Econ 240A

Econ 240A

Power 6

The Challenger Disaster

lhttp://onlineethics.org/moral/boi sjoly/RB-intro.html

The Challenger

lThe issue is whether o-ring failure onprior 24 prior launches istemperature dependent

lThey were considering launchingChallenger at about 32 degrees

lWhat were the temperatures of priorlaunches?

Challenger

Launch

Only 4 launches

Between 50 and

64 degrees

Challenger

lDivide the data into two groups

•12 low temperature launches, 53-70 degrees

•12 high temperature launches, 70-81 degrees

Temperature

O-Ring Failure

Yes

Temperature

O-Ring Failure

Yes

Probability of O-RingFailure Conditional OnTemperature, P/T

lP/T=#of Yeses/# of Launches atlow temperature

•P/T=#of O-Ring Failures/# ofLaunches at low temperature

•Pˆ = k(low)/n(low) = 5/12 = 0.41

lP/T=#of Yeses/# of Launches athigh temperature

•Pˆ = k(high)/n(high) = 2/12 = 0.17

Are these two ratessignificantly different?

lDispersion: p*(1-p)/n

•Low: [p*(1-p)/n]1/2 = [0.41*0.59/12]1/2=0.14

•High: [p*(1-p)/n]1/2 = [0.17*0.83/12]1/2=0.11

lSo .41 - .17 = .24 is 1.7 to 2.2standard deviations apart? Is thatenough to be statisticallysignificant?

Interval Estimation andHypothesis Testing

Outline

lInterval Estimation

lHypothesis Testing

lDecision Theory

-1.645

0.050

a Z value

of 1.96 leads

to an area of

0.475, leaving

0.025 in the

Upper tail

Interval Estimation

lThe conventional approach is tochoose a probability for the intervalsuch as 95% or 99%

So z values

of -1.96 and

1.96 leave

2.5% in each

tail

-1.96

2.5%

1.96

http://www.sfgate.com/election/races/2003/10/07/map.shtml

TwoCalifornias

Interval Estimation

lSample mean example: MonthlyRate of Return, UC Stock IndexFund, Sept. 1995 - Aug. 2004

•number of observations: 108

•sample mean: 0.842

•sample standard deviation: 4.29

•Student’s t-statistic

•degrees of freedom: 107

Sample

Mean

0.842

Appendix B

Table 4

p. B-9

2.5 % in the

upper tail

Interval Estimation

l95% confidence interval

lsubstituting for t

Interval Estimation

lMultiplying all 3 parts of theinequality by 0.413

lsubtracting .842 from all 3 partsof the inequality,

Interval EstimationAn Inference about E(r)

lAnd multiplying all 3 parts of theinequality by -1, which changes the signof the inequality

lSo, the population annual rate of returnon the UC Stock index lies between19.9% and 0.2% with probability 0.95,assuming this rate is not time varying

Hypothesis Testing

Hypothesis Testing: 4 Steps

lFormulate all the hypotheses

lIdentify a test statistic

lIf the null hypothesis were true, whatis the probability of getting a teststatistic this large?

lCompare this probability to a chosencritical level of significance, e.g. 5%

-1.645

5 % lower tail

Step #4: compare the

probability for the test

statistic(z= -1.33) to the

chosen critical level

(z=-1.645)

Decision Theory

Decision Theory

lInference about unknown populationparameters from calculated samplestatistics are informed guesses. Soit is possible to make mistakes. Theobjective is to follow a process thatminimizes the expected cost ofthose mistakes.

lTypes of errors involved in acceptingor rejecting the null hypothesisdepends on the true state of naturewhich we do not know at the timewe are making guesses about it.

Decision Theory

lFor example, consider a possibleproposition for bonds to financedams, and the null hypothesis thatthe proportion that would vote yeswould be 0.4999 (or less), i.e. p ~0.5. The alternative hypothesiswas that this proposition wouldwin i.e., p >= 0.5.

Decision Theory

lIf we accept the null hypothesiswhen it is true, there is no error.If we reject the null hypothesiswhen it is false there is noerror.

Decision theory

lIf we reject the null hypothesiswhen it is true, we commit atype I error. If we accept thenull when it is false, we commita type II error.

Decision

Accept null

Reject null

True State of Nature

p = 0.4999

H2o Bonds lose

P >= 0.5

Bonds win

No Error

Type I error

No Error

Type II error

Decision Theory

The size of the type I error isthe significance level orprobability of making a type Ierror, 

The size of the type II error isthe probability of making a typeII error, 

Decision Theory

We could choose to make thesize of the type I error smallerby reducing for example from5 % to 1 %. But, then whatwould that do to the type IIerror?

Decision

Accept null

Reject null

True State of Nature

p = 0.4999

P >= 0.5

No Error

1 - 

Type I error



No Error

1 - 

Type II error



Decision Theory

lThere is a tradeoff between thesize of the type I error and thetype II error.

Decision Theory

lThis tradeoff depends on thetrue state of nature, the value ofthe population parameter weare guessing about. Todemonstrate this tradeoff, weneed to play what if gamesabout this unknown populationparameter.

What is at stake?

lSuppose you are for WaterBonds.

lWhat does the water bondscamp want to believe about thetrue population proportion p?

•they want to reject the nullhypothesis, p=0.4999

•they want to accept thealternative hypothesis, p>=0.5

Cost of Type I and Type II Errors

lThe best thing for the waterbonds camp is to lean the otherway from what they want

lThe cost to them of a type Ierror, rejecting the null when itis true (i.e believing the bondswill pass) is high: over-confidence at the wrong time.

lExpected Cost E(C) = CIhigh (typeI error)*P(type I error) + CIIlow(type II error)*P(type II error)

Costs in Water Bond Camp

lExpected Cost E(C) = CIhigh (type Ierror)*P(type I error) + CIIlow (type IIerror)*P(type II error)

E(C) = CI high(type I error)* CIIlow(type II error)* 

lRecommended Action: makeprobability of type I error small, i.e.run scared so chances of losing staysmall

Decision

Accept null

Reject null

True State of Nature

p = 0.499

Bonds lose

P >= 0.5

Bonds win

No Error

1 - 

Type I error 

C(I)

No Error

1 - 

Type II error 

C(II)

E[C] = C(I)* + C(II)* 

Bonds: C(I) is large so make small

How About Costs to theBond Opponents Camp ?

lWhat do they want?

lThey want to accept the null,p=0.499 i.e. Bonds lose

lThe opponents camp shouldlean against what they want

lThe cost of accepting the nullwhen it is false is high to them,so C(II) is high

Costs in the opponentsCamp

lExpected Cost E(C) = Clow(type Ierror)*P(type I error) + Chigh(type IIerror)*P(type II error)

E(C) = Clow(type I error)* Chigh(type II error)* 

lRecommended Action: makeprobability of type II error small, i.e.make the probability of acceptingthe null when it is false small

Decision

Accept null

Reject null

True State of Nature

p = 0.499

Bonds lose

P >=0.5

Bonds win

No Error

1 - 

Type I error 

C(I)

No Error

1 - 

Type II error 

C(II)

E[C] = C(I)* + C(II)* 

Opponents: C(II) is large so make small

Decision Theory Example

If we set the type I error, to 1%, thenfrom the normal distribution (Table 3),the standardized normal variate z willequal 2.33 for 1% in the upper tail. For asample size of 1000, where p~0.5 fromnull

Decision theory example

So for this poll size of 1000, withp=0.5 under the null hypothesis,given our choice of the type I errorof size 1%, which determines thevalue of z of 2.33, we can solve fora

2.33

1 %

Decision Theory Example

lSo if 53.7% of the polling sample,or 0.5368*1000=537 say they willvote for water bonds, then wereject the null of p=0.499, i.e thenull that the bond proposition willlose

Decision Theory Example

lBut suppose the true value of pis 0.54, and we use thisdecision rule to reject the null if537 voters are for the bonds,but accept the null (of p=0.499,false if p=0.54) if this number isless than 537. What is the sizeof the type II error?