Hypothesis Testing1

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Hypothesis testing



The Hypothesis Testing Approach

C onjectures ( )Hypotheses

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What is a Hypothesis?

A hypothesis is an assumption about thepopulation parameter.

o A parameter is a characteristic of the

population, like its mean or variance.o The parameter must be identified before

analysis.•



Motivation for Hypothesis Testing

• The intent of hypothesis testing is formally examine twoopposing conjectures (hypotheses), H0 and HA.

• These two hypotheses are mutually exclusive andexhaustive so that one is true to the exclusion of theother.

• We accumulate evidence - collect and analyze sampleinformation - for the purpose of determining which of the two hypotheses is true and which of the twohypotheses is false.

• Beyond the issue of truth, addressed statistically, is theissue of justice. Justice is beyond the scope of statistical investigation.

•



Hypothesis Testing Process

Population

Assume the

population

mean age is 50.

(Null Hypothesis)

REJECT

The Sample

Mean Is 20

SampleNull Hypothesis

, !o not likely



Errors in Making Decisions

• Type I Error

• Reject True Null Hypothesis (“False Positive”)

• Has Serious Consequences

• Probability of Type I Error Is α• Called Level of Significance

• Set by researcher

• Type II Error

• Do Not Reject False Null Hypothesis (“False

Negative”)• Probability of Type II Error Is β (Beta)

•



Example-The Trial System

Correct Decision Incorrect Decision

Innocent Individual Guilty IndividualGoes Free Goes Free

Incorrect Decision Correct Decision

Innocent Individual Guilty Individual

Is Disciplined Is Disciplined

Innocent

Guilty

V e

r d i

c

t

In Truth, the Defendant is:H0: Innocent H A : Guilty



Hypothesis Testing & Justice System

• State the Opposing Conjectures, H0 andHA.

• Determine the amount of evidencerequired, n, and the risk of committing a “type I error”,

• What sort of evaluation of the evidenceis required and what is the justification for this? (type of test)

• What are the conditions which proclaimguilt and those which proclaiminnocence? (Decision Rule)

• Gather & evaluate the evidence.• What is the verdict? (H0 or HA?)

• Determine “Zone of Belief”: ConfidenceInterval.

• What is appropriate justice? ---

Conclusions•



True, But Unknown State of the World

H0is True H A is True

Ho is True

Decision

H A is True

Correct Decision Incorrect Decision Type II Error Probability = β

Incorrect Decision Correct DecisionType I Error Probability = α



α & β Have an Inverse Relationship



Hypothesis Testing Algorithm

1.State H0

2.State H1

3.Choose α

4.Choose n

5.Choose Test:6. Set Up Critical Value(s)

7. Collect Data

8. Compute Test Statistic

9. Make Statistical Decision

10. Express Decision



Rejection Region

Z 0

α

eject H0

Z 0

ejectH 0

α

H 0: µ ≥

0 H 1 : H 1 :

µ < 0

H 0: µ

≤ H 1 :

µ > 0

ust Be S i g n i f i c a n t l y

Below µ =0



Z-test & C.I. for µ

Ø H0: µ = < > µ 0 vs. HA: µ ≠ > < µ 0

Ø n = _______ α = _______

Ø Testing a Hypothesis About a Mean;

Ø Process Performance Measure is Approximately

Normally Distributed;Ø We “Know” σ

Ø Therefore this is a “Z-test” - Use the Normal Distribution.

Ø Zcalc = (X - µ 0)/(σ / /n)

Ø The Confidence Interval for µ is Given by: X+ Zα /2(σ / n )

Ø Interpretation

Ø



t-test and Confidence Interval

s H0: µ = < > µ 0 vs. HA: µ > < µ 0

s n = _______ α = _______

• Testing a Hypothesis About a Mean;

• Process Performance Measure is Approximately Normally Distributedor We Have a “Large” Sample;

• We Do Not Know σ Which Must be Estimated by S.

Ø Therefore this is a “t-test” - Use Student’s T Distribution.

