LSSG Black Belt Training Hypothesis Testing Introduction Always

LSSG Black Belt Training Hypothesis Testing

Introduction Always about a population parameter Attempt to prove (or disprove) some assumption Setup: alternate hypothesis: What you wish to prove Example: Change in Y after LSS Project null hypothesis: Assume the opposite of what is to be proven. The null is always stated as an equality. Example: Y after project is same as before 2

The test 1. Take a sample, compute statistic of interest. Standardized mean customer satisfaction score 2. How likely is it that if the null were true, you would get such a statistic? (the p-value) How likely is it that the sample would show (by random chance) the difference that we see after the LSS project, if in fact there was no improvement? 3. 4. If very unlikely, then null must be false, hence alternate is proven beyond reasonable doubt. If quite likely, then null may be true, so not enough evidence to discard it in favor of the alternate. 3

Types of Errors Null is really True reject null, Type I Error assume alternate is proven Believe in improvement when none occured do not reject null, Good Decision evidence for alternate not strong enough Null is really False Good Decision Type II Error Cannot show improvement when in fact it occured 4

The Testing Process 1. 2. 3. 4. 5. Set up Hypotheses (Null and Alternate ) Pick a significance level (alpha) Compute critical value (based on alpha) Compute test statistic (from sample) Conclude: If |test statistic| > |critical value|, then Alternate Hypothesis proven at alpha level of significance. 5

Hypothesis Testing Roadmap Hypothesis Testing Continuous Normal Attribute Non-Normal c 2 Contingency Tables Means Variance Medians Variance Correlation Z-tests c 2 Correlation Levene’s t-tests F-test Sign Test Same tests as Non-Normal Medians ANOVA Bartlett’s Wilcoxon Correlation Kruskal. Wallis Regression Mood’s Friedman’s 6

Parametric Tests Use parametric tests when: 1. 2. 3. The data are normally distributed The variances of populations (if more than one is sampled from) are equal The data are at least interval scaled 7

One sample z - test A gap between two parts should be 15 microns. A sample of 25 measurements shows a mean of 17 microns. Test whether this is significantly different from 15, assuming the population standard deviation is known to be 4. One-Sample Z Test of mu = 15 vs not = 15 The assumed standard deviation = 4 N Mean SE Mean 95% CI Z P 25 17. 0000 0. 8000 (15. 4320, 18. 5680) 2. 50 0. 012 8

Z-test for proportions You wish to test the hypothesis that at least two-thirds (66%) of people in the population prefer your brand over Brand X. Of the 200 customers surveyed, 140 say they prefer your brand. Is this statistically significant? Test and CI for One Proportion Test of p = 0. 66 vs p > 0. 66 Sample X N Sample p 1 140 200 0. 700000 95% Lower Bound Z-Value P-Value 0. 646701 1. 19 0. 116 9

One sample t-test Error Reduction % 10 12 9 8 7 12 14 13 15 16 18 12 18 19 20 17 15 The data show reductions in percentage of errors made by claims processors after undergoing a training course. The target was a reduction of at least 13%. Was it achieved? 10

One Sample t-test – Minitab results One-Sample T: Error Reduction Test of mu = 13 vs > 13 95% Lower Variable N Mean St. Dev SE Mean Bound T P Error Reduction 17 13. 8235 3. 9248 0. 9519 12. 1616 0. 87 0. 200 The p-value of 0. 20 indicates that the reduction in Errors could not be proven to be greater than 13%. P-value of 0. 20 shows that the probability is greater than alpha (0. 05) that the difference may be 13% or less. 11

Two Sample t-test You realize that though the overall reduction is not proven to be more than 13%, there seems to be a difference between how men and women react to the training. You separate the 17 observations by gender, and wish to test whethere is in fact a significant difference between genders in error reduction. M 10 12 9 8 7 12 14 13 F 15 16 18 12 18 19 20 17 15 12

Two Sample t-test The test for equal variances shows that they are not different for the 2 samples. Thus a 2 -sample t test may be conducted. The results are shown below. The p-value indicates there is a significant difference between the genders in their error reduction due to the training. Two-sample T for Error Reduction M vs Error Reduction F N Mean St. Dev SE Mean Error Red M 8 10. 63 2. 50 0. 89 Error Red F 9 16. 67 2. 45 0. 82 Difference = mu (Error Red M) - mu (Error Red F) Estimate for difference: -6. 04167 95% CI for difference: (-8. 60489, -3. 47844) T-Test of difference = 0 (vs not =): T-Value = -5. 02 P-Value = 0. 000 DF = 15 Both use Pooled St. Dev = 2. 4749 13

