Homework Solutions to Chapter 36

The average of the k sample standard deviations is σ

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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.
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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.
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern