统计应用软体StatisticalComputationsandAnalysis000001.ppt
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1、統計應用軟體 Statistical Computations and Analysis,Lecture 7: Hypothesis Testing (II),常見的假說檢定,Case 1: Test on proportions Single population Two populations Case 2: Test on means with variance known or sample size 30 Single population Two populations Case 3: Test on means of Normal distributions with varia
2、nce unknown Single population Two populations More than two populations Case 4: Test on variances of Normal distributions Single population Two populations More than two populations,常見的假說檢定,Case 1: Test on proportions: z-test Z-test (prop.test) Case 2: Test on means with variance known or sample siz
3、e 30 Z-test (no available function) Case 3: Test on means of Normal distributions with variance unknown Single population: t-test (t.test) Two populations: t-test (t.test) More than two populations: ANOVA (lm or aov) Case 4: Test on variances of Normal distributions Single population: Chi-squared te
4、st (no available function) Two populations: F-test (var.test) More than one populations: Bartlett Test (bartlett.test),Case 3,Single population: t-test (t.test) Two populations: Dependent samples: paired t-test (t.test) Independent samples: : pooled t-test (t.test) : Welchs t-test (t.test) More than
5、 two populations: ANOVA,單一樣本均值的顯著性檢定 t-test,正常成人血中平均膽固醇為180mg/dL (假設符合常態分布)。今調查某地區 16 位成人平均膽固醇為200mg/dL, 變方為 2500; 問此地區平均膽固醇是否與180mg/dL有差異? H0: = 180 Ha: 180 However,未知! 需用樣本變方估計之。,See Corollary in Lecture 5.,單一樣本均值的顯著性檢定 t-test,H0: = 180 Ha: 180 = 200, S2 = 2500, n = 16 T = p-value = P( 200 | H0) x
6、 2 = 2*(1-pt(1.6,15) = 0.1304,單一樣本均值的顯著性檢定 t-test,Test by t.test(x, y = NULL, alternative = c(“two.sided“, “less“, “greater“), mu = 0, paired = FALSE, var.equal = FALSE, conf.level = 0.95, .),單一樣本均值的顯著性檢定 t-test,EX. 膽固醇檢測範例:,兩樣本均值顯著性檢定,例1: 今欲比較洗腎病人透析前後之體重是否不同,6位病人腎臟透析前後體重如下表。,兩樣本均值顯著性檢定,A, B兩種嬰兒奶粉,A
7、 奶粉試用於 9 個初生男嬰,B 奶粉試用於 10 個初生男嬰。一個月後嬰兒體重增加如下:,兩樣本均值顯著性檢定,成對樣本: 洗腎病人透析先後體重變化:同一位病人透析先後之體重均測量自同一病人, 因此, 同一病人二次體重是配對的 (Paired) 觀測值。 同一試驗單位在不同環境所得之觀測值為成對樣本! 非成對樣本: A.B 兩種嬰兒奶粉: 吃 A 奶粉的 9 位男嬰與吃 B 奶粉的10 位男嬰是不同; 吃 A 奶粉 9 位男嬰體重增加之觀測值與吃 B 奶粉 10 位男嬰體重增加之觀測值無關獨立樣本 由不同試驗單位所得之觀測值為非成對樣本!,成對樣本均值顯著性檢定,針對每一試驗單位, 計算其成
8、對觀測值差: Di = Xi1 Xi2 將 Di 視為一組新的樣本。欲檢驗兩成對樣本均值是否相同, 等於檢驗 Di 的平均值是否為0! 回到單一樣本均值顯著性檢定! Di N(D , D2),單一樣本均值的顯著性檢定,H0: =0 當 (i) 觀測值來自常態分布且變方已知, 或 (ii) 樣本大小 n30, 則採用 z-test (no R function) 當樣本大小 n 30, 則採用 t-test (t.test)。,成對樣本均值顯著性檢定,洗腎病人透析前後體重範例: H0: D = 0 Ha: D 0 統計值: ( Di 的平均值) 當 n 30, 且D 未知, Under H0 ,
9、成對樣本均值顯著性檢定,洗腎病人透析前後體重範例: = 3.22, SD2 = 2.4737, n = 6 T = p-value = P( 3.22|H0) x 2 = 2*(1-pt(5.015,6-1) = 0.004,成對樣本均值顯著性檢定,Test by t.test(x, y = NULL, alternative = c(“two.sided“, “less“, “greater“), mu = 0, paired = FALSE, var.equal = FALSE, conf.level = 0.95, .),paired = FALSE,paired = TRUE,成對樣本
10、均值顯著性檢定,EX.洗腎病人透析前後體重範例:,非成對樣本均值顯著性檢定,由不同試驗單位所得觀測值為非成對樣本! X1 N(1 , 12) X2 N(2 , 22) By Theorem 1(i): N(1-2 , 12/n1 +22/n2),非成對樣本均值顯著性檢定,若 12, 22 已知, 可利用z-test來檢驗兩族群均值是否相同。 若 12, 22 未知: (1) if 12 = 22 = 2, V( ) = 12/n1 +22/n2 = 2(1/n1+1/n2) 則 其中,非成對樣本均值顯著性檢定,(2) 若12 22 , 則無法直接套用 t 分布! 修正方法有以下兩種 (任選其一
11、即可) Welchs t-test: (1) 加權 T 值 (2) 加權 t 分布之自由度,非成對樣本均值顯著性檢定,How to know if 12 = 22 ? 假說檢定: H0: 12 = 22 Ha: 12 22 整理: 非成對樣本均值顯著性檢定 Step1: Test H0: 12 = 22 Step2: Test H0: 1 = 2,非成對樣本均值顯著性檢定,嬰兒奶粉範例: Step1: Test H0: 12 = 22,非成對樣本均值顯著性檢定,嬰兒奶粉範例: Step2: Test H0: 1 = 2 t.test(x, y = NULL, alternative = c(“
12、two.sided“, “less“, “greater“), mu = 0, paired = FALSE, var.equal = FALSE, conf.level = 0.95, .),var.equal = FALSE,var.equal = TRUE,非成對樣本均值顯著性檢定,非成對樣本均值顯著性檢定,例:痛風病人與正常人血中尿酸量之變異,非成對樣本均值顯著性檢定,痛風病人與正常人血中尿酸量範例: Step1: Test H0: 12 = 22,非成對樣本均值顯著性檢定,痛風病人與正常人血中尿酸量範例: Step2: Test H0: 1 = 2,Two Populations 均
13、值比較,例: 欲比較兩種飼料對天竺鼠體重之影響, 將8隻天竺鼠隨機指派食用兩種飼料(4隻飼料A; 4隻飼料B),兩星期後體重增加之觀測值(g),Three Populations 均值比較,例: 欲比較三種飼料對天竺鼠體重之影響, 將12隻天竺鼠隨機指派食用兩種飼料(4隻飼料A; 4隻飼料B; 4隻飼料C),兩星期後體重增加之觀測值(g),均值比較,H0: A = B vs Ha: A B 1 = 0.05: probability of type I error for 1st test 1 1: probability of no type I error for 1st test H0:
14、 B = C vs Ha: B C 2 = 0.05: probability of type I error for 2nd test 1 2: probability of no type I error for 2nd test H0: A = C vs Ha: A C 3 = 0.05: probability of type I error for 3rd test 1 3: probability of no type I error for 3rd test,均值比較,Overall type I error = 至少有一次 test 犯 type I error = 1 P(3
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