A random sample of 84 eighth grade students' scores on a national mathematics assessment test has a mean score of 292. This test result prompts a state school administrator to declare that the mean score for the state's eighth graders on this exam is more than 285. Assume that the population standard deviation is 40. At $\alpha=0.12$, is there enough evidence to support the administrator's claim? Complete parts (a) through (e). (a) Write the claim mathematically and identify $\mathrm{H}_{0}$ and $\mathrm{H}_{\mathrm{a}}$. Choose the correct answer below. A. \[ \begin{array}{l} H_{0}: \mu=285 \\ H_{a}: \mu>285(\text { claim) } \end{array} \] D \[ \begin{array}{l} H_{0}: \mu=285 \text { (claim) } \\ H_{a} \cdot \mu>285 \end{array} \] B. \[ \begin{array}{l} H_{0}: \mu \leq 285 \\ H_{a}: \mu>285 \text { (cläm) } \end{array} \] E. \[ \begin{array}{l} H_{0}: \mu<285 \\ H_{a}: \mu \geq 285 \text { (claim) } \end{array} \] C. $H_{0}: \mu \geq 285$ (claim) $H_{a} \cdot \mu<285$ F. $H_{0}: \mu \leq 285$ (claim) $H_{a}: \mu>285$ (b) Find the standardized test statistic $z$ $z=$ (Round to two decimal places as needed.)

Step 1 ：The null hypothesis is usually a statement of no effect or no difference. The alternative hypothesis is what we are trying to prove. In this case, the administrator's claim is that the mean score is more than 285. So, the null hypothesis should be that the mean score is less than or equal to 285 and the alternative hypothesis should be that the mean score is more than 285. Therefore, the correct answer is B. \[H_{0}: \mu \leq 285\] \[H_{a}: \mu>285 \text { (claim) }\]

Step 2 ：We can calculate the z-score using the formula: \[z = \frac{X - \mu}{\sigma}\] where: X is the sample mean, \(\mu\) is the population mean, and \(\sigma\) is the standard deviation of the population. In this case, X = 292, \(\mu\) = 285, and \(\sigma\) = 40. We can substitute these values into the formula to find the z-score.

Step 3 ：Substituting the given values into the formula, we get: \[z = \frac{292 - 285}{40} = 1.60\]

Step 4 ：Now, we need to compare this z-score with the critical z-score for a significance level of 0.12. The critical z-score is the z-score that corresponds to the cut-off point for the significance level in the standard normal distribution. If the calculated z-score is greater than the critical z-score, we reject the null hypothesis. If the calculated z-score is less than or equal to the critical z-score, we fail to reject the null hypothesis.

Step 5 ：The critical z-score is approximately 1.17. This means that the cut-off point for the significance level of 0.12 is 1.17 standard deviations above the mean.

Step 6 ：Since the calculated z-score (1.60) is greater than the critical z-score (1.17), we reject the null hypothesis. This means that there is enough evidence to support the administrator's claim that the mean score for the state's eighth graders on this exam is more than 285.

Step 7 ：Final Answer: \[\boxed{(a) \text{The correct answer is B.} \begin{array}{l} H_{0}: \mu \leq 285 \\ H_{a}: \mu>285 \text { (claim) } \end{array}}\] \[\boxed{(b) \text{The standardized test statistic } z \text{ is approximately 1.60.}}\] \[\boxed{\text{There is enough evidence to support the administrator's claim that the mean score for the state's eighth graders on this exam is more than 285 at } \alpha=0.12.}\]