1

Econ 240A

Power 6

2

The Challenger Disaster

http://www.msnbc.msn.com/id/11031097/

3

The Challenger

The issue is whether o-ring failure on prior 24 prior launches is temperature dependent

They were considering launching Challenger at about 32 degrees

What were the temperatures of prior launches?

4

Histogram

0

1

2

3

4

5

6

7

34 39 44 49 54 59 64 69 74 79 84 89

Temperature Bin In 50 Ranges

Fre

qu

en

cy

ChallengerLaunch

Only 4 launches Between 50 and64 degrees

5

Challenger

Divide the data into two groups• 12 low temperature launches, 53-70

degrees• 12 high temperature launches, 70-81

degrees

Temperature O-Ring Failure

53 Yes

57 Yes

58 Yes

63 Yes

66 No

67 NO

67 No

67 No

68 No

69 No

70 No

70 Yes

Temperature O-Ring Failure

70 Yes

70 No

72 No

73 No

75 Yes

75 No

76 No

76 No

78 No

79 No

80 No

81 No

8

Probability of O-Ring Failure Conditional On

Temperature, P/T P/T=#of Yesses/# of Launches at

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

Launches at low temperature• Pˆ = k(low)/n(low) = 5/12 = 0.41

P/T=#of Yesses/# of Launches at high temperature• Pˆ = k(high)/n(high) = 2/12 = 0.17

9

Are these two rates significantly different?

Dispersion: 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 So .41 - .17 = .24 is 1.7 to 2.2

standard deviations apart? Is that enough to be statistically significant?

10

Interval Estimation and Hypothesis Testing

11

Outline

Interval Estimation Hypothesis Testing Decision Theory

Density Function for the Standardized Normal Variate

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

-5 -4 -3 -2 -1 0 1 2 3 4 5

Standard Deviations

Den

sity

2]1/)0[(2/1*]2/1[)( zezf

0

Cumulative Distribution Function for a Standardized Normal Variate

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

-5 -4 -3 -2 -1 0 1 2 3 4 5

Standard Deviations

Pro

ba

bilt

y

0

Density Function for the Standardized Normal Variate

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

-5 -4 -3 -2 -1 0 1 2 3 4 5

Standard Deviations

Den

sity

2]1/)0[(2/1*]2/1[)( zezf

-1.645

0.050

15

a Z valueof 1.96 leadsto an area of0.475, leaving0.025 in the Upper tail

16

Interval Estimation

The conventional approach is to choose a probability for the interval such as 95% or 99%

17

So z values of -1.96 and1.96 leave2.5% in eachtail

Density Function for the Standardized Normal Variate

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

-5 -4 -3 -2 -1 0 1 2 3 4 5

Standard Deviations

Den

sity

2]1/)0[(2/1*]2/1[)( zezf

-1.96

2.5% 2.5%

1.96

19

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

Two Californias

20

Times to Produce a cell Phone: Sample of 50

Problem 10.35 Data set Xr10-35

Times

25.9

29.4

26.3

28.2

26

27.8

25.2

27.7

30.2

26.5

27.2

27.2

26.9

28.8

26.3

27.3

27.2

29.1

27.2

26.3

28.4

26.8

24.7

25.6

26.3

25.7

28.3

28.2

28.1

26.6

27.7

24.8

25.1

26.5

27.4

24.4

26.3

25.3

26.5

25.8

26.8

25.6

26.2

29.2

27.1

26.4

25.9

25.9

26.9

25.3

22

Sample statistics

Times

Mean 26.81

Standard Error 0.183085795

Median 26.55

Mode 26.3

Standard Deviation 1.294612069

Sample Variance 1.676020408

Kurtosis -0.053059002

Skewness 0.493039805

Range 5.8

Minimum 24.4

Maximum 30.2

Sum 1340.5

Count 50

23

Mean Time to Assemble

Prob[?<=(x^-)/(s/n)<=?] = 0.95

24

Student’s t-Distribution

What does it look like? EViews What are the critical values for

2.5% in each tail? Text

25

@ctdist(x,v)@dtdist(x,v)@qtdist(p,v)@rtdist(v) t-distribution for , and v>0. Note that v=1 is the Cauchy

@ctdist(x,v)@dtdist(x,v)@qtdist(p,v)@rtdist(v) t-distribution

Gen tran=@rtdist(48)Gen tdens=@dtdist(tran,48)

0.0

0.1

0.2

0.3

0.4

-3 -2 -1 0 1 2 3

TRAN

TD

EN

S

Student's t-distribution, simulated, 48 dof

27

Appendix BTable 4p. B-9

28

Mean Time to Assemble

Prob[?<=(x^-)/(s/n)<=?] = 0.95 Prob[-2.01<=(26.81-)/(1.29/50)<=2.01] =

0.95 Prob[-2.01<=(26.81-)/0.182)<=2.01] = 0.95 Prob[-0.366<=(26.81-)<=0.366] =0.95 Prob[0.37>=(-26.81>=- 0.37] =0.95 Prob[26.81+0.37>=>=26.81-0.37] = 0.95 Prob[27.18>= >=26.44] =0.95

29

Hypothesis Testing

30

Hypothesis Testing: 4 Steps

Formulate all the hypotheses Identify a test statistic If the null hypothesis were true, what is

the probability of getting a test statistic this large?

