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# LSSG Black Belt Training Hypothesis Testing - PowerPoint PPT Presentation

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: Person is guilty of crime

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### LSSG Black Belt Training

Hypothesis Testing

Always about a population parameter

Attempt to prove (or disprove) some assumption

Setup:

alternate hypothesis: What you wish to prove

Example:Person is guilty of crime

null hypothesis: Assume the opposite of what is to be proven. The null is always stated as an equality.

Example:Person is innocent

• Take a sample, compute statistic of interest.

The evidence gathered against defendant

• How likely is it that if the null were true, you would get such a statistic? (the p-value)

How likely is it that an innocent person would be found at the scene of crime, with gun in hand, etc.

• 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.

• 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.

Continuous

Attribute

Normal

Non-Normal

c2 Contingency Tables

Correlation

Variance

Variance

Means

Medians

Same tests as Non-Normal Medians

Z-tests

Levene’s

Correlation

c2

Sign Test

t-tests

F-test

ANOVA

Bartlett’s

Wilcoxon

Correlation

Kruskal-Wallis

Regression

Mood’s

Friedman’s

Hypothesis Testing Roadmap

Use parametric tests when:

• 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

Used when testing to see if sample comes from a known population. A sample of 25 measurements shows a mean of 16. Test whether this is significantly different from a the hypothesized mean of 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

70% of 200 customers surveyed say they prefer the taste of Brand X over competitors. Test the hypothesis that more than 66% of people in the population prefer Brand X.

Test and CI for One Proportion

Test of p = 0.66 vs p > 0.66

95%

Lower

Sample X N Sample p Bound Z-Value P-Value

1 140 200 0.700000 0.646701 1.19 0.116

BP Reduction%

10

12

9

8

7

12

14

13

15

16

18

12

18

19

20

17

15

The data show reductions in Blood Pressure in a sample of 17 people after a certain treatment. We wish to test whether the average reduction in BP was at least 13%, a benchmark set by some other treatment that we wish to match or better.

One-Sample T: BP Reduction

Test of mu = 13 vs > 13

95%

Lower

Variable N Mean StDev SE Mean Bound T P

BP 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 BP could not be proven to be greater than 13%. There is a 0.20 probability that it is not greater than 13%.

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 treatment. You separate the 17 observations by gender, and wish to test whether there is in fact a significant difference between genders.

M F

10 15

12 16

9 18

8 12

7 18

12 19

14 20

13 17

15

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 reaction to the treatment.

Two-sample T for BP Reduction M vs BP Reduction F

N Mean StDev SE Mean

BP Red M 8 10.63 2.50 0.89

BP Red F 9 16.67 2.45 0.82

Difference = mu (BP Red M) - mu (BP 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 StDev = 2.4749

From the book “Doing Data Analysis with Minitab 14” by Robert Carver:

• 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]

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 theSum of Squares. 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.

See www.statsoft.com for more details.

One-way ANOVA: Type A, Type B

Source DF SS MS F P

Factor 1 24.00 24.00 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%

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).

Minitab: Stat/ANOVA/One-Way (unstacked)

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.

Strength Temp Speed

20.0 Low Slow

22.0 Low Slow

21.5 Low Slow

23.0 Low Fast

24.0 Low Fast

22.0 Low Fast

25.0 High Slow

24.0 High Slow

24.5 High Slow

17.0 High Fast

18.0 High Fast

17.5 High Fast

Two-way ANOVA: Strength versus Temp, Speed

Source DF SS MS F P

Temp 1 3.5208 3.5208 5.45 0.048

Speed 1 20.0208 20.0208 31.00 0.001

Interaction 1 58.5208 58.5208 90.61 0.000

Error 8 5.1667 0.6458

Total 11 87.2292

S = 0.8036 R-Sq = 94.08% R-Sq(adj) = 91.86%

The results show significant main effects as well as an interaction effect.

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.

From the book “Doing Data Analysis with Minitab 14” by Robert Carver:

• 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]

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.

A 2F 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 –

• temperature,

• baking time, and

• amount of butter,

each at two levels – HIGH and LOW.

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.

Minitab: Stat/DOE/Create Factorial Design/Display Factorial Designs

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.

