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Using samples of 197 credit card statements, an auditor found the following: Det

ID: 353915 • Letter: U

Question

Using samples of 197 credit card statements, an auditor found the following:

  

  

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

  

  

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.)

   

  

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.)

  

  

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.)


What alpha risk would control limits of .0038 and .0470 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.)

  

  

Using control limits of .0038 and .0470, is the process in control?

  

  

Suppose that the long-term fraction defective of the process is known to be 2 percent. What are the values of the mean and standard deviation of the sampling distribution? (Round your intermediate calculations and final answers to 2 decimal places.)

  

  

Construct a control chart for the process, assuming a fraction defective of 2 percent, using two-sigma control limits. Is the process in control?

  

Using samples of 197 credit card statements, an auditor found the following:

Use Table-A.

Explanation / Answer

a)

Fraction defective in each sample = Number with errors / 197 sample size

b) Estimate of true fraction defectivew = AVERAGE(0.0152, 0.0152, 0.0254, 0.0457) = 0.0254

c) Mean = 0.0254

Standard deviation of Sampling distribution = STDEV(0.0152, 0.0152, 0.0254, 0.0457) = 0.01436

d) For alpha risk of 0.03,

Z = NORMSDIST(1-0.03) = 0.834

Lower limit = Mean - Z*Std dev = 0.0254 - 0.834*0.01436 = 0.0134

Upper limit = Mean + Z*Std dev = 0.0254 + 0.834*0.01436 = 0.0374

Sample Number with Errors Fraction Defective 1 3 0.0152 2 3 0.0152 3 5 0.0254 4 9 0.0457
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