Chi-squared test of independence For tabulated count data. Two types of glass sheets are manufactured, and the defects found on 111 sheets are tabulated based on the type of glass (Type A and Type B) and the location of the defect on each sheet (Zone 1 and Zone 2). You wish to test whether the two variables (Type of glass and Location of error on the glass) are statistically independent of each other. Intro to LSS 14

Chi Square Test Results Tabulated statistics: Glass Type, Location Rows: Glass Type Columns: Location Zone 1 Zone 2 All Type A 29 29. 98 23 22. 02 52 52. 00 Type B 35 34. 02 24 24. 98 59 59. 00 All 64 64. 00 47 47. 00 111. 00 Cell Contents: Count Expected count Pearson Chi-Square = 0. 143, DF = 1, P-Value = 0. 705 Likelihood Ratio Chi-Square = 0. 143, DF = 1, P-Value = 0. 705 15

Assignment From the book “Doing Data Analysis with Minitab 14” by Robert Carver: 1. 2. 3. Pages 138 – 142: Choose any 3 of the datasets mentioned on those pages and answer the related questions. [1 -sample means] Pages 148 -151: Choose any 3 of the datasets mentioned on those pages and answer the related questions. [1 -sample proportions] Pages 165 -168: Choose any 3 of the datasets mentioned on those pages and answer the related questions. [2 -sample tests] 16

Basics of ANOVA Analysis of Variance, or ANOVA is a technique used to test the hypothesis that there is a difference between the means of two or more populations. It is used in Regression, as well as to analyze a factorial experiment design, and in Gauge R&R studies. The basic premise of ANOVA is that differences in the means of 2 or more groups can be seen by partitioning the Sum of Squares (SS) is simply the sum of the squared deviations of the observations from their means. Consider the following example with two groups. The measurements show the thumb lengths in centimeters of two types of primates. Total variation (SS) is 28, of which only 4 (2+2) is within the two groups. Thus 24 of the 28 is due to the differences between the groups. This partitioning of SS into ‘between’ and ‘within’ is used to test the hypothesis that the groups are in fact different from each other. Obs. Type A Type B 1 2 3 4 6 7 8 Mean SS 3 2 7 2 Overall Mean = 5 SS = 28 See www. statsoft. com for more details. 17

Results of ANOVA The results of running an ANOVA on the sample data from the previous slide are shown here. The hypothesis test computes the F-value as the ratio of MS ‘Between’ to MS ‘Within’. The greater the value of F, the greater the likelihood that there is in fact a difference between the groups. looking it up in an F-distribution table shows a p-value of 0. 008, indicating a 99. 2% confidence that the difference is real (exists in the Population, not just in the sample). One-way ANOVA: Type A, Type B Source DF SS MS F P Factor 1 24. 00 0. 008 Error 4 4. 00 1. 00 Total 5 28. 00 __________________ S = 1 R-Sq = 85. 71% R-Sq(adj) = 82. 14% Minitab: Stat/ANOVA/One-Way (unstacked) 18

Two-Way ANOVA Strength 20. 0 22. 0 21. 5 23. 0 24. 0 22. 0 25. 0 24. 5 17. 0 18. 0 17. 5 Temp Low Low Low High High Speed Slow Slow Fast Fast The results show significant main effects as well as an interaction effect. Is the strength of steel produced different for different temperatures to which it is heated and the speed with which it is cooled? Here 2 factors (speed and temp) are varied at 2 levels each, and strengths of 3 parts produced at each combination are measured as the response variable. Two-way ANOVA: Strength versus Temp, Speed Source DF SS Temp 1 3. 5208 Speed 1 20. 0208 Interaction 1 58. 5208 Error 8 5. 1667 Total 11 87. 2292 MS F P 3. 5208 5. 45 0. 048 20. 0208 31. 00 0. 001 58. 5208 90. 61 0. 000 0. 6458 S = 0. 8036 R-Sq = 94. 08% R-Sq(adj) = 91. 86% 19

Two-Way ANOVA The box plots give an indication of the interaction effect. The effect of speed on the response is different for different levels of temperature. Thus, there is an interaction effect between temperature and speed. 20

Assignment From the book “Doing Data Analysis with Minitab 14” by Robert Carver: 1. 2. Pages 192 – 194: Choose any 3 of the datasets mentioned on those pages and answer the related questions. [1 -way ANOVA] Pages 204 – 206: Choose any 3 of the datasets mentioned on those pages and answer the related questions. [2 -way ANOVA] 21

