Analyze階段考古題解說.doc

6 Sigma DMAIC GB Analyze階段題庫本題庫由選擇題/計(jì)算題/簡(jiǎn)答題三部份組成,所有題目全部附答案,題目來自歷次考試.一. 選擇題:( A )1.下面選項(xiàng)何者不屬於Noisy Data AnalysisA. Regression Analysis(迴歸分析) B. Outliers (異常點(diǎn)分析)C. Autocorrelation (自我相關(guān)性分析) D. Normality (常態(tài)分佈分析)( A )2.下面那一選項(xiàng)不是用來評(píng)估兩種物料某一計(jì)量值品質(zhì)特性是否有差異的Minitab操作選項(xiàng).A 2 Proportions. B. 2 Variances C. Power and Sample Size D. Normality Test ( B )3.下面各選項(xiàng)中對(duì)Paired Sample的實(shí)驗(yàn)策略敘述不正確的是A. 可用於比較同一組樣本被兩個(gè)不同量測(cè)設(shè)備量測(cè)的結(jié)果B. 計(jì)算Sample Size時(shí)Paired Sample的Sample Size計(jì)算方法與2 Sample t相同 C. 並不要求檢定標(biāo)準(zhǔn)差是否一致. D. 數(shù)據(jù)收集方式為成對(duì)的數(shù)據(jù)收集方式( B )4.下面各選項(xiàng)中關(guān)於使用One Way ANOVA分析的敘述不正確的是A. P-Value0.05時(shí),平均值的95%信賴區(qū)間兩兩交錯(cuò) B. 計(jì)量/計(jì)數(shù)型都適用 C. 分析前需檢定標(biāo)準(zhǔn)差一致 D. 可用來檢定3組或3組以上數(shù)據(jù)的平均值有無差異( D )5工程師希望評(píng)估某産品不同溫度熱處理後硬度有無差異，先後在四種不同的溫(600, 700, 800和900度)實(shí)驗(yàn)並測(cè)量出硬度(假定是常態(tài)分佈),請(qǐng)選擇最適合的分析工具：A. 1-Sampel Z-Test B. Regression analysisC. Regression and Correlation analysis D. One Way ANOVA( B )6在執(zhí)行One Way ANOVA分析時(shí)，下列哪一個(gè)是作爲(wèi)基本的條件之一A Mean一致 B Variance一致 C 非Normal數(shù)據(jù) D Median一致(BC )7 選用無母數(shù)(Nonparametrics)方法作爲(wèi)檢定工具，必須先決定：A. 記數(shù)型或計(jì)量型 B. Variance是否相等 C. 是否常態(tài)分佈 D.隨機(jī)化二.計(jì)算題 1.A生產(chǎn)線和B生產(chǎn)線生產(chǎn)加工同樣類型的半導(dǎo)體產(chǎn)品,為了驗(yàn)證兩個(gè)生產(chǎn)線所生產(chǎn)的產(chǎn)品的電阻是否相同,收集數(shù)據(jù)如下,我們可以認(rèn)為A線和B線生產(chǎn)的產(chǎn)品電阻相同嗎?原題數(shù)據(jù)a.請(qǐng)?jiān)O(shè)立虛無假設(shè): H0: A=B ; A =Bb.請(qǐng)?jiān)O(shè)立對(duì)立假設(shè): H1: AB ; A Bc.正態(tài)性判定: Minitab: Graph Probability Plot MultipleANS:A生產(chǎn)線 : P-Value = 0.905 0.05 , 資料服從常態(tài)B生產(chǎn)線 : P-Value = 0.396 0.05 , 資料服從常態(tài)d.等標(biāo)準(zhǔn)差的判定: Minitab: Stat Basic Statistics 2 Variances ANS:P-Value = 0.487 0.05 , A,B兩生產(chǎn)線標(biāo)準(zhǔn)差無顯著差異 e.使用的驗(yàn)證方法(: 2-Sample t f.驗(yàn)證結(jié)論:Minitab: Stat Basic Statistics 2-Sample t Two-sample T for A vs B N Mean StDev SE MeanA 10 12.922 0.248 0.078B 10 12.354 0.315 0.10Difference = mu (A) - mu (B)Estimate for difference: 0.56800095% CI for difference: (0.300553, 0.835447)T-Test of difference = 0 (vs not =): T-Value = 4.48 P-Value = 0.000 DF = 17ANS:P-Value = 0.000 Probability Plot MultipleANS:A設(shè)備 : P-Value = 0.782 0.05 , 資料服從常態(tài)B設(shè)備 : P-Value = 0.205 0.05 , 資料服從常態(tài)C設(shè)備 : P-Value = 0.172 0.05 , 資料服從常態(tài)D設(shè)備 : P-Value = 0.977 0.05 , 資料服從常態(tài)d.等標(biāo)準(zhǔn)差的判定: Minitab: Data Stack ColumnsMinitab: Stat ANOVA Test for Equal Variances.ANS:P-Value = 0.27 0.05 , A,B,C,D 4臺(tái)設(shè)備長(zhǎng)度標(biāo)準(zhǔn)差無顯著差異e.