Statistical Tests

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# Statistical Tests - PowerPoint PPT Presentation

Statistical Tests . Statistical tests. Descriptive statistics : Description of observations Probability theory : Theoretical considerations ( no empirical data ) Statistical tests : Combination of descriptive statistics and probability theory

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### Statistical Tests

Statistical tests
• Descriptivestatistics:
• Description ofobservations
• Probabilitytheory:
• Theoreticalconsiderations(noempiricaldata)
• Statistical tests:
• Combinationofdescriptivestatisticsandprobabilitytheory
• Basedon observations (samples) conclusionsaredrawn
• Caution: Conclusionsbased on statisticalinferenceareneverdeterministic
• Conclusioncanalwaysbe „wrong“
• Minimal requirement: giveprobabilityforerror.
Statistical tests
• General procedureofstatisticaltests:
• Formulationofreasearchhypothesis (alternative hypothesis H1) andofformulationofthe null hypothesis H0 (hypothesisthatthereisnoeffect, typicallytheoppositeoftheresearchhypothesis).
• Choice of an appropriatestatisticaltestandteststatistic
• Choice ofrequiredlevelofsignficance. Oftenα =0.05 ischosen. This is an incrediblyweakcondition(personal opinion Tim Becker).
• Computationofteststatisticfromdataandresultingdecision (rejectionoracceptanceof null hypothesis).
Statistical tests
• Decidebetweentwo alternative hypothesesbased on information
• Example:
• Accusedatcourt
• Information: witnesses, evidence, expert opinion, ...
• Decision: guilty – not guilty
• The judgehastwohypothesisforhisverdict:
• H0: The accusedis not guilty (null hypothesis)
• H1: The accusedisguilty (alternative hypothesis)
• Whatcan happen?

1) Decision correct: Guilty gets acquittal

2) Decision correct: A guilty is convicted

3) Wrong decision: Innocent is convicted

Example:

Accusedatcourt

4) Wrong decision: Guilty gets acquittal

Statistical tests

H0: Accused is not guilty (Null hypothesis)

H1: Accusedisguilty (Alternative hypothesis)

Verdict

Unknown truth

H0 not guilty

H1guilty

H0 not guilty

H1 guilty

Verdict

Unknown truth

H0

not guilty

H1guilty

H0 not guilty

Type II error

H1 guilty

Type I error

Situation 3: Type I error

Situation 4: Type II error

Statistical tests

Errors of different kind and impact

Verdict

Unknown truth

H0

not guilty

H1guilty

H0 not guilty

Type II error

H1 guilty

Type I error

Statistical tests

Terminology:

Probability of type I error: 

Probability of type II error: 

Question: Which probability shall be made small?

Strategy:

• Do not sentence innocents („minimize “)
• Always decide not guilty
•  = 0
• Many guilties get acquittal
•  
• Sentence all guilties („minimize “)
• Always decide guilty
•  = 0
• Many innocents are sentenced
•  

Decision

Unknown truth

H0

H1

H0

Type II error ()

H1

Type I error ()

Statistical tests
• Whicherror, or , ist worse?
• can‘tbejudged in general,but depends on consideredhypotheses.

Example 1: Case at court

 has to be minimized.

„In dubio pro reo“ (innocent until proven guilty)

Decision

Unknown truth

old drug

H0

New drug H1

H0 -old drug

Type II error ()

H1 – new drug

Type I error ()

Statistical tests

Example 2: Is a new drug better than an established drug?

H0: new drug is not betterH1: new, more expensive drug is better

Type I error ():New drug will be given, although not better

• Consequences:
• Patient getunnecessary treatment
• Treatment becomesmore expensive

Type II error ():It is overlooked that the new drug is better

• Consequences:
• Patient does not getoptimaltreatment
• Loss ofresearchcost

Decision

Unknown truth

H0

Type II error ()

Type I error ()

Statistical tests

Type I error ():Adverse reaction is assumed, although not present

Type II error ():Severe adverse reaction is not recognized

Consequence:

Type II error () has to be minimized

### Statistical tests

Formulationofhypotheses

Blood pressure is lowered

Great drug ???

Statistical tests

Example: Drug to lower blood pressure

H0: Drug does not workH1: Drug works

1) How to measure the effect?

 Definetargetvalue X

X: systolic blood pressure in mmHg

2) Investigation

Patient: Measurement before drug intake: 190 mmHg Measurement after drug intake: 185 mmHg

 H1 correct???

Difference may be a chance effect !

