Data from the federal Occupational Safety and Health (OSHA) shows that the numbe

Question

Data from the federal Occupational Safety and Health (OSHA) shows that the number of workplace injuries for a given manufacturing process in the years from 2000 to 2003 were 5.2, 4.6, 4.3, and 4.8 per one thousand workers. a. Use these data to give an interval that will, with 90% confidence, contain the number of workplace injuries in 2004. b. An industrial leader states that the average workplace injury rate is more per one thousand workers. Should you believe the industrial leader? What assumptions are you making?

Explanation / Answer

Solution:

Part a

First of all we have to find the confidence interval for the population mean.

Confidence interval = Xbar -/+ t*S/sqrt(n)

From the given data, we have

Sample mean = Xbar = 4.725

Sample standard deviation = S = 0.377491722

Sample size = n = 4

Degrees of freedom = n - 1 = 4 – 1 = 3

Confidence level = 90%

Critical t value = 2.3534

Lower limit = Xbar - t*S/sqrt(n)

Lower limit = 4.725 – 2.3534*0.377491722/sqrt(4) = 4.28

Upper limit = Xbar + t*S/sqrt(n)

Upper limit = 4.725 + 2.3534*0.377491722/sqrt(4) = 5.17

Confidence interval = (4.28, 5.17)

Part b

We should not believe the industrial leader because the value 4.2 does not contain in the given Confidence interval = (4.28, 5.17). We assumed that the sample is came from the normally distributed population.

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