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Observations on shear strength for 26 subgroups of test spot welds, each consisting of six welds, yield \(\Sigma {\bar x_i} = 10,980,\Sigma {s_i} = 402\), and \(\Sigma {r_i} = 1074\). Calculate control limits for any relevant control charts.

Short Answer

Expert verified

Compute \(LCL\)and \(UCL\) for four different charts.

Step by step solution

01

Step 1:To Find the Corresponding of definition

There could be four total relevant charts\( - S\)chart,\(R\)chart,\(\bar X\)chart based on\(\bar s\)and\(\bar X\)chart based on\(\bar r\). See the definition of corresponding charts in the book.

\(S\)chart:

\(LCL = 15.4615 - \frac{{3 \times 15.4615\sqrt {1 - 0.95{2^2}} }}{{0.962}} = 15.4615 - 14.9141 = 0.55\)\(UCL = 15.4615 + \frac{{3 \times 15.4615\sqrt {1 - 0.95{2^2}} }}{{0.962}} = 15.4615 + 14.9141 = 30.37\)

\(R\)chart:

\(\begin{array}{l}LCL = 41.31 - \frac{{3 \times 0.848 \times 41.31}}{{2.536}} = 41.31 - 41.44\mathop = \limits^{(1)} 0\\UCL = 41.31 + \frac{{3 \times 0.848 \times 41.31}}{{2.536}} = 41.31 + 41.44 = 82.75\end{array}\)

\(\bar X\)chart based on\(\bar s\):

\(\begin{array}{l}LCL = 422.31 - \frac{{3 \cdot 15.4615}}{{0.952 \cdot \sqrt 6 }} = 402.42\\UCL = 422.31 + \frac{{3 \cdot 15.4615}}{{0.952 \cdot \sqrt 6 }} = 442.20\end{array}\)

\(\bar X\)chart based on\(\bar r\):

\(\begin{array}{l}LCL = 422.31 - \frac{{3 \cdot 41.3077}}{{2.536 \cdot \sqrt 6 }} = 402.36\\UCL = 422.31 + \frac{{3 \cdot 41.3077}}{{2.536 \cdot \sqrt 6 }} = 442.26.\end{array}\)

(1): when \( \le 0\) set it to zero.

02

Step 2:Final proof

Compute \(LCL\)and \(UCL\) for four different charts.

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Most popular questions from this chapter

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.)

The standard deviation of a certain dimension on an aircraft part is \(.005\;cm\). What CUSUM procedure will give an in-control ARL of 600 and an out-of-control ARL of 4 when the mean value of the dimension shifts by \(.004\;cm\)?

Consider the single-sample plan that utilizes\(n = 50\)and\(c = 1\)when\(N = 2000\). Determine the values of AOQ and ATI for selected values of\(p\), and graph each of these against\(p\). Also determine the value of\(AOQL\).

In the case of known 饾泹 and 饾洈, what control limits are necessary for the probability of a single point being outside the limits for an in-control process to be .005?

Consider the control chart based on control limits\({\mu _0} \pm 2.81\sigma /\sqrt n \)

a. What is the ARL when the process is in control?

b. What is the ARL when\(n = 4\)and the process mean has shifted to\(\mu = {\mu _0} + \sigma \)?

c. How do the values of parts (a) and (b) compare to the corresponding values for a\(3 - \sigma \)chart?

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