1.1 Descriptive statistics

Used to describe quality characteristics and relationships. This includes statistics such as the mean,standard,range and measures of distribution of data.

1.2 Statistical process control

Involves inspecting a random sample of the output from a process and deciding whether the process is producing products with characteristics that fall within a predetermined range.

1.3 Acceptance sampling

This is a process of randomly inspecting a sample of goods and deciding whether to accept or reject entire lot based on the results.

All three of these statistical quality control categories are helpful in measuring and evaluating the quality of products or services. However, statistical process control (SPC) tools are used most frequently because they identify quality problems during the production process. For this reason, we will devote most of the chapter to this category of tools. The quality control tools we will be learning about do not only measure the value of a quality characteristic. They also help us identify a change or variation in some quality characteristic of the product or process. We will first see what types of variation we can observe when measuring quality. Then we will be able to identify specific tools used for measuring this variation.

1.4 Two general approaches to the management of quality

1.5 Two sources of variation

1.6 Definition of Key Terms}

1.7 Benefits of Quality Control

• Better uniformity of product quality
  
• Improved utilization of raw materials
  
• More efficient use of equipment through early detection of faults
  
• Reduction in scrap and rework
  
• Stronger producer–consumer relations

1.8 Specification, Production and Inspection

Before production starts and decision has to be made as to what is to be designed and made.Net comes the actual manufacturing of the product. Finally it must be determined whether the product manufactured is what was intended. The quality of a manufactured product must be looked at in terms of these three functions of specification, production and inspection.

1.9 Statistical Control Process

1.10 Type I and Type II Errors

2.1 Control Charts for Variable

These charts may be applied to any quality characteristic that is measurable. In order to control measurable characteristics, we have to exercise control on the measure of location as well as the measure of dispersion

2.2 Control limits

When \(\mu\) and \(\sigma\) are Unknown

Let \(X_{ij}\), $ i=1,2,…n $ and $ j=1,2,..,k $ be the measurement on the $ i^ {th} $ sample.The sample mean$ X_i$, the range \(R_i\) and the standard deviation \(S_i\) for the $ i^ {th} $ sample are given by \[ \overline{X_i} = \sum_{i=1}^{n} \frac{X_{ij}}{n} \] \[R_i=max X_{ij}-min X_{ij} \] \[ S_i^2 = \frac{1}{n-1}\sum_{j=1}^{n} {(X_{ij}-\overline{X_i.})^2} \]

The average of sample means, sample ranges and sample standard deviations are respectively.

\[ \overline{\overline{X_i}} = \frac{1}{k}\sum_{i=1}^{k} {\overline{X_i}} \]

\[ \overline{R} = \frac{1}{k}\sum_{i=1}^{k} {R_i} \]

\[ \overline{S} = \frac{1}{k}\sum_{i=1}^{k} {S_i} \]

In practice , however \(\sigma\) is estimated by using sample ranges and sample standard deviation respectively.An unbiased estimate of the population mean \(\mu\) is X-double bar while an unbiased estimated of standard deviation using the sample range is

\(\hat{\sigma} = \dfrac{\overline{R}}{d_2}\)

where \({d_2}\) is a constant depending on the sample size n.Using these estimate of the population mean\(|mu\) and standard deviation \(\sigma\), the \(3-\sigma\) control limits are given by \[\overline{\overline{X}}\pm \frac{\overline{R}}{d_{2}\sqrt{n}}\]

\[\overline{\overline{X}} \pm A_{2}\overline{R}\]

where \(A_2=\frac{3}{d_{2}\sqrt{n}}\) is also a constant depending on the sample sizes. we have tables which give the values of $ A_{2}$ for various sample sizes.

Example

To determine whether a machine manufacturing springs is in control. 25 samples of size 4 were taken after every 2 hours and their strength measured. The measurement of the 25 samples are shown below by setting control units using ranges. Determine whether the process is in control.

Solution

\[ \overline{\overline{X_i}} = \frac{1}{k}\sum_{i=1}^{k} {\overline{X_i}}=\frac{1}{25}\sum_{i=1}^{25} {\overline{X_i}}=500.38 \]

\[ \overline{R} = \frac{1}{k}\sum_{i=1}^{k} {R_i}= \frac{1}{25}\sum_{i=1}^{25} {R_i}=3.934 \]

The control limit are given by

\[\overline{\overline{X}} \pm A_{2}\overline{R}\]

For \(n=4\), we have \(A_{2} = 0.739\) Using tables.

\[UCL={500.380+0.73{(3.934)}}=503.25\] \[CL={\mu}=500.3802.5\] \[LCL={500.380-0.73{(3.934)}}=497.51\]

Conclusion

All the means fall within the control limits. we conclude that the process is in control.

