Lean与六西格玛培训第8课：Hypothesis Test

Hypothesis Testing,May 2015,Kenny,Operational ExcellencePanyu / China,Objectives of Module:,Understand the importance of Hypothesis testing. Understand the application of Hypothesis Testing as the decision making 1 sample t-test 2 samples t-test Pair t-test Normality Test Equal Variance Test 1 proportion test 2 proportions test 1-way ANOVA,Population,Data Collection : Sample,Process,Measure,Analysis,Sample Statistics are used to estimate population attributes,Generating Basic Statistics for Information,Why Learn Hypothesis Testing,Hypothesis testing is the tool that help us to make the right decision onBuying the new machineChange for better processChange for better supplierChange for better cost consciousAs the convincing tool Convince people (mostly management) with the statistic proof Prove to customer that the quality isnt deteriorated or getting betterProve to supplier that the quality is already deteriorated or getting worse,Warm Up Example,The batteries life are claimed by supplier is normal distribution with m=120 hours and s=2.5 hours. 12 batteries being tested in IQC as following :-117.5118.1  126.3  119.7  123.7  119.5119.2113.8117.4121.3122.4116.5Can we suspect the claimed by this supplier ?,Descriptive Statistic,Descriptive Statistics: Batteries Variable     Mean  StDev  Variance  Minimum  Median  Maximum  Range   IQRBatteries  119.62   3.40     11.58   113.80  119.35   126.30  12.50  4.70,Any decision can be made ?,Graphical Plot,Any Conclusion ? And is it final? Or still subject to argument?,Hypothesis Test,One-Sample T: Batteries Test of mu = 120 vs < 120                  95%                                          UpperVariable    N     Mean  StDev  SE Mean    Bound      T      PBatteries  12  119.617  3.403    0.982  121.381  -0.39  0.352,In fact, by Hypothesis test (1-sample t-test), in statistically, we cannot claim the average of batteries life span is less than 120 hours.,P-value > 0.05, concluded that statistically we dont have sufficient data to show the average of batteries life span is less than 120 hours.,How Do Hypothesis Tests Work?,This Hypothesis, Ho (Null Hypothesis, always equal) is tested against statistics computed from data.If the “test” rejects the Null Hypothesis, a pre-defined Alternate Hypothesis, Ha (not equal, greater than, less than) is selected.Ho : m1=m2Ha : m1=m2 or m1m2 Previous Example:Ho : Supplier is given us the average of life span equal to 120 hoursHa : Supplier is given us the average of life span less than 120 hoursIn Statistical TermHo : m = 120 hoursHa : m < 120 hours,Collect Data, in order for us to prove that either we should accept Ho or reject Ho (hence accept Ha).,Risk for Decision Making,Rejecting the null hypothesis Ho when it is true is defined as a Type I error.The probability of a Type I error is denoted by (0< <1).Accepting the null hypothesis when it is false is defined as a Type II error.Is frequently due to sample sizes being too small.The probability of a Type II error is denoted by (0< <1).,any 12="" 120="" decision="" making="" is="" involved="" risk="" only="" god="" for="" a="" and="" p-value="" 1.="" considered="" the="" then="" by="" sample="" size="" center="" limit="" theorem="" not="" cover="" mean="" population="" can="" be="" described="" in="" standard="" normal="" curve="" or="" with="" m="120" as="" 2.="" average="" of="" samples="" shown="" 119.6="" from="" previous="" 3.="" question="" how="" far="" tested="" value="" that="" we="" claim="" batteries="" span="" less="" than="" 4.="" statistic="" calculated="" probability="" infinity="" to="" 119.6.="" 0.352.="" this="" 5.="" if="" are="" willing="" reject="" clearly="" should="" accept="" 6.whenever="" on="" left="" beyond="" 7.="" concluded="" rules="" as:="" whenever="" p="">a => Accept HoWhenever p value  Reject Ho,Setting a risk,In Hypothesis testing we decide on the alpha risk we can tolerate. 