Ø tcalc = (X - µ 0)/(s/ /n )

Ø The Confidence Interval for µ is Given by:

Ø X + tα /2(s/ n )

Ø Interpretation•



Z-test & C.I. for p

s H0: p = < > p0 vs. HA: p > < p 0

s n = _______ α = _______

• Testing a Hypothesis About a Proportion;

• We have a “large” sample, that is, both np0 and n(1-p0)

> 5• Therefore this is a “Z-test” - Use the Normal Distribution.

s Zcalc = (p - p0)/( p0(1-p0)/n )

s _____ Reject H0 in Favor of HA. _______ FTR H0.

s The Confidence Interval for pis Given by:

p + Zα /2( p(1-p)/n )

s Interpretation

•



, .Advance Inc

Integrated CircuitManufacturing

&Methods Materials



Interested in increasing productivity rating in the

, .integrated circuit division Advance Inc determined that a methods review course would be of value to employees

.in the IC division

To determine the impact of this measure they reviewed historical productivity records for the division and determined that the average level was 100 with a

.standard deviation of 10

Fifty IC division employees participated in the course and

-the post course productivity of these employees was, , .measured on average to be 105

Assume that productivity ratings are approximately. .distributed Did the course have a beneficial effect

Test the appropriate hypothesis at the α = .05 level.of significance

- & : “Z Test Confidence Interval Training Effect”Example



Training Effect Example

•H0: µ < 100 HA: µ > 100

• =n 50 α = .05

•( ) ( ) ( )i testing a mean ii normal distribution iii σ = 10 is-known so that this is a Z test

•Z

calc = ( -X

µ 0)/(σ / ) = ( - )/ ( / ) = / . =n 105 100 10 50 5 1 414

.3 536

• % :The 95 Confidence Interval is Given by X + Zα /2 (σ / )n which is 105 + . ( . ) =1 96 1 414 105 + . .2 77 or 102 23 < µ < .107 77

• Thus the course appears to have helped improve IC division employee productivity from an average level of 100 to a level that is at

. . .least 102 23 and at most 107 77• - : “ ”A follow up question is this increase worth the investment?



Loan Application Processing



’First People s Bank

’First People s Bank would like to improve their loan.application process In particular currently the

amount of time required to process loan applications is approximately normally distributed with a mean of

.18 days Measures intended to simplify and speed the process.have been identified and implemented Were they

effective? Test the appropriate hypothesis at the α = .05 level of significance if a sample of 25

applications submitted after the measures were

.implemented gave an average processing time of 15 2. .days and a standard deviation of 2 0 days



First People’s Bank

sH0: µ > 18 HA: µ < 18

s =n 25 α = .05s( ) ( ) ( )i testing a mean ii normal distribution iii σ is

-unknown and must be estimated so that this is a t test

stcalc = ( -X µ 0)/( / √ ) = ( . - )/ ( / √ ) = - . /. =s n 15 2 18 2 25 2 8 4- .7 00

s % :The 95 Confidence Interval is Given by X + tα /2 ( /√ )s n which.is 15 2 + . (. ) = .2 064 4 15 2 +. .83 or 14 37 < µ

< .16 03

s Thus the course appears to have helped decrease the average time required to process a loan application from 18 days to a level

. . .that is at least 14 37 days and at most 16 03 days



χ 2-test & C.I. for σ H0: σ = < > σ 0 .vs HA: σ = > < σ 0

=n _______ α = _______

Testing a Hypothesis About a Standard Deviation (or Variance);The Measured Trait (e.g. the PPM) is Approximately Normal;

Therefore this is a “χ 2-test” - Use the Chi-Square Distribution. χ 2calc = ( - )n 1 s2/(σ 2

0 )

The Confidence Intervals for σ 2 and σ :are Given by

( - )n 1 s2/χ 2,large α /2 < σ 2 < ( - )n 1 s2/χ 2

,small α /2

and

√( - )n 1 s2/χ 2 ,large α /2 < σ < √( - )n 1 s2/χ 2 ,small α /2

Interpretation

v



, .Fast Facts Financial Inc

( ), .Fast Facts Financial FFF Inc provides credit reports to lending institutions that evaluate applicants for home, , , .mortgages vehicle home equity and other loans

.A pressure faced by FFF Inc is that several competing credit reporting companies provide reports in about the same

