HW-03New

HW-03New - HOMEWORK FOR EXST 7036 (Chapter 3) Student Name:...

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HOMEWORK FOR EXST 7036 (Chapter 3) Student Name: Tanza Erlambang 3.2) In the 2000 US Presidential election, Palm Beach County Florida was the focus of unusual voting patterns caused by a confusing “butterfly bailout”. Many voters claimed they voted mistakenly for Pat Buchanan, when they intended to vote for Al Gore. Figure 3.8 shows the total number of votes for Buchanan plotted against number of votes for Ross Perot in 1996, by county in Florida (See A.Agresti and B Presnell, Statist.Sci ., 17: 436-440, 2003) a. In county i, let π i denote the proportion of vote for Buchanan and let x i for proportion vote of Perot in 1996.The Linear probability model fitted to all counties except Palm Beach, π ^ i =- 0.0003+0.0304x i .Give the value of p in interpretation:The estimated proportion vote for Buchanan in 2000 was roughly p% of that for Perot in 1996. Answer: Let say x i = 1000, thus: π ^ i =- 0.0003+0.0304x i → π ^ i =- 0.0003+0.0304(10000) = 303.9997 = 304 Then, P. Buchanan = 304/10000 (100%) = 3.04% of Perot b . For Palm Beach County, , π i = 0.0079 and x i = 0.0774. Does this result appar to be an outlier?. Finding π i ^ i and π i - π ^ i (Analyses conducted by statisticians predicted that fewer than 900 votes were truly intended for Buchanan, compared with the 3407 he received. George W Bush won the state by 537 votes and, with it, the Electoral College and the election). Answer: Estimated proportion: π ^ i =- 0.0003+0.0304(0.0774) = 0.0021 π i - π ^ i = 0.0079 – 0.0021 = 0.0058 π i ^ i = 0.0079/0.0021 = 3.76 → the actual value is 3.76 times the estimated value → suggested that it is an outlier 1
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SAS Output: The GENMOD Procedure Model Information Data Set WORK.CRAB Distribution Binomial Link Function Probit Response Variable (Events) y Response Variable (Trials) n Number of Observations Read 173 Number of Observations Used 173 Number of Events 111 Number of Trials 173 Criteria For Assessing Goodness Of Fit Criterion DF Value Value/DF Deviance 171 195.4621 1.1431 Scaled Deviance 171 195.4621 1.1431 Pearson Chi-Square 171 168.2119 0.9837 Scaled Pearson X2 171 168.2119 0.9837 Log Likelihood -97.7311 Algorithm converged. Analysis Of Parameter Estimates Standard Wald 95% Confidence Chi- Parameter DF Estimate Error Limits Square Pr > ChiSq Intercept 1 -2.2383 0.5155 -3.2487 -1.2279 18.85 <.0001 weight 1 1.0990 0.2171 0.6736 1.5245 25.63 <.0001 Scale 0 1.0000 0.0000 1.0000 1.0000 NOTE: The scale parameter was held fixed. a. Report the fit for the probit model, with weight predictor Answer: Probit (π ^ i ) = -2.238 + 1.099 (weight) b . Find π ^ at the highest observed weight, 5.20 kg. 2
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HW-03New - HOMEWORK FOR EXST 7036 (Chapter 3) Student Name:...

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