Determine the value of the P-value using the given data and test statistic: Given in the table are the BMI statistics for random samples of men and women. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. Use a 0.01 significance level for both parts. \begin{tabular}{|c|c|c|} \hline & Male BMI & Female BMI \\ \hline $\boldsymbol{\mu}$ & $\mu_{1}$ & $\mu_{2}$ \\ \hline $\mathbf{n}$ & 40 & 40 \\ \hline$\overline{\mathbf{x}}$ & 28.2204 & 24.0101 \\ \hline $\mathbf{s}$ & 7.249005 & 4.208679 \\ \hline \end{tabular} The test statistic, $t$, is 3.18 . (Round to two decimal places as needed.) The P-value is $\square$. (Round to three decimal places as needed.)

Step 1 ：The problem provides us with the following data: sample sizes \(n_1 = 40\) and \(n_2 = 40\), sample standard deviations \(s_1 = 7.249005\) and \(s_2 = 4.208679\), and the test statistic \(t = 3.18\).

Step 2 ：We are asked to find the P-value for a two-sample t-test with unequal variances. The degrees of freedom for this test is calculated using the formula: \[df = \frac{(s1^2/n1 + s2^2/n2)^2}{(s1^2/n1)^2/(n1-1) + (s2^2/n2)^2/(n2-1)}\]

Step 3 ：Substituting the given values into the formula, we find that the degrees of freedom \(df \approx 62.61\).

Step 4 ：The P-value is the area under the t-distribution curve to the right of the test statistic. Since we are doing a one-tailed test, we need to find 1 minus the cumulative distribution function (CDF) value at the test statistic \(t = 3.18\).

Step 5 ：Calculating this, we find that the P-value is approximately 0.001145383589568283.

Step 6 ：Rounding this to three decimal places, we get the final answer: The P-value is \(\boxed{0.001}\).