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Unformatted text preview: Prefinal for AMS310 SOLUTIONS I1 086 . 15 * 2 58 . 2 2 2 / = = = n Z E I2 48 . 15 10 60 * 58 . 2 2 / = = = n Z E I3 (a) I4 (d) I5 2 . 4 2 44 . 1 * 841 . 5 2 / = = = n s t E I6 (b) I7 (a) I8 (d) I9 20 10 * 2 2 = = = X Y I10 (a) I11 25 . 25 . 50 . ) 2 ( ) 3 ( ) 3 ( = = = = F F X P I12 323 . 667 . 1 ) 2 ( 1 ) 3 ( = =  = X P X P Longer problem: II1(a) p = 170/2835 =0.06. So point estimate of 100p =100x 0.06=6 % II1(b) First estimate SE of p = n n x n x ) 1 ( = 2835 ) 2835 170 1 ( 2835 170 =0.0045. Thus ) 100 ( p SE =0.45% II1(c). The estimate of this quantity is p =70/940=0.074 Thus and our estimate of the percentage of defiectives given an item is produced on the night shift equals 100x0.074= 7.4%. This is more than the overall percentage of defectives produced by workers (6%) in this sample. II1(d) H : p 1 = p 2= p 3. Here p 1, p 2, p 3 denote the proportion of defectives produced on Day shift, Evening shift and Night shift repectively. and Night shift repectively....
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This homework help was uploaded on 01/31/2008 for the course AMS 310.01 taught by Professor Mendell during the Fall '03 term at SUNY Stony Brook.
 Fall '03
 Mendell

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