Question

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.

average temperatures and snow accumulations

average temperature (°f)	45	29	24	39	45	17	25	21	35	37
snow accumulation (in.)	5	12	28	6	10	22	29	14	12	8

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.