12 SSGB Amity BSI Hypothesis

8/9/2019 12 SSGB Amity BSI Hypothesis

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Module-12

Hypothesis Testing


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Objectives of this module

To introduce a range of hypothesis tests suitablefor testing for differences between averages.

Specifically:

1 Sample t-test

2 Sample t-test

Paired t-test

ANOVA


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Tools for Verifying the Causes

Verify the CausesPlot - ScatterMatrix

Boxplot

Dotplot

CorrelationHypothesis Tests

ANOVA Regression

Design ofExperiments

Identify possible

relationships

Determine strengthof relationships

Quantifyrelationships

Identify and quantifycomplex relationships

Increasingcomplexit

y

ofre

lationship


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Hypothesis Tests for Average

Before a hypothesis test for averages can be selected,there are two preliminary questions:

1) Are the samples Normally distributed?

Each of the samples that are to be included in the test must

contain Normally distributed data in order for the results of thetest to be statistically valid.

2) How many samples do you want to compare?

The number of samples refers to the number of different

samples (subgroups) that are to be compared, not the amount ofdata that is in each of those samples (thats sample size).


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Are the

Samples

normally

distributed

Determinethe Numberof Samples

Transform theDatausing Box-Cox

CompareMedian

Values

YES

NO

NO

1 Sample t - Test

2 Sample t - Test

One-way Anova

1

2

2

3+

Paired t - Test

To compare the Averages of samples of data


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Are the

Samples

normally

distributed


Transform theData

using Box-Cox

CompareMedian

Values

Does theData have

Outliers

Use the

Moods

Median

Test

Use theKruskalWallis

Test

YES

YESNO

NO

NO

To compare the Medians of samples of data


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An Expensive Situation

You manage a warranty claims department. A

customer claims loss of earnings of $1,245 for an

item which usually is about $1,000. You examine 250 previous claims of the same

item to make a comparison and find the average

to indeed be $997 with a standard deviation $88. You want to know if the customer is over-

claiming or if it is reasonable


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Innocent or Guilty?

If the customer is not over-claiming (it is a

legitimate claim) then we would expect the

claim to fit with the pattern of data representedby the previous 250 claims

If the customer is over-claiming then we wouldexpect the claim to not fit the pattern of data

from the previous 250 claims


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Does the Claim fit the Pattern?

You remember from previous module that if the data isnormally distributed you can apply normal theory

997 1085 1173 1261

s = 88

1245

By using standard Normaltables we can calculate theprobability of a claim of$1245 based on the historicaldata. If the probability is low,then we can assume anIllegitimate claim

Z =X - X

s

Z = = 2.821245 - 99788From Standard Normal tables the P-value, the probability of being equal to orgreater than 1245 is 0.0024 or 0.24%. In other words we would expect such a claimto happen 1 in 417 claims


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P-values are Probabilities of Interest

P----value

Tail area Area under curve beyond value of interest

Probability of being at value of interest orbeyond

-4 -3 -2 -1 0 1 2 3 4 -4 -3 -2 -1 0 1 2 3 4 -4 -3 -2 -1 0 1 2 3 4

Value of

Interest

Value of

Interest

Value of

Interest

Value of

Interest


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Making the Decision

The Normal theory is telling us that based on the previous250 claims, we should expect a claim of $1245 or greaterevery 417 claims

Hence: This claim is legitimate it is that 1 in 417

This is not legitimate - it does not fit the previous data

Experience shows a small p-value (0 to 0.05) means The probability is small that the value of interest comes

from that distribution by chance therefore something elseis going on

Since our p-value 0.024 < 0.05 we can conclude that theclaim is not legitimate


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What Have we done?

In the previous example we have used the properties of theNormal distribution to test whether the occurrence of anevent could have happened by chance (the data fits the

expected pattern) or there is a real difference (the data doesnot fit the expected pattern)

This type of situation occurs frequently during Six Sigma

improvement projects, either in the

Analyze phase when we are looking for differences toidentify potential roots causes

Improve and Control phases when we are aiming todemonstrate that a real change has been made we havemade a difference


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Analyze: Is there a Difference?

A common question during the Analyze phase is is therereally a difference?

For example: We suspect the output of a process depends

upon the supplier

Is there a difference?

Process YA

Variation inprocess output

Sample yA and sA

Supplier A

Process YB


Sample yB and sB

Supplier B


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Improve: Have We Made a Difference?

A common question having improved a process is

have we really made a difference?

