Brian Cabrera_Statistics_Hw5

Brian Cabrera_Statistics_Hw5 - Brian Cabrera October 31,...

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Brian Cabrera October 31, 2011 Homework 5 Problem 1 YES; the mean is between 1.9771 and 2.0199 and my expected mean was 2. This is simply because an average of 3 and 1 is 2. Problem 2
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0.64 0.56 0.48 0.40 0.32 0.24 Median Mean 0.52 0.51 0.50 0.49 0.48 1st Quartile 0.44000 Median 0.48000 3rd Quartile 0.56000 Maximum 0.76000 0.48744 0.50456 0.48000 0.52000 0.09173 0.10387 A-Squared 3.62 P-Value < 0.005 Mean 0.49600 StDev 0.09742 Variance 0.00949 Skew ness 0.033424 Kurtosis -0.315310 N 500 Minimum 0.24000 Anderson-Darling Normality  Test 95%  Confidence I nterv al for Mean 95%  Confidence I nterv al for Median 95%  Confidence I nterv al for StDev 9 5 %  Confidence I nter vals Summary for C26 YES; this is because the 95% confidence interval for mean is 0.48744 and 0.50456 which is very close to my expected mean of 0.50 since it is a fair coin toss it is usually 50% probability of landing on one side. Problem 3
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This note was uploaded on 11/05/2011 for the course ECON UA.18.1-01 taught by Professor Romanfrydman during the Fall '11 term at NYU.

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Brian Cabrera_Statistics_Hw5 - Brian Cabrera October 31,...

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