使用的驗(yàn)證方法: ANOVAf.驗(yàn)證結(jié)論:Minitab: Stat ANOVA One Way One-way ANOVA: 長(zhǎng)度 versus 設(shè)備 Source DF SS MS F P設(shè)備 3 29.645 9.882 57.26 0.000Error 36 6.213 0.173Total 39 35.858S = 0.4154 R-Sq = 82.67% R-Sq(adj) = 81.23% Individual 95% CIs For Mean Based on Pooled StDevLevel N Mean StDev -+-+-+-+-A 10 15.460 0.568 (-*-)B 10 16.320 0.394 (-*-)C 10 17.160 0.337 (-*-)D 10 14.890 0.314 (-*-) -+-+-+-+- 15.20 16.00 16.80 17.60Pooled StDev = 0.415ANS:P-Value = 0.000 Basic Statistics 2Proportions.Test and CI for Two Proportions Sample X N Sample p1 129 2897 0.044529 148 3325 0.044511Difference = p (1) - p (2)Estimate for difference: 0.000017544795% CI for difference: (-0.0102563, 0.0102914)Test for difference = 0 (vs not = 0): Z = 0.00 P-Value = 0.997ANS:P-Value = 0.997 0.05, 表示A,B兩家公司不良率無顯著差異 4.用4臺(tái)設(shè)備生產(chǎn)同樣的產(chǎn)品,並可以把產(chǎn)品分為A,B,C 等級(jí),為了瞭解設(shè)備別生產(chǎn)的產(chǎn)品的 A,B,C等級(jí)比率是否相同,從全體生產(chǎn)的產(chǎn)品中抽取樣品,並區(qū)分為設(shè)備別,等級(jí)別如下,我們可 以認(rèn)為此4臺(tái)設(shè)別有差異嗎? a.請(qǐng)?jiān)O(shè)立虛無假設(shè): H0: P1=P2=P3=P4 b.請(qǐng)?jiān)O(shè)立對(duì)立假設(shè): H1: P1, P2,P3,P4至少有一個(gè)不等; c.使用的驗(yàn)證方法: Chi-Square Test(卡方分布)d.驗(yàn)證結(jié)論 Minitab: Stat Tables Chi-Square TestChi-Square Test: A, B, C Expected counts are printed below observed countsChi-Square contributions are printed below expected counts A B C Total 1 81 25 14 120 84.85 26.36 8.79 0.175 0.071 3.091 2 85 11 3 99 70.00 21.75 7.25 3.214 5.313 2.491 3 69 31 7 107 75.66 23.51 7.84 0.586 2.388 0.089 4 45 20 5 70 49.49 15.38 5.13 0.408 1.389 0.003Total 280 87 29 396Chi-Sq = 19.218, DF = 6, P-Value = 0.004ANS:P-Value = 0.004 Probability Plot Multiple p1=0.919 p2=0.14 p3=0.802 p4=0.314 p5=0.495 全部大於0.05, 所以全部具備常態(tài)性3) 標(biāo)準(zhǔn)差驗(yàn)證方法及結(jié)果: Test for Equal Variance Minitab : Data Stack ColumnsMinitab : Stat ANOVA Test Equal Variance ANS:P-Value = 0.249 0.05 所以可以認(rèn)為標(biāo)準(zhǔn)差相等4)平均值驗(yàn)證方法及結(jié)果: One way ANOVA Minitab: Stat ANOVA One Way One-way ANOVA: 平整度 versus 工人 Source DF SS MS F P工人 4 0.13960 0.03490 19.98 0.000Error 55 0.09609 0.00175Total 59 0.23569 Individual 95% CIs For Mean Based on Pooled StDevLevel N Mean StDev -+-+-+-+man1 12 5.0655 0.0558 (-*-)man2 12 5.0157 0.0377 (-*-)man3 12 4.9670 0.0306 (-*-)man4 12 4.9297 0.0330 (-*-)man5 12 5.0329 0.0467 (-*-) -+-+-+-+ 4.950 5.000 5.050 5.100ANS: P-Value = 0.000 Basic Statistics 2-ProportionsTest and CI for Two Proportions Sample X N Sample p1 293 550 0.5327272 341 708 0.481638Difference = p (1) - p (2)Estimate for difference: 0.051088995% CI for difference: (-0.00452818, 0.106706)Test for difference = 0 (vs