2.3 The Range Chart( R-Chart)

It is known that the sample range is related to the process standard deviation. Therefore process variability may be controlled by plotting values of R from successive samples on a control chart. This control chart will be called R-chart

For small samples n<10 the range is a reasonable measure of dispersion but it is inefficient if the sample are large \((n\geq 10).\)

2.4 Control chart for Standard deviation (\(\sigma-\) Chart)

\(\sigma-\) Known

In sampling from a normal distribution the control limits for $ $ charts are given by

\[UCL_{\sigma}=B_2{\sigma}\] \[CL_{\sigma}=\sigma \] \[LCL_{\sigma}=B_1{\sigma}\]

\(\sigma-\)- Unknown

\(\hat{\sigma}\) is estimated by \[\hat{\sigma}=\frac{\overline{S}}{C_2}\]

The control limits are given by

\[UCL_{\overline{S}}=B_4{\overline{S}}\] \[CL_{\overline{S}}=\overline{S}\] \[LCL_{\overline{S}}=B_3{\overline{S}}\]

NOTE

When the sample size n is moderately large, say \(n>10 or 12\), the range method for estimating \(\sigma\) loses statistical efficiency. In these cases, it is best to replace the usual \(\overline{X}\) and \(R\) charts where the process standard deviation is estimated directly instead of indirectly through the use of \(R\). Therefore for the control purpose, from each subgroup, we must calculate the sample means $ $ and the sample standard deviation \(S\) ie.

\(\sigma-Unknown\)

If \(n\leq 10\) , use \(\overline{X}\) and \(R- charts\)

If $ n> 10$ , use \(\overline{X}\) and $ S- charts$

Example

Sample of size \(n=6\) are collected from a process every half hour. After 50 samples have been collected, we calculated $=20.0 $ and $ =1.5 $. Assuming that both charts exhibit control and that the quality characteristics is normally distributed.

Required

• Estimate the process standard deviation

• Find the control limits on the \(\overline{X}\) and \(S\) charts

• Suppose an \(R\) chart were to be substituted for the \(S\) charts, what would be the appropriate parameter of the \(R\) charts

Solution

\[\hat{\sigma}=\frac{\overline{S}}{C_2}=\frac{1.5}{0.8686}=1.7269\]

The control limits of the \(\overline{X}\) charts are given by \[UCL_{\overline{X}}={20.0+1.41{(1.5)}}=22.115\] \[CL={\overline{\overline{X}}}=20.0\] \[LCL_{\overline{X}}={20.0-1.41{(1.5)}}=17.885\]

For the \(S\) chart control limits are given by

\[UCL_{\overline{S}}=B_4{\overline{S}}=1.97(1.5)=2.955\] \[CL_{\overline{S}}=\overline{S}=1.5\] \[LCL_{\overline{S}}=B_3{\overline{S}}=0.03(1.5)=0.045\]

We known that \(\hat{\sigma}=\frac{\overline{R}}{d_2}\)

\[UCL_R=D_4\overline{R}=2.003(4.38)=8.7731\] \[CL_R=\overline{R}=4.38\] \[LCL_R=D_3\overline{R}=0(4.38)=0\]

4.1 The control charts for fraction defective (P-charts)

Suppose that a production process is producing items such that the proportion of defective items in the population of manufactured item is \(p\). Then the process will be said to be under control if the proportion of defective items in samples is the same as the population proportion \(p\), so \(p\) is also the probability that a manufactured item is defective.

Let \(X\) be a random variable such that

\[X=\begin{Bmatrix} 1 ,& defective with probability ,p\\ 0 ,& Otherwise \end{Bmatrix}\]

The the p.d.f of \(X\) is

\[f(x)=\begin{Bmatrix} p^x(1-p)^{1-x} ,& x=0,1\\ 0 ,& Otherwise \end{Bmatrix}\]

This is Bernoulli distribution with \(E(p)=p\) and \(var(p)=pq\).

