value .12.00 points Problem 10-5 Using samples of 194 credit card statements, an

Question

value .12.00 points Problem 10-5 Using samples of 194 credit card statements, an auditor found the following: Use Table-A. Sample Number with errors 5 4 610 a. Determine the fraction defective in each sample. (Round your answers to 4 decimal places.) Sample Fraction defective 0258 0206 0309 0515 2 4 b. If the true fraction defective for this process is unknown, what is your estimate of it? (Enter your answer as a percentage rounded to 1 decimal place. Omit the "%" sign in your response.) Estimate 3.2

Explanation / Answer

a) Fraction Defective in each sample = No of defects in sample / Number of Credit card in sample = Nf of defective credit card/ 194

e.g For Sample 1, Fraction Defective = 5/194 =  0.0258

b) fraction defect of process, p bar = Total Defective credit card/ Total Credit card = Sum of defects /Sum of Card in sample = 25/194*4 =25/776 = 3.2%

c) Mean is as of same as calculated above , pbar =25/776 = 0.0322

Standard deviation = Sqrt(Pbar*(1-Pbar)/n) =Sqrt(0.0322*(1-0.0322)/194) = 0.0126

d) alpha = 0.03

P(p bar) = 1-0.03 = 0.97

Looking at Z table

Z = 1.88

UCL = Mean+Z*Standard Deviation

LCL = Mean - z*Standanrd Deviation

UCL = 0.0322+1.88*0.0126 =0.0560

LCL = 0.0322-1.88*0.0126 = 0.00837

e) LCL = 0.0174 = Mean - Z*Std. = 0.0322 -z*0.0126

=> z=1.17 Seeing Z table P (<1.17) = 0.8790

alpha = 1- 0.8790 = 0.121

f) if p value of any sample falls outside of UCL and LCL it is unstable

Sample 4 is outside of UCL , 0.0470 and hence proces is not in control

g) p bar = 0.02

Mean = 0.02

Std =  Sqrt(Pbar*(1-Pbar)/n) = sqrt(0.02*0.98/194) = 0.0100

h)

UCL = Mean+Z*Standard Deviation = 0.02+2*0.0100 = 0.0401

LCL = Mean - z*Standanrd Deviation =0.02-2*0.0100= 0

Sample 4 with Fraction = 0.0515 falls outside UCL

so process not control

Sample No. Of error Fraction defective in each sample 1 5 0.0258 2 4 0.0206 3 6 0.0309 4 10 0.0515

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