , w i ) would be S zw / S zz S ww where S zw = n i =1 ( z i - ¯ z )( w i - ¯ w ) = n i =1 z i w i = S xy . Also S zz = S xx and S ww = S yy . So the sample correlation is the same. 2. When you throw a coin 10 times, you observed 3 heads. Test if the coin is biased at α = 0 . 2 . Clearly state appropriate parameters, hypotheses, a test statistic. You may use the following R output (15pts). > pbinom(0:5,10,0.5) [1] 0.00 0.01 0.06 0.17 0.38 0.62 Solution Let p be the probability of getting head. Then H 0 : p = 0 . 5 vs. H 1 : p 6 = 0 . 5 . Assign X i = 1 if i -th tossing yields head and X i = 0 oth- erwise. Note that P ( X i = 1) = p, P ( X i = 0) = 1 - p . The point estimator for p would be ˆ p = 10 i =1 X i / 10 = ¯ X . We will take 10 i =1 X i rather than ¯ X as the test statistic because it will be much easier to figure out the distribution of the test statis- tic. Note that 10 i =1 X i Binomial (10 , p ) . Under H 0 , 10 i =1 X i Binomial (10 , 0 . 5) . We reject H 0 if 10 i =1 x i is either too small or too large. Note that P ( 10 i =1 X i 3) = P ( 10 i =1 X i 7) = 0 . 17 from symmetry. So the P -value is 0.34. We do not

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