CHANGE


Old Process Yold

Variation in

process output

Sample yold and sold

New Process Ynew


Sample ynew and snew


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10 11 12 13 14 15 16 17 18 19 20

0

10

20

Yold

Fre

quency

10 11 12 13 14 15 16 17 18 19

0

10

20

Ynew

Fre

quency

Looking at the histograms may give us the idea thatthere is a difference

But can we be sure?


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A Difference?

It is possible that we have taken two samples from the samedistribution

Oldsample

Newsample

While on face value they look different statistically theyare not!! The difference is due to chance (sampling)


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Hypothesis Testing

In the first example we calculated probabilities to help usdecide. This situation is common, where we are interested inmaking a decision about differences between two or more

sets of data that relate to real world situations Hypothesis Testing allows us to decide whether an observed

difference is real or has happened by chance

In simple terms we expect there to be a difference betweenexpected and observed outcomes due to random(chance/common cause variation) factors. Hypothesis testingis concerned with whether these differences are so large that

they cannot be explained by random (chance) effects Hypothesis tests are often based on sample data in which

case we use sampling distributions rather than thedistribution of individual values


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Hypothesis Testing Concept

In order to show that an apparent difference is real

or due to chance we start by assuming that there is

no difference (Null Hypothesis, H0). If the observed difference is within that expected by

chance then the Null Hypothesis of no difference iscorrect.

If the observed difference is greater than that expected bychance then the Null Hypothesis of no difference is not

correct.


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Greater Than Expected!

Mean x

-1x 1x-2x 2x-3x 3x

Distribution ofsample means -

+-Theoretically the

limits are

68.26%

95.45%

99.73%100%

We cannot be 100% certain that an actual sample meanfalls outside the distribution. There is always an

element of risk


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95% Confidence

Mean x

-1.96x 1.96x

Distribution ofsample means

95%

Experience has shownthat setting the decision

Boundaries at 95%provides a good balance

between being able tomake a decision and the

risk of making thewrong decision


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Risk

Mean x

95%

A point here could be dueto chance or there is a difference

If we decide they are different

hen they are not we have made atype I error

The risk of making this error

Is called the -risk

A point here could be due tochance or there is a difference.

If decide they are the same when

they are not we have made atype II error

The risk of making this error

is called the -risk


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Type I and II errors

Type I - Deciding they are different when they are not

Type II - Not detecting a difference when there is one

P-value = the actual probability of making a Type I error

The probability of a Type II can be calculated given anassumed true difference

Both are important Guarding too heavily against one increase the risk of the

other

Increasing the sample size Reduces Type II errors

Allows you to detect smaller differences

More Later


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Hypothesis Tests

Hypothesis tests can be used in a number of situations:

A sample is taken and found to have a mean x, is this value equal toa target value T?

Samples taken before and after a change, the two sample means aredifferent - but are they?

Samples from a process for different stratification factors, the twosample means are different - but are they?

Samples taken before and after a change, the standard deviations aredifferent - but are they?

Samples from a process for different stratification factors, the twosample standard deviations are different - but are they?

The proportion defective appears to vary with different factors butdoes it?


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Steps in Hypothesis Testing

Hypothesis tests follow the pattern

1. determine the null hypothesis H0

2. determine the alternative hypothesis H1 or Ha3. determine statistical test

4. state level of risk

5. establish sample size6. collect data

7. analyse data

8. accept or reject the null hypothesis9. determine conclusion


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1. The Null Hypothesis

The null hypothesis is always stated so as to

specify that there is no (null) difference

When comparing means or variances the null

hypothesis is:

H0: = Target

H0: 1 = 2H

0

: 1

= 2


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2. The Alternative Hypothesis

The alternative hypothesis opposes the null

hypothesis and can therefore has several options

H1: 1 < 2

orH1: 1 > 2

or

H1: 1 2

The choice dependsupon the problemMinitab gives theoption to select

these

For example


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3. Determine Tests

Compare two or more groupproportions.

Chi-square test

Compare two or more group

variances.

Test for equal variances (F-test,

Bartletts test, Levenes test)

Compare two or more groupaverages.

ANOVA

(Analysis Of Variance)

Compare two group averages when

data is matched.

paired t-test

Compare a group average against atarget.

Compare two group averages.

t-test

Can be used toHypothesis Test


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3. Determine Test

Hypothesis tests are used when the input or process (X)variable is discrete.

If the X data are continuous, use Regression analysis tojudge whether they are related to the output (Y)variable.

