1 Econ 240A Power 6. 2 The Challenger Disaster l 1031097/ 1031097.
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Transcript of 1 Econ 240A Power 6. 2 The Challenger Disaster l 1031097/ 1031097.
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
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
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