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Unformatted text preview: 920 (a) The critical values are given by 0 /2 / . Therefore, /2 / = 0.15. = ( < 4.85) + ( > 5.15) = 0 < 1 3500 0.25/8 4.85  5 + 0 > 0.25/8 5.15  5 IE300/GE331 HW9 /2 = 0.15 8/0.25 = 1.6971, = 2(1.6971) = 0.0897. Alternatively, 940 (a) = 0.0897 The parameter of interest is the true mean tensile strength 0 = 3500, = 0.01, 0 = /
x = 3450 (b) Solve the equation 0  = /2 = 2.8868 , we obtain = 2(2.8868) = 0.004. When 0.004, (c) = 0  /2 = we fail to reject the null hypothesis. Otherwise, we reject the null hypothesis. = 0.004 is the answer.
3455 .3853470 Reject the null hypothesis and conclude the true mean tensile strength is significantly different from 3500 at = 0.01. 0 + /2 , = 60, = 12, 0 =  0 . Reject 0 if 0  > /2 where 0.005 = 2.58  0 / = 2.8868 3544.6153470 = (3455.385 3544.615) (e) The 99% CI on the true mean strength is given by  /2 60 That is 3405.313 3494.687. Since 0 = 3500 does not fall within this interval, we reject the null hypothesis. 3450  0.005 12 = ( 4.3079)  ( 0.8438) = 0.8006 + /2 3450 + 0.005 12 60 ...
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This note was uploaded on 05/04/2010 for the course GE 331 taught by Professor Negarkayavash during the Spring '09 term at University of Illinois at Urbana–Champaign.
 Spring '09
 NegarKayavash

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