sol5 - 8-20.a 1 The parameter of interest is the true mean...

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Unformatted text preview: 8-20.a) 1) The parameter of interest is the true mean yield, μ.2) H: μ= 90 3) H1: μ≠904) α= 0.055)zxn=- μσ/6) Reject Hif z< -zα/2where -z0.025= -1.96 or z> zα/2where z0.025= 1.967) x=90 48., σ= 3z904890350 36=-=./.8) Since -1.96 < 0.36 < 1.96 do not reject Hand conclude the yield is not significantly different from 90% at α= 0.05. b) P-value = 2[1-=-=Φ( .)][.].0 362 10 640580 71884c) n = (29(29(29(29(29zzzzαβσδ/.....22220 0250 052222238590196165954 67+=+-=+-=n 22455.d)β =+- --+-ΦΦzz0 0250 025909235909235..//= Φ(1.96 + -1.491) -Φ(-1.96 + -1.491) = Φ(0.47) -Φ(-3.45) = Φ(0.47) -(1 -Φ(3.45))= 0.68082 -( 1 -0.99972)= 0.68054.e) For α= 0.05, zα/2= z0.025= 1.96xznxzn- ≤≤+0 0250 025..σμσ90 481963590 4819635....- ≤≤+μ87.85 ≤μ≤93.11With 95% confidence, we believe the true mean yield of the chemical process is between 87.85% and 93.11%.8-23.a) 1) The parameter of interest is the true mean bulb life, μ....
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This homework help was uploaded on 02/26/2008 for the course IE 121 taught by Professor Perevalov during the Spring '08 term at Lehigh University .

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sol5 - 8-20.a 1 The parameter of interest is the true mean...

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