Compare this probability to a chosen critical level of significance, e.g. 5%

31

Density Function for the Standardized Normal Variate

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

-5 -4 -3 -2 -1 0 1 2 3 4 5

Standard Deviations

Den

sity

2]1/)0[(2/1*]2/1[)( zezf

-1.645

5 % lower tail

Step #4: compare the probability for the teststatistic(z= -1.33) to the chosen critical level(z=-1.645)

33

Decision Theory

Decision Theory Inference about unknown population

parameters from calculated sample statistics are informed guesses. So it is possible to make mistakes. The objective is to follow a process that minimizes the expected cost of those mistakes.

Types of errors involved in accepting or rejecting the null hypothesis depends on the true state of nature which we do not know at the time we are making guesses about it.

Decision Theory For example, consider a possible

proposition for bonds to finance dams, and the null hypothesis that the proportion that would vote yes would be 0.4999 (or less), i.e. p ~ 0.5. The alternative hypothesis was that this proposition would win i.e., p >= 0.5.

Decision Theory If we accept the null hypothesis

when it is true, there is no error. If we reject the null hypothesis when it is false there is no error.

37

Decision theory

If we reject the null hypothesis when it is true, we commit a type I error. If we accept the null when it is false, we commit a type II error.

Decision

Accept null

Reject null

True State of Nature

p = 0.4999H2o 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 is the

significance level or probability of making a type I error,

The size of the type II error is the probability of making a type II error,

40

Decision Theory

We could choose to make the size of the type I error smaller by reducing for example from 5 % to 1 %. But, then what would that do to the type II error?

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 There is a tradeoff between the

size of the type I error and the type II error.

43

Decision Theory

This tradeoff depends on the true state of nature, the value of the population parameter we are guessing about. To demonstrate this tradeoff, we need to play what if games about this unknown population parameter.

44

What is at stake?

Suppose you are for Water Bonds. What does the water bonds camp

want to believe about the true population proportion p?• they want to reject the null

hypothesis, p=0.4999• they want to accept the alternative

hypothesis, p>=0.5

Cost of Type I and Type II Errors

The best thing for the water bonds camp is to lean the other way from what they want

The cost to them of a type I error, rejecting the null when it is true (i.e believing the bonds will pass) is high: over-confidence at the wrong time.

Expected Cost E(C) = CIhigh (type I

error)*P(type I error) + CIIlow (type II

error)*P(type II error)

46

Costs in Water Bond Camp

Expected Cost E(C) = CIhigh (type I

error)*P(type I error) + CIIlow (type II

error)*P(type II error) E(C) = CI high(type I error)* CII

low(type II error)* Recommended Action: make

probability of type I error small, i.e. run scared so chances of losing stay small

Decision

Accept null

Reject null

True State of Nature

p = 0.499Bonds lose

P >= 0.5Bonds 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

48

How About Costs to the Bond Opponents Camp ?

What do they want? They want to accept the null,

p=0.499 i.e. Bonds lose The opponents camp should lean

against what they want The cost of accepting the null

when it is false is high to them, so C(II) is high

49

Costs in the opponents Camp

Expected Cost E(C) = Clow(type I error)*P(type I error) + Chigh(type II error)*P(type II error)

E(C) = Clow(type I error)* Chigh(type II error)*

Recommended Action: make probability of type II error small, i.e. make the probability of accepting the null when it is false small

Decision

Accept null

Reject null

True State of Nature

p = 0.499Bonds lose

P >=0.5Bonds 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%, then

from the normal distribution (Table 3), the standardized normal variate z will equal 2.33 for 1% in the upper tail. For a sample size of 1000, where p~0.5 from null

0158.0/)5.0ˆ(1000/5.0*5.0/)5.0ˆ(33.2

/)1(*/)ˆ()ˆ(/)]ˆ(ˆ[

ppz

nppppppEpz

52

Decision theory example

So for this poll size of 1000, with p=0.5 under the null hypothesis, given our choice of the type I error of size 1%, which determines the value of z of 2.33, we can solve for a5368.0ˆ p

Density Function for the Standardized Normal Variate

0

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0.25

0.3

0.35

0.4

0.45

-5 -4 -3 -2 -1 0 1 2 3 4 5

Standard Deviations

Den

sity

2]1/)0[(2/1*]2/1[)( zezf

2.33

1 %

Decision Theory Example So if 53.7% of the polling sample, or

0.5368*1000=537 say they will vote for water bonds, then we reject the null of p=0.499, i.e the null that the bond proposition will lose

55

Decision Theory Example

But suppose the true value of p is 0.54, and we use this decision rule to reject the null if 537 voters are for the bonds, but accept the null (of p=0.499, false if p=0.54) if this number is less than 537. What is the size of the type II error?

