A random sample of 85 eighth grade students' scores on a national mathematics assessment test has a mean score of 284 . 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 275 . Assume that the population standard deviation is 31. At $\alpha=0.04$, 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. \[ \begin{array}{l} H_{0}: \mu<275 \\ H_{a}: \mu \geq 275 \text { (claim) } \end{array} \] D. \[ \begin{array}{l} H_{0}: \mu=275 \text { (claim) } \\ H_{a}: \mu>275 \end{array} \] B. \[ \begin{array}{l} H_{0}: \mu \leq 275 \\ H_{a}: \mu>275 \text { (claim) } \end{array} \] E. \[ \begin{array}{l} H_{0}: \mu \geq 275 \text { (claim) } \\ H_{a}: \mu<275 \end{array} \] c. \[ \begin{array}{l} H_{0}: \mu \leq 275 \text { (claim) } \\ H_{a}: \mu>275 \end{array} \] F. \[ \begin{array}{l} H_{0}: \mu=275 \\ H_{a}: \mu>275 \text { (claim) } \end{array} \] (b) Find the standardized test statistic $z$. $z=2.68$ (Round to two decimal places as needed.) (c) Find the P-value. P-value $=\square$ (Round to three decimal places as needed.)
Solution
Step 1 ：The claim is that the mean score for the state's eighth graders on this exam is more than 275. This can be written mathematically as \(\mu > 275\). The null hypothesis, \(\mathrm{H}_{0}\), is the statement that the claim is false, so \(\mathrm{H}_{0}: \mu \leq 275\). The alternative hypothesis, \(\mathrm{H}_{\mathrm{a}}\), is the claim, so \(\mathrm{H}_{\mathrm{a}}: \mu > 275\).
Step 2 ：The standardized test statistic z can be calculated using the formula: \[z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}\] where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean, \(\sigma\) is the population standard deviation, and \(n\) is the sample size. Plugging in the given values: \[z = \frac{284 - 275}{31 / \sqrt{85}} \approx 2.68\]
Step 3 ：The P-value is the probability that a z-score is more extreme than the observed z-score, assuming the null hypothesis is true. Because this is a right-tailed test (the alternative hypothesis is \(\mu > 275\)), the P-value is the area to the right of the observed z-score on the standard normal distribution. Using a standard normal table or a calculator, we find that the P-value is approximately 0.004.