Patient: Measurement before drug intake :190 mmHg Measurement after drug intake :140 mmHg

Difference may be an individual effect

 Sample / independentrepetition

Patient

xi before

xi after

difference di

1

190

185

5

2

185

175

10

3

180

185

-5

4

180

180

0

5

190

170

20

n

190

175

15

Statistical tests

 Decisiondepends on meanof „reduction in bloodpressure“

• Conclusion on expectationμd oftargetvalue X shallbemade
Statistical tests

Hypotheseswithrespecttotargetvalue X must beformulated in termsoftheexpectation

Twotypesofhypotheseshavetobedistinguished:

One-sidedhypotheses:

X: targetvalue: reductionofbloodpressure in mmHg

H0: Drug does not work: μd = 0 (μd ≤ 0)

H1: Drug works: μd > 0

Tow-sidedhypotheses:

X: targetvalue: reductionofbloodpressure in mmHg

H0: Drug does not work: μd = 0

H1: Drug works: μd 0

### Statistical tests

One-sidedtest

Sample

Patient

before

after

difference

1

190

185

5

2

180

185

-5

50

190

175

15

n=50

d

20

0

All we have is one sample with

Statistical tests

X: „reduction of blood pressure“ [mmHg]

H0: μd = 0

H1: μd > 0

 Make decision

H0

H1

20

0

Statistical tests

Decision:

 H0 is true

 H1 is true

Decision „changes“ at a limitthatneedspre-specification.

Sample

Patient

before

after

difference

1

190

185

5

2

180

185

-5

5%

n

190

175

15

0

Itisthe 95% - quantileof

Statistical tests

H0: μd = 0H1: μd> 0

One-sided question

α = 5 %

Limit has to be pre-specified!

Limit isimplicitlygivenby.

Whatisthelimitforα = 5 %?

Two-sidedtest

### Statistical tests

Test statistic – t test

Sample

Patient

before

after

difference

1

190

185

5

2

180

185

-5

n

190

175

15

t

0

• Mean of the differences

Test statistic t

Statistical tests

HypothesesH0: μd = 0H1: μd0

= 5 %

What shall decision depend on?

• Standard deviation s
• Sample size n

sample

Patient

before

after

difference

1

190

185

5

2

180

185

-5

n

190

175

15

0

t

Statistical tests
• T-test forpairedsamples

target: reduction in bloodpressure d

Hypotheses H0: μd= 0 vs. H1: μd 0(two-sidedtest) = 5 %

Test statistict (In R: t.test() ):

In case(!) diisnormallydistributed:

• tist - distributedwith (n-1) degreesoffreedom

„Student‘s t-Test“

Statistical tests

Quantiles tf;0.95and tf;0.975 ofthetf– distribution

sample

target: reduction in blood pressure d

H0: μd= 0 vs. H1: μd 0(two-sided)

= 5 %

Patient

before

after

difference

1

190

185

5

2

180

185

-5

15

190

175

15

2,5%

2,5%

0

-2,145

2,145

6,21

t

n=15  f=n-1=14

p=

f=

0,950

0,975

13

14

15

20

1.771

1.761

1.753

1.725

2.160

2.145

2.131

2.086

t = 6,21 > 2,145 = t14; 0.975

Statistical tests

Decision

Compare t = 6,21 with Quantile

Decision:

H0 is discardedH1 is accepted=5 %

Statistical tests
• Rejection criteria:
• One-sidedhypothesis:
• : vs. : : rejectioninterval :
• : vs. : : rejectioninterval:
• canberejected a
• Tow-sidedhypothesis:
• :canberejectedatif or , i.e. if
• : can berejectedatif is not located in the()100%-confidenceintervall:

Therapy A

Hypotheses:

H0:  = 0H1:  0

Therapy B

Patient

Patient

blood pr

blood pr

1

1

150

140

2

2

135

125

3

3

140

130

nA

nB

132

145

μA

μB

Statistical tests
• T-test forun-pairedsamples

Example: Blood pressure after 3 months, fortwo different therapies

Question:Isthere a differencebetweentherapy A andtherapy B?

Hypotheses:

H0: μA = μBH1: μAμB

Therapy A

Therapy B

Patient

Patient

Blood pr

Blood pr

1

1

150

140

2

2

125

135

Test statistic:

3

3

140

130

nB

nA

145

132

test statistic t is t – distributed with (nA + nB -2) degrees of freedom

Statistical tests

Hypotheses:

H0: μA = μBH1: μAμB

(two-sided)

= 5 %

Parameters for a test statistic t?

