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08 STAT HWB a - Dr V.R Bencivenga Economics 329 PRACTICE...

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Dr. V.R. Bencivenga Economics 329 PRACTICE ASSIGNMENT #8B: RIGHT ANSWERS 1. 2 i , N iid n ,..., 1 i , X n , N ~ X 2 because the population is normal (with known variance) 1 . 3 X 2 = .25 2 n = 25 = .05 1 n z X n z X C u u 95 . 25 25 . 96 . 1 1 . 3 25 25 . 96 . 1 1 . 3 C where 1.96 is the 97.5 th percentile of Z 2. 2 i , N iid n ,..., 1 i , X n , N ~ X 2 because the population is normal (with known variance) 5 . 110 X 2 = 9 n = 10 = .05 1 n z X n z X C u u where z u is the 2 1 quantile (the 2 1 100 percentile) of Z 95 . 10 3 96 . 1 110 10 3 96 . 1 5 . 110 C b. 1 n z X C where z is the 1 quantile of Z 95 . 645 . 1 10 3 5 . 110 C c. 1 n z X C where z is the 1 quantile of Z 95 . 10 3 645 . 1 110 C 3. 2 i , N iid n ,..., 1 i , X n , N ~ X 2 because the population is normal (with known variance) 2 = 2500 = .01 1 n z X n z X C u u
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2 The length of the confidence interval should not exceed 10, i.e., 10 n z 2 n z X n z X L u u u . This inequality may be solved for the minimum sample size: 664 n 10 2500 576 . 2 2 n 10 n z 2 u . 4. 2 i , N iid n ,..., 1 i , X 1 n t ~ n s X because the population is normal (with unknown variance) 38 . X s 2 = .06 2 n = 25 = .05 1 n s t X n s t X C u u 95 . 25 06 . 064 . 2 38 . 25 06 . 064 . 2 38 . C 95 . 405 . 355 . C where 2.064 is the 97.5 th percentile of the t 24 random variable 5.a. 2 i , N iid n ,..., 1 i , X 1 n t ~ n s X because the population is normal (with unknown variance) 268 . 1 X s 2 = .228 2 n = 31 = .05 1 n s t X n s t X C u u 95 . 31 228 . 042 . 2 268 . 1 31 228 . 042 . 2 268 . 1 C 95 . 352 . 1 184 . 1 C where 2.042 is the 97.5 th percentile of the t 30 random variable b. 2 1 n 2 2 ~ s 1 n because the population is normal 1 s 1 n s 1 n C 2 l 2 2 2 u 2 = .05 95 .
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