Let \(X_1, X_2, ... ,X_n\) be n independent observation from the production process, then the best estimate of \(p\) is given by

\[\hat{p}=\frac{\sum_{i=1}^{n}x_i}{n}\]

Let \(D\) be the number of defective items in a sample of size \(n\), then

\[d=\sum_{i=1}^{n}x_i=n\hat{p}\]

The random variable \(D\) has the binomial distribution with parameters \(n\) and \(p\).

\[f(d)=\begin{Bmatrix} \binom{n}{d}p^d(1-p)^{n-d} ,& d=0,1,2,...,n\\ 0 ,& Otherwise \end{Bmatrix}\]

\(E(D)=np\); \(Var(D)=npq\) ;\(\hat{p}=\frac{D}{n}\)

\(E(\hat{p})=E(\frac{D}{n})=\frac{1}{n}E(D)=\frac{1}{n}*np=p\)

\(Var(\hat{p})=Var(\frac{D}{n})=\frac{1}{n^2}Var(D)=\frac{1}{n^2}npq=\frac{pq}{n}\)

The \(3-\sigma\) control limit are given by

\(E(p)\pm 3S.E(\hat{p}=p\pm3\sqrt{\frac{pq}{n}})\)

The control limits when \(p\) is known are

\[ UCL=p+3\sqrt{\frac{pq}{n}}\] \[ CL=p\] \[ LCL=p-3\sqrt{\frac{pq}{n}}\]

The Control limits when \(p\) is not known

We take m samples each of size \(n\) and for each sample we find the number of defective items

The estimate of \(p\) is given by

\[\overline{p}=\frac{\sum_{i=1}^{m}\hat{p_i}}{m}\]

The control limits are given by;

\[ UCL=\overline{p}+3\sqrt{\frac{{\overline{p}}({1-\overline{p}})}{n}}\] \[ CL=\overline{p}\] \[ LCL=\overline{p}-3\sqrt{\frac{{\overline{p}}({1-\overline{p}})}{n}}\]

Example

The sample fraction defective for 27 samples of size 50 are given below

Required

Calculate the control limit for the \(p-\) chart

Solution

\[\overline{p}=\frac{\sum_{i=1}^{27}\hat{p_i}}{2}=0.2081\]

The control limits are given by;

\[ UCL=\overline{p}+3\sqrt{\frac{{\overline{p}}({1-\overline{p}})}{n}}=0.2081+3\frac{\sqrt{0.2081(1-0.2081)}}{50}=0.3804\] \[ CL=\overline{p}=0.2081\] \[ LCL=\overline{p}-3\sqrt{\frac{{\overline{p}}({1-\overline{p}})}{n}}=0.2081+3\frac{\sqrt{0.2081(1-0.2081)}}{50}=0.0358\]

4.2 Control Chart for numbers of defectives \(np\) or \(d\) chart

An \(np\) or\(d\) chart shows the actual number of defectives found in each sample. If \(p\) is the fraction defective, then the \(3-\sigma\) control limits for the \(d\) charts are given by \[ UCL=np+3\sqrt{np(1-p)}\] \[ CL=np\] \[ LCL=np-3\sqrt{np(1-p)}\]

If \(p\) is not known then, we use the mean fraction defective \(\overline{p}\) given by \[\overline{p}=\frac{\sum_{i=1}^{m}\hat{p_i}}{m}\]

The control limits in this chart are given by; \[ UCL=n\overline{p}+3\sqrt{n\overline{p}(1-\overline{p})}\] \[ CL=n\overline{p}\] \[ UCL=n\overline{p}+3\sqrt{n\overline{p}(1-\overline{p})}\]

Remarks on \(p\) and \(d\) -charts

• If the number of items inspected on each occasion is the same, then it is in material if one uses the \(p\) -chart of the \(d\) -chart. But it would be more convenient and meaningful to use the \(d\) -chart than the \(p\) -chart

• If the sample size varies from sample to sample then the central line as well as the control limits would also vary from sample to sample. In this case the \(p\)-chart is relatively simple and is preferred to the \(d\) -chart.

Example

control chart is used to control the fraction non-conforming of a process. 10 subgroups yield the following results

Required

• Set up a control chart for the number non-conforming in samples of size \(n=100\). If any point plots outside the control limits assume the assignable causes can be found and recalculate the control limit

• For the chart established in part(1), what is the probability of detecting a shift in the process fraction nonconforming to 0.30 on the first sample after the shift has occurred.

