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Chapter 16: Q23E (page 699)
When\(n = 150\), what is the smallest value of\(\bar p\)for which the LCL in a\(p\)chart is positive?
\(\bar p = \frac{3}{{53}}\)
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The target value for the diameter of a certain type of driveshaft is \(.75\) in. The size of the shift in the average diameter considered important to detect is \(.002\) in. Sample average diameters for successive groups of \(n = 4\)shafts are as follows: \(.7507,.7504,.7492,.7501,.7503\), \(.7510,.7490,.7497,.7488,.7504,.7516,.7472,.7489\), \(.7483,.7471,.7498,.7460,.7482,.7470,.7493,.7462\), \(.7481\). Use the computational form of the CUSUM procedure with \(h = .003\) to see whether the process mean remained on target throughout the time of observation.
The accompanying table gives sample means and standard deviations, each based on\(n = 6\)observations of the refractive index of fiber-optic cable. Construct a control chart, and comment on its appearance. (Hint:\(\Sigma {\bar x_i} = 2317.07\)and\(\left. {\Sigma {s_i} = 30.34.} \right)\)
\(\begin{aligned}{*{20}{l}}{Day\;\;\;\;\;\;\;bar\left( x \right)\;\;\;\;s\;\;\;\;\;\;\;\;\;\;\;\;Day\;\;\;\;\;\;\;bar\left( x \right)\;\;\;\;s}\\{1\;\;\;\;\;\;\;\;\;\;\;\;95.47\;\;\;\;1.30\;\;\;\;\;\;13\;\;\;\;\;\;\;\;\;\;97.02\;\;\;\;1.28}\\{2\;\;\;\;\;\;\;\;\;\;\;\;97.38\;\;\;\;.88\;\;\;\;\;\;\;\;14\;\;\;\;\;\;\;\;\;\;95.55\;\;\;\;1.14}\\{3\;\;\;\;\;\;\;\;\;\;\;\;96.85\;\;\;\;1.43\;\;\;\;\;\;15\;\;\;\;\;\;\;\;\;\;96.29\;\;\;\;1.37}\\{4\;\;\;\;\;\;\;\;\;\;\;\;96.64\;\;\;\;1.59\;\;\;\;\;\;16\;\;\;\;\;\;\;\;\;\;96.80\;\;\;\;1.40}\\{5\;\;\;\;\;\;\;\;\;\;\;\;96.87\;\;\;\;1.52\;\;\;\;\;\;17\;\;\;\;\;\;\;\;\;\;96.01\;\;\;\;1.58}\\{6\;\;\;\;\;\;\;\;\;\;\;\;96.52\;\;\;\;1.27\;\;\;\;\;\;18\;\;\;\;\;\;\;\;\;\;95.39\;\;\;\;.98}\\{7\;\;\;\;\;\;\;\;\;\;\;\;96.08\;\;\;\;1.16\;\;\;\;\;\;19\;\;\;\;\;\;\;\;\;\;96.58\;\;\;\;1.21}\\{8\;\;\;\;\;\;\;\;\;\;\;\;96.48\;\;\;\;.79\;\;\;\;\;\;\;\;20\;\;\;\;\;\;\;\;\;\;96.43\;\;\;\;.75}\\{9\;\;\;\;\;\;\;\;\;\;\;\;96.63\;\;\;\;1.48\;\;\;\;\;\;21\;\;\;\;\;\;\;\;\;\;97.06\;\;\;\;1.34}\\{10\;\;\;\;\;\;\;\;\;\;96.50\;\;\;\;.80\;\;\;\;\;\;\;\;22\;\;\;\;\;\;\;\;\;\;98.34\;\;\;\;1.60}\\{11\;\;\;\;\;\;\;\;\;\;97.22\;\;\;\;1.42\;\;\;\;\;\;23\;\;\;\;\;\;\;\;\;\;96.42\;\;\;\;1.22}\\{12\;\;\;\;\;\;\;\;\;\;96.55\;\;\;\;1.65\;\;\;\;\;\;24\;\;\;\;\;\;\;\;\;\;95.99\;\;\;\;1.18\;\;\;\;\;\;\;}\end{aligned}\)
The following numbers are observations on tensile strength of synthetic fabric specimens selected from a production process at equally spaced time intervals. Construct appropriate control charts, and comment (assume an assignable cause is identifiable for any out of-control observations).
\(\begin{array}{*{20}{r}}{ 1. }&{51.3}&{51.7}&{49.5}&{12.}&{49.6}&{48.4}&{50.0}\\{ 2. }&{51.0}&{50.0}&{49.3}&{13.}&{49.8}&{51.2}&{49.7}\\{ 3. }&{50.8}&{51.1}&{49.0}&{14.}&{50.4}&{49.9}&{50.7}\\{ 4. }&{50.6}&{51.1}&{49.0}&{15.}&{49.4}&{49.5}&{49.0}\\{ 5. }&{49.6}&{50.5}&{50.9}&{16.}&{50.7}&{49.0}&{50.0}\\{ 6. }&{51.3}&{52.0}&{50.3}&{17.}&{50.8}&{49.5}&{50.9}\\{ 7. }&{49.7}&{50.5}&{50.3}&{18.}&{48.5}&{50.3}&{49.3}\\{ 8. }&{51.8}&{50.3}&{50.0}&{19.}&{49.6}&{50.6}&{49.4}\\{9.}&{48.6}&{50.5}&{50.7}&{20.}&{50.9}&{49.4}&{49.7}\\{ 10. }&{49.6}&{49.8}&{50.5}&{21.}&{54.1}&{49.8}&{48.5}\\{ 11. }&{49.9}&{50.7}&{49.8}&{22.}&{50.2}&{49.6}&{51.5}\end{array}\)
When\({S^2}i\)s the sample variance of a normal random sample,\((n - 1){S^2}/{\sigma ^2}\)has a chi-squared distribution with\(n - 1df\), so
\(P\left( {\chi _{.999,n - 1}^2 < \frac{{(n - 1){S^2}}}{{{\sigma ^2}}} < \chi _{.001,n - 1}^2} \right) = .998\)
from which
\(P\left( {\frac{{{\sigma ^2}\chi _{.999,n - 1}^2}}{{n - 1}} < {S^2} < \frac{{{\sigma ^2}\chi _{.001,n - 1}^2}}{{n - 1}}} \right) = .998\)
This suggests that an alternative chart for controlling process variation involves plotting the sample variances and using the control limits
\(\begin{aligned}{l}LCL = \overline {{s^2}} {\chi _{ - 999_{ - n - 1}^2}}/(n - 1)\\UCL = \overline {{s^2}} \chi _{.001,n - 1}^2/(n - 1)\end{aligned}\)
Construct the corresponding chart for the data of Exercise\(11\). (Hint: The lower- and upper-tailed chi-squared critical values for\(5\)df are\(.210\)and\(20.515\), respectively.)
Develop a single-sample plan for which \(AQL = .02\) and LTPD \( = .07\) in the case \(\alpha = .05,\beta = .10\). Once values of \(n\) and \(c\) have been determined, calculate the achieved values of \(\alpha \) and \(\beta \) for the plan.
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