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Unformatted text preview: Problem 21. 1 2 3 4 145 153 150 147 y psi y psi y psi y psi = = = = 150 μ = 3 psi σ = a). Since the breaking strength of a fiber is required to be at least 150 psi, we consider two hypotheses: • H : the breaking strength is greater or equal to 150: 150 μ ≥ (will also accept μ =150). • H 1 : the breaking strength is smaller than 150: 150 μ < . We would accept the fiber only if the null hypothesis H fails to be rejected (i.e., is accepted) or H 1 could be rejected. b). ( 29 1 2 3 4 1 148.75 4 y y y y y = + + + = Test statistic: From table of Cumulative Standard Normal Distribution: ( 29 { } 0,05 arg 1 0.05 0.95 1.645 Z z = Φ =  = = Since 0,05 Z Z < , we fail to reject (accept) hypothesis H . c). ( 29 ( 29 ( 29 1 ( ) 1 0.798 0.202 P value Z P x Z P x Z P x Z = Φ = < = = =  <  =  = Since 0,05 0,202 P value α = < = we should accept hypothesis H . d). The 95% confidence interval is ( 29 95% , U∞ , where 95% 0,05 / 148.75 1.645*3/ 2 151.2175 U y Z n σ = + = + = ....
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This note was uploaded on 08/27/2011 for the course EIN 4905 taught by Professor Staff during the Spring '08 term at University of Florida.
 Spring '08
 Staff

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