Solution

\[\overline{p}=\frac{\sum_{i=1}^{10}\hat{d_i}}{10}=\frac{210}{1000}=0.21\]

The control limits in this chart are given by; \[ UCL=n\overline{p}+3\sqrt{n\overline{p}(1-\overline{p})}=100(0.21)+3\sqrt{100(0.21)(1-0.21)}=33.21\] \[ CL=100(0.21)=21\] \[ LCL=n\overline{p}-3\sqrt{n\overline{p}(1-\overline{p})}=100(0.21)-3\sqrt{100(0.21)(1-0.21)}=8.79\]

We drop sample number 3,6 and 7 which plot outside the control limits.

\[\overline{p}=\frac{210-35-5-45}{7*100}=\frac{125}{700}=0.18\]

\[UCL=100(0.18)+3\sqrt{100(0.18)(1-0.18)}=29.53\] \[CL=100(0.18)=18\] \[LCL=100(0.18)-3\sqrt{100(0.18)(1-0.18)}=6.47\]

Sample number 5 with 30 defective item is outside the control limits dropping this, we find that

\[\overline{p}=\frac{125-30}{6*100}=\frac{95}{600}=0.16\]

\[UCL=100(0.16)+3\sqrt{100(0.16)(1-0.16)}=27\] \[CL=100(0.16)=16\] \[LCL=100(0.16)-3\sqrt{100(0.16)(1-0.16)}=5\]

Conclusion

All the remaining points are within the control limits. We adopt these limits for future control.

Probability of not detecting a shift

\[P(5\leq D\leq 27/p)=0.30\]

\[P\left[\frac{(5-np)}{\sqrt{npq}}\leq \frac{(D-np)}{\sqrt{npq}}\leq \frac{(27-np)}{\sqrt{npq}}\right]\]

\[P\left[\frac{(27-30)}{\sqrt{21}}\leq Z\leq \frac{(27-30)}{\sqrt{21}}\right]\]

\[P\left[\frac{-25}{\sqrt{4.58}}\leq Z\leq \frac{-3}{\sqrt{4.58}}\right]\]

\[P(-5.45\leq Z\leq -0.66)=0.2576\]

Therefore, the probability of detecting on the first subsequent sample is \(1-0.2546=0.7454\)

4.3 The Control Chart for non-Conformity

The \(\overline{X}\) and \(R\) charts may be applied to any quality characteristics that is measurable. The \(p\) and \(d\) charts may be applied to the results of any inspection that accepts or reject individual items of product. Thus both these types of charts are broadly useful in any statistical quality control process. The control chart for non-conformity generally called the \(c\) -chart, has a much more restricted field of usefulness.

4.4 The \(u\) -chart}

For the \(c\) - chart that we have discussed that sample size is exactly one inspection unit. The inspection unit is chosen for operational or data- collection simplicity, we want to look at cases where the sample size is not necessarily one inspection unit.

• We redefine a new inspection unit that is equal to n time the old inspection unit. If \(\overline{c}\) is the observed mean number of non-conformity’s in the original inspection, than the new control limits are given by

\[UCL=n\overline{c}+3\sqrt{n\overline{c}}\] \[CL=n\overline{c}\] \[LCL=n\overline{c}-3\sqrt{n\overline{c}}\]

• The second approach involves setting up a control chart based on the average number of non- conformities per inspection uni, if we find n total non-conformities in a sample of n inspection units, then average number of non-conformities per inspection unit is \(u=\frac{c}{n}\)

The Parameter of the control chart are

\[UCL=\overline{u}+3\sqrt{\frac{\overline{u}}{n}}\] \[CL=\overline{u}\] \[LCL=\overline{u}-3\sqrt{\frac{\overline{u}}{n}}\]

When \(\overline{u}\) represent the average number of non-conformities per inspection unit in a preliminary set of data Control limits found would be regarded as trial control limits. The chart developed this way id called the control chart for non-conformities per unit or the \(u\)- charts.

Example

A personal computer manufacturer wishes to establish a control chart for non-conformities per unit on the final line. The sample size is selected as the computers. Data on the number of non-conformities is 20 samples of 5 computers each are shown below.

Required

Estimate the average number of non-conformities per unit

Solution

\[\overline{u}=\frac{\sum_{i=1}^{20}u_i}{20}=2.27\]

The Parameter of the control chart are; \[UCL=\overline{u}+3\sqrt{\frac{\overline{u}}{n}}=1.93+3\sqrt{\frac{1.93}{5}}=3.79\] \[CL=\overline{u}=1.93\] \[LCL=\overline{u}-3\sqrt{\frac{\overline{u}}{n}}=1.93-3\sqrt{\frac{1.93}{5}}=0.07\]

Conclusion

Therefore, all the point lies within the control limits, thus we can use the above trial control for current control.

