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Unformatted text preview: EXERCISES IN STATISTICS Series A, No. 9 1. The average length of a finger bone of 10 fossil skeletons of the proconsul hominid is 3.73cm, and the standard deviation is 0.34cm. Find 80% and 90% confidence intervals for the mean length of the bone in the species. Answer. We are told that ¯ x 10 = 3 . 73 and that s 10 = p P ( x i − ¯ x ) 2 /n = 0 . 34. On the assumption that x i ∼ N ( μ,σ 2 ); i = 1 ,...,n is a normally distributed random sample, there are ¯ x ∼ N µ μ, σ 2 n ∂ and ¯ x − μ σ √ n ∼ N (0 , 1) . Since σ 2 is unknown, we must use ˆ σ 2 = P ( x i − ¯ x ) 2 n − 1 = s 2 n n − 1 whence ˆ σ √ n = s √ n − 1 = . 34 3 . From the result that ¯ x − μ ˆ σ/ √ n = ( (¯ x − μ ) σ √ n ¡ s P ( x i − ¯ x ) 2 σ 2 ( n − 1) ) ∼ N (0 , 1) q χ 2 ( n − 1) ( n − 1) = t ( n − 1) , we may derive a probability statement of the form P µ ¯ x − b ˆ σ √ n < μ < ¯ x + b ˆ σ √ n ∂ = Q. The t (9) tables indicate that P { t (9) > b } = 0 . 95 = ⇒ b = 1 . 833 and P { t (9) > b } = 0 . 90 = ⇒ b = 1 . 383 . Therefore, the 80% confidence interval is 3 . 73 ± (1 . 833 × . 34 / 3) = [3 . 522 , 3 . 938] and the 90% confidence interval is 3 . 73 ± (1 . 383 × . 34 / 3) = [3 . 573 , 3 . 887] 2. Mr. Smith has been threatened with the loss of his job if he persists in arriving late at the oﬃce. Prior to this threat, his average arrival time over 10 dayslate at the oﬃce....
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This note was uploaded on 03/02/2012 for the course EC 2019 taught by Professor D.s.g.pollock during the Spring '12 term at Queen Mary, University of London.
 Spring '12
 D.S.G.Pollock

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