利用Minitab的基础统计29114

,附录,1.,利用,Minitab,的基础统计,LGENT SIX SIGMA TASK TEAM,基础统计,BeforeAfter,58.560.0,60.354.9,61.758.1,69.062.1,64.058.5,62.659.9,56.754.4,例题,1:,求基本的统计量,开发新产品的公司，为了知道它的效果，把7名主妇为对象:,实验结果他们的体重变化如右图,为了知道体重变化的程度？,Stat/basic statistics/display descriptive statistics,附录,1-,1,/41,LG Electronics/LGENT 6TASK TEAM,基础统计,Descriptive Statistics,Variable N Mean Median TrMean StDev SE Mean,Before 7 61.83 61.70 61.83 4.01 1.52,After 7 58.27 58.50 58.27 2.79 1.05,Variable Minimum Maximum Q1 Q3,Before 56.70 69.00 58.50 64.00,After 54.40 62.10 54.90 60.00,Trimmed Mean(,整理平均,),Minitab removes the smallest 5%and the largest 5%of the values(rounded the nearest integer),and then averages the remaining data,SE Mean(Standard Error of Mean)StDev/N,例题,1:,求基本的统计量,附录,1-,2,/41,LG Electronics/LGENT 6TASK TEAM,基础统计,calc/calculator,BeforeAfter,58.560.02.5641,60.354.9-8.9552,61.758.1-5.8347,69.062.1-10.0000,64.058.5-8.5938,62.659.9-4.3131,56.754.4-4.0564,附录,1:,求基本统计量,附录,1-,3,/41,LG Electronics/LGENT 6TASK TEAM,基础统计,Random,抽取工厂所生产的,21,个产品后，测定结果如下,.,求例题,1,中求的统计量外的多种的统计量？,calc/column statistics or row statistics,stat/basic statistics/store descriptive statistics,例题,2:,求多种统计量,附录,1-,4,/41,LG Electronics/LGENT 6TASK TEAM,基础统计,执行这样的过程，计算的,Data Window,以列别出现,.,例题,2:,求多种统计量,附录,1-,5,/41,LG Electronics/LGENT 6TASK TEAM,基础统计,例题,2:,求多种统计量,统计量要在,Session Window,上显示,?,Manip/display data,选择,所愿的,统计量,Data Display,Row size Mean1 SEMean1 StDev1 Variance1 Q1_1 Median1 Q3_1,1 27.5 28.2190 0.424467 1.94515 3.78362 26.9 28.4 29.65,2 27.6,3 27.6,4 30.3,5 28.8,6 22.9,7 26.6,8 31.8,9 28.4,10 26.9,11 30.0,12 31.2,13 29.4,14 28.0,15 26.8,16 28.8,17 28.5,18 26.3,19 29.9,20 26.9,21 28.4,Row IQR1 Sum1 Minimum1 Maximum1 Range1 N1,1 2.75 592.6 22.9 31.8 8.9 21,Data Display,Row Mean1 SEMean1 StDev1 Variance1 Q1_1 Median1 Q3_1 IQR1,1 28.2190 0.424467 1.94515 3.78362 26.9 28.4 29.65 2.75,Row Sum1 Minimum1 Maximum1 Range1 SSQ1,1 592.6 22.9 31.8 8.9 16798.3,附录,1-,6,/41,LG Electronics/LGENT 6TASK TEAM,基础统计,例题,3:,概率分布的活用,-2,项分布,(1),感染到某种传染病能恢复的概率为,0.4,。,15,名被传染时，,5,名恢复的概率为,?,Calc/probability distributions/binominal,试行,成功概率,成功次数,Probability Density Function,Binomial with n=15 and p=0.400000,x P(X=x),5.00 0.1859,附录,1-,7,/41,LG Electronics/LGENT 6TASK TEAM,基础统计,例题,3:,概率分布的活用,-2,项分布,(2),感染到某种传染病能恢复的概率为,0.4.,在,15,名当中至少,10,名能恢复的概率为？