5.1 Single Sampling Plans

If the decision about accepting or rejecting a lot is taken on the basis of one sample only, the acceptance sampling plan is described as Single sampling plan.

Let \(N\) be the lot size; \(n\) the sample size; \(c\) the acceptance number,

then the single sampling plan is described as follows;

• Select a random sample of \(n\) items from the lot of size \(N\)

• Inspect all the items in the sample.Let \(X\) be the number of defective items in the sample

• If \(X\leq c\) ; Accept the lot

• If \(X> c\) ; Reject the lot

The OC curve for the single sampling plan

In a lot of incoming quality \(p\), the number of defective items is \(NP\) and the number of non- defective items is \(N-NP=N(1-p)\). The probability of getting exactly \(x\) defective item in a sample of size \(n\) from this lot is given by the hyper-geometric distribution.

\[f(x:p)=\frac{\binom{Np}{x} \binom{N(1-p)}{n-x}}{\binom{N}{n}}\]

Therefore, the probability of accepting a lot of quality \(p\) is given by

\[P_a=\sum_{x=0}^{c}\frac{\binom{Np}{x} \binom{N(1-p)}{n-x}}{\binom{N}{n}}\]

Computing \(P_a\) for various \(p\) generated the OC curve. The computation of the hyper geometric probabilities is not easy, so in practices the binomial approximation is used, that is

\[P_a=\sum_{x=0}^{c}\binom{n}{x} p^{x}q^{n-x}\]

When we use the equation above for calculating \(P_a\), we only need the parameters \(n\) and \(c\), therefore at times we say that a single sampling plan is defined by these two parameters.

Example

If the lot fraction defective is \(p=0.01\),\(n=89\) and \(c=2\), then the \(P_a\) is given by \[P_a=P(x\leq 2)=P(x=0)+P(x=1)+P(x=2)\] \[P_a=\binom{89}{0}(0.01)^{0} (0.99)^{89}+\binom{89}{1}(0.01)^{1} (0.99)^{88}+\binom{89}{2}(0.01)^{2} (0.99)^{87}=0.9397\]

For various values of \(p\) the table below gives the probability of acceptance \(P_a\) when \(n=89\) and \(c=2\)

\[f(x)=\begin{Bmatrix}\binom{89}{x}p^{x} (1-p)^{89-x} &, x=0,1,...,89\\ 0 &, Elsewhere\end{Bmatrix}\]

Plotting \(P_a\) against \(p\) gives the OC curve

In the sampling plan \(n=89\), \(c=2\), if the lots are one percent defective the probability of acceptance is approximately 0.94. This means that if 100 lots from a process that manufactures one percent defective product are submitted to this sampling plan, we would expect to accept 94 of the lots and rejects 6 of them.

The ideal OC curve can be approached by increasing the sample size and keeping the acceptance number \(c\) proportional to \(n\).

5.2 Double Sampling Plan

If the decision is to accept or reject a lot on the basis of two samples plan is described as double sampling plan. It is defined by four parameters

• \(n_1=\) Sample size on first sample

• \(c_1=\) Acceptance number on first sample

• \(n_2=\) Sample size on second sample

• \(c_2=\) Acceptance number on second sample

Thus a random sample of size n, items is selected from the lot and the number of defectives in the sample\(X_1\)

• If \(X_1\leq c_1\), the lot is accepted on the first sample

• If \(X_1 > c_1\), the lot is rejected on the first sample

• If \(c_1<X_1< c_2\), a second sample of size \(n_2\) is drawn from the remaining items in the lot, and the number of defectives in the sample \(X_2\) observed.

