Sampling Distribution of $S^2$

• If a random sample of size $n$ is taken from a normal population with mean $\mu$ and variance $\sigma^2$, and the sample variance $S^2$ is computed.
• Consider the distribution of the statistics $\frac{(n-1)S^2}{\sigma^2}$
  $$\sum_{i=1}^n (X_i-\mu)^2=\sum_{i=1}^n \left[(X_i-\bar{X})+(\bar{X}-\mu) \right] ^2$$
  $$=\sum_{i=1}^n (X_i-\bar{X})^2+\sum_{i=1}^n (\bar{X}-\mu)^2+2(\bar{X}-\mu)\sum_{i=1}^n (X_i-\bar{X})$$
  $$=\sum_{i=1}^n (X_i-\bar{X})^2+n(\bar{X}-\mu)^2$$

• Dividing both sides by $\sigma^2$ and substituting $(n-1)S^2$ for $\sum_{i=1}^n (X_i-\bar{X})^2$, we obtain
  $$\frac{1}{\sigma^2}\sum_{i=1}^n (X_i-\mu)^2=$$
  chi-squared random variable with $n$ degrees of freedom
  $$\frac{(n-1)S^2}{\sigma^2}+$$
  $$\frac{(\bar{X}-\mu)^2}{\frac{\sigma^2}{n}}$$
  chi-squared random variable with $1$ degrees of freedom

• Theorem 8.4:
  
• It is customary to let $\chi_{\alpha}^2$ represent the $\chi^2$-value above which we find an area of $\alpha$.
• This is illustrated by the shaded region in Fig. 7.15.

  Figure 7.15: The chi-squared distribution.

  

  For $\nu=7$, $\chi_{0.05}^2=14.067$, and $\chi_{0.95}^2=2.1677$ (Table A.5)

• Example 8.10: A manufacturer of car batteries guarantees that his batteries will last, on the average, 3 years with a standard deviation of 1 year.
• If five of these batteries have lifetimes of 1.9, 2.4, 3.0, 3.5, and 4.2 years, is the manufacturer still convinced that his batteries have a standard deviation of 1 year?
• Assume that the battery lifetime follows a normal distribution.
• Solution:
  $$s^2=\frac{n\sum_{i=1}^n X_i^2-(\sum_{i=1}^n X_i)^2}{n(n-1)}=\frac{5*48.26-15^2}{5*4}=0.815$$
  $$\chi^2=\frac{(n-1)s^2}{\sigma^2}=\frac{4*0.815}{1}=3.26$$

• Since $n=5$, $\chi^2$ has $\nu=n-1=4$ degrees of freedom.
• From Table A.5 row $\nu=4$, wee see that
  $$\chi_{0.025}^2=11.143  and \chi_{0.975}^2=0.484$$

• Since 95% of the values with 4 degrees of freedom fall between $0.484$ and $11.143$, the computed value with $\sigma^2=1$ is reasonable (since our $\chi^2=3.26$ falls within this range).
• Therefore the manufacturer has no reason to suspect that the standard deviation is other than 1 year.