HW7Solutions

HW7Solutions - 4 a We use the sample mean x to estimate the...

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4. a. We use the sample mean, x to estimate the population mean μ . 1407 . 8 27 80 . 219 ˆ = = Σ = = n x x i b. We use the sample median, 7 . 7 ~ = x (the middle observation when arranged in ascending order). c. We use the sample standard deviation, () 660 . 1 26 94 . 1860 27 8 . 219 2 2 = = = s s d. With “success” = observation greater than 10, x = # of successes = 4, and 1481 . ˆ 27 4 = = = n x p e. We use the sample (std dev)/(mean), or 2039 . 1407 . 8 660 . 1 = = x s 5. a. ( ) 2 1 μμ = = Y E X E Y X E ; 434 . 575 . 8 141 . 8 = = y x b. 2 2 2 1 2 1 2 2 n n Y V X V Y X V Y X σσ + = + = + = ; 2 2 2 1 2 1 n n Y X V Y X σ + = = The estimate would be 5687 . 20 104 . 2 27 66 . 1 2 2 2 2 2 1 2 1 = + = + = n s n s s Y X . c. 7890 . 104 . 2 660 . 1 2 1 = = s s d. ( ) ( ) 1824 . 7 104 . 2 66 . 1 2 2 2 2 2 1 = + = + = + = Y V X V Y X V 6. a. , ) ( ) ( λ = = = X E X E so X is an unbiased estimator for the Poisson parameter ; since n = 150, , 317 ) 1 )( 7 ( ... ) 37 )( 1 ( ) 18 )( 0 ( = + + + = i x 11 . 2 150 317 ˆ = = = x . b. n n x λσ = = , so the estimated standard error is 119 . 150 11 . 2 ˆ = = n 7. a. min(x i ) = 202 and max(x i ) = 525, so the estimate of the number of planes manufactured is max(x i ) - min(x i ) + 1 = 525 – 202 + 1 = 324.
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b. The estimate will equal the true number of planes manufactured iff min(x i ) = α and max(x i ) = β , i.e., iff the smallest serial number in the population and the largest serial number in the
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HW7Solutions - 4 a We use the sample mean x to estimate the...

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