Problem
Sir R. A. Fisher, a famous statistician, showed that the critical values of a chi-square distribution can be approximated by the standard normal distribution \[ \chi_{k}^{2}=\frac{\left(z_{k}+\sqrt{2 v-1}\right)^{2}}{2} \] where $v$ is the degrees of freedom and $z_{k}$ is the $z$-score such that the area under the standard normal curve to the right of $z_{k}$ is $k$. Use Fisher's approximation to find $\chi_{0.975}^{2}$ and $\chi_{0.025}^{2}$ with 100 degrees of freedom. Compare the results with those found in the table of the chi-square distribution. In Fisher's approximation the value of $\chi_{0.975}^{2}$ is $\square$. (Type an integer or a decimal.) an example Get more help - Clear all Check answer
Solution
Step 1 ：Let's denote the degrees of freedom as \(v\) and the z-score as \(z_{0.975}\). Given that \(v = 100\) and \(z_{0.975} = 1.959963984540054\).
Step 2 ：Substitute these values into Fisher's approximation formula: \(\chi_{0.975}^{2} = \frac{(z_{0.975} + \sqrt{2v - 1})^2}{2}\).
Step 3 ：Calculate the value of \(\chi_{0.975}^{2}\) to get approximately 129.06942386990755.
Step 4 ：Round the result to two decimal places to get \(\boxed{129.07}\).
