Points: Test the claim about the population mean $\mu$ at the level of significance $\alpha$. Assume the population is normally distribut Claim: $\mu<4915 ; \alpha=0.01$ Sample statistics: $\bar{x}=5117, s=5725, n=54$ (Type integers or decimals. Do not round.) Find the standardized test statistic $t$. \[ \mathrm{t}=0.26 \] (Round to two decimal places as needed.) Find, the $P$-value. \[ \mathrm{P}= \] (Round to three decimal places as needed.)

Step 1 ：Given values are: population mean \(\mu = 4915\), sample mean \(\bar{x} = 5117\), standard deviation \(s = 5725\), sample size \(n = 54\), and level of significance \(\alpha = 0.01\).

Step 2 ：Calculate the t statistic using the formula: \(t = \frac{\bar{x} - \mu}{s / \sqrt{n}}\).

Step 3 ：Substitute the given values into the formula to get: \(t = \frac{5117 - 4915}{5725 / \sqrt{54}}\).

Step 4 ：The calculated t statistic is approximately 0.26.

Step 5 ：Calculate the p-value using the t statistic and the degrees of freedom (n-1).

Step 6 ：The calculated p-value is approximately 0.398.

Step 7 ：Since the p-value is greater than the level of significance (0.01), we fail to reject the null hypothesis.

Step 8 ：This means that we do not have enough evidence to support the claim that the population mean is less than 4915.

Step 9 ：The final answer is: The standardized test statistic \(t\) is approximately \(\boxed{0.26}\) and the \(P\)-value is approximately \(\boxed{0.398}\).