# Using Statistical Test Method to Establish a Decision Model of Performance Evaluation Matrix
## 4 Statistical Two-Tailed Test  
As noted above, confirming the effect upon the improvement is an important step in management. Therefore, we use a statistical two-tailed test to judge whether the satisfaction of all the CTQ service items needs to be enhanced in this section. We let set A be the set formed by the service items that need to be enhanced; then,  
\[A=\{i|(x_{i},y_{i})\in Q_{2}\cap Q_{3}\}. \tag{30}\]  
\(\theta_{bi}^{S}\) is the value of satisfaction index before improvement, \(\theta_{ai}^{S}\) is the value of satisfaction index after improvement, and \(i\in A\). If the satisfaction sample data are \(\big{(}X^{\prime}{}_{i,1},\cdots,\ X^{\prime}{}_{i,j},\cdots,\ X^{\prime}{}_{ i,n}\big{)}\) for \(i\in A\), then the estimator of \(\theta_{ai}^{S}\) can be shown as follows:  
\[\theta_{ai}^{S}=\frac{1}{n}\times\sum_{j=1}^{n}X^{\prime}{}_{i,j}. \tag{31}\]  
Similarly, the estimator of \(\theta_{bi}^{S}\) for \(i\in A\) as follows:  
\[\hat{\theta}_{bi}^{S}=\frac{1}{n}\times\sum_{j=1}^{n}X_{i,j}. \tag{32}\]  
Then,  
\[Z_{hi}=\frac{\sqrt{n}\big{(}\hat{\theta}_{hi}^{S}-\theta_{hi}^{S}\big{)}}{S_ {Xhi}},\ h=a,b\ \text{and}\ i\in A, \tag{33}\]  
where  
\[S_{Xai}=\sqrt{\frac{1}{n-1}\times\sum_{j=1}^{n}\big{(}X^{\prime}_{i,j}-\overline {X}^{\prime}_{i}\big{)}^{2}}\ \text{and}\ S_{Xhi}=\sqrt{\frac{1}{n-1}\times\sum_{j=1}^{n}\big{(}X_{i,j}- \overline{X}_{i}\big{)}^{2}}. \tag{34}\]  
Then, \(Z_{hi}\) is an approximate value, and it is distributed as a standard normal distribution when sample size \(n\) is large. Thus,  
\[\begin{array}{ll}1-\alpha&=p(-Z_{\alpha/2}\leq Z_{hi}\leq Z_{\alpha/2})\\ &=p\bigg{(}-Z_{\alpha/2}\leq\frac{\sqrt{n}\big{(}\hat{\theta}_{bi}^{S}-\theta_ {bi}^{S}\big{)}}{S_{Xhi}}\leq Z_{\alpha/2}\bigg{)}\\ &=p\Big{(}\hat{\theta}_{hi}^{S}-Z_{\alpha/2}\frac{S_{Xhi}}{\sqrt{n}}\leq\theta _{hi}^{S}\leq\hat{\theta}_{hi}^{S}+Z_{\alpha/2}\frac{S_{Xhi}}{\sqrt{n}}\Big{)}.\end{array} \tag{35}\]  
Then, the \(100(1-\alpha)\%\) confidence interval of \(\theta_{bi}^{S}\) and \(\theta_{ai}^{S}\) can be demonstrated as follows:  
\[\text{before improvement}:\ \ [L\theta_{bi},U\theta_{bi}]=\bigg{[}\hat{\theta}_{ bi}^{S}-Z_{\alpha/2}\frac{S_{Xhi}}{\sqrt{n}},\theta_{bi}^{S}+Z_{\alpha/2}\frac{S_{Xhi}}{ \sqrt{n}}\bigg{]}; \tag{36}\]  
\[\text{after improvement}:\ \ [L\theta_{ai},U\theta_{ai}]=\bigg{[}\hat{\theta}_{ ai}^{S}-Z_{\alpha/2}\frac{S_{Xai}}{\sqrt{n}},\hat{\theta}_{ai}^{S}+Z_{\alpha/2} \frac{S_{Xai}}{\sqrt{n}}\bigg{]}. \tag{37}\]  
Obviously, when the value of \(\theta_{ai}^{S}\) is bigger than that of \(\theta_{bi}^{S}\), it indicates that the improvement is effective. When \(\theta_{ai}^{S}\) is equal to \(\theta_{bi}^{S}\), it means that the improvement has no effect. When the value of \(\theta_{ai}^{S}\) is smaller than that of \(\theta_{bi}^{S}\), improvement does not occur; in fact, it is made worse. Accordingly, the statistical hypothesis test for validating the effectiveness of improvement is displayed as follows:  
\[\text{null hypothesis}\ H_{0}:\ \theta_{bi}^{S}=\theta_{ai}^{S}; \tag{38}\]  
\[\text{alternative hypothesis}\ H_{1}:\ \theta_{bi}^{S}\
eq\theta_{ai}^{S}. \tag{39}\]Based on Equations (36) and (37), we perform the two-tailed statistical test using two confidence intervals. The test rules are defined as follows:  
1. If \(U\theta_{bi}<L\theta_{ai}\), then the improvement of service item \(i\) works well.
2. If \([Lb_{bi},U\theta_{bi}]\cap[L\theta_{ai},U\theta_{ai}]\
eq\phi\), then service item \(i\) does not improve significantly.
3. If \(U\theta_{ai}<Lb_{bi}\), then it means that improvement does not occur; in fact, it is made worse.  
Based on the rules of the two-tailed statistical test, we can assist the system administrator of the teaching APP to verify the improvement effect upon all the service items that need to be improved and determine whether the improvements should continue so as to ensure the system quality of the teaching APP.