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H 0 : H 1 : α = Decision Rule: If then do not reject H 0 , otherwise reject H 0 . Test Statistic: Decision: Conclusion: We have found ________________ evidence at the _____ level of significance that. H 0 : H 1 : α = Decision Rule: If
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H0: H1: α = Decision Rule: If then do not reject H0, otherwise reject H0. Test Statistic: Decision: Conclusion: We have found ________________ evidence at the _____ level of significance that
H0: H1: α = Decision Rule: If then do not reject H0, otherwise reject H0. Test Statistic: Decision: Conclusion: We have found ________________ evidence at the _____ level of significance that .05 .05
H0: H1: α = Decision Rule: If then do not reject H0, otherwise reject H0. Test Statistic: Decision: Conclusion: We have found ________________ evidence at the _____ level of significance that the true average diameter of the ball bearings produced differs from .25 inches. μ = .25 μ ≠ .25 .05 .05
Jaggia and Kelly (1stedition) Critical Value(s) Table 1 Group Flowchart * means coverage is different from text. yes Wald-Wolfowitz One-Sample Runs Test for Randomness pp. 634-638 Z using σ pp. 277-284 yes Z-table 2 1 Normal population ? yes no n > 30 ? Wilcoxon Signed-Ranks *pp. 610-614 (assumes population is symmetric) no WSR Table 3 no at least interval known ? no Normal population ? no n > 30 ? mean or median level of data ? yes t with df = n-1 pp. 288-294 t-table 4 yes ordinal Sign Test *pp. 631-634 Sign Table 5 Z pp. 294-298 yes Z-table 6 np> 5 and n(1-p) > 5 ? proportion Parameter ? no Binomial Table Binomial/ Hypergeometric variance or standard deviation yes chi-square (df = n-1) pp. 336-344 Chi-square Table 7 Normal population ? no ? Default case
H0: H1: α = Decision Rule: If then do not reject H0, otherwise reject H0. Test Statistic: Decision: Conclusion: We have found ________________ evidence at the _____ level of significance that the true average diameter of the ball bearings produced differs from .25 inches. μ = .25 μ ≠ .25 .05 .05
H0: H1: α = Decision Rule: If then do not reject H0, otherwise reject H0. Test Statistic: Decision: Conclusion: We have found ________________ evidence at the _____ level of significance that the true average diameter of the ball bearings produced differs from .25 inches. μ = .25 μ ≠ .25 .05 df = n – 1 = 60 Do not reject H0 Reject H0 Reject H0 .025 .025 t = -2.0003 0 t = 2.0003 .05
H0: H1: α = Decision Rule: If -2.0003 < tcomputed < 2.0003 then do not reject H0, otherwise reject H0. Test Statistic: Decision: Conclusion: We have found ________________ evidence at the _____ level of significance that the true average diameter of the ball bearings produced differs from .25 inches. μ = .25 μ ≠ .25 .05 df = n – 1 = 60 Do not reject H0 Reject H0 Reject H0 .025 .025 t = -2.0003 0 t = 2.0003 .05
3 Steps to standard deviation 1. Calculate the variation of the sample, SS. 2. Calculate the variance of the sample, S2. 3. Calculate the standard deviation of the sample, S.
H0: H1: α = Decision Rule: If -2.0003 < tcomputed < 2.0003 then do not reject H0, otherwise reject H0. Test Statistic: Decision: Conclusion: We have found ________________ evidence at the _____ level of significance that the true average diameter of the ball bearings produced differs from .25 inches. μ = .25 µ ≠ .25 .05 df = n – 1 = 60 Do not reject H0 Do not reject H0 Reject H0 Reject H0 .025 .025 t = -2.0003 0 t = 2.0003 Do not reject H0. .05 insufficient