10% Chance we make the wrong decision = 0.10 5 % Chance we make the wrong decision = 0.05 Typical 1 % Chance we make the wrong decision = 0.01,Beta Risk,Beta Risk is the probability of failing to reject the null hypothesis when a difference exists.,Another Example (2 samples t-test),New type of fuel being developed for better distance drive against the current existing one. 20 cars being tested for distance drive for 20 liters gasoline, as data shown. (Car distance drive worksheet) You are being call to make the decision whether change to new type of fuel will save us more energy or money (considered the processing cost of each liter are same), in other word, give us more distance drive.,Descriptive Statistics: Current, New Variable    Mean  StDev  Minimum  Median  MaximumCurrent   102.40   4.61    93.30  102.95   109.60New       106.61   3.79   101.30  106.30   116.30,Either Box or Dot plots, it look like they are really significant difference, but people is still able to argue since the graphical analysis is a subjective.,1. State the objectives,2. Establish Ho & Ha in English and statistical words.,3. Select the appropriate statistical test (assumed probability distribution, z, t, f or Chi-Squared),4. Select the a level (usually 5%),6. Collect the Data,Hypothesis Test Steps,7. Test for assumption,8. Using Minitab, determine p-valueIf p-value > 0.05, then accept HoIf p-value < 0.05, then reject Ho,9. Translate the statistic conclusion to human language,5. Select b level and calculate the sample size n,Is the new fuel will give us more distance drive?,Ho : new fuel = current fuel : mn = mcHa : new fuel > current fuel : mn > mc,In this case, use 2 samples t-test (will discuss in more detail later in this module),Set a = 0.05,Forget it for the time being,Done, as data shown.,Mostly is normality test, but for the time being, assume both data are normal distributed, also equal variance.,More detail next page, however p-value = 0.002 from MinitabThus, reject Ho, and accept Ha,Conclusion : Its significantly the new fuel will give us more distance drive if compare to current fuel type.,Important word in the conclusion since in statistic, everything is difference (or variation never equal to zero), but hypothesis test proof significantly difference or greatar than or less than.,Minitab for Hypothesis Test,Minitab Output,Two-Sample T-Test and CI: New, Current Two-sample T for New vs Current          N    Mean  StDev  SE MeanNew      20  106.61   3.79     0.85Current  20  102.40   4.61      1.0Difference = mu (New) - mu (Current)Estimate for difference:  4.2150095% lower bound for difference:  1.96561T-Test of difference = 0 (vs >): T-Value = 3.16  P-Value = 0.002  DF = 38,Another Example,12 samples of batteries being tester by IQC engineer for both suppliers, current (A) and new (B). We want to shift to supplier B (new) due to the cheaper price by 10%. Base on the data, can we conclude that both suppliers are having the no significant difference of quality of life span?,1. State the objectives,2. Establish Ho & Ha in English and statistical words.,3. Select the appropriate statistical test (assumed probability distribution, z, t, f or Chi-Squared),4. Select the a level (usually 5%),6. Collect the Data,7. Test for assumption,5. Select b level and calculate the sample size n,Is both suppliers are having the same life span ?,Ho : Supplier A = Supplier B : mA = mBHa : Supplier A = Supplier B : mA = mB,Use 2 samples t-test,Set a = 0.05,Forget it for the time being,Done, as data shown.,Assumed both are normal distributed and equal variance,8. Using Minitab, determine p-valueIf p-value > 0.05, then accept HoIf p-value < 0.05, then reject Ho,Next Example (continue),Two-Sample T-Test and CI: Supplier A, Supplier B Two-sample T for Supplier A vs Supplier B             N    Mean  StDev  SE MeanSupplier A  12  119.23   2.87     0.83Supplier B  12  118.28   4.04      1.2Difference = mu (Supplier A) - mu (Supplier B)Estimate for difference:  0.95000095% CI for difference:  (-2.017047, 3.917047)T-Test of difference = 0 (vs not =): T-Value = 0.66  P-Value = 0.514  DF = 22,9. Translate the statistic conclusion to human language,Conclusion : Its not enough evident to show that both suppliers are having the difference average of batteries life span. Thus, we can shift to supplier B (new) for better price.,Since p > a, We should accept Ho, meant they are not significant difference.,Important wording in the conclusion (can be “not sufficient data”, but never say “they are the same”), since we are only concluded they are not significantly difference base on our samples that being measured in this test. Or fail to reject Ho.,Always draw graphic, this is the common sense,Exercise 1 (data in Exercise 1 worksheet),Q1 :  A machine produces metal rods used in an automobile suspension system. A random sample of 15 rods is selected, and the diameter is measured. The resulting data are shown below. 