Question

Installing a muffler at Barut Garage, Inc. involves a five work tasks. Azra has timed workers performing these tasks seven times, with the results (in minutes) shown in the following table.

No	Tasks	1	2	3	4	5	6	7	PR
1	Select correct mufflers	5	4	6	3	4	5	5	120
2	Remove old muffler	8	8	8	6	8	7	6	85
3	Install new muffler	15	13	15	14	12	15	13	85
4	Inspect the work	4	4	3	3	3	3	3	110
5	Complete paperwork	9	8	9	8	10	7	10	110

  
      
Given that Azra does not want an error of more than 0.07 and a confidence level less than 0.85 answer the following:
  
Provide the z-score:  Round to the nearest hundredth, for example 4.22
  
Does the number of observations for the following tasks warrant such confidence level?  
Enter YES or NO.
  
Task # 3 Install new muffler      :  
Task # 5 Complete paperwork   :  
  
What are the required numbers of observations for these tasks? Enter the number.
  
Task # 3 Install new muffler      :  
Task # 5 Complete paperwork   :  
  
Assume that the number of observations for the following tasks justifies the confidence level and accuracy mentioned above. Also note that Azra factors in a total of fatigue and personal allowance of 0.15 applied to job time. Provide the following results from your solution:

Task	Normal Time*	Standard Time*
# 2 Remove old muffler	min	min
# 4 Inspect the work	min	min

  
Provide the total standard time* for the whole job:   minutes.
*Round to the nearest hundredth, for example 8.45 minutes

Answer 1

Task	1	2	3	4	5	6	7	PR	mean	NT
1	5	4	6	3	4	5	5	120	4.571429	5.485714
2	8	8	8	6	8	7	6	85	7.285714	6.192857
3	15	13	15	14	12	15	13	85	13.85714	11.77857
4	4	4	3	3	3	3	3	110	3.285714	3.614286
5	9	8	9	8	10	7	10	110	8.714286	9.585714

for 0.85 confidence level, z score will be 0.935

Number of observations needed for a task =[ z xSD / h x Mean]²

where h is the tolerance level

for task 3 N= [ 0.935 x1.2149 / 0.07x13.857]² = 1.37 =2

hence the process does not warrant such high confidence level,since current number of observations is 7

For task 5 N = [ 0.935 x 1.1126 / 0.07x8.7142]² = 2.9

Hence the process does not warrant such high confidence level, since current number of observations is 7

Task 2 normal time = rating * mean time = 6.1928

Task 2 standard time = 6.1928 /1- fatigue allowance = 6.1928 /1-0.15 =7.2856

Task 4 normal time = 3.6142

Task 4 standard time = 3,6142 / 1- 0.15 =4.252

Normal time for complete task = 36.657

Standard time for complete task = 36.657 /1-0.15 = 43.1258