相互独立的两组正态总体

1、 相互独立的两组正态总体相互独立的两组正态总体Independent samples T testThere are two independent samples of resistors. Their resistant values are as follows: A: 140, 138, 143, 142, 144, 137B: 135, 140, 142, 136, 138, 140Is there a significant difference between the means of the two samples? Paired samples T test配对样本假设检验

2、两个样本之间不相互独立：配对数据是指对两个总体X, Y进行联合观测取得的数据，用(X1,Y1)， (X2,Y2), , (Xn, Yn) 表示。D = (X-Y) N(d , 2)d = 1-2 为比较两种轮胎的耐磨性，随机地从A牌与B牌中个抽取一个安装在同一辆车上，共安装5辆车进行轮胎的寿命试验，记录下每个轮胎磨损到不能用为止的耐用里程，得如下数据：(10.6, 10.2), (9.8, 9.4), (12.3, 11.8), (9.7, 9.1), (8.8, 8.3). 试以 0.05的显著水平，检验这两种轮胎的耐磨性能是否存在显著差异。解：设A牌轮胎的耐用里程为X，B牌的耐用里程为Y。

3、 d = X-Y d: 0.4, 0.4, 0.5, 0.6, 0.5 SPSSAnalyze| Compare means | Paired samples T test 某校从一年级学生中抽取16名女生，按智商（IQ)、家庭的社会经济情况、是否独生子女、环境等条件进行配对，每对中一名上过幼儿园，另一名未上过。对这16名学生进行某种心理学中的敏度测验，分数如下。问这一调查是否证明幼儿园在给一年级作预备上有益？12345678上过8374676470678164未上7874636668637765 解：上过幼儿园者分数为X, 未上过幼儿园者分数为Y。 配对差异为d，其真值为d。d = X-YH

4、ypothesis H0: d = 0, H1: d 0One way anova (analysis of variance) 方差分析 12344148404044495039484944464349484642455041Four samples (groups) of 5 wing length of houseflies randomly selected are as follows. Are the data homogeneous?One way anova 12344148404044495039484944464349484642455041Grand mean43.648

5、.046.442.445.129.212.075.245.2161.619.64meansquareGroup#Sum of sqaure within groups: 161.6df=16Sun of sqaure among groups: 19.64 * 5 = 98.2df=3square 某灯泡厂用四种不同配方制成的灯丝生产了四批灯泡，在每批灯泡中随机地抽取若干个测得其使用寿命（单位：小时），所得数据如表。试问这四种灯丝生产的灯泡使用寿命有无显著差异。灯泡使用寿命数据表灯泡使用寿命数据表灯丝12345678甲1600161016501680170017201800乙158016401

6、64017001750丙14601550160016201640174016601820丁151015201530157016801600anova Analysis of variance. The purpose of it is to determine whether there is significant difference among groups of data through the comparison of the variance within groups and the variance among groups. One way anovaA geneticis

7、t recorded the following measurements taken on two-week-old mice of a particular strain. Is there evidence that the variance among mice in different litters is larger than one would expect on the basis of the variability found within each litter? Litter #1 19.49 20.62 19.51 18.09 22.75 #222.94 22.15

8、 19.16 20.98 23.13 #323.06 20.05 21.47 14.90 19.70 #415.09 21.48 22.48 18.79 19.70 #516.72 19.22 16.62 20.74 21.82 #620.00 19.79 21.15 14.88 19.79 #721.52 20.37 21.93 20.14 22.28 有5种油菜品种，分别在4块实验田上种植，所得亩产量如表所示（kg 单位）。试问（1）不同油菜品种对平均亩产影响是否显著？（2）不同块田地的亩产量是否有显著差异。品种 田块1234A1256222280298A2244300290275A325

9、0277230322A4288280315259A5206212220212连续型随机变量的分布The distribution of continuous variables1. 正态分布, the normal distribution2. 2 分布, chi-square distribution3. t分布, t-distribution4. F分布, F-distribution表的构成总是把数值较大的一个S2 作为分子, 对应于n1。故 F值总是大于1，Appendix of TablesContinuous variables1. The normal distribution

10、function2. T-distribution critical values3. Chi-square distribution critical values4. F-distribution critical valuesCalculations1.Calculate probability2.Calculate confidence interval3.Calculate necessary sample size4.Hypothesis test:Given significant value: compare critical valueDirectly calculate s

11、ignificant level: compare with 0.05It is important to realize that small probability may occur sometimes, that is a reason that we have two types of errors.Statistics Large sample hypothesis test Increasing the sample size increases the sensitivity of a statistical test. Small sample inference If to