3 value: 12.00 points Problem 10-5 Using samples of 194 credit card statements,

Question

3 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 1 2 3 4 5 4 6 10 a. Determine the fraction defective in each sample. (Round your answers to 4 decimal places.) Sample Fraction defective 0258 0206 0309 0515 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 % 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.)

Explanation / Answer

a) Fraction defective = Number of defects / Sample size

b) True fraction defective = p? = ?np / (n*k) = 25/(4*194) = 0.0322 or 3.2 %

c) Estimate of mean, p? = ?np / (n*k) = 0.0322  

Standard deviation, ? = ?[ p? (1-p?)/n] = 0.0127

d) Z = NORMSINV(1-0.03/2) = 2.17

Lower limit, LCL = p? - z*? = 0.0322 - 2.17*0.0127 = 0.0046

Upper limit, UCL = p? + z*? = 0.0322 + 2.17*0.0127 = 0.0597

e)

For LCL = 0.0174 , z = ( p? - LCL)/? = (0.0322 - 0.0174)/0.0127 = 1.17 (rounded off)

For UCL = 0.0470 , z = (UCL - p?)/? = (0.0470 - 0.0322)/0.0127 = 1.17 (rounded off)

Alpha risk = 2*(1 - NORMSDIST(z)) = 2*(1 - NORMSDIST(1.17)) = 0.2420

f) Observation 4 (0.0515) lies outside the UCL. Therefore, process is not in control

g) Value of mean = 0.02

Standard deviation = ?0.02*(1-0.02)/194 = 0.01

Number of Errors (np) Fraction defective (p) 5 0.0258 4 0.0206 6 0.0309 10 0.0515

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