Assignment 5-Chapter 10- Quality Control s Have A Nominal Time Of 80 Min Using s

Question

Assignment 5-Chapter 10- Quality Control s Have A Nominal Time Of 80 Min Using samples of 194 credit card statements, an audioor found the following: Use Tablo-A Sample Number with errors 5 3 7 10 a. Determine the fraction defective in each sample. (Round your answers to 4 decimal places. Sample defective 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 G. What is your estimate of the mean and standard deviation of the sampling distibution of fractions defective for samples of this size? (Round your intermediate calculations and final answers to 4 decimal places.) Mean Standard deviation d. What control limits would give an alpha risk of 03 for this process? (Round your intermediate calculations to 4 decimal places. Round your "" value to 2 decimal places and other answers to 4 decimal places.) Lower limit Upper limit

Explanation / Answer

a) Fraction defective:

Sample 1 = 5/194 = 0.0258

Sample 2 = 3/194 = 0.0155

Sample 3 = 7/194 = 0.0361

Sample 3 = 10/194 = 0.0515

b) True % defective = Total defectives in all four samples/(Number of samples)*100 = (5+3+7+10)/(4*194)*100 = 3.2%

c) Mean (p) = Total number of defectives/Total number of samples = (5+3+7+10)/(4*194) = 0.0322

n =194

Standard deviation (s) = sqrt(p*(1-p)/n) = sqrt(0.0322*(1-0.0322)/194) = 0.0127

d) Alpha risk = 0.03, Probability of non defective = 1-Alpha risk = 1-0.03 = 0.97

z value for 0.97 probability is 1.88 (Refer Z table)

Upper limit for fraction defective = p + z*s = 0.0322 + 1.88*0.0127 = 0.0561 (In number = 0.0561*194 = 10.88)

Lower limit for fraction defective = p - z*s = 0.0322 - 1.88*0.0127 = 0.0084 (In number = 0.0127*194 = 1.62)

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