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Unformatted text preview: Julie, Tung and Wei Professor Rakhlin STAT 101 HW9 Solutions December 9, 2011 64 points in total Question. Chap 15, # 46 (5 points) Show the 95% zinterval or tinterval for the indicated parameter. (a) (1 point) μ given ¯ x = 45 ,s = 80 ,n = 33 ¯ x ± t . 05 / 2 s √ n = 45 ± t . 05 / 2 80 √ 33 which turns out to be 73.36677 and 16.63323. (1 point, 0.5 for some work, 0.5 for the correct answer. Reasonable rounding suffice) (b) (1 point) μ given ¯ x = 255 ,s = 16 ,n = 21 ¯ x ± t . 05 / 2 s √ n = 255 ± t . 05 / 2 16 √ 21 which turns out to be 247.8881 and 262.1119 (1 point, 0.5 for some work, 0.5 for the correct answer. Reasonable rounding suffice) (c) (1 point) p given ˆ p = 0 . 25 ,n = 48 ˆ p ± z . 05 / 2 √ ˆ p (1 ˆ p ) √ n = 0 . 25 ± z . 05 / 2 √ . 25(1 . 25) √ 48 which turns out to be 0.1275023 and 0.3724977 (1 point, 0.5 for some work, 0.5 for the correct answer. Reasonable rounding suffice) (d) (2 point) p given ˆ p = 0 . 9 ,n = 52 We must compute Wilson’s intervals because (1 ˆ p )52 = 5 . 2 < 10 (1 point). Therefore, the intervals are ˆ p ± z . 05 / 2 √ ˆ p (1 ˆ p ) √ n +4 where ˆ p = 52(0 . 9)+2 52+4 = 0 . 8714286. Then, our interval becomes 0.7837603 and 0.9590968 (1 point, 0.5 for some work, 0.5 for the correct answer. Reasonable rounding suffice) Question. Chap 15, # 49 (3 points) A package of light bulbs promises an average life of more than 750 hours per bulb. A consumer group did not believe the claim and tested a sample of 40 bulbs was 740 hours with s = 30 hours. The manufacturer responded that its claim was based on testing hundreds of bulbs....
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This note was uploaded on 04/04/2012 for the course STAT 101 taught by Professor Heller during the Fall '08 term at UPenn.
 Fall '08
 Heller
 Statistics

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