2005-Spring-Final_s

# 2005-Spring-Final_s - 1(3 1 3 3 2 3 n1 = n2 = 25 X1 = 115 X2 = 98 s1 = 30 s2 = 25(a H0 Ha 2 2 = 80(or > 80 2 80 test statistic t= X2 p = 3:6 > 1:71

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1. (3+1+3+3+2+3) n 1 = n 2 = 25 ; X 1 = 115 ; X 2 = 98 ;s 1 = 30 ;s 2 = 25 (a) H 0 : 2 = 80( or 2 80) H a : 2 > 80 test statistic: t = X 2 s 2 = p n 2 = 3 : 6 > 1 : 71 = t 24 ; 0 : 05 Reject H 0 : There is evidence that the mean bill in county II is above \$80 at 0.05 level of (b) Data for county II is normally distributed. (c) p-value = P ( T 24 > 3 : 6) < 0 : 005 (d) H 0 : 1 = 2 H a : 1 2 t = X 1 ± X 2 s p q 1 n 1 + 1 n 2 = 2 : 177 > 1 : 6772 = t 48 ; 0 : 05 Reject H 0 : There is evidence that the mean monthly bill is higher in county I than in county II (e) (i) Data for both counties are normal. (ii) ± 2 1 = ± 2 2 (f) X 1 ± X 2 ² t 48 ; 0 : 025 s p q 1 n 1 + 1 n 2 : i.e. (1.297, 32.703). 2. (2+2+3+2+3) n 1 = 1090 ;n 2 = 2065 ; ^ p 1 = 283 = 1090 = 0 : 2596 ; ^ p 2 = 1053 = 2065 = 0 : 5099 : ^ p = (283 + 1053) = (1090 + 2065) = 0 : 4235 (a) H 0 : p 1 = p 2 H a : p 1 < p 2 (b) z = ^ p 2 ^ p 1 r ^ p (1 ^ p ) 1 n 1 + 1 n

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## This note was uploaded on 02/27/2010 for the course ISOM ISOM111 taught by Professor Anthonychan during the Spring '09 term at HKUST.

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2005-Spring-Final_s - 1(3 1 3 3 2 3 n1 = n2 = 25 X1 = 115 X2 = 98 s1 = 30 s2 = 25(a H0 Ha 2 2 = 80(or > 80 2 80 test statistic t= X2 p = 3:6 > 1:71

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