JUST NUMBER 4 PLEASE. On Excel. Computer upgrade times (in minutes) are being ev
ID: 421195 • Letter: J
Question
JUST NUMBER 4 PLEASE. On Excel.
Computer upgrade times (in minutes) are being evaluated. Samples of five observations each have been taken, and the results are as listed. Using factors from Table 10.3, determine upper and lower control limits for mean and range charts, and decide if the process is in control. SAMPLE 9.2 80.5 79.6 78.9 80.5 79.7 78.8 78.7 79.6 79.4 79.6 80.6 80.0 81.0 80.479.7 80.4 80.5 78.4 80.4 80.3 79.4 80.8 80.0 81.0 80.1 80.8 80.6 78.8 81.1 s Using samples of 200 credit card statements, an auditor found the following: Sample Number with errors 4 2 5 a. Determine the fraction defective in each sample b. If the true fraction defective for this process is unknown, what is your estimate of it? c. What is your estimate of the mean and standard deviation of the sampling distribution of fractions defective for samples of this size? d. What control limits would give an alpha risk of .03 for this process? e. What alpha risk would control limits of .047 and.003 provide? f. Using control limits of 047 and.003, is the process in control? g. Suppose that the long-term fraction defective of the process is known to be 2 percent. What are the values of the mean and standard deviation of the sampling distribution? h. Construct a control chart for the process, assuming a fraction defective of 2 percent, using two-sigma control limits. Is the process in control? ional a test ields inconclusive results andExplanation / Answer
PLEASE FIND BELOW ANSWER TO QUESTION # 4:
Following table calculates mean and Range for each sample :
Sample
Value 1
Value 2
Value 3
Value 4
Value 5
Mean
Range
1
79.2
78.8
80
78.4
81
79.48
2.6
2
80.5
78.7
81
80.4
80.1
80.14
2.3
3
79.6
79.6
80.4
80.3
80.8
80.14
1.2
4
78.9
79.4
79.7
79.4
80.6
79.6
1.7
5
80.5
79.6
80.4
80.8
78.8
80.02
2
6
79.7
80.6
80.5
80
81.1
80.38
1.4
Accordingly ,
Xbar = Mean of Mean values = Sum of all mean values / 6 ( i.e. number of samples) = 479.76/6 = 79.96
Rbar = Average of Range values = Sum of Range values / Number of samples = 11.2/6 = 1.87
To be noted is that sample size = n = 5
Following are the value of relevant constants as derived from standard table for Xbar and Range chartfor n = 5 :
A2 = 0.577
D4= 2.114
D3 = 0
Control Limit for Xbar chart :
Upper Control Limit ( UCLx ) = Xbar + A2.Rbar = 79.96 + 0.577 x 1.87 = 79.96 + 1.078 = 81.038
Lower Control Limit ( LCLx ) = Xbar – A2.Rbar = 79.96 – 0.577x 1.87 = 79.96 – 1.078 = 78.882
Control limits for Range chart :
Upper Control Limit = UCLr = D4.Rbar = 2.114 x 1.87 = 3.953
Lower Control Limit = LCLr = 0
Since all the values Mean and range as calculated for each of 6 samples are within X bar control limit 78.882 – 81.038 and Range control limit 0 – 3.953 respectively, the process is in control
Sample
Value 1
Value 2
Value 3
Value 4
Value 5
Mean
Range
1
79.2
78.8
80
78.4
81
79.48
2.6
2
80.5
78.7
81
80.4
80.1
80.14
2.3
3
79.6
79.6
80.4
80.3
80.8
80.14
1.2
4
78.9
79.4
79.7
79.4
80.6
79.6
1.7
5
80.5
79.6
80.4
80.8
78.8
80.02
2
6
79.7
80.6
80.5
80
81.1
80.38
1.4
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