,average amount of time but are able to promise a lower time-than FFF Inc the reason being that the variation in time required to compile and summarize credit data is smaller

.than the time required by FFF

&FFF has identified implemented procedures which they

.believe will reduce this variation If the historic standard. ,deviation is 2 3 days and the standard deviation for a.sample of 25 credit reports under the new procedures is 1 8

, = .days then test the appropriate hypothesis at the 05.level of significance Assume that the time factor is

.approximately normally distributed



FFF Example•H0: σ = < > σ 0 .vs HA:σ > < σ 0 where σ 0 = .2 3

• =n 25 α = . .05•Testing a Hypothesis About a Standard Deviation (or Variance);•The Measured Trait (e.g. the PPM) is Approximately Normal;•Therefore this is a “χ 2-test” - Use the Chi-Square Distribution.

•χ 2calc = ( - )n 1 s2/σ 2

0 = ( )( .24 1 82 )/ ( .2 32) = . / . = .77 76 5 29 14 70

• .77 76/ .39 3641 < σ 2 < .77 76/ . .12 4011 or 1 975 < σ 2 < .6 27

• .so that 1 405 days < σ < .2 50 days

• . .Evidence is inconclusive Work should continue on this



Two Sample Testsand

Confidence Intervals



H0: μ1 – μ2 = ≥ ≤ μd

HA: μ1 – μ2 < > μd

n1= _____ n2= =_____ α 0

Comparison of Means from Two Processes Normality Can Be Reasonably Assumed

Are the two variances known or unknown?( )a -Known Z test( )b -Unknown but Similar in Value t test with n1+n2 – 2 df( )c - “ ”Unknown and Unequal t test with complicated df

-Critical Values and Decision Rules are the same as for any Z test- .or t test

Tests and Intervals for TwoMeans



C.I. for μ1

– μ2

X1 – X2 ZσX1-X2

or X1 – X2 tS X1-X2

Decisions – Same as any other Z or T test.Implications – Context Specific



(a)Z = [(X1 – X2) – μd]

σ√(1/n1 + 1/n2)

Z = [(X1 – X2) – μd] √(σ2

1/n1 + σ22/n2)

(b) t = [(X1 – X2) – μd] (assume equal variances)

Sp√(1/n

1+ 1/n

2) where df = n

1+n

2– 2

and Sp2 = (n1-1)S1

2 + (n2-1)S22

(c ) t = [(X1 – X2) – μd] (do not assume equal variances)

√(S12/n1 + S2

2/n2) where df = [(s12 /n1) + (s2

2/n2)]2

(s12 /n1)

2+ (s2

2/n2)2

n1 – 1 n2 – 1



: -Equality of Variances The F Test

H0: σ 1 = ≥ ≤ σ 2 vs. HA: σ 1 < > σ 2

n1= _____ n2= _____ α = _____

Test of equality of variances F-test



Fcalc = S12/S2

2

Make a decision.

Fcalc / Fn1-1,n 2-1, α /2 large ≤ σ 12/σ 2

2 ≤ Fcalc /Fn1-1,n 2-

1,α /2 small

C.I. for σ1/σ2 is obtained by taking squareroots of the endpoints of the above C.I. for

σ12/σ22

Conclusions / Implications – Context Specific.



&Tests Intervals for TwoProportions

H0: p1 – p2 = ≥ ≤ pd

HA: p1 – p2 < > pd

n1= _____ n2= _____ α = 0

Comparison of Proportions from Two Processesn1p1, n2p2, n1(1-p1) and n2(1-p2) all ≥ 5

Z-test

Critical Values and Decision Rules are the same as for any Z-test.



Z = [(p1 – p2)] IF pd = 0 √ p(1-p)(1/n1 + 1/n2) where p = (X1+X2)/(n1 + n2)

^ ^

Z = [(p1 – p2) – pd] IF pd 0

^ ^ ^ ^ √ (p1(1--p1)/n1 + p2(1-p2)/n2

^ ^ ^ ^ ^ ^C.I. for p1-p2is (p1 – p2) Z /2 √ (p1(1--p1)/n1 + p2(1-p2)/n2

^ ^



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