# Fuzzy Evaluation Model of Process Improvement Capability with Costs Consideration
## 4 An Application Example  
In this section, we use an application example to illustrate the fuzzy testing method outlined in Section 3. Since central Taiwan is the center of the tool machines and machining industry, the method was applied to an important component produced by a machinery processing factory in central Taiwan. It is presented as relative values due to business privacy considerations. As mentioned earlier, since the value \(R\) is usually less than 1.0 (\(R<1.0\)), at this time, \(\sqrt{\delta^{2}+R\gamma^{2}}<\sqrt{\delta^{2}+\gamma^{2}}\). If the process is capable (\(C_{PM}\geq 1.0\)), then \(C_{PIM}>1.0\). Obviously, usually \(C_{PIM}>C_{PM}\). Therefore, when the process is incapable (\(C_{PM}<1.0\)), then the value \(C_{PIM}\) can be requested to exceed 1.0. If the value of the process improvement capability index is at least 1.0 (\(VL=1.0\)), it means that the potential reaches the required level. Next, this study, based on the model established in Section 3, uses the following steps to complete the fuzzy test, as displayed below:  
Step 1: Establish the null hypothesis and the alternative hypothesis as follows:  
null hypothesis \(H_{0}\):\(C_{PIM}\geq 1.0\)  
versus  
alternative hypothesis \(H_{1}\):\(C_{PIM}<1.0\)  
Step 2: According to the calculation ratio of the improvement cost (\(C_{2}\)) of the process precision and the improvement cost (\(C_{1}\)) of the process accuracy, that is,  
\(R=\frac{C_{1}}{C_{2}}=0.4\)  
Step 3: Based on the sample data with a sample size \(n=36\), we calculate the observed values of \(\delta^{*}\) and \(\gamma^{*}\), respectively, as follows:  
\(\delta_{0}^{*}=\frac{1}{36}\sum_{i=1}^{36}y_{i}=0.31\)  
and  
\(\gamma_{0}^{*}=\sqrt{\frac{1}{n-1}\sum_{i=1}^{n}\left(y_{i}-\delta_{0}^{*} \right)^{2}}=\sqrt{\frac{1}{35}\sum_{i=1}^{36}\left(y_{i}-0.31\right)^{2}}=0.26\)  
Step 4: Based on Equations (16), (17) and (26), we calculate the values of \(E_{0}\) and \(CI\) with \(\alpha=\)0.05, respectively, as follows:  
\(E_{0}(0.05)=\frac{Z_{0.05}}{\sqrt{\chi_{0.97535}^{2}}}\times\sqrt{\frac{n-1}{n}}\)  
\(=\frac{1.96}{\sqrt{53.20}}\times\sqrt{\frac{35}{36}}\)\(=0.265\)  
\(CI=[C_{1},C_{2}]\)  
\(=[\delta_{0}^{*}-E_{0}(0.05)\times\gamma_{0}^{*},\delta_{0}^{*}+E_{0}(0.05)\times \gamma_{0}^{*}]\)  
\(=[0.31-0.265\times 0.26,0.31+0.265\times 0.26]\)  
\(=[0.241,0.379]\)  
Step 5: Based on step 4 and Equation (26), we calculate the values of \(C_{PIM}^{*}\) and \(UC_{PIM}\) with \(\alpha=\)0.05, respectively, as follows:  
\(C_{PIM}^{*}=\frac{1}{3\sqrt{\left(0.31\right)^{2}+0.4\times\left(0.26\right) ^{2}}}\)  
\(=0.95\)  
Therefore, we can obtain \(UC_{PIM}(\alpha)\) for \(CI_{1}>\)0.2410, as follows:  
\(UC_{PIM}(\alpha)=\left(3\gamma_{0}^{*}\sqrt{\left(\frac{\delta_{0}^{*}}{\gamma _{0}}-E_{0}(\alpha)\right)^{2}+\frac{\left(n-1\right)R}{\gamma_{0.5-\sqrt{1- \alpha}/2n-1}^{2}}}\right)^{-1}\)  
\(=\left(3\times 0.26\sqrt{\left(\frac{0.31}{0.26}-E_{0}(\alpha)\right)^{2}+ \frac{35\times 0.4}{\gamma_{0.5-\sqrt{1-\alpha}/2n-235}^{2}}}\right)^{-1}\)  
In particular, \(UC_{PIM}\) with \(\alpha=\)0.05 is  
\(UC_{PIM}=\left(3\times 0.26\sqrt{\left(\frac{0.31}{0.26}-0.265\right)^{2}+ \frac{35\times 0.4}{20.57}}\right)^{-1}=1.053\)  
Step 6: According to the Equations (29) and (30) we calculate the value, and use \(VL\) 1.000 to draw the membership function.