## 3 Experimental Results
### Effect of Vibration on DC Resistance
#### 3.2.1 Parameters Estimation Based on Samples before Vibration  
The resistance of the battery before the vibration is regarded as a sample of this batch of batteries, the mean value and the standard deviation of which are estimated by the method of parameter estimation. In the next section, the internal resistance of the samples after vibration is tested with the standard by which the parameters are estimated based on the samples before vibration.  
Figure 10: DC resistance distribution before and vibration test.  
Before the parameter estimation, it is necessary to test the population distribution of the internal resistance of samples before vibration in the same way as described in Section 3.1 for the capacity distribution test; that is to say, the frequency histogram is fitted at first, then the K-S hypothesis testing is carried out, and finally the distribution of the internal resistance based on the samples before vibration is obtained as normal distribution \(N\) (37.20, 1.056\({}^{2}\)).  
Estimation of Confidence Interval for the Population Mean \(\mu_{0}\)  
According to the theorem related to statistics, suppose the sample values obey the population normal distribution \(N\) (\(\mu,\ \sigma^{2}\)), applying the following formula:  
\[\frac{X-\mu}{S/\sqrt{n}}\ \sim\ t(n-1), \tag{10}\]  
where \(X\), \(S^{2}\) denotes the sample mean and sample variance, respectively, and \(t(n-1)\) does not depend on any unknown parameters.  
Using \(\frac{X-\mu}{S/\sqrt{n}}\) as the statistic, for a given confidence level \(1-\alpha\), the following formula applies:  
\[P\left\{-t_{\frac{\alpha}{2}}(n-1)<\frac{X-\mu}{S/\sqrt{n}}<t_{\frac{\alpha}{ 2}}(n-1)\right\}=1-\alpha. \tag{11}\]  
Thus, the confidence interval of the population mean \(\mu_{0}\) is obtained as  
\[X\pm\frac{S}{\sqrt{n}}t_{\frac{\alpha}{2}}(n-1). \tag{12}\]  
When the confidence level \(1-\alpha\) is set to be 0.95, by taking advantage of MATLAB R2012a, the confidence interval of the overall mean \(\mu_{0}\) is calculated, resulting in (36.81, 37.58) m\(\Omega\).  
Estimation of Confidence Interval for the Population Standard Variation \(\sigma_{0}\)  
According to the theorem related to statistics, suppose the sample values obey the population normal distribution \(N\) (\(\mu,\ \sigma^{2}\)), applying the following formula:  
\[\frac{(n-1)S^{2}}{\sigma^{2}}\ \sim\chi^{2}(n-1), \tag{13}\]  
where \(X\), \(S^{2}\) denotes the sample mean and the sample variance, respectively, and \(\chi^{2}(n-1)\) does not depend on any unknown parameters.  
Using \(\frac{(n-1)S^{2}}{\sigma^{2}}\) as the statistic, for a given confidence level \(1-\alpha\), the following formula applies:  
\[P\left\{\chi^{2}_{1-\frac{\alpha}{2}}(n-1)<\frac{(n-1)S^{2}}{\sigma^{2}}<\chi ^{2}_{\frac{\alpha}{2}}(n-1)\right\}=1-\alpha. \tag{14}\]  
Thus, the confidence interval of the population standard variation \(\sigma_{0}\) is obtained as  
\[\left[\frac{\sqrt{n-1}S}{\sqrt{\chi^{2}_{\frac{\alpha}{2}}(n-1)}},\frac{\sqrt {n-1}S}{\chi^{2}_{1-\frac{\alpha}{2}}(n-1)}\right]. \tag{15}\]  
When the confidence level \(1-\alpha\) is set to be 0.95, by taking advantage of MATLAB R2012a, the confidence interval of the population standard variation \(\sigma_{0}\) is calculated, resulting in (0.844, 1.411) m\(\Omega\)