Y

X

Continuou

s

Discre

te

(Proportions)

Discrete(Groups) Continuous

Chi-Square

t-testPaired t-test

ANOVA

Logisticregression

Regression


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4. State level of Risk

It is very important to specify before conducting the test thelevel of risk that the decision will be based on

The risk means that there is the chance that we will make

the wrong decision In assessing the risk level we need to consider the situation

and the business it is a question of significance vs.importance

Significant means there is a real difference as proven by ahypothesis test

Important means that the difference observed is important

to the business Typically the -risk is set at 5% which provides a 95%confidence


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"Important" vs. "Significant" Differences,

Important but not significant differences

Sometimes you cannot claim a difference is statisticallysignificant yet the observed difference is of importance to

the business Example: Production volumes

An increase of 1000 items produced per day is observed during the pilot.

An increase of 1000 is important to the business.

However, the difference is not statistically significant

Either the observed difference is due to random variation and no true

difference exists, or the the variation is too large (or sample size too

small) to detect the difference.

You (and your business leaders) need to decide if it is worth

the risk to go ahead and implement the new process


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5. Establish Sample Size

The ability to detect differences is related to the

sample size

A small sample means we cannot detect smalldifferences, but the cost of data collection is low

A large sample size will enable the detection of small

differences, but the cost of data collection could be

high

Selecting the right sample size depends upon

thePower = (1 - )


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6. Collect Data

Remember all the rules and guidelines about

collecting data

Sample size Good operational definition

Clear measurement procedure

Un-biased samples

Good Gauge R&R


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7. Analyse Data

All hypothesis tests can be conducted by hand

calculation BUT

Minitab Rules!


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8. Acceptance or Rejection

Method 1if the p-value is > or = to - fail to reject H0if the p-value is < - reject H0 and accept H1

Method 2

if 0 falls within the confidence interval - fail to reject H0if 0 falls outside the confidence interval - reject H0 and

accept H1

In practice we use Methods 2 and 3 from information providedby Minitab


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t -tests

i f i


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Mr Gossett (1876 1936) workedfor the Guinness company andpublished his work under the pen

name of Student, hence thestudent T-test.

The T-test measures thesignificance of shift in the average.average.

t- tests a bit of history



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Are the

Samples

normally

distributed


Transform the

Datausing Box-Cox

CompareMedianValues

YES

NO

NO

1 Sample t - Test

2 Sample t - Test

One-way Anova

1

2

2

3+

Paired t - Test


1 S l (1)


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1 Sample t-test (1)

1 Sample t-tests are used to compare the average

of one sample of data against a known average.

The known average may be a historical average

or an industry benchmark.

For the purpose of the test, the known average is

assumed to be exact.

Open data file: One-Sample-t.MPJ

1 S l t t t (2)


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1 Sample t-test (2)

A project team has implemented a new invoicingprocess, and wants to compare it against the

historical average of 16.5 days.

Hypothesis:

The Individual Value

Plot suggests that theaverage invoice timefor the new process is

lower (quicker) thanthe historical averageof 16.5

New

Process

22

20

18

16

14

12

10

Individual Value Plot of New Process

New process

16.5

1 S l t t t (3)


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1 Sample t-test (3)

A 1 Sample t-test can be used to test if the differencebetween the averages is statistically significant.

The hypotheses for the test would be:

Ho: There is no difference between the average invoice time

of the new process and 16.5 Ha: New process is lower than average invoice time of 16.5

Because our theory is that there is a difference, we

expect to reject the Null hypothesis.

1 S l t t t (4)


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1 Sample t-test (4)

MINITAB: Stat > Basic Statistics > 1-Sample t

There are two optionsfor entering data.

1st Option: Enter thecolumn that contains the

data here.

2nd Option: If you knowthe sample size, meanand standard deviation

of the data it can beentered here.

In both cases, the known averagethat the data is to be comparedagainst should be entered here.

1 Sample t tests (5)


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1 Sample t-tests (5)

Select the graph most appropriatefor the samples sizes involved. If indoubt, check all three.

Leave the confidence level of95%

Alternative: less than




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The p-value for the1 sample t-test isfound in the sessionwindow output. Inthis case, the p-valueis 0.040

Session Window

Output for 1 Sample

t-test

Since the p-value for this test is lessthan 0.05, we can be 95% confident

that there is a difference between theaverage invoicing time of the newprocess versus the historical average.We reject the Null Hypothesis

One-Sample T: New Process

Test of mu = 16.5 vs < 16.5

95%Upper

Variable N Mean StDev SE Mean Bound T P

New Process 15 14.9818 3.1218 0.8060 16.4015 -1.88 0.040



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New Process

22201816141210

_X

Ho

Individual Value Plot of New Process(with Ho and 95% t-confidence interval for the mean)

MINITAB adds a blue line to

the graphs that represents theconfidence interval of the

sample average, and alsoindicates the position of Ho(the known average).

