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sec_8_2_2_sol

# sec_8_2_2_sol - APPM 4/5570 Solutions to Problems from...

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APPM 4/5570 Solutions to Problems from Section 8.2 (continued) 25. n = 16, normal, σ = 0 . 3, x = 5 . 25 Everything will be done with Z ’s. a. Step One: Hypotheses: H 0 : μ = 5 . 5 H a : μ 6 = 5 . 5 Step Two: The Test Statistic Z = X - μ 0 σ/ n = 5 . 25 - 5 . 5 0 . 3 / 4 ≈ - 3 . 33 Step Three: The Rejection Region The rejection region is two-tailed and located on both the high and low ends of the Z curve, above z α/ 2 and below - z α/ 2 . We are not given an α , so we could “default” to α = 0 . 05, which makes the cutoffs - 1 . 645 and 1 . 645. In this case, we would reject H 0 , in favor of H a . Or, we could say that, since the test statistic is obviously quite extreme for a Z -statistic, we are likely to be in the rejection region for most “reasonable” or “typical” values of α . Step Four: Conclusion At 0 . 05 level of significance and most other reasonable levels, we will reject H 0 in favor of H a . This data strongly suggests that the true average percentage of SiO 2 is different from 5 . 5.

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