Homework Solutions to Chapter 36

# The average of the k sample standard deviations is σ

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each sample is called R. The average of the k sample standard deviations is σ σ . σ Xbar and σ R are the standard deviations of the distributions of the sample means and sample ranges, respectively. These distributions tend to be normal, regardless of the shape of the parent population (Shewhart's Law of Sample Statistics). So, a. σ is the standard deviation of the population while σ Xbar is the standard deviation of the distribution of sample average values b. σ is the standard deviation of the population while σ bar is the average standard deviation for the k samples of size n taken from the population c. σ Xbar , σ R and σ σ are the standard deviations of three distribution; - the standard deviation of the distribution of the k sample averages, - the standard deviation of the distribution of the k sample ranges, - and the standard deviation of the distribution of the k sample standard deviations, respectively. 27. No! USL is given by the designer. UCL is a statistical parameter. There is no relationship.

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28. The USL is 0.70 + 0.20 = 0.90. The LSL is 0.70 – 0.20 = 0.50 These are specifications on individual items. The charts reflect sample statistics and the numbers are based on “n” items. Problems: 6. For the Parent Population, the parent population mean is estimated by X and so is 0.72 the parent population standard deviation is σ’ = R bar /d 2 , and for n = 4, d 2 = 2.06 (Figure 36-13) and R bar = 0.16. σ’ = 0.16 / 2.06 = 0.078 Process Capability C p = (USL – LSL) / 6 σ’ C p = (0.9 – 0.5) / ( 6 (0.078)) = 0.855 NOTE: This value is much smaller than the 1.33 value suggested as the minimum value for good process capability.
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