Based on a careful work study in the Hofstetter Corp., the results shown in the following table have been observed: Observation (minutes per cycle) Performance Element Rating 1 2 3 4 5 Prepare daily reports 115% 36 40 33 43 40 Photocopy results 110% 13 9 36* 14 14 Label and package reports 95% 2 2 Distribute reports 90% 16 18 22 52 5 5 17 45^ * Photocopying machine broken; included as delay in the allowance factor ^ Power outage; included as delay in the allowance factor The allowance factor for this type of work is given to be 12%. To determine the average, one needs first to drop the outliers for which an exception exists. For the given data, the average time for the "prepare daily reports" operation = 38.40 minutes (round your response to two decimal places). Normal Time = Average Timex Performance Rating Factor The normal time for the "prepare daily reports" operation = places). 44.16 The average time for the "photocopy results" operation = = 12.50 minutes (round your response to two decimal minutes (round your response to two decimal places). Normal Time = Average Timex Performance Rating Factor The normal time for the "photocopy results" operation = 13.75 minutes (round your response to two decimal places). The average time for the "label and package reports" operation places). = 3.80 minutes (round your response to two decimal Normal Time = Average Timex Performance Rating Factor The normal time for the "label and package reports" operation = places). 3.61 minutes (round your response to two decimal The average time for the "distribute reports" operation = 18.25 minutes (round your response to two decimal places). Normal Time = Average Timex Performance Rating Factor The normal time for the "distribute reports" operation = 16.43 77.95 minutes (round your response to two decimal places). minutes (round your response to The normal time for the overall process is the sum of the four normal times = two decimal places). Standard time is determined using the relationship, Normal Time Standard Time = 1 - Allowance Factor minutes (round your response to two decimal places). The standard time for this process = 88.57 Suppose you need to determine the number of observations that are required for a 90% confidence level within 5% accuracy. To do this, first you will need to determine the number of observations for each operation and then the sample size for the entire task is provided by the biggest value across each operation. The formula for finding the appropriate sample size is: 2 ZS Required sample size = n = hx where h=accuracy level desired in percent of the job element, expressed as a decimal (5% = 0.05) z= number of standard deviations required for desired level of confidence (e.g. for 90% confidence, z = 1.65) s= standard deviation of the initial sample x= mean of the initial sample n = required sample size For the first operation, s = 3.91 (x-x) 2 S= n-1 minutes (round your response to two decimal places). x was found to be 38.40 minutes. h is given as 5%. The value of z for the given confidence level = 1.65. Using the above values, n for the first operation = whole number). 12 observations (round your response to the immediate higher One would be required to repeat the above steps for the remaining three operations and then find the biggest required number of the observations to be able to determine the number of observations for the overall process. The number of observations that are required for a 90% confidence level within 5% accuracy = response to the immediate higher whole number). 204 (round your