In this case, Ho (the known

average) is outside theconfidence interval of the

sample average andtherefore we can prove adifference between the two.

New Process

F

requency

22201816141210

5

4

3

2

1

0 _X

Ho

Histogram of New Process(with Ho and 95% t-confidence interval for the mean)

1 Sample t-test Whiskey or water?


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1 Sample t-test Whiskey or water?

Problem : A cocktail bar suspects that one of its suppliers ofwhiskey is adding water to increase profits. The freezingtemperature of whiskey has a normal distribution with a

mean of= - 0.5450 C. Adding water raises the freezing

temperature. The cocktail bar wants to know if its suspicionsare correct. Is the supplier adding water to its whiskey?

Procedure: A sample from each of ten bottles is obtained from

this supplier. Experimental Unit : A bottle.

Measurement : Freezing temperature of the sample.

Significance Level : = 0.05

Data Set : Whiskey.MPJ



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Are the

Samplesnormally

distributed


Transform the

Datausing Box-Cox

CompareMedianValues

YES

NO

NO

1 Sample t - Test

2 Sample t - Test

One-way Anova

1

2

2

3+

Paired t - Test

o co pa e t e ve ages o sa p es o data

The concept of Two Sample t-tests


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You take two samples ofdata from the sameprocess, but at different

times. Has the processmean moved?

Would you think the

process mean has movedif the results had been..

What did you look at to make your decisions?What did you look at to make your decisions?

The concept of Two Sample t tests



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2 Sample t-tests are used to compare the averages of twosamples of data. The two samples may represent:

two different suppliers

two different processes

two different teams

two different products etc.

The two samples can be of different sample sizes, since the 2

sample t-test takes account of both sample sizes.

The two samples can have different levels of variation(standard deviation) since the two sample t-test takes

account of this as well.Open data file: Two-Sample-T-Appeal.MPJ



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Process

Time(days)

OldNew

42.5

40.0

37.5

35.0

32.5

30.0

27.5

25.0

Boxplot of Time (days) vs Process

A project team has implemented a new appealsprocess, and the box plot below shows the cycle

time of the new and old processes.

Hypothesis:

The Box Plot suggests

that the average cycletime for the new

appeals process is lower(quicker) than the old

process.



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A 2 Sample t-test can be used to test if thedifference between the average cycle times is

statistically significant.

The hypotheses for the test would be:

Ho: There is no difference between the average cycle

times for the new and old processes.

Ha: The new cycle time is lower than the cycle times for

the old processes.

Because our theory is that the new is lower, we

expect to reject the Null hypothesis.



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There are several options forentering data.

1st Option: If the data is in asingle column, with a secondcolumn containing subgroupdata.

2nd

Option: If the two datasamples are in two separatecolumns.

3rd Option: If you know thesample size, mean and standarddeviation of both samples, thisinformation can be enteredhere.




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p ( )

Set the options as shown

above:

a confidence of 95%

a test difference of 0.0 Alternative: less than




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Two-Sample T-Test and CI: Time (days), Process

Two-sample T for Time (days)

Process N Mean StDev SE MeanNew 20 32.12 3.46 0.77

Old 25 34.35 3.12 0.62

Difference = mu (New) - mu (Old)

Estimate for difference: -2.2370095% upper bound for difference: -0.56083T-Test of difference = 0 (vs


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p ( )

Process

Time

(days)

OldNew

42.5

40.0

37.5

35.0

32.5

30.0

27.5

25.0

Boxplot of Time (days) by Process

The graphs provided by the 2 sample t-test are thesame as those provided by using the graph menu.

By default, MINITAB adds a symbol to show theaverage of each subgroup.

Process

Time

(days)

OldNew

42.5

40.0

37.5

35.0

32.5

30.0

27.5

25.0

Individual Value Plot of Time (days) vs Process



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p

Exercise: Using the data file: Two-Sample-T-Appeal.MPJ

2) Repeat the test using the samples in different

columns option.

3) Repeat the test using the summarised data

option.



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Are the

Samplesnormally

distributed


Transform the

Datausing Box-Cox

CompareMedianValues

YES

NO

NO

1 Sample t - Test

2 Sample t - Test

One-way Anova

1

2

2

3+

Paired t - Test

12 SSGB Amity BSI Hypothesis

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