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www.forcemotors.com Name of the Presentation I dd.mm.yyyy
Minitab Training Manual – Part II
04.05.2021
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Prepared By : Subhash Sirsat
'Shriram Kulkarni
Minitab®
Practice Session Module-II
Supported By: Makarand Kanade
Sr. VP : ( Corporate Quality , R & D)
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Histogram
Minitab®
Practice Session Module-II
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Hypothesis Testing X-Attribute
Y is Variable Y is attribute
Variation -σ Average-σ Proportion -P
One to Standard 1 – Variance test 1 –t Test 1-P test
One to one
2- Variance test (F test- Normal distribution) , Levens test (for Non normal test)
2- t Test 2- P test
One to many
Equal variance test a) Bartlett’s test (for Normal data)
b) Levens test (for non
normal data)
1 Way Anova ( if variances are Equal)
Chi- Square Test
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Hypothesis Testing :Key concepts
Purpose of hypothesis testing is not to question calculated values , but to make judgment about the difference in
the values of two data sets.
Next step , is to write hypothesis : null hypothesis is a statement of innocence. Says , there is no difference of
in the values of two data sets.
“alpha” is called significance level for hypothesis testing ( generally 0.05)
(1-alpha) is called confidence level for hypothesis testing (generally 95%)
We need a certain minimum confidence level for hypothesis testing
e.g : Ho : E (A) = E(B) “ Efficiencies of both coaches A & B are same.
Alternate Hypothesis : Is statement of guilty for the above null hypothesis
We have 3 possible alternative hypothesis.
H1: E(A) < E(B) ( efficiency of A is less than B) one tailed (a=0.05)
H1: E(A) >E(B) ( Efficiency of A is greater than B)one tailed (a=0.05)
H1: E(A) not equal to E(B) two tailed (a=0.025)
H0 : Null Hypothesis
H1: Alternative hypothesis
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It works very much like a court case
When we have a suspect ; we have to take a decision whether he / She is innocent or
guilty .
Null Hypothesis is statement of innocence (Ho)
A accused is said to be innocent unless proved guilty
We have collect sufficient proofs to take decision about the innocence .
Null hypothesis is statement of No change or No Difference.
Alternative hypothesis challenges Null hypothesis; If Null hypothesis is proven wrong
alternative hypothesis must be right.
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Hypothesis Testing :Key concepts
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Types of Errors : Type – I Error :- To react on outcome as it comes from special cause , when it actually comes from common cause of variation ( also called alpha error ) “over control = tampering=over adjustment ” . When variation due to common cause , changing the process thinking that it is due to abnormality and adjusting the process , increases the variation . Type II Error : - also called beta error . To react to an outcome , when actually it comes from special cause . Treating special cause , as a common cause . Not doing analysis , even in the presence of special cause leads to unstable process which is unpredictable and increased
variations due to presence of abnormality .
CORRECT DECISION
( CONFIDENCE : 1 - alpha) Error ( TYPE – II )
( Beta Error) : 10 %
ERROR ( TYPE – I )
( Alpha Error ): 5 % CORRECT DECISION
(Power : 1- beta)
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Hypothesis Risk P Value Conclusions
H0 Ha A <0.05
We conclude that there is
strong evidence for us to
Reject Null Hypotheses &
Accept Alternative
hypothesis
H0 Ha B >/=0.05
We have not proved that null
hypothesis is true
We didn’t have enough
evidence to reject H0
We typically will either collect
more data or accept null
hypothesis by default
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Hypothesis Testing : Drawing Conclusion
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General Rule
P Low , Null Go
P High , Null is Guy
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Used to compare standard Deviation of data with target value of Standard Deviation (one to Standard)
Hypothesis Statement
H0: б=target б
Ha: б < target б
: б > target б
: б≠ target б
Example: Ho: бv > 0.003
Ha : бv< 0.003
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One Variance Test
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Click here on “ hypothesis testing”
Step -1
One Variance Test
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Click here on “ 1 –Sample standard Deviation ”
Step -2
One Variance Test
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Select data from column
Put target value of Std Deviation
Select Hypothesis
Step -3
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Since p value is less than 0.05 we can conclude with 95 % confidence that std deviation of bore dia is less than
0.003
Output
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Graphical representation of data
Output
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Report card: Checks data integrity ;sample size, & test validity
Output
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This test is used to compare standard Deviation of two data .
Hypothesis Statement
Ho :б1 = б2 Null Hypothesis
Ha : б1< б2 Alternate hypothesis
: б1> б2
: б1 ≠б1
Example: Comparing variation in samples of two suppliers 1) Vikas Industries 2) Jai Bhavani Mata.
