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

http://slidepdf.com/reader/full/ise-362-probability-and-statistics-p-value-and-f-distribution-presentation 1/23

ISE 326

Prob & Stats II



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



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



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



Proportions

• The Test Statistic is:

• Reject H0 : p = p0 if

or



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




• The Manufacturer Pulls 200 Random Samples

and Finds That 4 Are Defective.

• Can The Manufacturer Demonstrate Process

Capability to The Customer?




• 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




• The Test Statistic

• Where x = 4, n = 200 and p0 = 0.05

• Reject H0 : p0 = 0.05 if = -1.645

•




• 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



Proportions

• Another Form of The Test Statistic

• Let

• Then

This Presents the Test Statistics in Terms of SampleProportion Rather Than in Terms of X



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



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



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



Two Proportions

• So then the Estimator For the Common

Parameter, is

= + +

• The Test Statistic Becomes:

0 = −

(1−)( 1 +

1)



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



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

=



F Distribution

• An F Distribution Looks Approximately Like :



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



F Distribution

• H0 : 1 = 2

• H1 : 1 ≠ 2

•

The Test Statistic is

• Rejection Criteria



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).



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



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

ISE 362 Probability and Statistics P value and F Distribution Presentation

Documents