Problems

1. A time study analyst is planning a work sampling study to estimate the percent of idle time in the packing department. The department consists of a foreman and three workers. The hourly wage of the foreman is $15 and the wage of the workers is $8. A time study technician has made 300 observations of these four employees at random times, producing the data below.

Idle observations 60
Working observations 240
Total observations 300

1. What is the estimate of the percent of idle time?
2. What is the standard error of the percent?
3. Estimate the weekly loss due to idle time in the packaging department. The department works a 40-hour week.
4. Suppose that an estimate of idle time within 2 percentage points, either way, is required. How many more observations should be taken?
5. What will be the point estimate of the percent of idle time after the additional observations are taken?

2. In the course of manufacturing an amplifier, an off-on switch must be attached to the front panel. A time study analyst wishes to determine how long it takes an experienced employee to perform this operation; he therefore times a worker during several repetitions, producing the data below in minutes. The sample standard deviation, s = 0.7277 min.

1. What is the true (population) mean time for attaching the off-on switch?
2. What is the observed time for attaching the off-on switch?
3. Suppose it was desired to estimate the mean time within 9 seconds either way. How many more observations would be required?

3. Use the data in Problem 2 to solve this problem.

1. The analyst has determined that the worker has a performance rating of 125 percent while being measured. What is the normal time?
2. Was this worker working rapidly or slowly?
3. The plant manager allows for these delays and interruptions during the working day.

Coffee breaks (2 per 8 hr. shift) 10 min. each

What is the allowance as a percent of job time?

4. What is the standard time for attaching the off-on switch?

4. Use your work in Problem 3 to solve this problem. The workers who install the off-on switch are paid $10.00 per hour, with fringe benefits of 35 percent.

1. Estimate the number of off-on switches which one worker should be able to install in one 8-hour shift.
2. Determine the standard labor cost of installing the switch.

5. This problem uses the data of problems #2 and #3 to illustrate the use of working time allowances based on a full day's time (rather than on job time).

1. What is the allowance as a percent of working time?
2. What is the standard time for attaching the switch?

1. What is the standard error?
2. What is the 95% confidence interval?
3. What is the interpretation of the confidence interval?

Solutions

1. a. <a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif::Image207::/sites/dl/free/0072443901/24520/Image207.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif">Image207 (1.0K)</a>Image207
b. <a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif::Image208::/sites/dl/free/0072443901/24520/Image208.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif">Image208 (1.0K)</a>Image208
c. The loss due to idle time will be the total of the loss due to the foreman's idle time and the loss due to the workers' idle time:
Foreman) .20(40 hrs.)($15.00) $120.00
Workers) .20(40 hrs.)($8.00)(3 workers) 192.00
Total weekly loss $312.00

d. <a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif::Image209::/sites/dl/free/0072443901/24520/Image209.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif">Image209 (1.0K)</a>Image209 . Therefore, 1,237 additional observations will be required.
e. The new value of <a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif::Image210::/sites/dl/free/0072443901/24520/Image210.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif">Image210 (0.0K)</a>Image210 cannot be calculated until the additional observations are obtained.

2. a. The population mean is unknown.

3. a. Normal Time (NT) = OT x PR = 8.82(1.25) = 11.025 min
b. Because a performance rating of 125% is greater than 100%, the worker was working rapidly.
c. The allowance percent (A) = 100(total allowance time)/(job time) = 100(20 + 20 + 30 + 20)/[(8(60) - 90] = 23.08% of job time.
d. Standard Time (ST) = NT(1 + A) = 11.025(1 + .2308)= 13.57 minutes per installation.

4. a. Number of installations = <a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif::Image213::/sites/dl/free/0072443901/24520/Image213.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif">Image213 (1.0K)</a>Image213 switches.
   b. The standard labor cost = <a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif::Image214::/sites/dl/free/0072443901/24520/Image214.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif">Image214 (1.0K)</a>Image214
.

5. a. The allowance percent (A) = 100(total allowance time)/(working time) = 100(90)/[8(60)] = 18.75%
b. Standard time (ST) = NT/(1 - A) = 11.025/(1 - .1875) = 13.57 minutes per installation. (Compare to #3-d).

6. a. <a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif::Image215::/sites/dl/free/0072443901/24520/Image215.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif">Image215 (1.0K)</a>Image215 min.
b The Lower Confidence Limit = <a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif::Image216::/sites/dl/free/0072443901/24520/Image216.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif">Image216 (1.0K)</a>Image216 n min.
The Upper Confidence Limit = <a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif::Image217::/sites/dl/free/0072443901/24520/Image217.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif">Image217 (1.0K)</a>Image217 min.

c. There is a .95 probability that the true (population) mean time for installing the switch lies between 8.46 min. and 9.18 min.