Ho : бVikas</=бJai Bhavani
Ha: бVikas >бJai Bhavani
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Two Variance Test
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Select data in 2 columns
Step -1
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Select -2 variance test
Step -2
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Compare std dev. Of Vikas Ind. & Jay Bhavani
Mata Ind.
Select Hypothesis
Click “OK ”
Step -3
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Conclusion : Since p values is more than 0.05 we can not conclude that std
deviation of Vikas is greater than Std Deviation of Jai Bhavani Mata.
Output
Graphical representation of data
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Graphical representation of data
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Conclusion: for Unusual data check one point ; check reason for unusual data.
Output
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Equal Variance test is used for comparing more than two standard deviations For Example : Comparing variations in 4 machines. Ho : бM/C1=бM/C2=бM/C3= бM/C 4 Ha: At least one б is different
Equal Variance Test
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Select “ hypothesis test”
Step -1
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Select “standard deviation test ”
Step -2
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Select data in required coloumns
Step -3
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Conclusion: Since P value is more than 0.05 there is no significant different in All machines б
CI for 4 Standard Deviation
Output
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Output
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Note : Look at, standard deviations of the M/c- 1 , M/c- 2 , M/c. - 3 & M/c – 4 CI for all 4 data are almost same
Output
Graphical representation of data
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One sample t test is used to compare mean of data with Standard. Example : Average height of Indian Men is less than 165 cm . In this Example we want to check Whether Mean Oil pump pressure is greater than 8 Bar Ho: цp </= 8 Ha: цp > 8
1 Sample Test
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Click “Here ”
Data from Example
Step -1
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Enter value of rows
Enter here “ mean value ”
Click “ hypothesis statement ” which is to be tested
Step -2
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Since P value is greater than 0.05 we can not conclude that mean oil pump pressure is grater than 8 bar
Output
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Output
Graphical representation of data
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Output
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2 Sample- t test is used to compare means ( Averages) of two data. Example: Average Surface finish of Vikas Ind. parts is greater than Average Surface finish of JBMI parts. Hypothesis Statement Ho : цv < цj Ha: цv > цj
2 Sample Test
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Select “Here ”Hypothesis test
Step -1
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Click “ 2 sample T test”
Step -2
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Select “ hypothesis statement, you want to test”
Step -3
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Conclusion: Since P value is less than 0.05 with 95% confidence we can conclude that Surface finish Vikas is greater than JBMI
Output
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One way ANOVA is used to compare means of more than 2 samples Example : Compare Mean Bore Sizes of 3 machines A ,B,C Hypothesis Statement : Ho : all means are Equal Ha : At least one mean is different.
One Way ANOVA Test
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Select Hypothesis Testing
Step -1
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Select one way ANOVA
Step -2
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Select Values in Separate columns
Step -3
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Conclusion : Since P value is less than 0.05 we can conclude that means size of Machines are not same . Mean of Machine “A “is different than “B” & “C “
Output
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Output
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Since CI of all three mean are not same we can conclude all means are statistically different
Output
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Output
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1 P test is used to compare percentage with respect to Standard value Example: Whether Rejection % in Castings is less than 7 %
1P Test
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Parts with dents
type of defect Quantity Checked parts defective for
dents Target
Painting defect 5500 15 0.5 % rejection
Hypothesis Statement : Rejection % is less than 0.5% Ho: % Rej >/= 0.5 Ha : % Rej </= 0.5
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Please enter data in 1 P test format
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Output
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Output
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Output
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Conclusion : Since P value is more than 0.05 we can not conclude that rejection % is less than 0.5%
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2-P test is used to compare % of two data
Example: % of Rejection Before improvement
% of Rejection after improvement
2P Test
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Check whether REJECTION % after taking action is significant
Sr # Rejection / total
production %
1 Rejection after taking action 7 /170 4.12%
2 Rejection before taking action 11/105 10.47%
Ho: Rejection % after action >/= Rejection before action Ha: Rejection % after action < Rejection before action
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Select “ Hypothesis testing”
Step -1
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Select “ 2 S” % defective Hypothesis testing
Step -2
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Enter “ Rejection % before improvement”
Enter “ Rejection % after improvement”
Click “ OK”
Step -3
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Conclusion : Since P value is less than 0.05 . We can conclude that % rejection before improvement is significantly greater than after improvement “
Output
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Output
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Output
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Chi – Square test is used to compare more two % defective test
Ho: All % Rejections are same
Ha: At least one % Rejection is different
CHI-Square Test
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Lot Qty Defective parts
500 15
400 20
2000 25
450 12
Data for test
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Select Hypothesis testing
Step -1
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Select Chi- square test % defective
Step -2
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Fill table for total no tested & No of defectives
Step -3
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Since P values is less than 0.05 with 95% we can conclude that difference among % defective are significant
Output
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Output
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Thank You
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