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Answers to Homework # 6 pages 226-228, #2, 6(a, c), 8(a, c), 9(a, c), 12, 15, 18b, 19(a, c), 20(a, b, c), 21, and 22. 2. The distribution of sample means will be normal (because n > 30), have an expected value of 100 and a standard error of σ / √n = 12 / √36 = 12/6 = 2 6. a) Calculate Standard Error: σ / √n = 10/ √4 = 10/2 = 5 b) σ / √n = 10 / √16 = 10/4 = 2.5 c) σ / √n = 10 / √25 = 10/5 = 2 8. a) σ M = σ / √n n = (σ / σ M) 2 (8 / 4) 2 = 2 2 = 4 c) σ M = σ / √n n = (σ / σ M) 2 (8 / 1) 2 = 64 9. a) σ M = σ / √n σ = σ M * √n 10 * √16 = 10*4 = 40 c) σ M = σ / √n σ = σ M * √n 2 * √16 = 2*4 = 8 12. a) The standard error would be 6:   /  n = 12 /  4 = 12/2 = 6 therefore the z- σ score would                          be 1: Z =      –      μ         55-50  = 5/6 = 0.83  A z-score of 0.83 is not an  extreme score. σ M 6 b) The standard error would be 2:   /  n = 12 /  36 = 12/6 = 2 therefore the z- σ score would be 1: Z =      –

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