Question

Using samples of 190 credit card statements, an auditor found the following:
Use Table-A.

Sample	1	2	3	4
Number with errors	5	3	5	12

a. Determine the fraction defective in each sample. (Round your answers to 4 decimal places.)

Sample	Fraction defective
1	.0263 .0263 Correct
2	.0158 .0158 Correct
3	.0263 .0263 Correct
4	.0632 .0632 Correct

b.If the true fraction defective for this process is unknown, what is your estimate of it? (Round your answer to 1 decimal place. Omit the "%" sign in your response.)

Estimate            3.3 3.3 Correct %

c. What is your estimate of the mean and standard deviation of the sampling distribution of fractions defective for samples of this size? (Round your intermediate calculations and final answers to 4 decimal places.)

Mean	.0329 .0329 Correct
Standard deviation	.0180 .0180 Incorrect

d.What control limits would give an alpha risk of .03 for this process? (Round your intermediate calculations to 4 decimal places. Round your "z" value to 2 decimal places and other answers to 4 decimal places.)

z = 2.17, ____ to ____

e.What alpha risk would control limits of .0470 and .0188 provide? (Round your intermediate calculations to 4 decimal places. Round your "z" value to 2 decimal places and "alpha risk" value to 4 decimal places.)

z =   alpha risk =

Answer 1

a) Sample 1 fraction defective =5/190 = 0.0263

Sample 2 fraction defective = 3/190 = 0.0158

Sample 3 fraction defective = 5/190 =0.0263

Sample 4 fraction defective = 12/190 = 0.0632

b) True sample defective = (0.0158+0.0263 + 0.0263+0.0632)*100/ 4 = 3.29 %

c)Mean = Mean = (0.0158+0.0263 + 0.0263+0.0632)/4 = 0.0329

SD=

Content	Statements Or observation	Difference From mean	Square of difference obtained
1	0.0158	-0.0171	0.00029241
2	0.0263	-0.0066	0.00004356
3	0.0263	-0.0066	0.00004356
4	0.0632	0.0303	0.00091809
Mean	0.0329		0.000324405

Now,

SD = ✓0.000324405 =0.0180

d) Alpha risk = 0.03, Z-Value from table = 2.17
Control limit = Mean +/- Z*Std. dev = 0.0329 + 2.17 * 0.0180 to 0.0329 -2.17 * 0.0180 = 0.0720 to 0 (value is negative)

e) Determine value z using formula Upper limits = mean value + z*(alpha)0.0470=0.0329+z*0.0180

Z=0.0470-0.0329/0.0180 = 0.78

Now, using normsdist, z= 0.78 it would be 0.78230

0.78230=1-Alpha/0.78

Alpha Risk = (1-0.78230)*0.78 = 0.1698