088.238.268.248.258.198.258.268.238.24i. Use hypothesis 1 sample t-test, prove that the rods diameter is significantly exceeds 8.20 mm. (use a = 0.05).Q2 :  The life of a battery used in a cardiac pacemaker is assumed to be normally distributed. A random sample of 10 batteries is subjected to an accelerated life test by running them continuously at an elevated temperature until failure, and the following lives are obtained. 25.5 h26.1h26.825.027.827.325.7i. The manufacturer wants to be certain that the mean battery life exceeds 25 h. What conclusions can be drawn from these data (use a = 0.05) ?,Exercise 1 (Continue),Q3 :  The Hotel Guest Surveys were conducted for two difference months, Data File (before and after the campaign) (0=totally not satisfaction, 10=perfect satisfaction). Assume normality and equal variance, is the campaign effective to increase the Customer Satisfaction levels ?Q4 : 2 printing machine (old and new) are tested for the performance, and parameters solder paste thickness is used for the evaluation. If the target of 7 mils is needed from the process. Base on the data, do you propose to buy a new printing machine in order to improve the process performance?,For all tests:p > 0.05 Fail to Reject Ho (null)p < 0.05 Reject Ho,Ho: P1=P2Ha: P1 P2Minitab:Stat -Bsc Stat -     1 or 2 Proportions,Ho:  m1 = m2Ha:  m1 m2Minitab:Stat - Basic Stats - 2-Sample T(Compares Means using pooled Std Dev)Check box to assume equal variances,Ho:  m1 = m2Ha:  m1 m2Minitab:Stat - Basic Stats - 2-Sample T(Compares Means using unpooled Std Dev)Check box to assume unequal variances,Quantitative Y (one factor only),Bartletts Test/ F Test,Normality Test,One Way ANOVA,Qualitative Y (2 factors only),ContingencyTable,Ho: Two factors are independentHa: Two factors are dependent Minitab:Stat -Tables - Chi-square Test,Ho:  m1 = m2 = m3 = .Ha:  at least one is differentMinitab:Stat - Anova- One-way(Be careful if Bartletts  p < 0.05)Assumes Equal Variances,Ho:  Data is NormalHa:  Data is NOT NormalMinitab:Stat - Basic Stat - Normality TestUse Anderson-Darling,Ho:  s1 = s2 = s3 = .Ha:  at leastone is differentMinitab:Stat - Anova - Homog of Variance(Requires stacked data),Non Normal,Normal,Chi-Squared,One Sample,Ho:  s1 = s targetHa:  s1 s targetMinitab:Stat - Basic Stat > Display Desc Stat Graphs > Graphical SummaryIf s target falls between CI,Then fail to reject Ho.,Two or More Samples,2 Sample T-Test( Variances Equal),2 Sample T-Test(Variances Not Equal),1 Sample T- Test,Ho:  m1 = m targetHa:  m1 m targetMinitab:Stat - Basic Stats - 1 Sample-T(This is also used for paired comparisons: Ho: m1 - m 2 = 0),Two or More Samples,Two  Samples,ProportionsTesting,Data,1 factor,2 factors,Hypothesis Test Guideline and Roadmap,Non- Parametric Test,Pair T-Test(If data are paired),Hypothesis Test Guideline and Roadmap,Normality Test,Ho : Distribution = NormalHa : Distribution = Normal,Normality test (from basic statistic) in fact is one of the hypothesis test, on which:,Using same data set from previous example of fuel distance drive.,There are not enough evident to prove that both distribution are not normal distribution,Equal Variance Test,Ho: 1 = 2Ho: 1 = 2,Equal variance test, the hypothesis statement are:,Using same data set from previous example of fuel distance drive.,Most of the time is tested of equal variance assumption before 2-samples t-test,Thats not enough evident to prove that both are difference variance.,Pair t-test,Difference/Delta,Machine A,Data is paired (Most of the time is “Before” and “after” study use same samples after certain process, to proof that certain parameters are not significantly impacted by the process, notes that the normality test should be perform on the delta before the pair t-test),Data is independent. Both machine use difference samples.,Y1Y2Y3Y4Y5Y6Y7Y8Y9Y10,Machine B,Pair Data,No Pair Data,Example,Re-flow Oven being tested for any changes shape of the board. The experiment is execution. The curvature of the board being measured for 20 