Question

An emergency service wishes to determine whether a relationship exists between the outside temperature and the number of emergency calls it receives for a 7-hour period. The data are shown. (25 points)
Temperature x 25 10 27 30 33
No. of calls y 7 4 8 10 11

a) Find the correlation coefficient r (If you use any software or online calculator, please take the screenshot.


b) Find the regression equation (If you use any software or online calculator, please take the screenshot. You will get extra points if you do by hand.)


c) Graph the regression equation. (If you use any software or online calculator, please take the screenshot.

Answer 1

Solution to a:

Correlation coefficient is given by:

$$r ={{n(\sum xy)- (\sum x) (\sum y)} \over { \sqrt{[n \sum x^{2} - {( \sum x)}^2] [n \sum y^{2} - {( \sum y)}^2]}}}$$

but;

$n=5, \sum x=125, \sum y=40$

$\sum x^2=3443, \sum xy=1094, \sum y^2=350$

therefore;

$$r = {{(5*1094)-(125*40)} \over {\sqrt{[5*3443-(125)^2][5*350 -(40)^2]}}}={470 \over 488.3646}=0.962396$$

Answer: 0.962396

Solution to b:

Regression equation is given by:

$$\hat{y} = a +b x$$

$$b={n( \sum xy) - ( \sum x)( \sum y) \over n( \sum x^2)- (\sum x)^2}$$

$$a = { \sum{y} - b \sum{x} \over n}$$

Therefore;

$$b = {(5*1094) - (40*125) \over (5*3443)- (125)^2 } = { 470 \over 1590} = 0.295597484$$

$$a= {40 - 0.295597484*125 \over 5} = {3.050314 \over 5} = 0.610062893$$

Answer: $\hat{y} = 0.6101+ 0.2956 x$

solution to c

Graph of the regression equation