The following table gives the data for the average temperature and the snow accumulation in several small towns for a single month. Determine the equation of the regression line, $\widehat{y}=b_{0}+b_{1} x$. Round the slope and $y$-intercept to the nearest thousandth. Then determine if the regression equation is appropriate for making predictions at the 0.05 level of significance.
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline \multicolumn{10}{|c|}{ Average Temperatures and Snow Accumulations } \\
\hline Average Temperature $\left({ }^{\circ} \mathrm{F}\right)$ & 36 & 34 & 20 & 40 & 45 & 15 & 34 & 21 & 31 & 39 \\
\hline Snow Accumulation (in.) & 5 & 12 & 28 & 9 & 14 & 26 & 26 & 14 & 13 & 8 \\
\hline
\end{tabular}

Step 1 ：Given the average temperatures and snow accumulations for several small towns for a single month, we are asked to determine the equation of the regression line, \(\widehat{y}=b_{0}+b_{1} x\), and round the slope and y-intercept to the nearest thousandth.

Step 2 ：We first calculate the slope (b1) using the formula: \[b_{1}=\frac{n(\sum x y)-(\sum x)(\sum y)}{n(\sum x^{2})-(\sum x)^{2}}\] where n is the number of observations, x is the average temperature, and y is the snow accumulation.

Step 3 ：We then calculate the y-intercept (b0) using the formula: \[b_{0}=\frac{\sum y-b_{1}(\sum x)}{n}\]

Step 4 ：Using these calculated values, we form the regression equation.

Step 5 ：To determine if the regression equation is appropriate for making predictions at the 0.05 level of significance, we calculate the correlation coefficient (r) using the formula: \[r=\frac{n(\sum x y)-(\sum x)(\sum y)}{\sqrt{[n \sum x^{2}-(\sum x)^{2}][n \sum y^{2}-(\sum y)^{2}]}}\]

Step 6 ：We compare the absolute value of r to the critical value for a two-tailed test at the 0.05 level of significance. If the absolute value of r is greater than the critical value, then the regression equation is appropriate for making predictions at the 0.05 level of significance.

Step 7 ：Given the data, we find that the slope of the regression line is approximately -0.538, and the y-intercept is approximately 32.433. This gives us the regression equation \(\widehat{y}=32.433-0.538x\).

Step 8 ：The correlation coefficient (r) is approximately -0.639, and the critical value for a two-tailed test at the 0.05 level of significance is approximately 2.306. Since the absolute value of r is less than the critical value, the regression equation is not appropriate for making predictions at the 0.05 level of significance.

Step 9 ：\(\boxed{\text{Final Answer: The equation of the regression line is } \widehat{y}=32.433-0.538x. \text{ The regression equation is not appropriate for making predictions at the 0.05 level of significance.}}\)