# The Fuzzy Process Quality Evaluation Model for the STB Quality Characteristic of Machining
## 2 Upper Confidence Limit  
In reality, process mean \(\mu\) and process standard deviation \(\sigma\) are usually undetected. Therefore, they must be estimated from subsamples taken when the process is thought to be in control [11]. As noted by [PERSON] [23], when \(n>10\) or 12, a process mean and process standard deviation \(\left(\overline{X}-S\right)\) control chart is often used and effectively reduces the standard deviation as well as the error of estimation. The estimation should usually be based on \(m\) subsamples, where \(m=20\) or 25. Firstly, each subsample has \(n\) observations of the STB quality characteristic; also, \(m\) subsamples are available. As a result, let \(\overline{X}_{h}\) and \(S_{h}^{2}\) respectively denote the subsample mean and subsample variance of the \(i\)th subsample below:  
\[\overline{X}_{h}=\frac{1}{n}{\sum_{j=1}^{n}{X_{hj}}} \tag{4}\]  
and  
\[S_{h}^{2}=\frac{1}{n-1}{\sum_{j=1}^{n}{\left({X_{hj}-\overline{X}_{h}}\right) ^{2}}}. \tag{5}\]  
Let \(N\) (\(N=m\times n\)) denote the total number of observations. We can use the overall sample mean and the pooled sample variance, which are the unbiased estimators, as follows:  
\[\overline{\bar{X}}=\frac{1}{m}{\sum_{h=1}^{m}{X_{h}}} \tag{6}\]  
and  
\[S^{2}=\frac{1}{m(n-1)}{\sum_{h=1}^{m}{(n-1)S_{h}^{2}}}=\frac{1}{m}{\sum_{h=1} ^{m}{S_{h}^{2}}}. \tag{7}\]  
Therefore, the estimator of \(Q_{\text{ISS}}\) can be shown as follows:  
\[\hat{Q}_{\text{ISS}}=\frac{USL-\frac{\overline{\bar{X}}}{X}}{S}. \tag{8}\]  
Under the assumption of normality, let  
\[K=\frac{m(n-1)S^{2}}{\sigma^{2}}=\sum_{h=1}^{m}\frac{(n-1)S_{h}^{2}}{\sigma^{ 2}}=\sum_{h=1}^{m}{K_{h}} \tag{9}\]  
where \(K_{h}\) represents the chi-square distribution with \(n-1\) degrees of freedom, and the characteristic function of \(K_{h}\) is \(\phi_{K_{h}}(t){=}(1-2 it)^{-(n-1)/2}\). Therefore, the characteristic function of \(K\) is  
\[ \begin{array}{c}\phi_{K}(t)=E\big{[}e^{itK}\big{]}=E\big{[}e^{it\Sigma_{h}^ {m}K_{h}}\big{]}=E\big{[}\prod_{h=1}^{m}e^{itK_{h}}\big{]}\\ \qquad=\prod_{h=1}^{m}\phi_{K_{h}}(t)=\prod_{h=1}^{m}(1-2 it)^{-(n-1)/2}\\ \qquad=(1-2 it)^{-m(n-1)/2}\end{array}  \tag{10}\]  
and the distribution of \(K\) is the chi-square distribution with \(m(n-1)\) degrees of freedom (i.e., \(\lambda_{m(n-1)}^{2}\)). Furthermore, the probability of \(K\) smaller than \(\chi_{1-\alpha/2,m(n-1)}^{2}\) is \(1{-}\alpha/2\) and equivalent to  
\[P\left(\frac{S}{\sigma}\leq\sqrt{\frac{\chi_{1-\alpha/2,m(n-1)}^{2}}{m(n-1)}} \right)=1-\frac{\alpha}{2} \tag{11}\]where \(\chi^{2}_{1-\alpha/2,m(n-1)}\) is the lower \(1-\alpha/2\) quantile of the chi-square distribution with \(m(n-1)\) degrees of freedom. Next, to obtain the upper confidence limit from quality index \(Q_{\text{ISS}}\), we let  
\[Z=\sqrt{mn}\chi\Big{[}Q_{\text{ISS}}-\hat{Q}_{\text{ISS}}\Big{(}\frac{S}{\sigma }\Big{)}\Big{]}=\sqrt{mn}\chi\Bigg{[}\frac{\overline{\hat{X}}-\mu}{\sigma} \Bigg{]} \tag{12}\]  
then \(Z\) is distributed as a standardized normal distribution (i.e., \(Z\)-\(N(0,1)\)). Therefore,  
\[p\Big{(}\sqrt{N}\Big{(}Q_{\text{ISS}}-\hat{Q}_{\text{ISS}}\times\frac{S}{\sigma }\Big{)}\leq Z_{\alpha/2}\Big{)}=1-\frac{\alpha}{2}. \tag{13}\]  
Equivalently,  
\[P\Big{(}Q_{\text{ISS}}\leq\hat{Q}_{\text{ISS}}\times\frac{S}{\sigma}+\frac{Z _{\alpha/2}}{\sqrt{mn}}\Big{)}=1-\frac{\alpha}{2} \tag{14}\]  
where \(Z_{\alpha/2}\) is the upper \(\alpha/2\) quantile of the standardized distribution. Both event A and event B are defined as follows:  
\[\text{A}=\Bigg{\{}\frac{S}{\sigma}\leq\sqrt{\frac{\chi^{2}_{1-\alpha/2,m(n-1 )}}{m(n-1)}}\Bigg{\}} \tag{15}\]  
and  
\[\text{B}=\Bigg{\{}Q_{ISS}\leq\hat{Q}_{ISS}\times\frac{S}{\sigma}+\frac{Z_{ \alpha/2}}{\sqrt{mn}}\Bigg{\}}. \tag{16}\]  
Obviously, the probability for both event A and event B is equal to \(1-\alpha/2\).