Based on the data shown below, calculate the correlation coefficient (rounded to three decimal places) \begin{tabular}{|c|r|} \hline$x$ & \multicolumn{1}{|c|}{$y$} \\ \hline 3 & 11.96 \\ \hline 4 & 12.3 \\ \hline 5 & 12.24 \\ \hline 6 & 9.88 \\ \hline 7 & 11.12 \\ \hline 8 & 8.46 \\ \hline 9 & 9.1 \\ \hline \end{tabular}

Step 1 ：Calculate the mean of x and y. The mean of x, denoted as \(\bar{x}\), is \(\frac{3+4+5+6+7+8+9}{7} = 6\). The mean of y, denoted as \(\bar{y}\), is \(\frac{11.96+12.3+12.24+9.88+11.12+8.46+9.1}{7} = 10.722857142857142\).

Step 2 ：Calculate the standard deviation of x and y. The standard deviation of x, denoted as \(\sigma_x\), is \(\sqrt{\frac{(3-6)^2+(4-6)^2+(5-6)^2+(6-6)^2+(7-6)^2+(8-6)^2+(9-6)^2}{6}} = 2\). The standard deviation of y, denoted as \(\sigma_y\), is \(\sqrt{\frac{(11.96-10.722857142857142)^2+(12.3-10.722857142857142)^2+(12.24-10.722857142857142)^2+(9.88-10.722857142857142)^2+(11.12-10.722857142857142)^2+(8.46-10.722857142857142)^2+(9.1-10.722857142857142)^2}{6}} = 1.412\).

Step 3 ：Calculate the covariance of x and y. The covariance of x and y, denoted as cov(x,y), is \(\frac{(3-6)*(11.96-10.722857142857142)+(4-6)*(12.3-10.722857142857142)+(5-6)*(12.24-10.722857142857142)+(6-6)*(9.88-10.722857142857142)+(7-6)*(11.12-10.722857142857142)+(8-6)*(8.46-10.722857142857142)+(9-6)*(9.1-10.722857142857142)}{6} = -1.12\).

Step 4 ：Calculate the correlation coefficient. The correlation coefficient, denoted as r, is \(\frac{cov(x,y)}{\sigma_x*\sigma_y} = \frac{-1.12}{2*1.412} = -0.396\).

Step 5 ：\(\boxed{-0.396}\) is the final answer.