prefinal_solutions - Prefinal for AMS310 SOLUTIONS Z/2 I-1...

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Prefinal for AMS310 SOLUTIONS I-1 086 . 0 15 * 2 58 . 2 2 2 / = = = n Z E α I-2 48 . 15 10 60 * 58 . 2 2 / = = = n Z E σ α I-3 (a) I-4 (d) I-5 2 . 4 2 44 . 1 * 841 . 5 2 / = = = n s t E α I-6 (b) I-7 (a) I-8 (d) I-9 20 10 * 2 2 = = = X Y σ σ I-10 (a) I-11 25 . 0 25 . 0 50 . 0 ) 2 ( ) 3 ( ) 3 ( = - = - = = F F X P I-12 323 . 0 667 . 0 1 ) 2 ( 1 ) 3 ( = - = - = X P X P Longer problem: II-1(a) p ˆ = 170/2835 =0.06. So point estimate of 100p =100x 0.06=6 % II-1(b) First estimate SE of p ˆ = n n x n x ) 1 ( - = 2835 ) 2835 170 1 ( 2835 170 - =0.0045. Thus ) ˆ 100 ( p SE =0.45% II-1-(c). The estimate of this quantity is p ˆ =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. II-1-(d) H 0 : 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. II-1(e) ANS P= P( χ 2 2 > 7.378)=0.04. Compare Pvalue to α , if P ≤α then reject Ho at α level ; if P> α then accept Ho. Since P= 0.04 < 0.05= α , we have P< α so we can reject Ho at the 0.05 level and conclude that the proportion of defectives is NOT the same for each shift.
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