Question

Using samples of 200 credit card statements, an Auditor found the following:

---Sample Number: 1    2     3     4     5     6     7   8 9    10    11   12   13   14   15

Number of Errors:    17  12   11   12   10    10   17   15   12   13    15    10     7     8      10

Determine lower and upper control limits for the fraction of errors using three-sigma limits.

Question 8 options:

	0 , .126
	.017 , .101
	.026 , .093
	.009 , .109
	None of the Above

The Quality Control Manager at a Credit Card Collection Center is concerned about how many complaints the company is getting from customers. The following data were collected for 15 days of operation:

-------Sample Number: 1     2     3     4    5    6    7    8    9    10    11    12    13    14    15

Number of Complaints: 10   4     1     0     3    3     4     5     10   10    0      8       2       0       7

Determine lower and upper control limits for the number of complaints they should expect each day using three-sigma limits.

Question 9 options:

	0 , 10.807
	.119 , 9.75
	.239 , 8.693
	0 , 12.92
	None of the Above

Answer 1

Total Number of Errors = 17+12+11+12+10+10+17+15+12+13+15+10+7+8+10 = 179

Total Number of Sample = 15*200 = 3000

p-bar = 179/3000 = 0.059667

Z = 3

UCLp = p-bar + Z*(p-bar*(1-p-bar)/n)^(1/2)

UCLp = 0.059667 + 3*(0.059667*(1-0.059667)/200)^(1/2)

UCLp = 0.10996

LCLp = p-bar - Z*(p-bar*(1-p-bar)/n)^(1/2)

LCLp = 0.059667 - 3*(0.059667*(1-0.059667)/200)^(1/2)

LCLp = 0.00942

Correct option will be .009 , .109

____________________________________________________________________________________________

C-bar = (10+4+1+0+3+3+4+5+10+10+0+8+2+0+7)/15 = 4.4667

UCLc = C-bar + 3*(C-bar)^(1/2)

UCLc = 4.4667 + 3*(4.4667)^(1/2)

UCLc = 10.80707

UCLc = 10.807

LCLc = C-bar - 3*(C-bar)^(1/2)

LCLc = 4.4667 - 3*(4.4667)^(1/2)

LCLc = -1.87367

LCLc = 0.00