Question 3 of 6: OC Function, an $\bar X$-$R$ Chart for Steel-Beam Tensile Strength, and Chart Design for a 1.2$\sigma$ Shift

Question 3: OC Function, an $\bar X$-$R$ Chart for Steel-Beam Tensile Strength, and Chart Design for a 1.2$\sigma$ Shift (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

(a) The operating characteristic function of an $\bar X$ chart

The operating characteristic (OC) function of a control chart, $\beta(\delta)$, gives the probability that a single sample statistic falls inside the control limits (i.e., fails to signal) as a function of the true size of a process shift $\delta$ (in $\sigma$ units). Plotted against $\delta$, it shows how quickly — and how reliably — the chart is expected to detect shifts of different sizes: a steep OC curve that drops rapidly from near 1 to near 0 as $\delta$ grows means the chart is very sensitive (small shifts are caught almost as reliably as large ones), while a flat, slowly-declining curve means the chart is sluggish and many samples must be taken before a moderate shift is likely to be caught.

Sample size $n$ controls the steepness of the OC curve directly: because the standard error of the plotted statistic scales as $\sigma/\sqrt n$, a larger $n$ narrows the control limits in $\sigma$-of-the-mean units without changing them in raw $\sigma_{\bar X}$ terms, so a shift of a given absolute size represents a larger number of standard errors and is detected far more reliably. Increasing $n$ therefore steepens the OC curve (lower $\beta$ for every non-zero $\delta$, i.e. better detection power) at the cost of a larger, more expensive sample every period; it leaves the in-control performance essentially unchanged, since that depends only on the number of $\sigma$ used for the limits ($k=3$), not on $n$.

When the process is exactly in statistical control ($\delta=0$), the OC function equals the probability that a single sample statistic falls inside the standard $3\sigma$ limits of a normal distribution, $\beta(0)=\Phi(3)-\Phi(-3)=0.9973$ — independent of $n$, since with no shift the plotted statistic is centred exactly on the control limits' own centre line regardless of how tightly $n$ has drawn those limits in.

(b) $\bar X$ and $R$ control limits, process mean/std. dev., and fraction below 240 kg/cm²

Given. $n=5$ per sample, $k=30$ samples, $\sum\bar X_i=7518$, $\sum R_i=375$; specification $250\pm10$ kg/cm² (LSL $=240$, USL $=260$); Appendix VI factors for $n=5$: $A_2=0.577$, $D_3=0$, $D_4=2.115$, $d_2=2.326$.

Find. The $\bar X$- and $R$-chart control limits; the in-control estimates $\hat\mu_0$, $\hat\sigma_0$; and the fraction of output below 240 kg/cm².

Approach. Compute the grand average and average range from the reported sums, apply the standard $\bar X$/$R$ control-limit formulas with the $n=5$ Appendix VI factors, estimate $\sigma$ from $\overline R/d_2$, then treat the assumed-in-control process as $N(\hat\mu_0,\hat\sigma_0^2)$ to get the tail probability below the 240 kg/cm² floor.

1. Grand average and average range. $$\bar{\bar X}=\frac{\sum\bar X_i}{k}=\frac{7518}{30}=\boxed{250.6\ \text{kg/cm}^2},\qquad \overline R=\frac{\sum R_i}{k}=\frac{375}{30}=\boxed{12.5\ \text{kg/cm}^2}.$$
2. $\bar X$-chart control limits. $$UCL_{\bar X}=\bar{\bar X}+A_2\overline R=250.6+0.577(12.5)=\boxed{257.81\ \text{kg/cm}^2},$$ $$LCL_{\bar X}=250.6-0.577(12.5)=\boxed{243.39\ \text{kg/cm}^2}.$$
3. $R$-chart control limits. $$UCL_R=D_4\overline R=2.115(12.5)=\boxed{26.44\ \text{kg/cm}^2},\qquad LCL_R=D_3\overline R=0(12.5)=\boxed{0}.$$
4. In-control mean and standard deviation. With the process assumed in control, $\hat\mu_0=\bar{\bar X}=\boxed{250.6\ \text{kg/cm}^2}$ and $\hat\sigma_0=\overline R/d_2=12.5/2.326=\boxed{5.374\ \text{kg/cm}^2}$.
5. Fraction below the 240 kg/cm² floor. Treating individual output as $N(250.6,\,5.374^2)$: $$z=\frac{240-250.6}{5.374}=-1.972,\qquad P(X<240)=\Phi(-1.972)\approx1-0.9757=\boxed{0.0243\ (2.43\%)}.$$ About 2.4% of output is expected to fall below the 240 kg/cm² minimum, even with the process running in control.

(c) Designing a chart to detect a $1.2\sigma$ downward shift within the first or second sample, power $\ge0.8$

Given. $\hat\mu_0=250.6$, $\hat\sigma_0=5.374$ kg/cm² from part (b); a downward mean shift of $1.2\sigma$; standard $k=3$ control limits; target: detect on the first or second sample after the shift with probability $\ge0.8$.

Find. The minimum constant sample size $n$ meeting the target, and the resulting $\bar X$-chart control limits.

Approach. For a shift of $\delta\sigma$, the probability that a single post-shift sample fails to signal is $\beta=\Phi(k-\delta\sqrt n)-\Phi(-k-\delta\sqrt n)\approx\Phi(k-\delta\sqrt n)$ (the second term is negligible for any $n$ worth considering). Two independent samples both failing to signal has probability $\beta^2$, so detecting within the first or second sample with probability $\ge0.8$ requires $1-\beta^2\ge0.8\Leftrightarrow\beta\le\sqrt{0.2}=0.4472$. Search over integer $n$ for the smallest one satisfying this.

1. Required single-sample miss probability. $$\beta^2\le0.2\ \Rightarrow\ \beta\le\sqrt{0.2}=\boxed{0.4472}.$$
2. Search over $n$. With $\delta=1.2$, $k=3$: $\beta(n)=\Phi(3-1.2\sqrt n)$. At $n=6$: $3-1.2\sqrt6=0.0606\Rightarrow\beta=\Phi(0.0606)=0.524>0.4472$ (fails). At $n=7$: $3-1.2\sqrt7=-0.1749\Rightarrow\beta=\Phi(-0.1749)=0.431\le0.4472$ (meets the target, with $1-\beta^2=1-0.431^2=0.815\ge0.8$). $n=7$ is therefore the minimum sample size. $$\boxed{n_{\min}=7}$$
3. Control limits for the redesigned chart. $$UCL=\mu_0+3\frac{\sigma}{\sqrt n}=250.6+3\frac{5.374}{\sqrt7}=250.6+6.09=\boxed{256.69\ \text{kg/cm}^2},$$ $$LCL=250.6-6.09=\boxed{244.51\ \text{kg/cm}^2}.$$