boards before and after the re-flow. Data as recorded. (pair t-test datasheet) Can we concluded that the curvature of the boards are increased after the re-flow oven?,Paired T-Test and CI: After, Before Paired T for After - Before             N     Mean    StDev  SE MeanAfter       20  16.8550   0.9747   0.2179Before      20  15.4850   1.2171   0.2722Difference  20  1.37000  1.61607  0.3613695% lower bound for mean difference: 0.74515T-Test of mean difference = 0 (vs > 0): T-Value = 3.79  P-Value = 0.001,The curvature after the reflow oven clearly significant greater than the before.,1-Proportion Test   (H0:   = 0),Compares the sample proportion against the target proportion.Use for discrete/attribute data. The normal distribution may be used as an approximation to the binomial when np > 5.,Example :After series of improvement in SMT process, QA Engineer is interested to know any significant reduction of yield loss in ICT station. 821 PCBA being tested, 46 of them are fail during the ICT.  Use a significance level (a) of 0.01 to test whether the yield loss is lower than the baseline of 7.8%.,Hypothesis StatementHo: p1 = 0.078Ho: p1 < 0.078,Since np > 5,Stat Basic Statistics 1 Proportions,Minitab Output,Session WindowTest and CI for One ProportionTest of p = 0.078 vs p < 0.078Sample   X    N   Sample p  99.0% Upper Bound  Z-Value  P-Value1 46  821  0.056029        0.074701      -2.35    0.009,What is your conclusion?,2 Proportions Test  (H0: 1 = 2),Stat Basic Statistics 2 Proportions,Extension from 1-Proportion test.Compares the sample proportion for 2 population.Use for discrete/attribute data.,Example:Operator A produced 23 defectives out of 624 items produced.  Operator B produced 14 defectives out of 703 produced.  Is Operator A producing higher percent defectives than Operator B at the 95% confidence level?What is your hypothesis statement?,Hypothesis StatementHo: pA = pB Ho: pA > pB,Session WindowTest and CI for Two Proportions Sample   X    N  Sample p1       23  624  0.0368592       14  703  0.019915Difference = p (1) - p (2)Estimate for difference:  0.016944395% lower bound for difference:  0.00181027Test for difference = 0 (vs > 0):  Z = 1.87  P-Value = 0.031,What is your conclusion?,Minitab Output,Test for non-normality distribution Nonparametric test,Also known as Distribution-free tests.Perform tests on the median. Hence does not make any assumption about the distribution of the sampled populations.,Roadmap of Nonparametric Tests,Quantitative Y,Quantitative Y (More than one factor),Quantitative Y (One factor only),Normal?- Yes or No,One Sample,Two Samples,Two Samples or More,1-Sample Sign Test1-Sample Wilcoxon  (Wilcoxon Signed   Rank Test),2-Sample Sign Test2-Sample Wilcoxon Sign Rank TestMann-Whitney,Kruskal-WallisMood MediansLevenesTest,Friedmans,Summary of Nonparametric Hypothesis Tests,Summary of Nonparametric Hypothesis Tests,Stat Nonparametric ,See your Mentor if you need more detail about nonparametric test.,Hypothesis Testing as the final step for Decision Making,First Guess - Engineering knowledge, Expected Results  Second Guess Collect Data, Calculate mean and sigma (and other Descriptive Statistic)  Third Guess - Plot the graphic like histogram, boxplot and dot-plot (Graphical Analysis)  Final Guess : Do hypothesis test, choosing alpha 0.05 (Inferential Statistic)  Decision Making - Make conclusion with certain degree of error (either a or b risk),Exercise 2,Q1 :  An engineer performs two assembly trials to determine the effect of oven atmosphere on the occurrence of tombstone defects on miniature discretes.  During the first trial, a nitrogen atmosphere is used in the reflow oven.  In the second trial, an air atmosphere is used in the reflow oven.  During the first trial, 26 defects are found on 645 parts.  During the second trial, 7 defects are found on 420 parts.  Is there a statistically significant difference between the two profile types and their impact on tombstone defects?Q2 :  Ten people have participated in a diet program. Both weights (before and after) in the program are measured (Data File). Is there evidence to support the claim that this particular diet-program is effective in producing a lean? Use a=0.05.Q3 :  New technology of the new design machine has claimed to be more consistency if co