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## Econ 240A

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**Econ 240A**Power 6**The Challenger Disaster**• http://onlineethics.org/moral/boisjoly/RB-intro.html**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?**Only 4 launches**Between 50 and 64 degrees Challenger Launch**Challenger**• Divide the data into two groups • 12 low temperature launches, 53-70 degrees • 12 high temperature launches, 70-81 degrees**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**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?**Outline**• Interval Estimation • Hypothesis Testing • Decision Theory**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**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**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.**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**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 =**Interval estimation**• Why can we use the normal distribution? • Where does the formula for z come from?**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:**And dividing by –0.015,**which changes the signs of the inequality: And using the standardized normal distribution, this probability equals ….0.5**0**-34.7**0**-34.7**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:**And dividing by –0.015,**which changes the signs of the inequality: And using the standardized normal distribution, this probability equals ….0.092**-1.33**-34.7**-1.33**-34.7**So a Z value**of 1.33 leads to an area of 0.408, leaving 0.092 in the Upper tail**Interval Estimation**• The conventional approach is to choose a probability for the interval such as 95% or 99%**So z values**of -1.96 and 1.96 leave 2.5% in each tail**1.96**-1.96 2.5% 2.5%**Substituting for z**And multiplying all three parts of the inequality by 0.015**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.**Two Californias**http://www.sfgate.com/election/races/2003/10/07/map.shtml**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**Sample**Mean 0.842**Appendix B**Table 4 p. B-9 2.5 % in the upper tail**Interval Estimation**• 95% confidence interval • substituting for t**Interval Estimation**• Multiplying all 3 parts of the inequality by 0.413 • subtracting .842 from all 3 parts of the inequality,**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**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%**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**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**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**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.**Also recall the 95% confidence interval on p which was**between 0.45 and 0.51, including the null that p= 0.5.**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.