HW1 - Problem 2-1. 1 2 3 4 145 153 150 147 y psi y psi y...

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Unformatted text preview: Problem 2-1. 1 2 3 4 145 153 150 147 y psi y psi y psi y psi = = = = 150 = 3 psi = 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 (will also accept =150). H 1 : the breaking strength is smaller than 150: 150 < . 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 = = - = = 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- = = < =- = = - < - = - = Since 0,05 0,202 P value = <- = 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 = + = + = ....
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HW1 - Problem 2-1. 1 2 3 4 145 153 150 147 y psi y psi y...

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