the following table gives the data for the average temperature and the snow accumulation in several small…

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, $hat{y}=b_0 + b_1x$. 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.\n\naverage temperatures and snow accumulations\n|average temperature (°f)|45|29|24|39|45|17|25|21|35|37|\n|snow accumulation (in.)|5|12|28|6|10|22|29|14|12|8|

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, $hat{y}=b_0 + b_1x$. 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.\n\naverage temperatures and snow accumulations\n|average temperature (°f)|45|29|24|39|45|17|25|21|35|37|\n|snow accumulation (in.)|5|12|28|6|10|22|29|14|12|8|

Answer

Explanation:

Step1: Calculate necessary sums

Let (x) be the average - temperature and (y) be the snow - accumulation. (n = 10) (number of data points). (\sum_{i = 1}^{n}x_{i}=45 + 29+24 + 39+45+17+25+21+35+37=317) (\sum_{i = 1}^{n}y_{i}=5 + 12+28+6+10+22+29+14+12+8=146) (\sum_{i = 1}^{n}x_{i}^{2}=45^{2}+29^{2}+24^{2}+39^{2}+45^{2}+17^{2}+25^{2}+21^{2}+35^{2}+37^{2}=45^{2}\times2 + 29^{2}+24^{2}+39^{2}+17^{2}+25^{2}+21^{2}+35^{2}+37^{2}=10329) (\sum_{i = 1}^{n}y_{i}^{2}=5^{2}+12^{2}+28^{2}+6^{2}+10^{2}+22^{2}+29^{2}+14^{2}+12^{2}+8^{2}=3136) (\sum_{i = 1}^{n}x_{i}y_{i}=45\times5+29\times12 + 24\times28+39\times6+45\times10+17\times22+25\times29+21\times14+35\times12+37\times8=3273)

Step2: Calculate the slope (b_1)

The formula for the slope (b_1) of the regression line is (b_1=\frac{n\sum_{i = 1}^{n}x_{i}y_{i}-\sum_{i = 1}^{n}x_{i}\sum_{i = 1}^{n}y_{i}}{n\sum_{i = 1}^{n}x_{i}^{2}-(\sum_{i = 1}^{n}x_{i})^{2}}) [ \begin{align*} b_1&=\frac{10\times3273 - 317\times146}{10\times10329-(317)^{2}}\ &=\frac{32730-46282}{103290 - 100489}\ &=\frac{- 13552}{2801}\ &\approx - 4.838 \end{align*} ]

Step3: Calculate the y - intercept (b_0)

The formula for the y - intercept (b_0) is (b_0=\bar{y}-b_1\bar{x}), where (\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}=\frac{317}{10}=31.7) and (\bar{y}=\frac{\sum_{i = 1}^{n}y_{i}}{n}=\frac{146}{10}=14.6) [ \begin{align*} b_0&=14.6-(-4.838)\times31.7\ &=14.6 + 4.838\times31.7\ &=14.6+153.365\ &=167.965 \end{align*} ]

Step4: Determine if the regression equation is appropriate

First, calculate the correlation coefficient (r). The formula for (r) is (r=\frac{n\sum_{i = 1}^{n}x_{i}y_{i}-\sum_{i = 1}^{n}x_{i}\sum_{i = 1}^{n}y_{i}}{\sqrt{(n\sum_{i = 1}^{n}x_{i}^{2}-(\sum_{i = 1}^{n}x_{i})^{2})(n\sum_{i = 1}^{n}y_{i}^{2}-(\sum_{i = 1}^{n}y_{i})^{2})}}) [ \begin{align*} r&=\frac{10\times3273 - 317\times146}{\sqrt{(10\times10329-(317)^{2})(10\times3136-(146)^{2})}}\ &=\frac{-13552}{\sqrt{2801\times9444}}\ &=\frac{-13552}{\sqrt{26452644}}\ &=\frac{-13552}{5143.21}\ &\approx - 0.264 \end{align*} ] For (n = 10) and a significance level of (\alpha=0.05), the critical - value of the Pearson correlation coefficient (two - tailed) is approximately (r_{c}=0.632). Since (|r| = 0.264<0.632), the regression equation is not appropriate for making predictions at the 0.05 level of significance.

Answer:

The regression line is (\hat{y}=167.965-4.838x), and the regression equation is not appropriate for making predictions at the 0.05 level of significance.