DOE Overview A design of experiment involves controlling specific inputs (factors) at various levels (typically 2 levels, like “High” and “Low” settings) to observe the change in output as a result, and analyzing the data to determine the significance and relative importance of factors. The simplest case would be to vary a single factor, say temperature, while baking cookies. Keeping all else constant, we can set temperature at 350 degrees and 400 degrees, and make several batches at those two temperatures, and measure the output desired – in this case it could be a rating by experts of crispiness of the cookies on a scale of 0 -10. 22

Full Factorial Designs A 2 F Factorial design implies that there are factors at 2 levels each. The case described on the previous slide with only one factor is the simplest. Having two factors at 2 levels would give us four combinations. Three factors yield 8 combinations, 4 would yield 16, and so forth. The following table shows the full factorial (all 8 combinations) design for 3 factors – l l l temperature, baking time, and amount of butter, each at two levels – HIGH and LOW. Temp Time Butter Low Low Low High Low High High 23

Fractional Factorials The previous example would require 8 different setups to bake the cookies. For each setup, one could bake several batches, say 4 batches, to get a measure of the internal variation. In practice, as the number of factors tested grows, it is difficult to even create all the setups needed, much less have replications within a setup. An alternative is to use fractional factorial designs, typically a ½ or ¼. As the name suggests, a ½ factorial design with 3 factors would only require 4 of the 8 combinations to be tested. This entails some loss of resolution, usually a confounding of interaction terms, which may be of no interest to the experimenter, and can be sacrificed. Temp Time Butter High Low High Low Minitab: Stat/DOE/Create Factorial Design/Display Factorial Designs 24

Running the Experiment – Outcome Values Once the settings to be used are determined, we can run the experiment and measure the values of the outcome variable. This table shows the values of the outcome variable “Crisp”, showing the crispiness index for the cookies, for each of the 8 settings of the full factorial experiment. Temp Time Butter Crisp Low Low 7 High Low 10 Low High Low 7 High Low 5 Low High 4 High Low High 9 Low High 8 High 8 25

Analysis of Data Analyzing the data in Minitab for the main effects and ignoring interaction terms, we get the following output: Factorial Fit: Crispiness versus Temp, Time, Butter Analysis of Variance for Crispiness (coded units) Estimated Effects and Coefficients for Crispiness (coded units) Source Main Effects Residual Error Total Term Effect Coef SE Coef Constant 7. 2500 0. 3750 Temp 3. 0000 1. 5000 0. 3750 Time 0. 5000 0. 2500 0. 3750 Butter 1. 5000 0. 7500 0. 3750 T P 19. 33 0. 000 4. 00 0. 016 0. 67 0. 541 2. 00 0. 116 S = 1. 06066 R-Sq = 83. 64% R-Sq(adj) = 71. 36% DF Seq SS Adj MS 3 23. 000 7. 667 4 4. 500 1. 125 7 27. 500 F P 6. 81 0. 047 Estimated Coefficients for Crispiness using data in uncoded units Term Constant Temp Time Butter Coef 7. 25000 1. 50000 0. 250000 0. 750000 Note that only the temperature is significant (p-value lower than 0. 05). The effect of temperature is 3. 00, which means that if temp. is set at HIGH, crispiness will increase by 3. 00 units on average, compared to the LOW setting. Minitab: Stat/DOE/Create Factorial Design/Analyze Factorial Design 26

Assignment From the book “Doing Data Analysis with Minitab 14” by Robert Carver: 1. Pages 309 – 310: Answer any 4 of the 7 questions on those pages. [DOE] 27

Hypothesis Testing Roadmap Hypothesis Testing Continuous Normal Attribute Non-Normal c 2 Contingency Tables Means Variance Medians Variance Correlation Z-tests c 2 Correlation Levene’s t-tests F-test Sign Test Same tests as Non-Normal Medians ANOVA Bartlett’s Wilcoxon Correlation Kruskal. Wallis Regression Mood’s Friedman’s 28

Non-Parametric Tests Use non-parametric tests: 1. 2. 3. 4. When data are obviously non-normal When the sample is too small for the central limit theorem to lead to normality of averages When the distribution is not known When the data are nominal or ordinal scaled Remember that even non-parametric tests have some assumptions about the data. 29

The sign test The story: A patient sign-in process at a hospital is being evaluated, and the time lapse between arrival and seeing a physician is recorded for a random sample of patients. You believe that currently the median time is over 20 minutes, and wish to test the hypothesis. 30

