In an attempt to reduce the number of person-hours lost as a result of industria

Question

In an attempt to reduce the number of person-hours lost as a result of industrial accidents, a large production plant installed new safety equipment. In a test of the effectiveness of the equipment, a random sample of 58 departments was chosen. The number of person-hours lost in the month prior to and the month after the installation of the safety equipment was recorded and the percentage change was calculated. Find the power of the test designed to determine if the the new safety equipment is effective when the mean percent reduction is actually 1.4%. Assume that the population standard deviation is 7% and that is 0.03.

Power = ???

Explanation / Answer

Ho: mean reduction = 0% (new safety equipment has no effect)
Ha: mean reduction > 0% (new safety equipment has positive effect)

where p represents the percent reduction in person-hours lost.

alpha = 0.03
standard deviation = 0.07

Since this is a one-tailed test (because of the directionality of the alternative hypothesis), the critical z-score, which defines the critical region, is z = 1.8807

Any test statistic greater than 1.8807 will lead to rejection of the null hypothesis and anything less than 1.8807 will lead to non-rejection of the null hypothesis.

1.8807 = (xbar – 0%) / [.07/sqrt(58)]

xbar = [.07/sqrt(58)](1.8807) = 0.0172

It means any sample mean proportion of less than 1.72% will lead to non-rejection.

But, it was given that the true mean is really 1.4%. The probability of getting a sample mean proportion of less than 1.72% when the true mean percent reduction is 1.4% is:

P(xbar < 1.72%)

= P{z < (0.0172- .0140) / [.07/sqrt(58)]}

= P{z < 0.351}

= 0.6372

Power = 1 - 0.6372 = 0.3628

Power = 0.3628

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