Miniab模塊功能培訓(xùn)

MINITAB培訓(xùn)(工程師級)

------BasicStatistics模塊此模塊有如下功能:模塊功能UseMinitab'sbasicstatisticscapabilitiesforcalculatingbasicstatisticsandforsimpleestimationandhypothesistestingwithoneortwosamples.Thebasicstatisticscapabilitiesincludeproceduresfor:CalculatingorstoringdescriptivestatisticsHypothesistestsandconfidenceintervalsofthemeanordifferenceinmeansHypothesistestsandconfidenceintervalsforaproportionorthedifferenceinproportionsHypothesistestforequalityofvarianceMeasuringassociationTestingfornormalityofadistribution打開一個(gè)軟件自帶的例子學(xué)習(xí)打開Quality文件數(shù)據(jù)如下選擇顯示描述性統(tǒng)計(jì)填完自變量和因變量后,點(diǎn)擊Graphs按鈕點(diǎn)擊OK后,顯示出箱線圖和統(tǒng)計(jì)數(shù)據(jù)解釋每個(gè)箱線線圖最低低點(diǎn)表示示當(dāng)天最最小值,最高點(diǎn)點(diǎn)表示最最大值,箱子高高,低點(diǎn)點(diǎn)分別表表示3/4,1/4數(shù)數(shù)字大小小,箱中中間一橫橫表示當(dāng)當(dāng)天幾個(gè)個(gè)數(shù)據(jù)的的中位數(shù)數(shù)在Session窗口口中具體體顯示MEAN平均值值,StDev標(biāo)準(zhǔn)準(zhǔn)差,Median中中位數(shù)等等信息.選擇存存貯描述述性統(tǒng)計(jì)計(jì)填完自變變量和因因變量后后,點(diǎn)擊擊OK多了些數(shù)數(shù)據(jù),自自動(dòng)將統(tǒng)統(tǒng)計(jì)的數(shù)數(shù)據(jù)如平平均值和和每日數(shù)數(shù)據(jù)量N=10存放在在C3-C5列列用圖形形概要方方式如圖,點(diǎn)點(diǎn)擊OKTheleftfourgraphs:histogramofdatawithanoverlaidnormalcurve(帶常態(tài)態(tài)曲線的的柱狀圖圖),boxplot箱線線圖),95%confidenceintervalsformean,and95%confidenceintervalsforthemedian.(平均均值和中中位數(shù)95%的的置信區(qū)區(qū)間范圍圍圖)右邊第一一塊數(shù)據(jù)據(jù)是正態(tài)態(tài)性檢驗(yàn)驗(yàn).顯示示P-Value=0.661,大于0.05,所以以數(shù)據(jù)符符合正態(tài)態(tài)分布.第二塊數(shù)數(shù)據(jù)是平平均值,偏態(tài)等等第三塊數(shù)數(shù)據(jù)是最最小最大大等第四塊數(shù)數(shù)據(jù)是平平均值,方差的的95%置信區(qū)區(qū)間等區(qū)間估計(jì)計(jì)與假設(shè)設(shè)檢驗(yàn)例假設(shè)設(shè)一個(gè)物物體的重重量未知知,為了了估計(jì)其其重量,用一個(gè)個(gè)天平稱稱5次,得到重重量5.525.485.645.515.45,假設(shè)設(shè)結(jié)果符符合標(biāo)準(zhǔn)差0.1的正態(tài)分分布,求求均值是否否是5.5和重量置置信水平平為95%的置置信區(qū)間間輸入數(shù)據(jù)據(jù)后,按按圖所示示按圖所示示將例子子中數(shù)據(jù)據(jù)填入后后點(diǎn)擊OK得到分析析結(jié)果見見Session窗口口分析One-SampleZ:C1Testofmu=5.5vsnot=5.5Theassumedstandarddeviation=0.1VariableNMeanStDevSEMean95%CIZPC155.520000.072460.04472(5.43235,5.60765)0.450.655由于假設(shè)設(shè)均值為為5.5,而置置信區(qū)間間(5.43235,5.60765)包含含了5.5,故故假設(shè)成成立,可可以認(rèn)為為重量為為5.5當(dāng)前一例例子中方方差未知知時(shí)(這這也是通通常的情情況),求重量量是否可可認(rèn)為是是5.5如圖填上上數(shù)據(jù),點(diǎn)擊OK結(jié)果在Session窗口中中顯示解釋One-SampleT:C1Testofmu=5.5vsnot=5.5VariableNMeanStDevSEMean95%CITPC155.520000.072460.03240(5.43003,5.60997)0.620.570同樣的,5.5在(5.43003,5.60997)范圍中中,可認(rèn)為假假設(shè)均值5.5成立.一個(gè)新例子公司購買了同同型號的兩臺(tái)臺(tái)機(jī)床(可認(rèn)認(rèn)為其加工的的零件尺寸服服從同方差的的正態(tài)分布),抽檢兩臺(tái)臺(tái)機(jī)床加工的的軸,分析兩兩臺(tái)機(jī)床加工工的軸有無明明顯差異機(jī)床甲:20.5/19.8/19.7/20.4/21.1/20/19/19.9機(jī)床乙:20.7/19.8/19.5/20.8/20.4/19.6/20.2輸入數(shù)據(jù)于兩兩列中Two-sampleTforC1vsC2NMeanStDevSEMeanC1820.0500.6260.22C2720.1430.5220.20Difference=mu(C1)-mu(C2)Estimatefordifference:-0.09285795%CIfordifference:(-0.738860,0.553146)T-Testofdifference=0(vsnot=):T-Value=-0.31P-Value=0.760DF=12由于假設(shè)兩均均值相等,差差為0,在(-0.738860,0.553146)中,故可認(rèn)認(rèn)為二者均值值無明顯差異異假設(shè)方差未知知,判斷兩個(gè)個(gè)總體的方差差是否相等,即尺寸離散散度是否相同同謝謝觀看