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H 0 : H 1 : α = Decision Rule: If then do not reject H 0 , otherwise reject H 0 . Test Statistic:

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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H 0 : H 1 : α = Decision Rule: If then do not reject H 0 , otherwise reject H 0 . Test Statistic:

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  1. 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

  2. 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

  3. 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 mean bar length of scale sets is less than 42 millimeters. μ = 42 μ < 42 .05 .05

  4. 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

  5. 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 mean bar length of scale sets is less than 42 millimeters. μ = 42 μ < 42 .05 .05

  6. Calculation of Wilcoxon Signed-Ranks Test Statistic di = Xi - M0 42 - 42 = X 39 - 42 = -3 42 - 42 = X 45 - 42 = +3 42 - 42 = X 43 - 42 = +1 43 - 42 = +1 40 - 42 = -2 40 - 42 = -2 42 - 42 = X 39 - 42 = -3 41 - 42 = -1 First, you calculate the differences. Any zero differences are discarded, because zero does not have a sign. di = Xi - M0 42 - 42 = 0 39 - 42 = -3 42 - 42 = 0 45 - 42 = +3 42 - 42 = 0 43 - 42 = +1 43 - 42 = +1 40 - 42 = -2 40 - 42 = -2 42 - 42 = 0 39 - 42 = -3 41 - 42 = -1 First, you calculate the differences. Any zero differences are discarded, because zero does not have a sign.

  7. Calculation of Wilcoxon Signed-Ranks Test Statistic di = Xi - M0 42 - 42 = X 39 - 42 = -3 42 - 42 = X 45 - 42 = +3 42 - 42 = X 43 - 42 = +1 43 - 42 = +1 40 - 42 = -2 40 - 42 = -2 42 - 42 = X 39 - 42 = -3 41 - 42 = -1 First, you calculate the differences. Any zero differences are discarded, because zero does not have a sign. di = Xi - M0 |di| 42 - 42 = 0 39 - 42 = -3 3 42 - 42 = 0 45 - 42 = +3 3 42 - 42 = 0 43 - 42 = +1 1 43 - 42 = +1 1 40 - 42 = -2 2 40 - 42 = -2 2 42 - 42 = 0 39 - 42 = -3 3 41 - 42 = -1 1 First, you calculate the differences. Any zero differences are discarded, because zero does not have a sign. Calculate the absolute differences.

  8. Calculation of Wilcoxon Signed-Ranks Test Statistic di = Xi - M0 |di| Ri 42 - 42 = 0 39 - 42 = -3 3 7 42 - 42 = 0 45 - 42 = +3 3 7 42 - 42 = 0 43 - 42 = +1 1 2 43 - 42 = +1 1 2 40 - 42 = -2 2 4.5 40 - 42 = -2 2 4.5 42 - 42 = 0 39 - 42 = -3 3 7 41 - 42 = -1 1 2 First, you calculate the differences. Any zero differences are discarded, because zero does not have a sign. Calculate the absolute differences. Rank the absolute differences. The smallest absolute difference gets a rank of 1, the second smallest absolute difference gets a rank of 2, and so on. If you have ties, you average the ranks. di = Xi - M0 42 - 42 = X 39 - 42 = -3 42 - 42 = X 45 - 42 = +3 42 - 42 = X 43 - 42 = +1 43 - 42 = +1 40 - 42 = -2 40 - 42 = -2 42 - 42 = X 39 - 42 = -3 41 - 42 = -1 First, you calculate the differences. Any zero differences are discarded, because zero does not have a sign.

  9. Calculation of Wilcoxon Signed-Ranks Test Statistic di = Xi - M0 |di| Ri Ri(+) W = 42 - 42 = 0 39 - 42 = -3 3 7 42 - 42 = 0 45 - 42 = +3 3 7 7 42 - 42 = 0 43 - 42 = +1 1 2 2 43 - 42 = +1 1 2 2 40 - 42 = -2 2 4.5 40 - 42 = -2 2 4.5 42 - 42 = 0 39 - 42 = -3 3 7 41 - 42 = -1 1 2 First, you calculate the differences. Any zero differences are discarded, because zero does not have a sign. Calculate the absolute differences. Rank the absolute differences. The smallest absolute difference gets a rank of 1, the second smallest absolute difference gets a rank of 2, and so on. If you have ties, you average the ranks. Finally, sum the ranks associated with the positive differences. di = Xi - M0 42 - 42 = X 39 - 42 = -3 42 - 42 = X 45 - 42 = +3 42 - 42 = X 43 - 42 = +1 43 - 42 = +1 40 - 42 = -2 40 - 42 = -2 42 - 42 = X 39 - 42 = -3 41 - 42 = -1 First, you calculate the differences. Any zero differences are discarded, because zero does not have a sign. 7 + 2 + 2 = 11 + + +

  10. Calculation of Wilcoxon Signed-Ranks Test Statistic di = Xi - M0 |di| Ri Ri(+) W = 7 + 2 + 2 = 11 42 - 42 = 0 39 - 42 = -3 3 7 n’ is the number of 42 - 42 = 0 nonzero differences. 45 - 42 = +3 3 7 7 You need it to get 42 - 42 = 0 the critical value(s). 43 - 42 = +1 1 2 2 Here n’ = 8. 43 - 42 = +1 1 2 2 40 - 42 = -2 2 4.5 40 - 42 = -2 2 4.5 42 - 42 = 0 39 - 42 = -3 3 7 41 - 42 = -1 1 2 First, you calculate the differences. Any zero differences are discarded, because zero does not have a sign. Calculate the absolute differences. Rank the absolute differences. The smallest absolute difference gets a rank of 1, the second smallest absolute difference gets a rank of 2, and so on. If you have ties, you average the ranks. Finally, sum the ranks associated with the positive differences. di = Xi - M0 42 - 42 = X 39 - 42 = -3 42 - 42 = X 45 - 42 = +3 42 - 42 = X 43 - 42 = +1 43 - 42 = +1 40 - 42 = -2 40 - 42 = -2 42 - 42 = X 39 - 42 = -3 41 - 42 = -1 First, you calculate the differences. Any zero differences are discarded, because zero does not have a sign.

  11. One-tail  = .05 n’ = 8 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 mean bar length of scale sets is less than 42 millimeters. μ = 42 or M = 42 μ < 42 M < 42 (5, 31) .05 Reject H0 Do not reject H0 .05 5 .05

  12. One-tail  = .05 n’ = 8 (5, 31) H0: H1: α = Decision Rule: If Wcomputed > 5 then do not reject H0, otherwise reject H0. Test Statistic: Decision: Conclusion: We have found ________________ evidence at the _____ level of significance that the mean bar length of scale sets is less than 42 millimeters. μ = 42 or M = 42 μ < 42 M < 42 .05 Reject H0 Do not reject H0 Do not reject H0 .05 5 Do not reject H0. insufficient .05

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