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, $hat{y}=b_{0}+b_{1}x$. 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. average temperatures and snow accumulations average temperature (f) 40 34 15 39 39 22 35 17 30 35 snow accumulation (in.) 6 16 21 6 12 28 23 15 15 5

Explanation:

Step1: Calculate necessary sums

Let \(x\) be the average temperature and \(y\) be the snow - accumulation.
\(n = 10\) (number of data points).
\(\sum_{i = 1}^{n}x_{i}=40 + 34+15+39+39+22+35+17+30+35 = 306\)
\(\sum_{i = 1}^{n}y_{i}=6 + 16+21+6+12+28+23+15+15+5 = 147\)
\(\sum_{i = 1}^{n}x_{i}^{2}=40^{2}+34^{2}+15^{2}+39^{2}+39^{2}+22^{2}+35^{2}+17^{2}+30^{2}+35^{2}\)
\(=1600 + 1156+225+1521+1521+484+1225+289+900+1225 = 10146\)
\(\sum_{i = 1}^{n}y_{i}^{2}=6^{2}+16^{2}+21^{2}+6^{2}+12^{2}+28^{2}+23^{2}+15^{2}+15^{2}+5^{2}\)
\(=36+256+441+36+144+784+529+225+225+25 = 2691\)
\(\sum_{i = 1}^{n}x_{i}y_{i}=40\times6+34\times16+15\times21+39\times6+39\times12+22\times28+35\times23+17\times15+30\times15+35\times5\)
\(=240+544+315+234+468+616+805+255+450+175 = 4102\)

Step2: Calculate the slope \(b_1\)

The formula for the slope \(b_1\) of the regression line is \(b_1=\frac{n\sum_{i = 1}^{n}x_{i}y_{i}-\sum_{i = 1}^{n}x_{i}\sum_{i = 1}^{n}y_{i}}{n\sum_{i = 1}^{n}x_{i}^{2}-(\sum_{i = 1}^{n}x_{i})^{2}}\)
Substitute the values:
\[

$$\begin{align*} b_1&=\frac{10\times4102 - 306\times147}{10\times10146-306^{2}}\\ &=\frac{41020-44982}{101460 - 93636}\\ &=\frac{- 3962}{7824}\\ &\approx - 0.506 \end{align*}$$

\]

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 \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}=\frac{306}{10} = 30.6\) and \(\bar{y}=\frac{\sum_{i = 1}^{n}y_{i}}{n}=\frac{147}{10}=14.7\)
\[

$$\begin{align*} b_0&=14.7-(-0.506)\times30.6\\ &=14.7 + 15.484\\ &\approx30.184 \end{align*}$$

\]

Step4: Determine the correlation coefficient \(r\)

The formula for the correlation coefficient \(r\) is \(r=\frac{n\sum_{i = 1}^{n}x_{i}y_{i}-\sum_{i = 1}^{n}x_{i}\sum_{i = 1}^{n}y_{i}}{\sqrt{(n\sum_{i = 1}^{n}x_{i}^{2}-(\sum_{i = 1}^{n}x_{i})^{2})(n\sum_{i = 1}^{n}y_{i}^{2}-(\sum_{i = 1}^{n}y_{i})^{2})}}\)
\[

$$\begin{align*} r&=\frac{10\times4102-306\times147}{\sqrt{(10\times10146 - 306^{2})(10\times2691-147^{2})}}\\ &=\frac{-3962}{\sqrt{7824\times4419}}\\ &=\frac{-3962}{\sqrt{34576256}}\\ &=\frac{-3962}{5880.16}\\ &\approx - 0.674 \end{align*}$$

\]

The degrees of freedom \(df=n - 2=10 - 2 = 8\). For a significance level of \(\alpha = 0.05\), the critical value of the Pearson correlation coefficient (from the table) is approximately \(r_{crit}=0.632\). Since \(|r|=0.674>0.632\), the regression equation is appropriate for making predictions.

The regression line equation is \(\hat{y}=30.184-0.506x\)

Answer:

The regression line equation is \(\hat{y}=30.184 - 0.506x\) and the regression equation is appropriate for making predictions at the 0.05 level of significance.