Assumption: XA N(μA; σA2) XB N(μB; σB2) mit σA2= σB2

Therapy A

-2,021

2,021

Therapy B

Patient

Patient

Blutdruck

Blutdruck

1

1

150

140

2

2

135

125

3

3

140

130

2,5%

2,5%

nA

nB

132

145

2,21

0

t

Statistical tests

Hypotheses:

H0: μA = μBH1: μAμB

(two-sided)

= 5 %

XA N(μA; σA2)XB N(μB; σB2) σA2= σB2

Decision:

Reject H0Accept H1=5 %

Compare t = 2,21 with the quantile of the t-distribution (nA=20, nB=22  f=nA+nB-2=40)

t = 2,21 > 2,021 = t40; 0.975

Statistical tests
• Rejectioncirteria:
• One-sidedhypothesis:
• : vs. : : canberejectedifatif
• :vs. : : can berejectedifatif
• Tow-sided hypothesis:
• : bzw. :can berejectedifatif 0is not locatcted in the()100%-confidenceinterval:

Test statistic:

Statistical tests
• T-test forun-pairedsampleswith different variancesσA2σB2

Hypothesesunchanged:H0: μA= μBvs. H1: μAμB

(two-sided)= 5 %

Parameters for Welch t test?

• If, then

teststatistic tist – distributedwith

degreesoffreedom

Statistical tests
• Rejectioncriterion:
• can be rejected at if

Assumption: XA N(μA; σA2) XB N(μB; σB2) mit σA2σB2

### Statistical tests

-test

Expected under H0

Observed

Post-op intricacies

OP-Type

Post-op intricacies

OP-Type

A

B

A

B

yes

55

yes

12

43

55

no

29

no

5

24

29

17

67

84

17

67

84

Statistical tests
• Testingcategorial variables

Here, „expected“ means that we assume that the op-type has no impact on whether post-op intircacies will occur or not

Expected

Observed

Post-op intricacies

OP-Type

Post-op intircacies

OP-Type

A

B

A

B

yes

55

yes

12

43

55

no

29

no

5

24

29

17

67

84

17

67

84

Test statistic (In R: chisq.test() ):

Statistical tests

Test statistic is 2-distributed with f = 1 degree of freedom

Expected

Observed

Post-op problems

OP-Type

Post-op problems

OP-Type

A

B

A

B

yes

55

yes

12

43

55

no

29

no

5

24

29

17

67

84

17

67

84

Test statistic:

Statistical tests

Hypotheses: H0: OP-Type and postoperative intricaciesareindependent H1: ..are not independent

= 5 % Two-sidedtest.

Compare 2 to quantile of 2 -distribution (degrees of freedom f=1)

2 = 0,26 < 3,814 = 2 1;0.95

Decision: H0 can not be rejected

Test statisitc:

Statistical tests
• Simplificationfor 2-by-2 tables

Quantile2 1;0.95 = 3,814

Degreesoffreedom: f = 1

Decision:

2  3,814: H0can not berejected.

2 > 3,814: H0canberejected.

Statistical tests
• General (k x m)-contingencytables:
• : Variable 1 and variable 2 areindependent vs. : variable 1 andvariable 2 are not independent
• observedfrequencies
• expectedfrequencies
Statistical tests
• Test statistic:
• is under asymptotically ²-distributed with (k-1)(m-1) degressoffreedom.
• Rejectioncriterion:
• canberejectedatif², where k isthenumberofrowsand m isthenumberofcolumns.
• The approximationoftheteststatisticwiththe² -distribution isacceptableiffor 80% ofthecellstheexpected(!) numberofcellcountsis >=5. OtherwiseuseFisher‘sexacttest(cf. Follwingpresentations).

### Statistical tests

Relative Riskand Odds Ratio

Statistical tests
• Relative Risk:
• A measureforriskdifferencebetweentwogroups.
• Itisthethefactorbywhichtheriskisincreased
• X=1: individual isexposedtoriskfactor
• X=0: individual is not exposed
• K=1: individual hasdisease
• K=0: individual does not havethedisease
• Conditionalprobabilitytobeill, underexposure : P(K=1|X=1)
• Conditionalprobabilitytobeillwithoutexposure : P(K=1|X=0)
Statistical tests

2x2 tableofprobabilities

P(K=0|X=0)=

P(K=0|X=1)=

P(K=1|X=0)=

P(K=1|X=1)=

• Relative risk
Statistical tests
• Odds ratio
• odds(A)ratioofprobabilityandprobabilityofoppositeevent
• Whentheprobabilites, theprobabilitesareestiamtedandthe OR iscomputedasfollows:

2 x 2 tableofobservedfrequencies

Statistical tests

Example: Retrospectivestudytodetectif Tonsillektomie increasesriskfor Morbus Hodgkin

=> Propabilitytogetthedisease in increasedunderexposure (Tonsillektomie)

Statistical tests
• Correspondingstatisticaltest
• canberejected, if 1 is not located in the()100%-confidenceintervallof OR:
Statistical tests
• Beispiel: Retrospektive Studie, ob Tonsillektomie das Risiko für Morbus Hodgkin erhöht
• ,
• ()100%-Konfidenzintervall für OR:
• kann zum Niveau verworfen werden, da 1 nicht im ()100%-Konfidenzintervall liegt.