The sign test – data Data for the test are as follows: Process Time 5 7 15 30 32 35 62 75 80 85 95 100 The histogram of the data shows that it is non-normal, and the sample size is too small for the central limit theorem to apply. The data are at least ordinal in nature (here they are ratio scaled), satisfying the assumption of the sign test. 31

Sign test - analysis Since the hypothesis is that the median is greater than 20, the test compares each value to 20. Those that are smaller get a negative sign, those that are larger than 20 get a positive one. The sign test then computes the probability that the number of negatives and positives observed would come about through random chance if the null were true (that the median time is 20 minutes). 32

Sign Test in Minitab - Results Sign Test for Median: Process Time Sign test of median = 20. 00 versus > 20. 00 Process Time N Below Equal Above P 12 3 0 9 0. 0730 Median 48. 50 In this data, there are 9 observations above 20, and 3 of them below. This can be shown to have a. 0730 probability of occurring, even if the median time for the population is in fact not greater than 20. Thus, there is insufficient evidence to prove the hypothesis (to reject the null) at the 5% level, but enough if you are willing to take a 7. 3% risk. 33

The sign test – other applications The sign test can also be used for testing the value of the median difference between paired samples, as illustrated in the following link. The difference between values in a paired sample can be treated as a single sample, so any 1 -sample hypothesis test can be applied. In such a case, the assumption is that the pairs are independent of each other. The equivalent parametric tests for the sign test are the 1 sample z test and the 1 -sample t-test. http: //davidmlane. com/hyperstat/B 135165. html 34

Wilcoxon Signed-Rank Test A test is conducted where customers are asked to rate two products based on various criteria, and come up with a score between 0 and 100 for each. The tester’s goal is to check whether product A, the new version, is perceived to be superior to product B. The null hypothesis would be that they are equal to each other. 35

Wilcoxon Signed-Rank Test The data Prod A 55 60 77 82 99 92 86 84 90 72 Prod B 50 62 70 78 90 95 90 80 86 71 Diff 5 -2 7 4 9 -3 -4 4 4 1 The measures are rankings by people, so the data are not necessarily interval scaled, and certainly not ratio scaled. Thus a paired sample t-test is not appropriate. A non-parametric equivalent of that is the Wicoxon Signed -Rank Test. This is similar to the sign test, but more sensitive. 36

Wilcoxon Signed-Rank Test Unlike the sign test, which only looks at whether something is larger or smaller, this tests uses the magnitudes of the differences, rank orders them, and then applies the sign of the difference and computes the sum of those ranks. This statistic (called W) has a sampling distribution that is approximately normal. For details on the technique, see the link below. Assumptions are: 1. Data are at least ordinal in nature 2. The pairs are independent of each other 3. Dependent variable is continuous in nature http: //faculty. vassar. edu/lowry/ch 12 a. html 37

Wilcoxon test in Minitab - Results Wilcoxon Signed Rank Test: Diff Test of median = 0. 000000 versus median > 0. 000000 N for N Test Diff 10 10 Wilcoxon Statistic 44. 5 P 0. 046 Estimated Median 2. 500 38

Mann-Whitney Test – Two Samples The Story: Customers were asked to rate a service in the past, and 10 people did so. After some improvements were made, data were collected again, with a new random set of customers. Twelve people responded this time. There is no pairing or matching of data, since the samples of customers for the old and the new processes are different. http: //faculty. vassar. edu/lowry/ch 11 a. html 39

Mann-Whitney Test – Two Samples Old 60 70 85 78 90 68 35 80 80 75 New 85 85 90 94 90 70 75 90 90 100 95 90 Note that the assumptions of a 2 -sample t-test are violated because the data are not interval scaled, and may not be normally distributed. The Mann-Whitney Test is the non-parametric alternative to the 2 -sample t-test. 40

Mann-Whitney Test – Two Samples The Mann-Whitney test rank orders all the data, with both columns combined into one. The ranks are then separated by group so the raw data is now converted to ranks. The sum of the ranks for each column is computed. The sums of ranks are expected to be in proportion to the sample sizes, if there is no difference between the groups. Based on this premise, the actual sum is compared to the expected sum, and the statistic is tested for significance. See details with another example on this link from Vassar Univ. : http: //faculty. vassar. edu/lowry/ch 11 a. html 41

Mann-Whitney Test in Minitab - Results Mann-Whitney Test and CI: Old, New N Median Old 10 76. 50 New 12 90. 00 Point estimate for ETA 1 -ETA 2 is -14. 00 95. 6 Percent CI for ETA 1 -ETA 2 is (-22. 00, -5. 00) W = 72. 5 Test of ETA 1 = ETA 2 vs ETA 1 < ETA 2 is significant at 0. 0028 The test is significant at 0. 0025 (adjusted for ties) 42

