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jgosset2 jgosset2
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7 years ago

Using samples of 194 credit card statements, an auditor found the following:
 
  Sample:                   1      2      3      4
  Number with errors:   4   6   5   8

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:_____   
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 "z" value to 2 decimal places and other answers to 4 decimal places.)
 
Z:_____   
Lower limit:_____
Upper limit:_____
 

e. What alpha risk would control limits of .0122 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.)
 
z =
alpha risk =
 
 
f. Using control limits of .0122 and .0470, is the process in control? Explain why.
 
yes or no
 
 
g. 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.)
 
Mean:_____   
Standard deviation: _____   
 

h.Construct a control chart for the process, assuming a fraction defective of 2 percent, using two-sigma control limits. Is the process in control? Explain why.
 
Yes or No
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14 Replies
Replies
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Educator
7 years ago
Hey! Welcome to the forum

Answered below Smiling Face with Open Mouth
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jgosset2 Author
wrote...
7 years ago
In part a., you gave a single fraction defective estimate, when my problem calls for a fraction defective estimate for samples 1,2,3,and 4. I felt that my answers for part a. and b. were correct since I followed the formulas I found, which is why I didn't include them in my question.
I got .0206, .0309, .0258, and .0412 for samples 1 through 4. For part b. I did (2.06+3.09+2.58+4.12) / (4) = 2.9625, which I rounded to 3.0 percent. For my mean on part c. I got .0296, which is just the answer from part b. not rounded. I got stuck at that point because I could not figure out how to get the standard deviation. None of what you have down for part a through b, aligns with what I have, so I'm not sure if you worked all them out wrong because I failed to include the parts that I had already answered or if I just did part b. wrong.
Answer accepted by topic starter
bio_manbio_man
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7 years ago
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