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)	45	34	24	43	35	16	27	20	35	39
snow accumulation (in.)	10	12	25	8	11	26	25	20	18	5

Explanation:

Step1: Calculate necessary sums

Let $x$ be the average - temperature and $y$ be the snow - accumulation.
We have $n = 10$.
$\sum_{i = 1}^{n}x_{i}=45 + 34+24 + 43+35+16+27+20+35+39=328$
$\sum_{i = 1}^{n}y_{i}=10 + 12+25+8+11+26+25+20+18+5=150$
$\sum_{i = 1}^{n}x_{i}^{2}=45^{2}+34^{2}+24^{2}+43^{2}+35^{2}+16^{2}+27^{2}+20^{2}+35^{2}+39^{2}$
$=2025+1156+576+1849+1225+256+729+400+1225+1521 = 10962$
$\sum_{i = 1}^{n}y_{i}^{2}=10^{2}+12^{2}+25^{2}+8^{2}+11^{2}+26^{2}+25^{2}+20^{2}+18^{2}+5^{2}$
$=100+144+625+64+121+676+625+400+324+25 = 3104$
$\sum_{i = 1}^{n}x_{i}y_{i}=45\times10+34\times12+24\times25+43\times8+35\times11+16\times26+27\times25+20\times20+35\times18+39\times5$
$=450+408+600+344+385+416+675+400+630+195 = 4503$

Step2: Calculate the slope $b_1$

The formula for the slope $b_1$ of the regression line is $b_1=\frac{n\sum_{i = 1}^{n}x_{i}y_{i}-\sum_{i = 1}^{n}x_{i}\sum_{i = 1}^{n}y_{i}}{n\sum_{i = 1}^{n}x_{i}^{2}-(\sum_{i = 1}^{n}x_{i})^{2}}$
$n\sum_{i = 1}^{n}x_{i}y_{i}-\sum_{i = 1}^{n}x_{i}\sum_{i = 1}^{n}y_{i}=10\times4503-328\times150=45030 - 49200=-4170$
$n\sum_{i = 1}^{n}x_{i}^{2}-(\sum_{i = 1}^{n}x_{i})^{2}=10\times10962-328^{2}=109620 - 107584 = 2036$
$b_1=\frac{-4170}{2036}\approx - 2.048$

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 $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}=\frac{328}{10}=32.8$ and $\bar{y}=\frac{\sum_{i = 1}^{n}y_{i}}{n}=\frac{150}{10}=15$
$b_0 = 15-(-2.048)\times32.8=15 + 67.1744\approx82.174$

The regression line is $\hat{y}=82.174-2.048x$

Step4: Calculate the correlation coefficient $r$

The formula for the correlation coefficient $r$ is $r=\frac{n\sum_{i = 1}^{n}x_{i}y_{i}-\sum_{i = 1}^{n}x_{i}\sum_{i = 1}^{n}y_{i}}{\sqrt{(n\sum_{i = 1}^{n}x_{i}^{2}-(\sum_{i = 1}^{n}x_{i})^{2})(n\sum_{i = 1}^{n}y_{i}^{2}-(\sum_{i = 1}^{n}y_{i})^{2})}}$
We already know $n\sum_{i = 1}^{n}x_{i}y_{i}-\sum_{i = 1}^{n}x_{i}\sum_{i = 1}^{n}y_{i}=-4170$
$n\sum_{i = 1}^{n}x_{i}^{2}-(\sum_{i = 1}^{n}x_{i})^{2}=2036$
$n\sum_{i = 1}^{n}y_{i}^{2}-(\sum_{i = 1}^{n}y_{i})^{2}=10\times3104 - 150^{2}=31040-22500 = 8540$
$r=\frac{-4170}{\sqrt{2036\times8540}}=\frac{-4170}{\sqrt{17387440}}\approx\frac{-4170}{4170.904}\approx - 0.999$

The degrees of freedom is $df=n - 2=10 - 2 = 8$. For $\alpha=0.05$ and $df = 8$, the critical - value of the Pearson correlation coefficient is approximately $\pm0.632$. Since $|r|\approx0.999>0.632$, the regression equation is appropriate for making predictions.

Answer:

The regression line is $\hat{y}=82.174 - 2.048x$ and the regression equation is appropriate for making predictions.