Download Six Sigma Class

1
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We first looked at this for the
regression analysis.
Introduction to (DOE) – one factor
test
Develops a model relating a
response to a factor - allows
prediction of responses at levels
other than where data was collected
◦ quantifies the strength of the model
◦ identifies outliers

Uses terminology involved in simple
linear regression
◦
◦
◦
◦
Intercept
Slope
Prediction Equation Y = f(X)
R-squared
X
Y
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Draw a “best fit” line through the data

Place the values you found for bo and
b1 in the following template.
y=

Where
+
X
bo
b1
bo = y-intercept
b1 = slope = Rise/Run
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xy
S xy  xy 

n
2



x
S x 2  x 2 

n
S xy
b1 

S x2


b 0  y  b1 x 
yˆ  b 0  b 1 x 
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How tall do we predict someone would be if their
shoe size is 9.5?
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IX-12
r
2
‘Correlation’ is a fancy word for how well the model
predicts the response from the factors.
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IX-18
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What is the r² best fit line?
S xy  xy 




S x 2  x 2
S y 2  y 2
r
xy

n
2

x 


n
2

y 


n
S xy
S x2 S y2

r2 
Same answer as slide 5?
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Did your
project
reduce
variation or
shift the
mean?
How do you
demonstrate
improvements
statistically
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IX-24
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The question one needs to ask
during a project is whether the
changes made as the result of a
Green Belt project have made a
difference in the process.
The question is, “Has the project
made a difference, and if so, how
confident are we that the
difference is statistically
significant?”
We strive for 95% confidence or
better (α=.05).
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
A hypothesis is:

Examples:
◦ a tentative explanation for an
observation that can be tested by further
investigation.
◦ taken to be true for the purpose of
investigation; an assumption.
◦ Average gas mileage differs depending
on type A or B gas.
◦ Probability of death in auto accidents
differs depending on seat belt usage.
◦ The type of aspirin determines the
amount of pain relief.
◦ A Six Sigma Green Belt project produced
significant results with regard to mean
and/or standard deviation improvement.
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
We set up 2 hypotheses
◦ H0 is called the null hypothesis
 what is being tested
 must contain an =
 assumed to be true
◦ H1 is called the alternate hypothesis

Based on the data we collect, we
must decide in favor of either H0
or H1. Which does the evidence
support?
◦ can only reject null, or fail to reject
null
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Define the problem and state objectives
State a “Null Hypothesis” (Ho)
State the “Alternative Hypothesis” (H1)
Establish significance level (a) and
determine the critical value
Collect sample data
Calculate test statistic and/or p-value
DECIDE:
What does the evidence suggest?
Reject Ho? or Fail to reject Ho?
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Based on a normal distribution
Z values from the Standard
Normal table
 Za / 2 (ˆ ) 
n

 errorlevel 
2
n = sample size
Za/2 = 1.96 for 95% confidence
= 2.576 for 99% confidence
̂ = estimated standard deviation
error level = amount of error allowed (i.e.
to estimate within +/-2 standard deviation,
the half interval width is 2)
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A test on cigarettes yielded an
average nicotine content of 15.6
milligrams and a standard deviation
of 2.1 milligrams.
If we want to be 99% confident in
detecting a 1 milligram change, what
sample size would be required?
What if we only want to be 95%
confident?
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Based on a binomial distribution
 ( Z a / 2 ) 2 p(1  p) 

n  
2
(p)


n = sample size
Za/2 = 1.96 for 95% confidence
= 2.576 for 99% confidence
p = average proportion defectives
1  p = q (proportion non-defectives)
 p = amount of error allowed
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You have just received a shipment of
computer memory chips.
Suppose we wanted to have 95%
confidence with detecting a change
of 1%. Assuming the historical
proportion of defectives in no more
than 2%, what sample size would be
required?
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Up to this point we have talked
about point estimators (mean,
standard deviation, etc) these
represent population parameters.
We can also construct an interval
that has a predetermined
probability of including the true
population parameter (). This is
a confidence interval.
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A test on a random sample of 9 cigarettes
yielded an average nicotine content of 15.6
milligrams and a standard deviation of 2.1
milligrams. Construct a 99% confidence
interval for the true average nicotine content.
What if we only want 95% confidence?
What happens to the 99% confidence interval
if we had tested 30 cigarettes originally?
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IX-35
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You have just received a shipment of
computer memory chips. From a sample of
100 chips, you find 6 to be defective. Find a
95% confidence interval for the true but
unknown proportion of defective chips in the
shipment.
What if we want 99% confidence?
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1.
2.
3.
4.
5.
6.
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Let’s set the Statapult up and take some data.
Set the Statapult up and reduce variation as
much as possible. Confirm with >25 shots.
Calculate and record average .
Choose a condition to test that you expect
will make a difference in the mean.
Use two inspectors. Record each inspectors
answers separately.
Fire 10 shots with the new setting.
Calculate sample average, X-bar and sample
standard deviation from one of the
inspectors.
Save the other inspector data for a later
example.
(1) Create a hypothesis test to determine if
the changed condition results in a statistical
difference.
(2) Determine if there is a difference between
the two inspectors.
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1.
2.
3.
4.
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Let’s set the Statapult up and take
some data.
Set the Statapult up and reduce
variation as much as possible.
Confirm with 10 shots. Calculate
and record sample standard
deviation s, and variance s².
Choose a condition to test that you
expect will make a difference in the
variation.
Shoot 10 shots with the new
condition and calculate sample
standard deviation s, and variance s².
Create a hypothesis test to
determine if the changed condition
results in a statistical difference.
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