# hw4sol~1 - know that the change in weight must be positive...

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Homework Four Solutions 7.12 a) 17/987 b) 987 ) 987 17 1 ( 987 17 c) ME=1.96 987 ) 987 17 1 ( 987 17 d) 17/987 1.96 987 ) 987 17 1 ( 987 17 =(.009, .025) Yes, because .025 <.1, we can conclude that fewer than 10% of all adults in the US were victims. 7.18 a) 1.645 b) 2.33 c) 3.29 7.27 a) 2.776 b) 2.145 c) 2.977 7.28 a) Slight skew in the boxplot, but since there are no outliers we can use the t-distribution to form a confidence interval. c) SE= 17 18 . 7 d) df=n-1=17-1=16. Remember for a 95% confidence interval we look down the 95% column and across the 16df row to get 2.12. e) 7.29 2.12 17 18 . 7 Because the confidence interval goes from a positive number to a positive number, we

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Unformatted text preview: know that the change in weight must be positive. However, since numbers between 3.6 and 11 are relatively small, the change in weight is probably small. 7.44ab a) 2 05 . ) 5 . 1 ( 5 . 96 . 1 =384.16…round to 385 b) 2 05 . ) 44 . 1 ( 44 . 96 . 1 =297.5….round to 298 7.48 First note from the hint that s 20000 6 120000 . Also, the Z-score corresponding to a 99% confidence interval is 2.58. 2 1000 ) 20000 ( 58 . 2 =2662.56. .round to 2663...
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## This note was uploaded on 02/04/2009 for the course STATISTIC 3011 taught by Professor Sally during the Spring '08 term at University of Minnesota Crookston.

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hw4sol~1 - know that the change in weight must be positive...

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