To determine the P-value for a left-tailed test with a t-statistic of \( t = 3.018 \) and \( n = 13 \), we need to find the degrees of freedom. The degrees of freedom for this test is calculated as:
\[ \text{df} = n - 1 = 13 - 1 = 12 \]
Next, we can consult the t-distribution table (Table A-3) to find the corresponding P-value for \( t = 3.018 \) with 12 degrees of freedom.
Looking up \( t = 3.018 \) in the t-table for 12 degrees of freedom, we find that:
- For \( t = 3.106 \), the one-tailed P-value is approximately 0.005.
- Since \( t = 3.018 \) is slightly less than \( t = 3.106 \), the P-value for \( t = 3.018 \) will be greater than 0.005 but less than 0.01.
Thus, we can conclude:
\[ 0.005 < P\text{-value} < 0.01 \]
Since our significance level is \( \alpha = 0.01 \) and the P-value falls within the range \( 0.005 < P\text{-value} < 0.01 \), we need to reject the null hypothesis. This is because the P-value is lower than the significance level.
Therefore, the correct option is:
D) 0.005 < P-value < 0.01; reject the null hypothesis.