0.0

0.1

0.2

0.3

0.4

0.5

440 460 480 500 520 540 560

Voters Supporting Bonds

DENSITY, p=0.50 DENSITY, p=0.54

p = 0.50 p = 0.54

alpha = 1 %

Figure: The Pobability of a Type II Error = 40%

Decision Rule: Reject Null if Voters>536

beta = 40%

Decision Theory Example What is the value of the type II error, if

the true population proportion is p = 0.54?

Recall our decision rule is based on a poll proportion of 0.536 or 536 for Bonds

z(beta) = (0.536 – p)/[p*(1-p)/n]1/2

Z(beta) = (0.536 – 0.54)/[.54*.46/1000]1/2

Z(beta) = -0.253

nppppbetaz /)1(*/)ˆ()(

Table 2: Probability of Type II Error Versus Population Proportion

True Population z value* F(z) = beta 1 - beta Proportion, p

0.51 1.64 0.950 0.050 0.52 1.01 0.844 0.156 0.53 0.38 0.648 0.352 0.54 -0.25 0.400 0.600 0.55 -0.89 0.187 0.813 0.56 -1.53 0.063 0.937 0.57 -2.17 0.015 0.985 0.58 -2.82 0.002 0.998 0.59 -3.47 0.000 1.000 0.6 -4.13 0.000 1.000

* z = (0.536 - p)/sqrt[p*(1-p)/1000]

Calculation of Beta

Figure 3: Operating Characteristic Curve

0.000

0.100

0.200

0.300

0.400

0.500

0.600

0.700

0.800

0.900

1.000

0.5 0.51 0.52 0.53 0.54 0.55 0.56 0.57 0.58 0.59 0.6 0.61

Presumed Population Proportion, p

Bet

aBeta Versus p (true)

p

Figure 4: Power Function of the Test

0.000

0.100

0.200

0.300

0.400

0.500

0.600

0.700

0.800

0.900

1.000

0.5 0.51 0.52 0.53 0.54 0.55 0.56 0.57 0.58 0.59 0.6 0.61

Supposed Population Proportion, p

1 -

be

ta

Power of the Test

p

1 -

Figure 4: Power Function of the Test

0.000

0.100

0.200

0.300

0.400

0.500

0.600

0.700

0.800

0.900

1.000

0.5 0.51 0.52 0.53 0.54 0.55 0.56 0.57 0.58 0.59 0.6 0.61

Supposed Population Proportion, p

1 -

be

ta

Power of the Test

p

1 - Ideal

Decision

Accept null

Reject null

True State of Nature

p = 0.499 P >= 0.5

No Error 1 -

Type I error

No Error 1 -

Type II error

63

Tradeoff

Between and Suppose the type I error is 5%

instead of 1%; what happens to the type II error?

64

Tradeoff

If then the Z value in our example is 1.645 instead of 2.33 and the decision rule is reject the null if 526 voters are for water bonds.

0.0

0.1

0.2

0.3

0.4

0.5

440 460 480 500 520 540 560

Voters Supporting Bonds

DENSITY, p=0.50 DENSITY, p=0.54

p = 0.50 p = 0.54

alpha = 1 %

Figure: The Pobability of a Type II Error = 40%

Decision Rule: Reject Null if Voters>502

beta = 40%

66

Interval Estimation Example

67

Interval Estimation

Sample mean example: Monthly Rate of Return, UC Stock Index Fund, 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

)//()( nsxt

Monthly Rates of Return On the UC Stock Index Fund

-15

-10

-5

0

5

10

Sep-9

5

Jan-

96

May

-96

Sep-9

6

Jan-

97

May

-97

Sep-9

7

Jan-

98

May

-98

Sep-9

8

Jan-

99

May

-99

Sep-9

9

Jan-

00

May

-00

Sep-0

0

Jan-

01

May

-01

Sep-0

1

Jan-

02

May

-02

Sep-0

2

Jan-

03

May

-03

Sep-0

3

Jan-

04

May

-04

Date

Rat

e

SampleMean0.842

69

Appendix BTable 4p. B-9

2.5 % in the upper tail

70

Interval Estimation

95% confidence interval

substituting for t

)108/29.4/()842(.)//()(

95.0)983.1983.1(

nsxt

tp

95.)983.1413./)842(.983.1( p

71

Interval Estimation

Multiplying all 3 parts of the inequality by 0.413

subtracting .842 from all 3 parts of the inequality,

95.)819.842.819.( p

95.)023.661.1( p

72

Interval EstimationAn Inference about E(r)

And multiplying all 3 parts of the inequality by -1, which changes the sign of the inequality

So, the population annual rate of return on the UC Stock index lies between 19.9% and 0.2% with probability 0.95, assuming this rate is not time varying

95.0)02.66.1( p