MidtermReviewSolutions 2010

MidtermReviewSolutions 2010 - Midterm Review- Fall 2010...

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1 Midterm Review- Fall 2010 Exercises from previous years’ midterms – solutions (except for 2008 and 2009) 2003 #3. (5 points) In a poll of 1000 persons, you find that 44% of the respondents state that they are in favor of a land tax. Build a 95% confidence interval for the percentage of the population that is in favor of the land tax. Let p be the true percentage in the population. We draw a random sample of 1000 persons. Let x be a dummy variable equal to 1 if the person is in favor of the tax, and to 0 if it is not. We estimate p with ˆ p x = . ( ) ( ) 1 var p p x n = which is estimated by ( )( ) .44 .56 .2464 .0002464 1000 1000 = = . Thus ( ) .0002464 .0157 se x = = The 95% confidence interval is ( ) .44 1.96 .0157 ± or ( ) ( ) .44 .031;.44 .031 .41;.47 + = 2003 #5 (15 points) In a study of turnover in the US labor market, annual data for the period 1950 to 1982 were used to estimate the following model: ( ) ( ) 0 1 2 log log quit unemp year u β = + + + where quit is the quit rate in manufacturing, defined as the number of people leaving their job voluntarily per 100 employees, unemp is the unemployment rate, and year is the year of the observation. The estimated model is: ( ) s ( ) log .08 0.34log 0.012 (.03) (0.12) (0.007) quit unemp year = 2 0.54 33 observations R n = = (iii) Test the hypothesis 2 0 = against 2 0 < at the 5% significance level. Interpret the result. 0 2 1 2 : 0 : 0 H H = < 30 .012 1.71 .007 t = = − The critical value for a 5% significance level, one-tailed test at n-k-1=33-2-1=30 degrees of freedom is 1.697. We thus reject that 2 0 = with a 5% chance of error. The parameter 2 measures the percentage change in quit rate for any additional year. This is also called the time trend. There is evidence that the quit rate has been declining over time. 2004: #1. (5 points) Denote by X the number of miles per gallon achieved by cars of a particular model. You are told that ( ) 2 20, 4 X N μ σ = = . What is the probability that for a random sample of 25 cars, the average miles per gallon X will be between 19 and 21?
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2 1 1 n i i X X n = = follows a Normal distribution of mean 20 μ = and variance 2 4 .16 25 n σ = = Therefore ( ) 20 0,1 .4 X z N = And the probability that average miles per gallon will be between 19 and 21 miles/gallon is:
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This note was uploaded on 02/10/2011 for the course EEP 118 taught by Professor Elizabeth during the Spring '11 term at University of California, Berkeley.

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MidtermReviewSolutions 2010 - Midterm Review- Fall 2010...

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