OHSU OGI Class Steve Brainerd Basic Statistics …myplace.frontier.com/~stevebrainerd1/STATISTICS/ECE-580...1/29/03 Class 2 Non-Normal Data Steve Brainerd 4 OHSU OGI Class ECE-580-DOE

1/29/03 Class 2 Non-Normal Data Steve Brainerd

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OHSU OGI Class ECE-580-DOE :Statistical Process Control and Design of Experiments Steve Brainerd

Basic Statistics :Non-Normal data and Distribution free tests

• Parametric data = Normal distribution • Non-Parametric data = Distribution Free

• 1. INTRODUCTION: Also called Distribution free tests!!

• Most processes, particularly those involving life data and reliability are not normally distributed.

• Most process capability tools, however, assume normality. Only through verifying data normality and selecting the appropriate data analysis method will the results be accurate.

• Reference: Experimental Statistics Handbook 91, United States Department of Commerce, Mary Gibbons Natrella Oct 1966, Library of Congress Catalog Card Number : 63-60072 Chapter 16 “ Some Tests Which are Inependent of the form of the Distribution


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• Some years ago, some statisticians held the belief that when processes were not normally distributed, there was something "wrong" with the process, or even that the process was "out of control." In their view, the purpose of the control chart was to determine when processes were non-normal so they could be "corrected," and returned to normality.


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• Most of the efforts done in the interest of quality improvement lead to non-normal processes, since they try to narrow the distribution using process stops. (Limits or adjustments to tool.)

• Similarly, nature itself can impose stops to a process, such as a service process whose waiting time is physically bounded at the lower end by zero. The design of a waiting process would move the process as close zero as economically possible, causing the process mode, median and average to move towards zero. This process would tend towards non-normality, regardless of whether it is stable or non-stable.


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• 1. INTRODUCTION.• Many processes do not follow the normal distributions. Some examples of

non-normal distributions would include:• Cycle time

• Calls per hour • Customer waiting time • Straightness • Perpendicularity

• Shrinkage • Defect counts• Can you name any others??


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• Normal distribution Non-Normal ( Cycletime data):


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• Cycle time data Example N =207


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• CYCLE TIME EXAMPLE

Common methods for handling non-normal data are:

•Sub group averaging

•Segmenting data

•Transforming data

•Using different distributions

•Non-parametric statistics

Descriptive Statistics


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Using different distributions: Cycletime data: Which non-normal distribution is a best fit. Various distributions overlaid on the data. We can see that the Weibull distribution is the best fit for the data.


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Using different distributions: Cycletime data:Weibull distribution is the best fit for the data.


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When performing statistical tests on data, it is important to realize that many statistical tests assume normality. If you have non-normal data, there are parametric equivalent statistical tests that should be employed. Table 2 below summarizes the statistical tests to use with normal process data, as well as the and non-parametric statistical test equivalents


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Why not just always use these non-parametric tests?Since non-parametric tests do not make an assumption about the distribution of the data, they have less information to use to determine significance. They are less powerful then the parametric tests, such that they have a more difficult time finding statistical significance.If the parametric test is not applicable , then use the appropriate non-parametric test


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Basic Statistics :Non-Normal data and Distribution free testsEXAMPLE 1: Using Wilcoxon test

Student t test: If one assumes Normally distributed data: Now let's consider a case where we would like to study whether the mean cycle time of this process is at 20 days. If we assume data normality and run a one sample student t test to confirm this hypothesis, Table 5 displays the results. Need to compare calculated t to t value in table

Ho: µ1 =20 Hi: µ1 = 20N total sample size: 207

Mean µ1 : 21.787 Standard deviation: 12.135 Standard Error of mean: 0.845T test results show significance with a P value of 0.035:Meaning: If I reject the Null Hypothesis µ1 = 20 I have a 3.5% chance of being wrong!


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Basic Statistics :Non-Normal data and Distribution free tests EXAMPLE 1: Using Wilcoxon test

Based on the above statistics, one would pronounce at an alpha risk of 5% that the mean of the data set is different than 20. If we were to verify the fact that the data is not normal, we would have run the one sample Wilcoxon test which is based on the medians rather than the means, and we would have obtained the results found in Table 6.

