You can calculate the P-value for a chi-square test using technology. After calculating the standardized test statistic, use the cumulative distribution function (CDF) to calculate the area under the curve. Use the P-value method to test the claim. A school administrator claims that the standard deviation for eighth-grade students on a test is greater than 30 points. A random sample of 19 eighth-grade students has a standard deviation of 30.4 points. At $\alpha=0.05$, is there enough evidence to support the administrator's claim? Identify the null and alternative hypotheses. A. \[ \begin{array}{l} H_{0}: \sigma \leq 30 \\ H_{a}: \sigma>30 \end{array} \] c. \[ \begin{array}{l} H_{0}: \sigma<30 \\ H_{a}: \sigma \geq 30 \end{array} \] B. \[ \begin{array}{l} H_{0}: \sigma \geq 30 \\ H_{a}: \sigma<30 \end{array} \] D. \[ \begin{array}{l} H_{0}: \sigma>30 \\ H_{a}: \sigma \leq 30 \end{array} \] Identify the standardized test statistic. $\chi^{2}=\square$ (Round to three decimal places as needed.)

Step 1 ：Identify the null and alternative hypotheses. The null hypothesis is usually a statement of no effect or status quo, and the alternative hypothesis is what we are trying to prove. In this case, the school administrator is claiming that the standard deviation is greater than 30, so this is our alternative hypothesis. The null hypothesis is the opposite of the alternative hypothesis, so it would be that the standard deviation is less than or equal to 30. Therefore, the correct hypotheses are: \[\begin{array}{l}H_{0}: \sigma \leq 30 \H_{a}: \sigma>30\end{array}\]

Step 2 ：Identify the standardized test statistic. The test statistic for a chi-square test is calculated as: \[\chi^{2} = \frac{(n-1)s^{2}}{\sigma^{2}}\] where n is the sample size, s is the sample standard deviation, and σ is the population standard deviation. In this case, n=19, s=30.4, and σ=30. We can plug these values into the formula to calculate the test statistic.

Step 3 ：Calculate the test statistic: n = 19, s = 30.4, sigma = 30, chi_square = 18.4832

Step 4 ：The null and alternative hypotheses are: \[\begin{array}{l}H_{0}: \sigma \leq 30 \H_{a}: \sigma>30\end{array}\]

Step 5 ：The standardized test statistic is \(\chi^{2} = \boxed{18.483}\)