Wax-com Electronics Inc.claims that a certain circuit board has a median operating life of less than 20,000 hours.A random sample of 25 such circuit boards showed 8 circuit boards failing before 20,000 hours and 17 circuit boards failing after 20,000 hours.Assume the useful life of the circuit board is not normally distributed.At ? = .05,can it be concluded that the sample contradicts the company's claim? Assume the useful life of the circuit board is not normally distributed.
A)Reject the null hypothesis.
B)Do not reject the null hypothesis.
$$Z = \frac { 8.5 - 25 ( .5 ) } { \sqrt { ( .5 ) ( 1 - .5 ) ( 25 } } = \frac { - 4 } { 2.5 } = - 1.6$$

Question

Quizplus · Accepted Answer

The calculated Z value of -1.6 does not fall in the critical region for a one-tailed test at the 0.05 significance level. Therefore, we do not reject the null hypothesis, meaning the sample does not provide sufficient evidence to contradict the company's claim.

Wax-Com Electronics Inc Z=8.5−25(.5)(.5)(1−.5)(25=−42.5=−1.6Z = \frac { 8.5 - 25 ( .5 ) } { \sqrt { ( .5 ) ( 1 - .5 ) ( 25 } } = \frac { - 4 } { 2.5 } = - 1.6Z=(.5)(1−.5)(25​8.5−25(.5)​=2.5−4​=−1.6

Wax-Com Electronics Inc $Z = \frac { 8.5 - 25 ( .5 ) } { \sqrt { ( .5 ) ( 1 - .5 ) ( 25 } } = \frac { - 4 } { 2.5 } = - 1.6$