Kruskal-Wallis Test – 3 or more samples Here the data would be similar to the Mann-Whitney test, except for having more than 2 samples. For parametric data, one would conduct an ANOVA to test for differences between 3 or more populations. The Kruskal-Wallis test is thus a non-parametric equivalent of ANOVA. 43

Kruskal-Wallis Test – Data Adults Teens Children 7 5 6 4 2 6 5 9 9 8 5 9 10 7 8 3 4 3 5 10 2 The data show ratings of some product by three different groups. The same data are shown stacked on the right to perform the test in Minitab. Rating 7 5 6 4 2 6 5 9 9 8 5 9 10 7 8 3 4 3 5 10 2 Factor Adults Adults Teens Teens Children Children 44

Kruskal-Wallis Test The Kruskal-Wallis test proceeds very similarly to the Mann. Whitney test. The data are all ranked from low to high values, and the ranks then separated by group. For each group, the ranks are summed and averaged. Each group average is compared to the overall average, and the deviation measured, weighted by the number of observations in each group. If the groups were identical, the deviations from the grand mean would be a small number (not 0, as one might intuitively think) that can be computed. The actual difference is compared to the expected one (H statistic computed) to complete the test. See the link below for details of the computation, if interested. http: //faculty. vassar. edu/lowry/ch 14 a. html 45

Kruskal-Wallis Test Minitab Results Kruskal-Wallis Test: Rating versus Factor Kruskal-Wallis Test on Rating Factor N Adults 7 Children 6 Teens 8 Overall 21 Median Ave Rank 5. 000 8. 6 3. 500 7. 2 8. 500 15. 9 11. 0 Z -1. 23 -1. 79 2. 86 H = 8. 37 DF = 2 P = 0. 015 H = 8. 48 DF = 2 P = 0. 014 (adjusted for ties) 46

Mood’s Median Test Mood median test for Rating Chi-Square = 10. 52 DF = 2 P = 0. 005 Individual 95. 0% CIs Factor N<= N> Median Q 3 -Q 1 --------+-----+-------Adults 6 1 5. 00 2. 00 (-------*-----) Children 5 1 3. 50 (-----*-----------) Teens 1 7 8. 50 1. 75 (----*-) --------+-----+-------4. 0 6. 0 8. 0 Overall median = 6. 00 The Mood’s median test is an alternative to Kruskal-Wallis. It is generally more robust against violations of assumptions, but less powerful. 47

Friedman’s Test is the non-parametric equivalent to a randomized block design in an ANOVA. In other words, there are 3 or more groups, but each row of values across the groups are matched. The story A person’s performance is rated in a normal state, rated again after introducing noise in the environment, and finally with the introduction of classical music in the background. This is done for a sample of 7 employees. 48

Friedman’s Test – Data Normal Noise 7 8 6 9 5 7 8 5 4 6 5 5 4 4 Music 8 8 7 9 9 The data show the ratings of performance by person in each of 3 conditions. The same data are stacked in the table to the right, for doing the test in Minitab. Each person represents a block of data, since the 3 numbers for that person are related. Perform 7 8 6 9 5 7 8 5 4 6 5 5 4 4 8 8 7 9 9 Group Normal Normal Noise Noise Music Music Block 1 2 3 4 5 6 7 49

Friedman’s Test - Analysis Friedman’s test also ranks the ratings, but this time the ranking is done internally within each row – the three scores for each person are ranked 1, 2, and 3. These ranks are then summed and averaged. If the groups are identical, then one would expect no difference in the sum or mean of rankings for each group. In other words, if the conditions did not affect the performance rating, the rankings would either be the same, or vary randomly across people to yield equal sums. The sums are compared to this expectation to test the hypothesis. See the following link for more details. http: //faculty. vassar. edu/lowry/ch 15 a. html 50

Friedman’s Test in Minitab – Results. Friedman Test: Perform versus Group blocked by Block S = 9. 50 DF = 2 P = 0. 009 S = 10. 64 DF = 2 P = 0. 005 (adjusted for ties) Group Music Noise Normal N 7 7 7 Est Median 8. 000 4. 667 7. 333 Sum of Ranks 19. 5 8. 0 14. 5 Grand median = 6. 667 51

Assignment From the book “Doing Data Analysis with Minitab 14” by Robert Carver: 1. Pages 293 – 294: Choose any 3 datasets on those pages and answer the related questions. [Nonparametric tests] 52