In a study, researchers wanted to measure the effect of alcohol on the hippocampal region, the portion of the brain responsible for long-term memory storage, in adolescents. The researchers randomly selected 18 adolescents with alcohol use disorders to determine whether the hippocampal volumes in the alcoholic adolescents were less than the normal volume of $9.02 \mathrm{~cm}^{3}$. An analysis of the sample data revealed that the hippocampal volume is approximately normal with no outliers and $\bar{x}=8.16 \mathrm{~cm}^{3}$ and $s=0.8 \mathrm{~cm}^{3}$. Conduct the appropriate test at the $\alpha=0.01$ level of significance.

Identify the t-statistic.
$t_{0}=\square$ (Round to two decimal places as needed.)
Identify the P-value
P-value $=\square$ (Round to three decimal places as needed.)

Step 1 ：Given the sample mean \(\bar{x} = 8.16\) cm³, the population mean \(\mu = 9.02\) cm³, the sample standard deviation \(s = 0.8\) cm³, and the sample size \(n = 18\).

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

Step 3 ：Substituting the given values into the formula, we get \(t = \frac{8.16 - 9.02}{0.8 / \sqrt{18}} = -0.86 / (0.8 / \sqrt{18}) = -0.86 / 0.18898 = -4.55\) (rounded to two decimal places).

Step 4 ：So, the t-statistic, \(t_0 = -4.55\).

Step 5 ：Next, we need to find the P-value. The P-value is the probability of obtaining a result as extreme as the observed data, assuming the null hypothesis is true.

Step 6 ：Since we are conducting a one-tailed test, we will look up the t-statistic in a t-distribution table with degrees of freedom \(df = n - 1 = 18 - 1 = 17\).

Step 7 ：Looking up a t-statistic of -4.55 in a t-distribution table with \(df = 17\), we find that the P-value is less than 0.001.

Step 8 ：So, the P-value = <0.001 (rounded to three decimal places).

Step 9 ：Therefore, the t-statistic is \(\boxed{-4.55}\) and the P-value is \(\boxed{<0.001}\).