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

Estimated Effects and Coefficients for Crispiness (coded units)

Term Effect Coef SE Coef T P

Constant 7.2500 0.3750 19.33 0.000

Temp 3.0000 1.5000 0.3750 4.00 0.016

Time 0.5000 0.2500 0.3750 0.67 0.541

Butter 1.5000 0.7500 0.3750 2.00 0.116

S = 1.06066 R-Sq = 83.64% R-Sq(adj) = 71.36%

Analysis of Variance for Crispiness (coded units)

Source DF Seq SS Adj SS Adj MS F P

Main Effects 3 23.000 23.000 7.667 6.81 0.047

Residual Error 4 4.500 4.500 1.125

Total 7 27.500

Estimated Coefficients for Crispiness using data in uncoded units

Term Coef

Constant 7.25000

Temp 1.50000

Time 0.250000

Butter 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

From the book “Doing Data Analysis with Minitab 14” by Robert Carver:

• Pages 309 – 310: Answer any 4 of the 7 questions on those pages. [DOE]

Continuous

Attribute

Normal

Non-Normal

c2 Contingency Tables

Correlation

Variance

Variance

Means

Medians

Same tests as Non-Normal Medians

Z-tests

Levene’s

Correlation

c2

Sign Test

t-tests

F-test

ANOVA

Bartlett’s

Wilcoxon

Correlation

Kruskal-Wallis

Regression

Mood’s

Friedman’s

Hypothesis Testing Roadmap

Use non-parametric tests:

• 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.

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.

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.

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).

Sign Test for Median: Process Time

Sign test of median = 20.00 versus > 20.00

N Below Equal Above P Median

Process Time 12 3 0 9 0.0730 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.

The sign test can also be used for testing the value of the median difference betweenpaired 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/B135165.html

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.

The data

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.

Prod A Prod B Diff

55 50 5

60 62 -2

77 70 7

82 78 4

99 90 9

92 95 -3

86 90 -4

84 80 4

90 86 4

72 71 1

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:

• Data are at least ordinal in nature

• The pairs are independent of each other

• Dependent variable is continuous in nature

http://faculty.vassar.edu/lowry/ch12a.html

Wilcoxon Signed Rank Test: Diff

Test of median = 0.000000 versus median > 0.000000

N

for Wilcoxon Estimated

N Test Statistic P Median

Diff 10 10 44.5 0.046 2.500

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/ch11a.html

Old New

60 85

70 85

85 90

78 94

90 90

68 70

35 75

80 90

80 90

75 100

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.

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/ch11a.html

Mann-Whitney Test and CI: Old, New

N Median

Old 10 76.50

New 12 90.00

Point estimate for ETA1-ETA2 is -14.00

95.6 Percent CI for ETA1-ETA2 is (-22.00,-5.00)

W = 72.5

Test of ETA1 = ETA2 vs ETA1 < ETA2 is significant at 0.0028

The test is significant at 0.0025 (adjusted for ties)

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.

Rating Factor

7 Adults

5 Adults

6 Adults

4 Adults

2 Adults

6 Adults

5 Adults

9 Teens

9 Teens

8 Teens

5 Teens

9 Teens

10 Teens

7 Teens

8 Teens

3 Children

4 Children

3 Children

5 Children

10 Children

2 Children

Adults Teens Children

7 9 3

5 9 4

6 8 3

4 5 5

2 9 10

6 10 2

5 7

8

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.

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/ch14a.html

Kruskal-Wallis Test: Rating versus Factor

Kruskal-Wallis Test on Rating

Factor N Median Ave Rank Z

Adults 7 5.000 8.6 -1.23

Children 6 3.500 7.2 -1.79

Teens 8 8.500 15.9 2.86

Overall 21 11.0

H = 8.37 DF = 2 P = 0.015

H = 8.48 DF = 2 P = 0.014 (adjusted for ties)

Mood median test for Rating

Chi-Square = 10.52 DF = 2 P = 0.005

Individual 95.0% CIs

Factor N<= N> Median Q3-Q1 --------+---------+---------+--------

Adults 6 1 5.00 2.00 (-------*-----)

Children 5 1 3.50 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.

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.

Perform Group Block

7 Normal 1

8 Normal 2

6 Normal 3

9 Normal 4

5 Normal 5

7 Normal 6

8 Normal 7

5 Noise 1

4 Noise 2

6 Noise 3

5 Noise 4

5 Noise 5

4 Noise 6

4 Noise 7

8 Music 1

8 Music 2

8 Music 3

8 Music 4

7 Music 5

9 Music 6

9 Music 7

Normal Noise Music

7 5 8

8 4 8

6 6 8

9 5 8

5 5 7

7 4 9

8 4 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.

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/ch15a.html

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)

Sum

Est of

Group N Median Ranks

Music 7 8.000 19.5

Noise 7 4.667 8.0

Normal 7 7.333 14.5

Grand median = 6.667

From the book “Doing Data Analysis with Minitab 14” by Robert Carver:

• Pages 293 – 294: Choose any 3 datasets on those pages and answer the related questions. [Non-parametric tests]