z检验与t检验的比较分析

1、本资料来源z检验与检验与t检验检验z test & t test Contentsn1. z test z检验检验n2. One-sample t test 单样本单样本t检验检验n3. Paired-samples t test 配对样本配对样本t检验检验n4. Two independent-samples t test 两独立样本两独立样本t检验检验n5. test When variances of the two samples are heterogeneous 方差不齐时两样本均数方差不齐时两样本均数 检验检验n6. Two type error in hypothesi

2、s test 假设检验中的两类错误假设检验中的两类错误tt1. z testpopulationsample 1sample 2sample rrXXX21(3 13)XXXzsampleXznobservation ：mean ：,1XXtdfnnS is unknown mean ：2,X Nnz-distribution versus t-distributiontFor very large samples, the t-test and z-test are identicalA sampling distribution for H0 showing the region of r

3、ejection for a = .05 in a 2-tailed z-test（双侧z检验）Xzn00:100H10:100HA sampling distribution for H0 showing the region of rejection for a = .05 in a 1-tailed z-test （单侧z检验）Xzn00:100H10:100H1. One-sample z testA sampling distribution for H0 showing the region of rejection for a = .05 in a 2-tailed z-test

4、（双侧z检验）0074.2723.38/6.5/ 100Xzn0:72H1:72H723.38,2*(1 normsdist (3.38)0.0007zP6.595%CI 74.2 1.96=100：(72.93，75.47)2. Two independent-samples z test两独立样本均数比较的两独立样本均数比较的z检验检验例例5-4 5-4 研究正常人与高血压患者胆固醇含量研究正常人与高血压患者胆固醇含量(mg%)(mg%)的资料如下的资料如下, ,试比较两组血清试比较两组血清胆固醇含量有无差别。胆固醇含量有无差别。正常人组正常人组 高血压组高血压组111506,180.6,

5、34.2nXS222142,223.6,45.8nXSn1. 1. 建立检验假设建立检验假设, , 确定检验水平确定检验水平n , ,即正常人与高血压患者血清胆固醇值总体均数相同即正常人与高血压患者血清胆固醇值总体均数相同; ;n , ,即正常人与高血压患者血清胆固醇值总体均数不同即正常人与高血压患者血清胆固醇值总体均数不同; ;na a = 0.05= 0.05n2. 2. 计算统计量计算统计量z z 值值012:H112:H22|180.6223.6|10.4034.2 /50645.8 /142zn3. 3. 确定确定P P 值值, , 作出推断结论作出推断结论n本例本例z z =10.

6、402.58,=10.402.58,故故P P 0.01,Descriptive StatisticsExplore)P值0.05W法与 D法我国国标GB4882-85推荐方法，Shapiro-Wilk W法（1965年）适用于Kolmogorov-Smirnov D法（？）适用于503 n501000n小于3或大于1000怎么办？H0: 样本对应的总体总体服从正态分布脂肪酸含量脂肪酸含量编号编号哥特里哥特里- -罗紫法罗紫法脂肪酸水解法脂肪酸水解法差值差值 d d1 10.8400.8400.5800.5800.2600.2602 20.5910.5910.5090.5090.0820.08

7、23 30.6740.6740.5000.5000.1740.1744 40.6320.6320.3160.3160.3160.3165 50.6870.6870.3370.3370.3500.3506 60.9780.9780.5170.5170.4610.4617 70.7500.7500.4540.4540.2960.2968 80.7300.7300.5120.5120.2180.2189 91.2001.2000.9970.9970.2030.20310100.8700.8700.5060.5060.3640.3640.2724.10 087ddS表 两种方法对乳酸饮料中的脂肪含量测

8、量结果H0：d d0 092. 7101087. 02724. 0nSdSdtdd262. 29 , 2/05. 0t应用条件n两变量对子之差来自随机、独立的正态分布总体2(0,)NEffect size(效应大小)How large is the effect (the difference)?nCohens dnSquared correlation coefficient (r2)Effect size: Cohens dnCohens d:nRepresents mean difference in standard deviation units 3.25=1.3052.49dddS

9、Effect size: Cohens dnSame guidelines for interpreting Cohens dEffect sizedSmall .20Medium.20 .80Large .80nPercent of variance explainednSymbol: r2 2224.52=0.2914.52+11trtdfEffect size: r2Effect size: r2nSame guidelines for interpreting r2Effect sizer2Small .254. Two independent-samples t test 两独立样本

10、t 检验应用于：Experimentaltreatment versus control实验研究 完全随机设计完全随机设计(completely random design)(completely random design)Existing groupsmales versus females观察研究 Goal ：Assess whether the means of the populations that two different samples came from are the same or different (or whether one is greater than th

11、e other, in a directional test)目的:比较两总体均数是否有差异。应用条件nRandom and independent samplesnNormalitynHomogeneity of variance两组变量值分别来自随机、独立的正态分布总体221222212(,)(,)NN 和且方差相等，即两独立样本两独立样本t t检验计算公式检验计算公式n 称为合并方差称为合并方差(combined/pooled variance(combined/pooled variance）1122121212121222121221122221(1)(1)(2()()11)2XXX

12、XCCXXXXXXtnnSSSSSnnnSnnnSnn2CS1212122XXXXtnnS表 5-2 25 名糖尿病患者两种疗法治疗后二个月血糖值(mmol/L) 编号 甲组血糖值(X2) 编号 乙组血糖值(X2) 1 8.4 1 5.4 2 10.5 2 6.4 3 12.0 3 6.4 4 12.0 4 7.5 5 13.9 5 7.6 6 15.3 6 8.1 7 16.7 7 11.6 8 18.0 8 12.0 9 18.7 9 13.4 10 20.7 10 13.5 11 21.1 11 14.8 12 15.2 12 15.6 13 18.7 例例5.3 255.3 25例糖尿

