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Econ 240A

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  1. Econ 240A Power 6

  2. The Challenger Disaster • http://onlineethics.org/moral/boisjoly/RB-intro.html

  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. Only 4 launches Between 50 and 64 degrees Challenger Launch

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

  6. Probability of O-Ring Failure Conditional On Temperature, P/T • P/T=#of Yeses/# 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 Yeses/# of Launches at high temperature • Pˆ = k(high)/n(high) = 2/12 = 0.17

  7. 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?

  8. Interval Estimation and Hypothesis Testing

  9. Outline • Interval Estimation • Hypothesis Testing • Decision Theory

  10. The Field Poll, 10-7-’04 • In a sample of approximately 1135 likely voters, 48% indicate they will vote for Senator Boxer Vs. 32% for Jones • If the poll is an accurate reflection or subset of the population of voters Nov. 2, what is the expected proportion that will vote for Boxer? • How much uncertainty is in that expectation? Power 4

  11. Power 4 Field Poll • The estimated proportion, from the sample, that will vote for recall is: • where is 0.48 or 48% • k is the number of “successes”, the number of people sampled who are for Boxer, approximately 545 • n is the size of the sample, 1135

  12. Field Poll Power 4 • What is the expected proportion of voters Nov. 2 that will vote for Boxer? • = E(k)/n = np/n = p, where from the binomial distribution, E(k) = np • So if the sample is representative of voters and their preferences, 48% should vote for Boxer in three weeks.

  13. Power 4 Field Poll • How much dispersion is in this estimate, i.e. as reported in newspapers, what is the margin of sampling error? • The margin of sampling error is calculated as the standard deviation or square root of the variance in • = VAR(k)/n2 = np(1-p)/n2 =p(1-p)/n • and using 0.48 as an estimate of p, • = 0.48*0.52/1135 =0.00022

  14. Interval Estimation • Based on the Poll of 48% for Boxer, what was the probability that the fraction, p, voting for Boxer would exceed 50%, i.e. lie between 0.5 and 1.0? • The standardized normal variate, z =

  15. Interval estimation • Why can we use the normal distribution? • Where does the formula for z come from?

  16. Solving for p: .015*z = 0.48 - p p = 0.48 -.015*z and substituting for p: and subtracting 0.48 from each of the 3 parts of this inequality:

  17. And dividing by –0.015, which changes the signs of the inequality: And using the standardized normal distribution, this probability equals ….0.5

  18. 0 -34.7

  19. 0 -34.7

  20. Solving for p: .015*z = 0.48 - p p = 0.48 -.015*z and substituting for p: and subtracting 0.48 from each of the 3 parts of this inequality:

  21. And dividing by –0.015, which changes the signs of the inequality: And using the standardized normal distribution, this probability equals ….0.092

  22. -1.33 -34.7

  23. -1.33 -34.7

  24. So a Z value of 1.33 leads to an area of 0.408, leaving 0.092 in the Upper tail

  25. Interval Estimation • The conventional approach is to choose a probability for the interval such as 95% or 99%

  26. So z values of -1.96 and 1.96 leave 2.5% in each tail

  27. 1.96 -1.96 2.5% 2.5%

  28. Substituting for z And multiplying all three parts of the inequality by 0.015

  29. And subtracting 0.48 from all three parts of the inequality And multiplying by -1, which changes the signs of the inequality: So a 95% confidence interval based on the poll, predicted a vote for Boxer of between 45% and 51%, an inference about the unknown parameter p. Z values of -2.575 and 2.575 leave 1/2% in each tail. You might calculate a 99% confidence interval for the poll.

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

  31. 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

  32. Sample Mean 0.842

  33. Appendix B Table 4 p. B-9 2.5 % in the upper tail

  34. Interval Estimation • 95% confidence interval • substituting for t

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

  36. 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

  37. Hypothesis Testing

  38. 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%

  39. Hypothesis Test Example • Field Poll on Boxer • Step #1: null, i.e. the maintained, hypothesis: true proportion for Boxer is 50% H0 : p = 0.5; the alternative hypothesis is that the true population proportion supporting Boxer is greater than 50%, Ha : p>0.5

  40. Hypothesis Test Example • Step #2: test statistic: standardized normal variate z • Step #3: Critical level for rejecting the null hypothesis: e.g. 5% in upper tail; alternative 1% in upper tail

  41. Step #4: compare the probability for the test statistic(z= -1.33) to the chosen critical level(z=1.645) Sample statistic 1.645 5 % upper tail 6

  42. Hypothesis Test Example • So, since –1.33 is not above the critical value of 1.645, I.e. not extreme and not in the upper tail, do not reject the null hypothesis that p=0.5 . • In terms of common sense, the sample proportion of 0.48 means p=0.5 is more likely than the alternative of p>0.5.

  43. Also recall the 95% confidence interval on p which was between 0.45 and 0.51, including the null that p= 0.5.

  44. And subtracting 0.48 from all three parts of the inequality And multiplying by -1, which changes the signs of the inequality: So a 95% confidence interval based on the poll, predicted a vote for Boxer of between 45% and 51%, an inference about the unknown parameter p. Z values of -2.575 and 2.575 leave 1/2% in each tail. You might calculate a 99% confidence interval for the poll.