Ch 8 stats hw - n = 1000 the z-score for a 95% CI is 1.96...

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Ch 8-2 #5: For large sample confidence intervals for the proportion in this situation you have: pHat ± z * sqrt( (pHat * (1-pHat)) / n) where pHat is the sample proportion z is the zscore for having α% of the data in the tails, i.e., P( |Z| > z) = α n is the sample size in this question we have pHat = 410/1000
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Unformatted text preview: n = 1000 the z-score for a 95% CI is 1.96 the 95% CI is 0.41 1.96 * sqrt( 0.41 * (1-0.41) / 1000 ) (0.3795159 , 0.4404841) b) about 38% is a good estimate for the minimum proportion of first - year defaults that were approved on the basis of falsified applications....
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This note was uploaded on 02/24/2012 for the course ECON 101 taught by Professor Smith during the Spring '11 term at University of Phoenix.

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