13、病患者随机分成两组，甲组单纯用药物治疗，乙组采用药物治例糖尿病患者随机分成两组，甲组单纯用药物治疗，乙组采用药物治疗合并饮食疗法，二个月后测空腹血糖疗合并饮食疗法，二个月后测空腹血糖(mmol/L)(mmol/L)如表如表5-2 5-2 所示，问两种疗法治所示，问两种疗法治疗后患者血糖值是否不同？疗后患者血糖值是否不同？由 原 始 数 据 算 得 :n1=12，X1=182.5，X12=2953.43，n2=13，X2=141.0， X22=1743.16，1X=X1/n1=182.5/12=15.21，2X=X2/n2=14.16/13=10.85 代入公式，得代入公式，得: :H0： 1=

14、 2，H1： 1 2，aa0.0515.21 10.852.639=12+13-2=231.652t，有统计学差异拒绝，023, 2/05. 0,05. 0069. 2HPttEffect size: Cohens dnCohens d:nRepresents mean difference in standard deviation units 1224.36218=1.05717.03311CXXdSEffect size: Cohens dnSame guidelines for interpreting Cohens dEffect sizedSmall .20Medium.20 .8

15、0Large .80nPercent of variance explainednSymbol: r2 2222.64=0.1032.6423trtdfEffect size: r2Effect size: r2nSame guidelines for interpreting r2Effect sizer2Small .25Excel计算方法Excel计算方法及结果SPSS计算方法SPSS计算结果方差齐性检验652. 1069. 2362. 4%CI95：5. test When variances of the two samples are heterogeneous 方差不齐时两样本均

16、数比较方差不齐时两样本均数比较 检验检验tt两样本方差齐性检验（homogeneity of variance test）1. 两样本方差齐性检验2. 2. LEVENE方差齐性检验（LEVENES TEST OF HOMOGENEITY OF VARIANCE）3. 将原样本观察值作离均差变换，或离均差平方变换然后进行完全随机设计的方差分析，其检验结果用于判断方差是否齐性。4. 因为后者对原数据是否为正态不灵敏，所以比较稳健，且该方法可用于后面章节方差分析的齐性检验。目前均推荐采用LEVENE方差齐性检验222201211221112222:()()1,1HHSFnnS双侧较大，（较小）22

17、220121122222112112220.05/2,(12,11):()12,16.1736;13,17.8210()13 112,12 1113.430,0.05,HHnSnSSFSFFP 双侧已知： 较大17.8210=1.1018，（较小） 16.1736查附表3， =1.1018方差齐同例例5.3 的方差齐性检验的方差齐性检验=FINV(0.025,12,11)例例5.4 5.4 两组小白鼠分别饲以高蛋白和低蛋白饲料，两组小白鼠分别饲以高蛋白和低蛋白饲料，4 4周后记录小白鼠体重增加周后记录小白鼠体重增加量量(g)(g)如表如表5-35-3所示，问两组动物体重增加量的均数是否相等？所

18、示，问两组动物体重增加量的均数是否相等？ 高蛋白高蛋白50504747434339395151434348485151424250504343低蛋白低蛋白36363838 3 33838363639393737353533333737393934343636H0: 12 22，H1: 12 22，a a0.05 402. 5269. 3659.172221SSF 1n n-1= 12-1=11-1= 12-1=11，2 2 = = n n-1= 13-1 =12-1= 13-1 =12， 查附表查附表3 3F F界值表，界值表， F F0.05/2,(11,12)0.05/2,(11,12)

19、3.34 =3.32153.34 =3.3215F F F F0.05/2,(11,12)0.05/2,(11,12)P P 0.05 0.05差别有统计学意义，即方差差别有统计学意义，即方差不齐同不齐同=FINV(0.05/2,11,12)=2*FDIST(5.402,11,12)=0.007Excel方差齐性检验Excel方差齐性检验t t 检验实例（表检验实例（表5.45.4数据）数据）22221122222222112212(/)(17.659/12 3.269/13)14.6874(/ )(/)(17.659/12)(3.269/13)12 113 111SnSnSnSnnn1222

20、121245.7536.5387.01717.6593.2691213XXtSSnn H H0 0： 1 1 2 2，H H1 1： 1 1 2 2，a a 0.050.05=TDIST(7.017,15,2)P=4.16E-06差异有统计学意义，高蛋白喂养的动物体重高于低蛋白组Excel做方差不齐的t 检验Excel做方差不齐的t 检验SPSS计算方法SPSS计算结果方差不齐的结果读取正态检验结果Tests of NormalityTests of Normalityb b.24412.048.88812.111VAR00001StatisticdfSig.StatisticdfSig.Ko

21、lmogorov-SmirnovaShapiro-WilkLilliefors Significance Correctiona. VAR00002 = 1.00b. Tests of NormalityTests of Normalityb b.15213.200*.95113.606VAR00001StatisticdfSig.StatisticdfSig.Kolmogorov-SmirnovaShapiro-WilkThis is a lower bound of the true significance.*. Lilliefors Significance Correctiona. VAR00002 = 2.00b. The t Test formula12121212122221122121212()()2(1)(1)1111(2)XXCXXXXtnnSnSnSSSnnnnnnDifference in the means1standard errorXXtnsOn