• If \(X_1+X_2\leq c_2\), the lot is accepted

• If \(X_1+X_2 > c_2\), the lot is rejected

The OC curve of the double Sampling plan

For a double sampling plan the lot will be accepted if and only if

• \(X_1 \leq c_1\)

• \(c_1 < X_1 \leq c_2\) and \(X_1 + X_2 \leq c_2\)

For any incoming quality \(p\) the probability of acceptance is given by

\[P_a=P(X_1 \leq c_1)+P(X_1 + X_2 \leq c_2)\]

Let \(f_1(x_1)\) be the probability of getting \(X_1\) defective item in the first sample , then

\[f_1(x_1)=\begin{Bmatrix}\binom{n_1}{x_1}p^{x_1} (1-p)^{n_1-x_1} &, x_1=0,1,...,n_1\\ 0 &, Elsewhere\end{Bmatrix}\]

If \(f_2(x_2 /x_1\), denotes the conditional probability of finding \(X_2\) defective in the second sample given that \(X_1\) defectives have already been found in the first sample, then

\[f_2(x_2/x_1)=\binom{n_2}{x_2}p^{x_2} (1-p)^{n_2-x_2}=f_2(x_2)\]

Let \(P_{a}^{'}\) denote the probability of acceptance on the first sample and \(P_{a}^{''}\) denote the probability of acceptance on the second sample, then

\[P_a=P_{a}^{'}+P_{a}^{''}\]

The joint p.d.f of \(X_1\) and \(X_2\) is given by

\[f(x_1,x_2)=f_1(x_1)*f_2(x_2/x_1)=\binom{n_1}{x_1}p^{x_1} (1-p)^{n_1-x_1} * \binom{n_2}{x_2}p^{x_2} (1-p)^{n_2-x_2}\]

\(P_{a}^{'}=P(X_1 \leq c_1)=\sum_{x_{1}=0}^{c_1}f_1(x_1)\)

\(P_{a}^{''}= P(c_1 < X_1 \leq c_2\) and \(X_1 + X_2 \leq c_2)\)

\[P_{a}^{''}= \sum_{(x_1,x_2)c_1 < X_1 \leq c_2 and X_1 + X_2 \leq c_2}f(x_1,x_2)\]

Example

For a double sampling plan with \(n_1=50;c_1=2;n_2=100\) and \(c_2=6\). Find the probability of acceptance \(P_a\). if the incoming lots have fraction defective is 0.05.

Solution

\[P_a=P_{a}^{'}+P_{a}^{''}\]

We accept on the first sample if \(X_1\leq 2\)

\(P_{a}^{'}=P(X_1 \leq 2)=\sum_{x_{1}=0}^{2}f_1(x_1)=\sum_{x_{1}=0}^{2}\binom{50}{x_1}(0.05)^{x_1} (0.95)^{50-x_1}\)

\[P_a=\binom{50}{0}(0.05)^{0} (0.95)^{50}+\binom{50}{1}(0.05)^{1} (0.95)^{49}+\binom{50}{2}(0.05)^{2} (0.95)^{48}=0.5405\]

We accept the lot on the second sample if \(2<X_1 \leq 6\) and \(X_1 + X_2 \leq 6\), the pair of points \((x_1, x_2)\) satisfying these are

\[P_{a}^{''}= \sum_{(x_1,x_2)c_1 < X_1 \leq c_2 and X_1 + X_2 \leq c_2}f(x_1,x_2)\]

\[f(x_1,x_2)=f_1(x_1)*f_2(x_2/x_1)=\binom{50}{x_1}0.05^{x_1} (095)^{50-x_1} * \binom{100}{x_2}0.05^{x_2} (0.95)^{100-x_2}\]

\[f_1(3)=\binom{50}{3}(0.05)^{3} (0.95)^{47}=0.2198\] \[f_1(4)=\binom{50}{4}(0.05)^{4} (0.95)^{46}=0.1360\] \[f_1(5)=\binom{50}{5}(0.05)^{5} (0.95)^{45}=0.0658\] \[f_1(6)=\binom{50}{6}(0.05)^{6} (0.95)^{44}=0.026\]

\[f_2(0)=\binom{100}{0}(0.05)^{0} (0.95)^{100}=0.0059\] \[f_2(1)=\binom{100}{1}(0.05)^{1} (0.95)^{99}=0.0312\] \[f_2(2)=\binom{100}{2}(0.05)^{2} (0.95)^{98}=0.0812\] \[f_2(3)=\binom{100}{3}(0.05)^{3} (0.95)^{97}=0.1396\]

\[P_{a}^{''}= 0.0754\] \[P_a=P_{a}^{'}+P_{a}^{''}=0.5405+0.0754=0.6159\]

6.1 Introduction

Process capability refers to the ability of a production process to meet or exceed preset specification. A critical aspect of statistical quality control is evaluating the ability of a production process to meet or exceed preset specification. To understand what this mean, lets look more closely at the term specification.

