# Statistical Hypothesis Testing for Asymmetric Tolerance Index
## 4 An Illustrative Example  
The quality characteristics of many machinery products have asymmetric tolerances [1; 2; 3]. Say, for instance, that the asymmetric tolerance of a quality characteristic for a machine part is \((LSL,T,USL)=(4.03,4.05,4.10)\). This gives us the following parameters:  
upper tolerance \(D_{u}=USL-T=4.10-4.05=0.05\),  
lower tolerance \(D_{l}=T-LSL=4.05-4.03=0.02\),  
\[d=Min\{D_{u},D_{l}\}=0.02,\]  
\[d_{u}=\frac{d}{D_{u}}=0.4,\]  
\[d_{l}=\frac{d}{D_{l}}=1.0.\]  
Figure 2: Flowchart of statistical hypothesis testing.  
According to the evaluation procedure provided in Section 3, we perform the following steps:  
Step 1: Based on the capability level required by customers, we calculate the following:  
Required capability: \(C=1.0\),  
Significance level: \(\alpha=0.05\),  
Sample size: \(n=36\),  
Null hypothesis: \(H_{0}\):\(C_{pk}^{\prime\prime}=1.0\),  
Alternative hypothesis: \(H_{1}\):\(C_{pk}^{\prime\prime}\
eq 1.0\).  
Step 2: We then calculate the value of sample mean \(\overline{x}\), sample standard deviation \(s\), and \(e_{t0}\) with sample size \(n=36\), as follows:  
\[\overline{x}=\frac{1}{36}{\sum\limits_{i=1}^{36}{x_{i}}}=4.07,\]  
\[s=\sqrt{\frac{1}{36}{\sum\limits_{i=1}^{36}{(x_{i}-\overline{x})}^{2}}}=0.005,\]  
\[e_{t0}=t_{\alpha/4\mu-1}{\times}{\left({\frac{s}{\sqrt{n}}}\right)}=t_{0.025 35}\times{\left({\frac{0.005}{\sqrt{36}}}\right)}=\left({\frac{0.005}{\sqrt{36 }}}\right)=0.0017.\]  
Obviously, the estimated error item is less than or equal to 0.0017 when \(n=36\) (\(|\overline{x}-\mu|\leq 0.0017\)). If \(e_{t0}\) is required to be smaller than 0.0017, the sample size of \(n\) needs to be larger. The value of \(n\) can be calculated by the inequality as shown below:  
\[n\geq{\left({\frac{s}{e_{t0}}t_{\alpha/4\pi-1}}\right)}^{2}.\]  
Step 3: Next, we calculate the values of \([\overline{x}-e_{t0},\overline{x}+e_{t0}]\) as follows:  
\[[\overline{x}-e_{t0},\overline{x}+e_{t0}]=[4.07-0.0017,4.07+0.0017]=[4.0683,4.0717].\]  
Since \(\overline{x}-e_{t0}=\)4.0683 \(>4.05\), we calculate the values of \(\hat{C}_{pk0}^{\prime\prime}\), \(LC_{pk}^{\prime\prime}\), and \(UC_{pk}^{\prime\prime}\) for case (2) as follows:  
\[C_{pk0}^{\prime\prime}=\frac{d-d_{u}(\overline{x}-T)}{3s}=\frac{0.02-0.4 \times(4.07-4.05)}{3\times 0.005}=0.8,\]  
\[LC_{pk}^{\prime\prime}=\left({\hat{C}_{pk0}^{\prime\prime}-\frac{d_{u}\times t _{0.02535}}{3\sqrt{n}}}\right)\times\sqrt{\frac{\chi_{0.025,35}^{2}}{n-1}}= \left({0.8-\frac{0.4\times 2.03}{3\sqrt{36}}}\right)\times\sqrt{\frac{20.569}{35}}=0.579,\]  
\[UC_{pk}^{\prime\prime}=\left({\hat{C}_{pk0}^{\prime\prime}+\frac{d_{u}t_{ \alpha/4\mu-1}}{3\sqrt{n}}}\right)\times\sqrt{\frac{\chi_{1-\alpha/4\mu-1}^{2} }{n-1}}=\left({0.8+\frac{0.4\times 2.03}{3\sqrt{36}}}\right)\times\sqrt{\frac{53.203}{35}}=0.919.\]  
Step 4: We use \(LC_{pk}^{\prime\prime}=0.579\) and \(UC_{pk}^{\prime\prime}=0.919\) to conduct statistical verification as follows:  
Since \(UC_{pk}^{\prime\prime}=0.919<C=1.0\), we reject \(H_{0}\) and conclude that the process is inadequate (\(C_{pk}^{\prime\prime}<1.0\)) and therefore in need of improvement.