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). \[ \mathrm{H}_{\mathrm{a}}: \mu>285 \text { (claim) } \] \[ \mathrm{H}_{\mathrm{a}}: \mu>285 \text { (claim) } \] D. \[ \begin{array}{l} H_{0}: \mu=285 \text { (claim) } \\ H_{a}: \mu>285 \end{array} \] E. \[ \begin{array}{l} H_{0}: \mu<285 \\ H_{a}: \mu \geq 285 \text { (claim) } \end{array} \] \[ H_{a}: \mu<285 \] F. \[ \begin{array}{l} H_{0}: \mu \leq 285 \text { (claim) } \\ H_{a}: \mu>285 \end{array} \] (b) Find the standardized test statistic $z$. $z=1.60$ (Round to two decimal places as needed.) (c) Find the P-value. $P$-value $=$ (Round to three decimal places as needed.)

Step 1 ：State the hypotheses. The first step is to state the null hypothesis and the alternative hypothesis.

Step 2 ：\(H_{0}: \mu=285\)

Step 3 ：\(H_{a}: \mu>285\)

Step 4 ：Calculate the test statistic. The formula for the z-score is \(z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean, \(\sigma\) is the population standard deviation, and \(n\) is the sample size.

Step 5 ：\(z = \frac{292 - 285}{40 / \sqrt{84}} = 1.60\)

Step 6 ：Find the P-value. The P-value is the probability that we would observe a result as extreme as the test statistic, assuming the null hypothesis is true. Since we are testing the claim that the mean score is greater than 285, we want to find the probability that the z-score is greater than the calculated test statistic. This is a one-tailed test, so we can find the P-value by looking up the z-score in a standard normal distribution table and subtracting the value from 1.

Step 7 ：\(P\)-value = 0.054

Step 8 ：Compare the P-value with the significance level \(\alpha = 0.12\). If the P-value is less than \(\alpha\), we reject the null hypothesis. If the P-value is greater than \(\alpha\), we fail to reject the null hypothesis.

Step 9 ：Since the P-value (0.054) is less than the significance level (0.12), we reject the null hypothesis.

Step 10 ：Therefore, 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 11 ：\(\boxed{\text{Yes}}\)