Econ 240A

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Econ 240A

• Power 6

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The Challenger Disaster

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

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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?

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Only 4 launches Between 50 and 64 degrees
Challenger Launch
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Challenger

• Divide the data into two groups
• 12 low temperature launches, 53-70 degrees
• 12 high temperature launches, 70-81 degrees

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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
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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
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Probability of O-Ring Failure Conditional On
Temperature, P/T

• P/Tof Yesses/ of Launches at low temperature
• P/Tof O-Ring Failures/ of Launches at low
  temperature
• Pˆ k(low)/n(low) 5/12 0.41
• P/Tof Yesses/ of Launches at high temperature
• Pˆ k(high)/n(high) 2/12 0.17

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Are these two rates significantly different?

• Dispersion p(1-p)/n
• Low p(1-p)/n1/2 0.410.59/121/2 0.14
• High p(1-p)/n1/2 0.170.83/121/2 0.11
• So .41 - .17 .24 is 1.7 to 2.2 standard
  deviations apart? Is that enough to be
  statistically significant?

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Interval Estimation and Hypothesis Testing
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Outline

• Interval Estimation
• Hypothesis Testing
• Decision Theory

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0
13
0
14
0.050
-1.645
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a Z value of 1.96 leads to an area of 0.475,
leaving 0.025 in the Upper tail
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Interval Estimation

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

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So z values of -1.96 and 1.96 leave 2.5 in
each tail
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1.96
-1.96
2.5
2.5
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Two Californias
http//www.sfgate.com/election/races/2003/10/07/ma
p.shtml
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Times to Produce a cell Phone Sample of 50

• Problem 10.35
• Data set Xr10-35

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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
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Sample statistics
Times 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
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Mean Time to Assemble

• Prob?lt(x-?)/(s/?n)lt? 0.95

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Students t-Distribution

• What does it look like? EViews
• What are the critical values for 2.5 in each
  tail? Text

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_at_ctdist(x,v)_at_dtdist(x,v)_at_qtdist(p,v)_at_rtdist(v) t-d
istribution for , and vgt0. Note that v1 is the
Cauchy
_at_ctdist(x,v) _at_dtdist(x,v) _at_qtdist(p,v) _at_rtdist(v)
t-distribution
Gen tran_at_rtdist(48) Gen tdens_at_dtdist(tran,48)
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Appendix B Table 4 p. B-9
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Mean Time to Assemble

• Prob?lt(x-?)/(s/?n)lt? 0.95
• Prob-2.01lt(26.81-?)/(1.29/?50)lt2.01 0.95
• Prob-2.01lt(26.81-?)/0.182)lt2.01 0.95
• Prob-0.366lt(26.81-?)lt0.366 0.95
• Prob0.37gt(?-26.81gt- 0.37 0.95
• Prob26.810.37gt?gt26.81-0.37 0.95
• Prob27.18gt ? gt26.44 0.95

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Hypothesis Testing
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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

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Step 4 compare the probability for the
test statistic(z -1.33) to the chosen critical
level (z-1.645)
-1.645
5 lower tail
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Decision Theory
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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.

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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 gt 0.5.

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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.

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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.

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True State of Nature
p 0.4999 H2o Bonds lose
P gt 0.5 Bonds win
Type II error
No Error
Accept null
Decision
Type I error
No Error
Reject null
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Decision Theory

• The size of the type I error is the significance
  level or probability of making a type I error, a.
• The size of the type II error is the probability
  of making a type II error, b.

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Decision Theory

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

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True State of Nature
p 0.4999
P gt 0.5
Type II error b
No Error 1 - a
Accept null
Decision
Type I error a
No Error 1 - b
Reject null
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Decision Theory

• There is a tradeoff between the size of the type
  I error and the type II error.

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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.

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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, p0.4999
• they want to accept the alternative hypothesis,
  pgt0.5

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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)

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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) a CII low(type II
  error) b
• Recommended Action make probability of type I
  error small, i.e. run scared so chances of losing
  stay small

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True State of Nature
P gt 0.5 Bonds win
p 0.499 Bonds lose
No Error 1 - a
Type II error b C(II)
Accept null
Decision
Type I error a C(I)
No Error 1 - b
Reject null
EC C(I) a C(II) b
Bonds C(I) is large so make a small
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How About Costs to the Bond Opponents Camp ?

• What do they want?
• They want to accept the null, p0.499 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

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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) a Chigh(type II
  error) b
• Recommended Action make probability of type II
  error small, i.e. make the probability of
  accepting the null when it is false small

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True State of Nature
p 0.499 Bonds lose
P gt0.5 Bonds win
No Error 1 - a
Type II error b C(II)
Accept null
Decision
Type I error a C(I)
No Error 1 - b
Reject null
EC C(I) a C(II) b
Opponents C(II) is large so make b small
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Decision Theory Example

• If we set the type I error, a, 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 p0.5 from null

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Decision theory example

• So for this poll size of 1000, with p0.5 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 a

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2.33
1
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Decision Theory Example

• So if 53.7 of the polling sample, or
  0.53681000537 say they will vote for water
  bonds, then we reject the null of p0.499, i.e
  the null that the bond proposition will lose

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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
  p0.499, false if p0.54) if this number is less
  than 537. What is the size of the type II error?

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Decision Theory Example

• What is the value of the type II error, b, 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)/n1/2
• Z(beta) (0.536 0.54)/.54.46/10001/2
• Z(beta) -0.253

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Calculation of Beta
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Beta Versus p (true)
b
p
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Power of the Test
1 - b
p
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Power of the Test
Ideal
1 - b
p
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True State of Nature
p 0.499
P gt 0.5
Type II error b
No Error 1 - a
Accept null
Decision
Type I error a
No Error 1 - b
Reject null
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Tradeoff

• Between a and b
• Suppose the type I error is 5 instead of 1
  what happens to the type II error?

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Tradeoff

• If a 5, 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.

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Interval Estimation Example
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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
• Students t-statistic
• degrees of freedom 107

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Sample Mean 0.842
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Appendix B Table 4 p. B-9
2.5 in the upper tail
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Interval Estimation

• 95 confidence interval
• substituting for t

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Interval Estimation

• Multiplying all 3 parts of the inequality by
  0.413
• subtracting .842 from all 3 parts of the
  inequality,

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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