,Calc/probability distributions/binominal,执行次数,成功概率,成功次数,Data Display,K2 0.0338333,*,在,Minitab,上，为了求,P(Xx),求,P(Xx),的值后,计算,1-,P(Xx),，求概率,.,附录,1-,8,/41,LG Electronics/LGENT 6TASK TEAM,基础统计,例题,3:,概率分布的活用,-,泊松分布,显示器的某种特性服从平均为,10,的泊松分布时，求,P(X14),?,Calc/probability distributions/Poisson,Cumulative Distribution Function,Poisson with mu=10.0000,x P(X=x),14.00 0.9165,附录,1-,9,/41,LG Electronics/LGENT 6TASK TEAM,基础统计,例题,3:,概率分布的活用,-,正态分布,显示器的某种特性为正态分布时，求,P(X-0.28),?,Calc/probability distributions/normal,Cumulative Distribution Function,Normal with mean=0 and standard deviation=1.00000,x P(X=x),-0.2800 0.3897,附录,1-,10,/41,LG Electronics/LGENT 6TASK TEAM,基础统计,例题,4:,母平均的置信区间,-,知道母分散时,从分散为,225,的正态分布母总体取出,25,个样本，得到如下资料,样本的平均为,64.32,。,求出母平均的,95%,的置信区间,Stat/basic statistics/1 Sample Z,556978,474952,857333,717089,765551,5352,7187,3870,6781,7363,Z Confidence Intervals,The assumed sigma=15.0,Variable N Mean StDev SE Mean 95.0%CI,C1 25 64.32 15.11 3.00 (58.44,70.20),附录,1-,11,/41,LG Electronics/LGENT 6TASK TEAM,基础统计,例题,4:,母平均的置信区间,-,不知道母分散的情况,推定新显示器的寿命时间，抽取,9,台，寿命时间以小时为单位测定的结果如下,假设寿命时间为正态分布时，求母平均寿命的,90%,的置信区间？,Stat/basic statistics/1 Sample t,50000,51000,54000,52000,54000,50000,53000,52000,52000,T Confidence Intervals,Variable N Mean StDev SE Mean 90.0%CI,C2 9 52000 1500 500 (51070,52930),附录,1-,12,/41,LG Electronics/LGENT 6TASK TEAM,基础统计,例题,5:,母平均差异的置信区间,-,两个母分散相同情况,对两个公司的,Hinge Force,差异的分析，取出各,20,个样本的结果如下,.,求两个公司,Hinge Force,差异的,95%,置信区间,(,假设两个母分散是一样的,),Stat/basic statistics/2 Sample t,Two Sample T-Test and Confidence Interval,Two sample T for PK vs RPM,N Mean StDev SE Mean,PK 20 2.1220 0.0387 0.0087,RPM 20 2.1395 0.0233 0.0052,95%CI for mu PK-mu RPM:(-0.0380,0.0030),T-Test mu PK=mu RPM(vs not=):T=-1.73 P=0.091 DF=38,Both use Pooled StDev=0.0320,假设,母分散,是一样的,附录,1-,13,/41,LG Electronics/LGENT 6TASK TEAM,基础统计,例题,5:,母平均差异的置信区间,-,两个母分散不同的情况,为了知道对两个公司的,Hinge Force,的差异，推出各,20,个标本的结果,.,对两个公司,Hinge Force,的差异，求出,95%,的置信区间,Stat/basic statistics/2 Sample t,Two Sample T-Test and Confidence Interval,Two sample T for PK vs RPM,N Mean StDev SE Mean,PK 20 2.1220 0.0387 0.0087,RPM 20 2.1395 0.0233 0.0052,95%CI for mu PK-mu RPM:(-0.0381,0.0031),T-Test mu PK=mu RPM(vs not=):T=-1.73 P=0.093 DF=31,假设,母分散,不一样的,附录,1-,14,/41,LG Electronics/LGENT 6TASK TEAM,基础统计,例题,6:,确认利用母平均的改善结果,运用英语分数的向上,Program，,比较,Program,进行前和进行后的英语分数，研讨向上,Program,在实际运用上,是否有用,.,Program,进行前,/,后的分数如下时，研讨,Program,对英语分数是否有用,(,各随机抽取,10,个样本,),Stat/basic statistics/paired t,beforeafter,7681,6052,8587,5870,9186,7577,8290,6463,7985,8883,Paired T-Test and Confidence Interval,Paired T for before-after,N Mean StDev SE Mean,before 10 75.80 11.64 3.68,after 10 77.40 12.18 3.85,Difference 10 -1.60 6.38 2.02,95%CI for mean difference:(-6.16,2.96),T-Test of mea