Interval Estimate with Specified Precision

• Previous section discusses for a given set of replications to calcualte the confidence interval and error. Sometimes we need to do the inverse, given a level of error and confidence, how many replications are needed?

• The half-length (h.i.) of a $100(1-\alpha)\%$ confidence interval for a mean $\theta$, based on the t distribution, is
  
  $$h.i. = t_{\alpha/2,R-1} * \hat{\sigma}(\hat{\theta}) (*)$$
  
  
  where $\hat{\sigma}(\hat{\theta}) = S/\sqrt{R}$, S is the sample standard deviation, R is the number of replications.

• Assume an error criterion $\epsilon$ is specified with a confidence level $1 - \alpha$, it is desired that a sufficiently large sample size R be taken such that
  
  $$P(\vert\hat{\theta} - \theta\vert < \epsilon) \ge 1 - \alpha$$
  
  

• Since we have the relation (*), the desired the error control condition can be written as
  
  $$h.i. = \frac{t_{\alpha/2,R-1} S_0} {\sqrt{R}} \le \epsilon$$
  
  

• Solve the above relation, we have
  
  $$R \ge \left( \frac{t_{\alpha/2,R-1} S_0} {\epsilon} \right)^2$$
  
  

  since $t_{\alpha/2,R-1} \ge z_{\alpha/2}$ the above relation can be written
  

  $$R \ge \left( \frac{z_{\alpha/2} S_0} {\epsilon} \right)^2$$
  
  

  For $R \ge 50, t_{\alpha/2,R-1} \approx z_{\alpha/2}$ the inequality with standard normal distribution holds.

  This says we need to run that many (R) replications to satisfy the error requirement.

• The true value of $\theta$ is in the following range with probability of $100(1-\alpha)\%$
  
  $$\hat{\theta} - \frac{t_{\alpha/2,R-1} S}{\sqrt{R}} \le \theta \le \hat{\theta} + \frac{t_{\alpha/2,R-1} S}{\sqrt{R}}$$
  
  

• Example 12.12 on page 449