Production specification often called tolerances, are preset ranges of acceptable quality characteristics such as product dimensions. For a product to be considered acceptable, its characteristics must fall within this preset range. Product specification are usually established by design engineers or product design specialists. For instance, the specification for the width of a machine part may be specified as 15 inches \(\pm 0.3\). This means that the width of the part should be 15 inches, though it is acceptable if it falls within the limits of 14.7 inches and 15.3 inches.

Specification for a product are preset on the basis of how the product is going to be used or what customer expectation are. As we have learned, any production process has a certain amount of natural variation associated with it. To be capable of producing an acceptable product, the process variation cannot exceed the preset specification. process capability thus involves evaluating process variability relative to preset product specification in order to determine whether the process is capable of producing an acceptable product.

Measuring Process Capability

To produce an acceptable product, the process must be capable and in control before production begins. Process capability is measured by the process capability index \(C_p\) which is computed as the ratio of the specification width to the width of the process variability.

\[C_p=\frac{USL-LSL}{6\sigma}\]

Where,

Specification width is the difference between the Upper Specification Limit (\(USL\)) and Lower Specification Limit (\(LSL\)) of the process

Process width computer as 6 standard deviation of the process being monitored

NB

The reason we use \(6\sigma\) is that most of the process measurement 99.74 percent falls within \(\pm 3\) standard deviation which is a total 6 standard deviation.

There are three possible ranges of values for \(C_p\) that also help us interpret its value.

• A value of \(C_p=1\) means that the process variability just meets specification. We would then say that the process is minimally capable.

• A value of \(C_p\leq 1\) means that the process is outside the range of specifications. This means that the process is not capable of producing within specification and the process must be improved.

• A value of \(C_p\geq 1\) means that the process variability is tighter than specifications and the process exceeds minimal capability

Example

Three bottling machine of Cococola are being evaluated for their capability

If specification are set between 15.8 and 16.2 litres, determine which of the machine are capable of producing within specification.

Solution

Comment

The results indicate that only machine A is capable of filling bottle within specification because it is the only machine that has s \(C_p\) value at or above one (1).

NB:

\(C_p\) valuable in measuring process capability. However, it has one shortcomings: It assumes that process variability is centered on the specification range. Unfortunately, This is not always the case. For instance, the specification limit are set between 15.8 and 16.2 litres, with a mean of 16 litres. However, the process variation is not centered, it has a mean of 15.9 liters. Because of this, certain proportion of products will fall outside the specification range.

The problem illustrated in the above example is not un-common, but it can lead to mistakes in the computation of the \(C_p\) measure. Because of this another measure for process capability is used more frequently.

\[C_pk=Min\left[{\frac{USL-\mu}{3\sigma}},{\frac{\mu-LSL}{3\sigma}}\right]\]

where \(\mu\) =mean of the process and \(\sigma\)= standard deviation of the process

This measure of process capability helps us address a possible lack of centering of the process over the specification range. To use this measure, the process capability of each half of the normal distribution is computed and minimum of the two is used.

Example

Compute the \(C_pk\) measure of process capability for the following machine and interpret the finding given the following information below.

\(USL=110\) ,\(LSL=50\) ,\(\mu=60\) and \(\sigma=10\)

\[C_pk=Min\left[{\frac{USL-\mu}{3\sigma}},{\frac{\mu-LSL}{3\sigma}}\right]\]

\[C_pk=Min\left[{\frac{110-60}{3\sigma}},{\frac{60-50}{3\sigma}}\right]\]

\[C_pk=Min\left[{\frac{110-60}{3(10)}},{\frac{60-50}{3(10)}}\right]\]

\[C_pk=Min\left[1.67,0.33\right]=0.33\]

This means that the process is not capable. The \(C_p\) measure of process capability gives us the following measure.

\[C_p=\frac{60}{6(10)}=1\]

Leading us to believe that the process is capable. The reason for the difference in the measures is that the process is not centered on the specification range.

Exercise

Q1.Three Ice cream packing machine at the creamy treat company are being evaluated for this capability. The following data are recorded.

If specification are set between 15.8 and 16.2 units, determine which of the machine are capable of producing within specification.

Q2. Compute the \(C_pk\) measure of process capability for the following machine and interpret the findings. What value would you have obtained with the \(C_p\) measure? Given the information below \(USL=100\) ,\(LSL=70\) ,\(\mu=80\) and \(\sigma=5\)