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单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,MINITAB培训,(工程师级),-Basic Statistics 模块,此模块有如下功能:,模块功能,Use Minitabs basic statistics capabilities for calculating basic statistics and for simple estimation and hypothesis testing with one or two samples.The basic statistics capabilities include procedures for:,Calculating or storing descriptive statistics,Hypothesis tests and confidence intervals of the mean or difference in means,Hypothesis tests and confidence intervals for a proportion or the difference in proportions,Hypothesis test for equality of variance,Measuring association,Testing for normality of a distribution,打开一个软件自带的例子学习,打开Quality文件,数据如下,选择 显示描述性统计,填完自变量和因变量后,点击 Graphs按钮,点击 OK 后,显示出 箱线图 和 统计数据,解释,每个箱,线,线图最,低,低点表,示,示当天,最,最小值,最高,点,点表示,最,最大值,箱子,高,高,低,点,点分别,表,表示3/4,1/4,数,数字大,小,小,箱,中,中间一,横,横表示,当,当天几,个,个数据,的,的中位,数,数,在Session窗,口,口中具,体,体显示MEAN平均,值,值,StDev标,准,准差,Median,中,中位数,等,等信息.,选择,存,存贮描,述,述性统,计,计,填完自,变,变量和,因,因变量,后,后,点,击,击 OK,多了些,数,数据,自,自动将,统,统计的,数,数据如,平,平均值,和,和每日,数,数据量N=10存放,在,在C3-C5,列,列,用 图,形,形概要,方,方式,如图,点,点击OK,Theleftfour graphs:histogram of datawithanoverlaidnormalcurve(带常,态,态曲线,的,的柱状,图,图),boxplot箱,线,线图),95%confidence intervalsformean,and 95%confidence intervalsforthe median.(平,均,均值和,中,中位数95%,的,的置信,区,区间范,围,围图),右边第,一,一块数,据,据是正,态,态性检,验,验.显,示,示 P-Value=0.661,大于0.05,所,以,以数据,符,符合正,态,态分布.,第二块,数,数据是,平,平均值,偏态,等,等,第三块,数,数据是,最,最小最,大,大等,第四块,数,数据是,平,平均值,方差,的,的95%置信,区,区间等,区间估,计,计与假,设,设检验,例 假,设,设一个,物,物体的,重,重量未,知,知,为,了,了估计,其,其重量,用一,个,个天平,称,称5次,得到,重,重量5.525.485.645.515.45,假,设,设结果,符,符合,标准差0.1,的正态,分,分布,求,求,均值是,否,否是5.5,和重量,置,置信水,平,平为95%的,置,置信区,间,间,输入数,据,据后,按,按图所,示,示,按图所,示,示将例,子,子中数,据,据填入,后,后点击OK,得到分,析,析结果,见,见Session窗,口,口,分析,One-Sample Z:C1,Test of mu=5.5 vs not=5.5,Theassumedstandarddeviation=0.1,VariableNMeanStDevSEMean95%CIZP,C155.520000.072460.04472(5.43235,5.60765)0.450.655,由于假,设,设均值,为,为5.5,而,置,置信区,间,间(5.43235,5.60765)包,含,含了5.5,故,故假设,成,成立,可,可以认,为,为重量,为,为5.5,当前一,例,例子中,方,方差未,知,知时(,这,这也是,通,通常的,情,情况),求重,量,量是否,可,可认为,是,是5.5,如图填,上,上数据,点击OK,结果在Session窗口,中,中显示,解释,One-Sample T:C1,Testof mu=5.5 vsnot=5.5,VariableNMeanStDevSE Mean95%CITP,C15 5.520000.072460.03240(5.43003,5.60997)0.620.570,同样的,5.5在,(5.43003,5.60997)范围,中,中,可认为,假,假设均值5.5成立.,一个新例子,公司购买了,同,同型号的两,台,台机床(可,认,认为其加工,的,的零件尺寸,服,服从同方差,的,的正态分布),抽检两,台,台机床加工,的,的轴,分析,两,两台机床加,工,工的轴有无,明,明显差异,机床甲:20.5/19.8/19.7/20.4/21.1/20/19/19.9,机床乙:20.7/19.8/19.5/20.8/20.4/19.6/20.2,输入数据于,两,两列中,Two-sampleT for C1vs C2,NMeanStDev SEMean,C1 820.0500.6260.22,C2 720.1430.5220.20,Difference=mu(C1)-mu(C2),Estimate for difference:-0.092857,95%CI fordifference:(-0.738860,0.553146),T-Test ofdifference=0(vs not=):T-Value=-0.31 P-Value=0.760DF=12,由于假设两,均,均值相等,差,差为0,在(-0.738860,0.553146)中,故可,认,认为二者均,值,值无明显差,异,异,假设方差未,知,知,判断两,个,个总体的方,差,差是否相等,即尺寸离,散,散度是否相,同,同,谢谢观看,/,欢迎下载,BY FAITH IMEANA VISIONOF GOOD ONE CHERISHES ANDTHEENTHUSIASMTHATPUSHES ONE TOSEEKITS FULFILLMENTREGARDLESS OFOBSTACLES.BY FAITHI BYFAITH,
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