# 兩獨立母體成功比例差- Z檢定(大樣本)：說明

## Presentation on theme: "兩獨立母體成功比例差- Z檢定(大樣本)：說明"— Presentation transcript:

10 - 7

n1 - 1 = degrees of freedom = Variance of Sample 2 n2 - 1 = degrees of freedom F

Two Sets of Degrees of Freedom df1 = n1 - 1; df2 = n2 - 1

Two Sets of Degrees of Freedom df1 = n1 - 1; df2 = n2 - 1 臨界值: FL( ) and FU( ) FL = 1/FU* (*degrees of freedom switched) n1 -1, n2 -1 n1 -1 , n2 -1

Two Sets of Degrees of Freedom df1 = n1 - 1; df2 = n2 - 1 臨界值: FL( ) and FU( ) FL = 1/FU* (*degrees of freedom switched) Reject H0 Reject H0 Do Not Reject a/2 a/2 FL FU F n1 -1, n2 -1 n1 -1 , n2 -1

Df1 = df2 = 24 臨界值: Reject Reject .025 .025 F 0.415 2.33 1.25

Df1 = df2 = 24 臨界值: Reject Reject .025 .025 F 0.415 2.33 1.25

H0: s12 = s22 H1: s12 ¹ s22  = 0.05 Df1 = df2 = 24 臨界值: Reject Reject Do not reject at a = 0.05 .025 .025 F 0.415 2.33 1.25

H0: s12 = s22 H1: s12 ¹ s22  = 0.05 Df1 = df2 = 24 臨界值: Reject Reject Do not reject at a = 0.05 .025 .025 沒有足夠證據證明兩母體變異數有顯著差異。 F 0.415 2.33 1.25

F Test in PHStat

F 檢定:單尾檢定(左尾) H0: s12 ³ s22 H1: s12 < s22 a = .05
Degrees of freedom switched Reject a = .05 F

F 檢定:單尾檢定（右尾） H0: s12 £ s22 H1: s12 > s22 a = .05 F Reject a = .05

2.若母體分配為常態分配且兩變異數皆已知，則不論樣本大小為多少皆用常態分配處理。

Reject a =.05 D = .084 a =.05 臨界值= df = n - 1 = 9 1.8331 3.15

Reject a =.05 D = .084 a =.05 臨界值= df = n - 1 = 9 1.8331 3.15 檢定統計量觀察值

Reject a =.05 D = .084 a =.05 臨界值= df = n - 1 = 9 1.8331 3.15 決策: Reject H0 檢定統計量觀察值

Reject a =.05 D = .084 a =.05 臨界值= df = n - 1 = 9 1.8331 3.15 決策: Reject H0 檢定統計量觀察值 結論: The new software package is faster.

Wilcoxon等級合檢定：兩母體分配（中位數）是否相同？

Wilcoxon等級合檢定：兩母體分配（中位數）是否相同？

Wilcoxon等級合檢定：程序 分別由母體1隨機抽取 n1個樣本，母體2抽取n2 個樣本

Wilcoxon等級合檢定：程序 將n1 + n2 個樣本觀察值，混合指派等級順序 Ri 加總各個樣本之等級Tj

Wilcoxon等級合檢定：H0、H1之設定
Two -Tail Test 雙尾檢定 H0: M1 = M2 H1: M1 ¹ M2 Do Not Reject Reject Reject T1L T1U M1 = median of population 1 M2 = median of population 2

Wilcoxon等級合檢定：H0、H1之設定

Wilcoxon等級合檢定：H0、H1之設定
Right -Tail Test 單尾檢定： H0: M1 £ M2 H1: M1 > M2 右尾檢定 Do Not Reject Reject T1U M1 = median of population 1 M2 = median of population 2

Wilcoxon等級合檢定： 例題7 你是一位生產企劃人員，你想要了解兩間工廠之生產作業速率(% of capacity)中位數是否相同？第一間工廠之作業速率觀察值為71, 82, 77, 92, 88，第二間工廠之作業速率為 85, 82, 94 & 97。 在 0.10 顯著水準之下，兩 間工廠之作業速率中位數是否 有顯著差異?

