ISE 362 Probability and Statistics P value and F Distribution Presentation
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Transcript of ISE 362 Probability and Statistics P value and F Distribution Presentation
7/27/2019 ISE 362 Probability and Statistics P value and F Distribution Presentation
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ISE 326
Prob & Stats II
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P Value
• P is the small Critical Value That Would Allow
H0 to Be Rejected Based on The Test Statistic.
• Using the Calculated Test Statistic and the DoF,
Interpolate a Test α.
• If it is Smaller Than Given α
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Proportions• Inferences Concerning a Difference Between
Populations Proportions
– Many Engineering Problems Are Concerned With a
Random Variable That Follows The Binomial Distribution.
• i.e.: Manufactured Items are Classified as Either Acceptable or
Defective.
– It is Usually Reasonable To Model The Occurrence of
Defective Parts, Where p Represents The Proportion of
Defective Items Produced.
– Hence The Hypothesis Testing is:
H0 : p = p0
H1 : p ≠ p0
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Proportions
• We Can Approximate a Normal Distribution
Testing As Along As p is Not Extremely Close
To Zero.
• Let X be The Number of Observations in a
Random Sample of Size n That Belong to The
Class Associated With p
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Proportions
• The Test Statistic is:
• Reject H0 : p = p0 if
or
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Proportions - Example
• A Semiconductor Manufacturer Produces
Controllers Used In Automobile Engine
Application.
• The Customer Requires That The Defective
Rate at a Not Exceed 0.05 and That the
Manufacturer Demonstrate Process Capability
at This Level of Quality Using α = 0.05
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Proportions - Example
• The Manufacturer Pulls 200 Random Samples
and Finds That 4 Are Defective.
• Can The Manufacturer Demonstrate Process
Capability to The Customer?
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Proportions - Example
• H0 : p = 0.05
• H1 : p < 0.05
• If The Manufacturer Can Make A Strong Claim About
The Process Capability if The Null Hypothesis Can BeRejected
• α = 0.05
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Proportions - Example
• The Test Statistic
• Where x = 4, n = 200 and p0 = 0.05
• Reject H0 : p0 = 0.05 if = -1.645
•
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Proportions - Example
• Z0 = -1.95
• We Should be Able to Reject H0.
•
Check to See if The P-value for Z0 is FurtherOut On The Tails
• For P = 0.0256 (Interpolated) Which is Smaller
Than α = 0.05
• Therefore The Process is Capable of Meeting
The Customer’s Quality Requirement
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Proportions
• Another Form of The Test Statistic
• Let
• Then
This Presents the Test Statistics in Terms of SampleProportion Rather Than in Terms of X
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Two Proportions
• The Previous Test Can Be Extended To The
Case Where There Are Two Binomial
Parameters of Interest
H0 : p1 = p2
H1 : p1 ≠ p2
• We Shall Look at Large Sample Procedure
based on a Normal Approximation
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Two Proportions
• Assume Two Random Sample Sizes and
are Taken From two Populations
• Let
and
Represent the Number of
Observations That Belong to The Class of
Interest in Samples 1 and 2, Respectively.
• Suppose That the Normal Approximation to
the Binomial is Applied to Each Population
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Two Proportions
• The Estimators of The Population Proportions
= and =
Have Approximate
Normal Distributions as Well
• The Test Statistic Becomes
0 = −
(1−)( 1 + 1)
Where p = p1 = p2
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Two Proportions
• So then the Estimator For the Common
Parameter, is
= + +
• The Test Statistic Becomes:
0 = −
(1−)( 1 +
1)
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Two Proportions
• Example
• Two Different Types of Polishing Solutions are BeingEvaluated as Possible Use in a Tumble Polish Operationfor the Manufacture of Interocular Lenses in the
Human Eye.• 300 Lenses Were Polished Using Fluid A, of These 253
Were Acceptable
• 300 Lenses Were Polished Using Fluid B, of These 196Were Acceptable
• Is There Any Reason To Believe The 2 Polishing FluidsDiffer?
• Use α = 0.01
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F Distribution
• Occasionally One Has to Compare the Variance Between 2Populations
• The F Distribution Has 2 Parameters, 1 and 2 Also Know AsDegrees of Freedoms (DoF)
• The Density Function is Complicated and is Not Generally Used
Explicitly.• There is a Connection Between an F Variable and the Chi-Squared
Variables
• If X1 and X2 are Independent Chi-Squared Random Variables with1 and 2 DoF, then the Ratio of the Variables Divided by Their DoF
Has an F distribution
=
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F Distribution
• An F Distribution Looks Approximately Like :
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F Distribution
• Let X1, X2, X3,….Xm be a Random Sample from aNormal Distribution with a Variance of 1
• Let Y1, Y2, Y3,….Yn be a Random Sample from aNormal Distribution with a Variance of
2
,
Independent of X.• Let and Represent the Sample Variances
•
Then =
• Has an F distribution with 1 = m-1 and 2 = n-1
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F Distribution
• H0 : 1 = 2
• H1 : 1 ≠ 2
•
The Test Statistic is
• Rejection Criteria
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F Distribution
• Example 9.14
• Testing the Ferritin Distribution BetweenOlder Men and Younger Men
• 28 (m) Older Men Were Sampled and WereFound to Have a Std Deviation of 52.6 mg/L(S1).
•26 (n) Younger Men Were Sampled and WereFound to Have a Std Deviation of 84.2 mg/L(S2).
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F Distribution
• Let 1 and 2 Be the Variance For the 2
Populations
• H0
:
1
= 2
• H1 : 1 < 2
• Let α = 0.01
•
Reject H0 if f ≤ F 0.99, 27, 25
• Only Have F 0.01, 25, 27 in the Table in the Book
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F Distribution
• F 0.99, 27, 25 = 1/ F 0.01, 25, 27
• F 0.01, 25, 27 = 2.54 so F 0.99, 27, 25 = 0.394
• = = (5.6)(84.)= 0.390
• Since 0.390 < 0.394, We Can Reject H0