N sample size test statistic T P and Median


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Basic Statistics :Non-Normal data and Distribution free testsEXAMPLE 1: Using Wilcoxon test

The Wilcoxon test indicates that the null hypothesis (test median is equal to 20) is accepted, and there is no statistical evidence that the median is different than 20.The above example illustrates the fact that assuming that the data is normal and applying statistical tests is dangerous. As abetter strategy in data analysis, it is better to verify the normality assumption and then -- based on the results -- use an appropriate data analysis method.


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The Wilcoxon T Statistic table

( Do not confuse with Student t)


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Basic Statistics :Non-Normal data and Distribution free tests EXAMPLE 2: Wilcoxon Signed-Ranks Test

Wilcoxon Signed-Ranks Test takes the magnitudes (how large the numbers are) of the differences into account

Applicable (only) to:data sets involving two (paired) dependent samples

---- To test whether the two samples come from populations with the same distribution or not


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Basic Statistics :Non-Normal data and Distribution free testsEXAMPLE 2: Wilcoxon Signed-Ranks Test

Assumption

In using the Wilcoxon signed-ranks test for two dependent (paired) samples, we assume that the population of differences (found from the pairs of data) has a distribution that is approximately symmetric.


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Wilcoxon Signed-Ranks Test

H0: The two samples come from populations with the same distribution.

H1: The two samples come from populations with different distributions.


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Wilcoxon Signed-Ranks Test : TEST STATISTIC T

(DO NOT CONFUSE WITH Student t)

Define Test statistic T = the smaller of the following two sums:

1. The sum of the absolute values of the negative ranks

2. The sum of the positive ranks

In particular, we follow the following 1 -5 steps to find the value of the test statistic T


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Step 1: For each pair of data, find the difference d by subtracting the second score from the first. Keep signs, but discard any pairs for which

d = 0.

Step 2: Ignore the signs of the differences, then rank the differences from lowest to highest. When differences have the same numerical value, assign to them the mean of the ranks involved in the tie.

Step 3: Attach to each rank the sign of the difference from which it came. That is, insert those signs that were ignored in step 2.

Step 4: Find the sum of the absolute values of the negative ranks. Also find the sum of the positive ranks.

Step 5: Let T be the smaller of the two sums found in step 4. This is the test statistic T


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Example: Weight Training:

Runner A B C D E FPre-Marathon weights (kg) 99 62 74 59 70 73Post-Marathon weights (kg) 94 62 66 58 70 76Step 1: d 5 0 8 1 0 -3Step 2: rank 3 * 4 1 * 2Step 3: signed rank 3 * 4 1 * -2Step 4: sum of Positive ranks = 3 + 4 + 1 = 8

sum of Negative ranks = 2Step 5: T = 2 (T is the smaller of the two sums )


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The Wilcoxon T Statistic table

( Do not confuse with Student t) for this example there are 6 pairs n = 6 thus T critical = 1


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•Procedure for Wilcoxon Signed -Ranks TestStep 1: Find the value of the test statistic T. Step 2: Let n be the number of pairs of data for which the difference d is not 0.

Based on the sample size n, find the critical values from Table.

Step 3: Reject the null hypothesis if the test T is less than or equal to the critical value. Otherwise, fail to reject the null hypothesis.

In this example test T = 2 and critical T from table = 1((from Table α = 0.05 two tail and n = 6)

Thus we accept the Null Hypothesis: H0: The two samples come from populations with the same distribution.STATEMENT: For the sample of marathon runners studied we could not detect any different in Pre and Post race weights.


