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Unformatted text preview: Estimators, Estimates, and Sampling distributions – Examples (with solutions) 1. Suppose that we use the Normal distribution to model the heights of females in the United states. Let’s assume that this Normal distribution has mean μ = 65 inches and standard deviation σ = 3 inches. (a) If we plan to take a random sample of size n , specify the probability distribution of the sample mean ¯ X , (give the distribution name and specify its parameters) Solution: From Theorem A, μ ¯ X = μ = 65 and σ ¯ X = σ/ √ n = 3 / √ n From the first part of theorem B, since ¯ X is based on a random sample from a Normal distribution, ¯ X ∼ N ( μ ¯ X ,σ ¯ X ). (b) If we plan to take a random sample of size n = 20, what is the probability that we will observe a sample mean within 2 inches of the mean μ = 65 inches. Solution: Using the fact that ¯ X ∼ N ( μ ¯ X ,σ ¯ X ), with μ ¯ X = μ = 65 and σ ¯ X = σ/ √ n = 3 / √ 20, we have: P (63 < ¯ X < 67) = P 63 μ ¯ X σ ¯ X < X μ ¯ X σ ¯ X < 67 μ ¯ X σ ¯ X = P 63 65 3 / √ 20 < Z < 67 65 3 / √ 20 = P ( 2 . 981424 < Z < 2 . 981424) = P ( Z < 2 . 981424) P ( Z < 2 . 981424) = pnorm (2 . 981424) pnorm ( 2 . 981424) = 0 . 9971309 (c) Compute and compare the standard deviation of the sample mean ¯ X for the cases n = 10 and n = 50. Solution: When n = 10 is σ ¯ X = 3 / √ 10 = 0 . 9486833 which is larger than when n = 50, σ ¯ X = 3 / √ 50 = 0 . 4242641 1 (d) Compute and compare the standard deviation of the sample mean based on a random sample of size n = 10 for the cases when the standard deviation of the Normal distribution used to model the heights of females is σ = 3 and σ = 8. Solution: When σ = 3 and n = 10, σ ¯ X = 3 / √ 10 = 0 . 9486833 which is smaller than when σ = 8 and n = 10, σ ¯ X = 8 / √ 10 = 2 . 529822 2. The following observations: x 1 ,x 2 ,...,x 24 are a realization of a random sample X 1 ,X 2 ,...,X 24 from a Normal distribution (generated using R’s random number gen erator). 56.51047 69.72663 72.32081 64.86753 64.56595 67.53100 66.64958 66.99755 66.58817 63.99538 60.96581 58.72800 60.78952 60.52094 70.59555 74.93141 65.59604 61.40853 63.00884 61.78672 65.56646 70.35814 69.93231 60.92663 (a) The observed sample mean is ¯ x = 65 . 20283, for which quantity is this value an estimate? Solution: The mean μ of the Normal distribution from which x 1 ,x 2 ,...,x 24 are a realization of a random sample. (b) If the actual mean and standard deviation of the distribution from which these data are a realization of a random sample is μ = 65 and σ = 4, specify the probability distribution (its name and the values for its parameters) from which ¯ x = 65 . 20283 was a realization....
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This note was uploaded on 05/06/2011 for the course STAT 5021 taught by Professor Staff during the Spring '08 term at Minnesota.
 Spring '08
 Staff
 Normal Distribution

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