A study of nonfatal occupational injuries in the United States found that about
ID: 3022541 • Letter: A
Question
A study of nonfatal occupational injuries in the United States found that about 31% of all injuries in the service sector involved the back. The National Institute for Occupational Safety and Health (NIOSH) recommended conducting a comprehensive ergonomics assessment of jobs and work stations. In response to this information, Mark Glassmeyer developed a unique erogonic handcard to help field service engineers be more productive and also to reduce back injuries from lifting parts and equipment during service calls. Using a sample of 382 field service engineers who were provided with these carts, Mark collected the following data:
Year 1 (without Cart) Year 2 (with cart)
Average call time 8.27 hours 7.98 hours
Standard deviation call time 1.36 hours 1.21 hours
Proportion of back injuries 0.018 0.010
Find 95% confidence intervals for the average call times and proportion of back injuries in each year. What conclusions would you reach based on your results?
Explanation / Answer
FOR YEAR 1, AVERAGE CALL TIME:
Note that
Margin of Error E = z(alpha/2) * s / sqrt(n)
Lower Bound = X - z(alpha/2) * s / sqrt(n)
Upper Bound = X + z(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.025
X = sample mean = 8.27
z(alpha/2) = critical z for the confidence interval = 1.959963985
s = sample standard deviation = 1.36
n = sample size = 382
Thus,
Margin of Error E = 0.136381455
Lower bound = 8.133618545
Upper bound = 8.406381455
Thus, the confidence interval is
( 8.133618545 , 8.406381455 )
***************************
FOR YEAR 2, AVERAGE CALL TIME:
Note that
Margin of Error E = z(alpha/2) * s / sqrt(n)
Lower Bound = X - z(alpha/2) * s / sqrt(n)
Upper Bound = X + z(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.025
X = sample mean = 7.98
z(alpha/2) = critical z for the confidence interval = 1.959963985
s = sample standard deviation = 1.21
n = sample size = 382
Thus,
Margin of Error E = 0.121339382
Lower bound = 7.858660618
Upper bound = 8.101339382
Thus, the confidence interval is
( 7.858660618 , 8.101339382 )
******************************
FOR YEAR 1, PROPORTION OF BACK INJURIES:
Note that
p^ = point estimate of the population proportion = x / n = 0.018
Also, we get the standard error of p, sp:
sp = sqrt[p^ (1 - p^) / n] = 0.006802371
Now, for the critical z,
alpha/2 = 0.025
Thus, z(alpha/2) = 1.959963985
Thus,
Margin of error = z(alpha/2)*sp = 0.013332402
lower bound = p^ - z(alpha/2) * sp = 0.004667598
upper bound = p^ + z(alpha/2) * sp = 0.031332402
Thus, the confidence interval is
( 0.004667598 , 0.031332402 )
************************************
FOR YEAR 2, PROPORTION OF BACK INJURIES:
Note that
p^ = point estimate of the population proportion = x / n = 0.01
Also, we get the standard error of p, sp:
sp = sqrt[p^ (1 - p^) / n] = 0.005090799
Now, for the critical z,
alpha/2 = 0.025
Thus, z(alpha/2) = 1.959963985
Thus,
Margin of error = z(alpha/2)*sp = 0.009977782
lower bound = p^ - z(alpha/2) * sp = 2.22181E-05
upper bound = p^ + z(alpha/2) * sp = 0.019977782
Thus, the confidence interval is
( 2.22181E-05 , 0.019977782 )
*****************************************
CONCLUSION:
We can see that there is a significant reduction in average call time, as the confidence intervals did not intersect. However, this reduction did not yield a significant reduction in the proportion of back injuries, as the proportion confidence intervals intersected.
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