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Basic Statistics :Non-Normal data and Distribution free testsWilcoxon Signed -Ranks Test

Wilcoxon Signed -Ranks Test

IDEAIf the null hypothesis (from the same distribution) is true, the sum of the positive ranks and the sum of the absolute values of negative ranks should be approximately the same. Both of them should be larger than the corresponding critical T value found in the table.---


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For n ≤ 30: T as defined

Test Statistic for theWilcoxon Signed-Ranks Test for Two Dependent Samples


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Test Statistics for theWilcoxon Signed-Ranks Test for Two Dependent Samples : Sample size dependence

For n ≤ 30: T

For n > 30:Distribution approximates a normal GuassionNote T calculated = smaller

sum)

n(n + 1)

n(n +1) (2n +1)z = 4

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T –


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For n ≤ 30: T

n(n + 1)

n(n +1) (2n +1)z =For n > 30: 4

T –

24Critical values:

For n ≤ 30, critical T values are in Wilcoxon Table

For n > 30, critical z values are in Z table


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Right Left

19197

116165116129171155112102188158121133

224171191207196165177165140188155219177174

Reaction times (in thousandths of a second) of 14 Subjects

Example:Based on the reaction timesprovided on the left, use theWilcoxon Signed-RanksTest to test the claim of no difference between reaction times with the rightand left hands. (α = 0.05)


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SignedRanks

Ranks ofDifferences

Differences dRight Left

19197

116165116129171155112102188158121133

–4.5–11–12

–8–13

–6–1–2–3

–14+4.5

–10–9–7

224171191207196165177165140188155219177174

4.51112

813

6123

144.5

109 7

–33–74–75–42–80–36–6

–10–28–86+33–61–56–41

EXAMPLE DATA:Reaction times for left and right hand (in thousandths of a second) of 14 Subjects


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The Wilcoxon T Statistic table n = 14 pairs


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Example: Wilcoxon Signed-Ranks Test Solution:

α = 0.05 (significance level)Step 1: (based on the table on the previous page)

sum of the positive ranks = 4.5sum of the negative ranks = 4.5 + 11 + … + 7 = 100.5 so, T = 4.5

Step 2: n = 14 < 30Critical Value C.V. = 21 (from Table α = 0.05 two tail and n = 14) )

Step 3: Since critical T value (21) > Calculated T value (4.5),we reject the null hypothesis. Conclusion: There is a difference between reaction times with the right and the left hands.


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Basic Statistics :Non-Normal data and Distribution free tests:Shainin B Vs C test also called “6 pack test”

Shainin’s Six Pack Test

Many current weathering studies use parametric statistical tools to analyze results. Researchersshould not overlook non-parametric tools for weathering data analysis. Non-parametric analysiscan be especially useful in weathering studies involving appearance issues and visualevaluations for acceptability ranking. Also, often non-parametric analysis does not requirenormally distributed data to be effective. Dorian Shainin has developed a system of uniquetechniques for process analysis in manufacturing environments. Shainin developed the conceptknown as "The Red X" source of variations in manufacturing processes. Shainin’s logicalapproaches to production processes may also represent valuable tools for investigatingweathering processes. One such tool includes Shainin’s "6 pack test." Results are rank ordered.The ranking is then associated with input variables. The probabilities associated with a specificorder are then evaluated.

For example, two different lots of automotive paint were obtained from a single supplier with asingle formulation. The research question was to see if lot to lot differences could result indifferent weathering characteristics. Three randomly selected specimens from each lot wereexposed side by side, backed in Florida at 5° South for six months. After weathering, thespecimens were measured for color. The ∆ b* values were ranked. The ranking andcorresponding batch numbers associated with each ∆ b* values are as follows:


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Basic Statistics :Non-Normal data and Distribution free tests:Shainin B Vs C test

Rank Order of Six Specimens

∆ b* value after 6 mos., Florida, backed, 5°

Batch Numbers

1 0.29 881 "C"

2 0.28 881 "C"

3 0.19 881 "C"

4 -0.08 921 "T"

5 -0.18 921 "T"

6 -0.19 921 "T"

There are 20 different equally likely rank orders for these 6 specimens. Similarly there is only a one in twenty chance that the three "C" specimens would all rank above the "T" specimens. The likelihood of this particular order from chance alone is 1/20 or 0.05. This handy little tool is an effective way to identify vital variations with out requiring normality or distribution tables. Readers are highly encouraged to seek out additional Shainin techniques for weathering processes, as well as, manufacturing process solutions.

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OHSU OGI Class Steve Brainerd Basic Statistics …myplace.frontier.com/~stevebrainerd1/STATISTICS/ECE-580...1/29/03 Class 2 Non-Normal Data Steve Brainerd 4 OHSU OGI Class ECE-580-DOE