第十章非参数统计.pptVIP

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第十章非参数统计.ppt

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第十章非参数统计

111 112 115 112 As a result of this class, you will be able to ... As a result of this class, you will be able to ... 12 Distribution has different shapes. Example 1: If inspecting 5 items the Probability of a defect is 0.1 (10%), the Probability of finding 0 defective item is about 0.6 (60%). If inspecting 5 items the Probability of a defect is 0.1 (10%), the Probability of finding 1 defective items is about .35 (35%). Example 2: If inspecting 5 items the Probability of a defect is 0.5 (50%), the Probability of finding 1 defective items is about .18 (18%). Note: Could use formula or tables at end of text to get Probabilities. Assume that the population is normally distributed. Allow students about 10 minutes to solve this. Note: More than 5 have been sold (6.4), but not enough to be significant. 51 n’ is the number of non-zero difference scores. Look up n’ = 5 in table; not 6! Rejection region does not include cut-off. Look up n’ = 5 in table; not 6! Rejection region does not include cut-off. Look up n’ = 5 in table; not 6! Rejection region does not include cut-off. 符号秩检验的解决 H0: 相同分布. Ha: 当前分布向右移动 ? = .05 n = 5 (not 6; 1 elim.) 关键值): 检验统计量: 判定: 结论: T0 拒绝H0 不拒绝H0 符号秩检验计算表 X1i X2i Di |Di | Ri Sign Sign Ri 9.98 9.88 9.88 9.86 9.90 9.83 9.99 9.80 9.94 9.87 9.84 9.84 +0.10 +0.02 +0.07 +0.19 +0.07 0.00 0.10 0.02 0.07 0.19 0.07 0.00 + + + + + ... +4 +1 +2.5 +5 +2.5 Discard Total T+= 15, T–= 0 4 1 2 2.5 5 3 2.5 ... 符号秩检验的解决 H0: 相同分布. Ha: 当前分布向右移动 ? = .05 n = 5 (not 6; 1 elim.) 关键值): 检验统计量: 判定: 结论: T0 拒绝H0 不拒绝H0 Wilcoxon Signed Rank Table (Portion) One-Tailed Two-Tailed n = 5 n = 6 n = 7 .. a = .05 a = .10 1 2 4 .. a = .025 a = .05 1 2 .. a = .01 a = .02 0 .. a = .005 a = .01 .. n = 11 n = 12 n = 13 : : : : 符号秩检验结果 H0:相同分布. Ha:当前分布向右移动 ? = .05 n’ = 5 (not 6; 1 elim.) Critical Value(s): 检验统计量: 判定: 结论: 1 T0 T – = 0 ? = .05拒绝原假设有证据表明新的方法更快拒绝H0 不拒绝H0 思考某机器制造厂在车削某个关键零件中，为了提高合格品的产量，对16名操作工脱产培训了4周，表记录了他们培训前后的平均日产量。试用符号秩检验法检验培训是否对产量的提高起作用？（显著水平0.0