homework5-solutions

# homework5-solutions - Statistics 5021 – Homework 5...

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Unformatted text preview: Statistics 5021 – Homework 5 (solutions) There are 20 total points (0.5 points for each part of each question and 8 points for handing in the assignment). One question will be graded for correctness. This homework is due Thursday, March 24 in your lab section. 1. James is conducting a hypothesis test for μ , the mean of the distribution of heights, H : μ = 65 inches H a : μ 6 = 65 inches He develops a decision rule where he will fail to reject H if ¯ x , the observed sample mean height, is between 64 and 66 inches, otherwise he will reject. Notice that he is ignoring the sample size and observed sample standard deviation. (a) Compute the probability of a type I error of James’ decision rule if it were to be conducted with a random sample of size n = 100 from N ( μ = 65 ,σ = 4) Solution: The type I error occurs when H is rejected when it is actually true. Since the test is conducted with a random sample from the null model (with standard deviation σ = 4), by Theorem A and the first part of Theorem B, the sample mean ¯ X ∼ N ( μ ¯ X ,σ ¯ X ), where μ ¯ X = μ = 65 and σ ¯ X = σ/ √ n = 4 / √ 100. P (Type I error) = P [(64 < ¯ X < 66) c ] = 1- P (64 < ¯ X < 66) = 1- P 64- μ ¯ X σ ¯ X < ¯ X- μ ¯ X σ ¯ X < 66- μ ¯ X σ ¯ X = 1- P 64- 65 4 / √ 100 < Z < 66- 65 4 / √ 100 = 1- P (- 2 . 5 < Z < 2 . 5) = 1- [ P ( Z < 2 . 5)- P ( Z <- 2 . 5)] = 1- [ pnorm (2 . 5)- pnorm (- 2 . 5)] = 1- . 9875807 = 0 . 0124193 (b) Compute the probability of a type II error of James’ decision rule if it were to be conducted with a random sample of size n = 100 from N ( μ = 66 ,σ = 4) Solution: The type II error occurs when we fail to reject H when H a is actually true. Since the test is conducted with a random sample from a specific 1 alternative model, by Theorem A and the first part of Theorem B, the sample mean ¯ X ∼ N ( μ ¯ X ,σ ¯ X ), where μ ¯ X = μ = 66 and σ ¯ X = σ/ √ n = 4 / √ 100. P (Type II error) = P (64 < ¯ X < 66) = P 64- μ ¯ X σ ¯ X < ¯ X- μ ¯ X σ ¯ X < 66- μ ¯ X σ ¯ X = P 64- 66 4 / √ 100 < Z < 66- 66 4 / √ 100 = 1- P (- 5 < Z < 0) = P ( Z < 0)- P ( Z <- 5) = pnorm (0)- pnorm (- 5) = 0 . 4999997 2. The data are 35 measured systolic blood pressures in mmHg 159 92 102 92 90 91 96 91 118 136 91 90 111 92 90 91 90 151 92 90 146 94 90 90 91 90 176 95 90 90 159 92 109 90 106 The observed sample mean is ¯ x = 104 . 6571 mmHg and the observed sample standard deviation is s = 24 . 53082 mmHg. A researcher feels that using the Normal distribution to model systolic blood pressures for the population is problematic. Is there statistical evidence at the α = 5% level that μ , the mean of the distribution of systolic blood pressure for the population is not equal to 110 mmHg?...
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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.

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homework5-solutions - Statistics 5021 – Homework 5...

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