To test $\mathrm{H}_{0}: \sigma=2.3$ versus $\mathrm{H}_{1}: \sigma \neq 2.3$, a random sample of size $n=17$ is obtained from a population that is known to be normally distributed.
(a) If the sample standard deviation is determined to be $s=1.2$, compute the test statistic.
$\chi_{0}^{2}=4.355$ (Round to the
(b) If the researcher decides to test this hypothesis at the $\alpha=0.05$ level of significance, determine the critical values.
The critical values are $\chi_{0.025}^{2}=\square$ and $\chi_{0.975}^{2}=$
(Round to three decimal places as needed.)

Step 1 ：Given a random sample of size \(n=17\) obtained from a population that is known to be normally distributed, we are testing the null hypothesis \(H_{0}: \sigma=2.3\) against the alternative hypothesis \(H_{1}: \sigma \neq 2.3\). The sample standard deviation is determined to be \(s=1.2\).

Step 2 ：The test statistic for a hypothesis test about a population standard deviation or variance is a chi-square statistic. The formula for the test statistic when testing a single population variance is given by: \[\chi^{2} = \frac{(n - 1)s^{2}}{\sigma^{2}}\] where \(n\) is the sample size, \(s\) is the sample standard deviation, and \(\sigma\) is the hypothesized population standard deviation.

Step 3 ：Substituting the given values into the formula, we get: \[\chi^{2} = \frac{(17 - 1)1.2^{2}}{2.3^{2}}\]

Step 4 ：Calculating the above expression, we find that the test statistic is \(\boxed{4.355}\) (rounded to three decimal places).

Step 5 ：If the researcher decides to test this hypothesis at the \(\alpha=0.05\) level of significance, we need to determine the critical values. These are the values of chi-square distribution that cut off the upper and lower 0.025 (since \(\alpha/2 = 0.05/2 = 0.025\)) in a chi-square distribution with \(n - 1 = 16\) degrees of freedom.

Step 6 ：Calculating these values, we find that the critical values are \(\boxed{6.908}\) and \(\boxed{28.845}\) (rounded to three decimal places).