Wilcoxon等級合檢定：例題7題解 Factory 1 Factory 2 Rate Rank Rate Rank 71 1 85 5
82 Tie 3 3.5 82 Tie 4 3.5 77 2 94 8 92 7 97 9 88 6 Rank Sum T2=19.5 T1=25.5

Lower and Upper Critical Values T1 of Wilcoxon Rank Sum Test
One-Tailed Two-Tailed 4 5 .05 .10 12, 28 19, 36 .025 11, 29 17, 38 .01 .02 10, 30 16, 39 .005 --, -- 15, 40 6

Wilcoxon等級合檢定：例題7題解 H0: M1 = M2 H1: M1 ¹ M2

Wilcoxon等級合檢定：例題7題解 H0: M1 = M2 H1: M1 ¹ M2 a = .10 n1 = 4 n2 = 5 臨界值:
Do Not Reject Reject Reject 12 28

Wilcoxon等級合檢定：例題7題解 檢定統計量:
H0: M1 = M2 H1: M1 ¹ M2 a = .10 n1 = 4 n2 = 5 臨界值: 檢定統計量: T1 = = 25.5 (Smallest Sample) Do Not Reject Reject Reject 12 28

Wilcoxon等級合檢定：例題7題解 檢定統計量:
H0: M1 = M2 H1: M1 ¹ M2 a = .10 n1 = 4 n2 = 5 臨界值: 檢定統計量: 決策: T1 = = 25.5 (Smallest Sample) Do not reject at a = 0.10 Do Not Reject Reject Reject 12 28

Wilcoxon等級合檢定：例題7題解 檢定統計量:
H0: M1 = M2 H1: M1 ¹ M2 a = .10 n1 = 4 n2 = 5 臨界值: 檢定統計量: 決策: 結論: T1 = = 25.5 (Smallest Sample) Do not reject at a = 0.10 Do Not Reject There is no evidence medians are not equal. Reject Reject 12 28

Wilcoxon等級合檢定： 常態分配近似法(Large Sample)

Wilcoxon符號等級檢定：成對母體分配是否相同？
 符號等級檢定 符號等級檢定不僅考慮差值的正負符號，同時考慮差值大小的等級的檢定方法。

Wilcoxon符號等級檢定：小樣本  符號等級檢定統計量(小樣本)
R+或R-：R+為正的D值的等級和， R-為負的D值的等級和。R=Min.{R+, R-}  決策法則 1.雙尾檢定 當RR/2時，則拒絕H0，否則接受H0。 R/2為臨界值。 2.單尾檢定 當RR時，則拒絕H0 ，否則接受H0 。 R為臨界值。

Wilcoxon符號等級檢定：大樣本  符號等級檢定統計量(大樣本n30)  決策法則 1.雙尾檢定
Z-Z/2時拒絕H0 ， Z-Z/2則接受H0 。 2.單尾檢定 Z-Z時拒絕H0 ， Z-Z則接受H0

Wilcoxon符號等級檢定：例題8 泡泡清潔用品公司宣稱其所生產的沐浴乳每瓶重量的中位數為300公克，今隨機抽查12瓶，得重量如下：

Wilcoxon符號等級檢定： 例題8題解 每瓶的重量與300之差的絕對值由小排到大，給予等級：

Wilcoxon符號等級檢定： 例題8題解 每瓶的重量與300之差的絕對值由小排到大，給予等級：

Wilcoxon符號等級檢定： 例題8題解 每瓶的重量與300之差的絕對值由小排到大，給予等級：

Wilcoxon符號等級檢定： 例題8題解 每瓶的重量與300之差的絕對值由小排到大，給予等級：