Question

the following table regarding traàffic trapast's) and the number of arrests Led th the traps(y . #Of traps |6| 5|714 # of atrests | 8 | 6 | 9 17 REMEMBER A TABLE MUST BE SHOwN TO CLARIFY wORK AND ANSWERS). Do the following a) Find the correlation coefficient between the varlables to the nearest hundredth' is the calculated correlation coefficient significant to the 95% level? alven reason b) e least squares equation(y-predict equation/linear regression equation)round to values to 2 decimal places In writing the answer.

Answer 1

(a)

From the given data, the following Table is calculated:

X	Y	XY	X2	Y2
6	8	48	36	64
5	6	30	25	36
7	9	63	49	81
4	7	28	16	49
Total = 22	30	169	126	230

$r=\frac{n\sum XY-\sum X\sum Y}{\sqrt{n\sum X^{2}-(\sum X)^{2}}\sqrt{n\sum Y^{2}-(\sum Y)^{2}}}$

(4 x 169) (22 x 30) = 0.80 (4 × 126) _ 222 V/(4 × 230)-302

(b)

Test statistic is given by:

1.1314 = 1.8857 0.80 × V4-2 rvn-2 0.6 V1- 0.802

$\alpha$ = 0.05

ndf= n - 2 = 4 - 2 = 2

From Table, critical values of t = $\pm$ 4.3027

Since the calculated value of t is less than critical value of t, we conclude that the calculated correlation coefficient is not significant to the 95% level.

(c)

From the given data, the following Table is calculated:

X	Y	XY	X2	Y2
6	8	48	36	64
5	6	30	25	36
7	9	63	49	81
4	7	28	16	49
Total = 22	30	169	126	230

$a=\frac{\sum Y\sum X^{2}-\sum X\sum XY}{n\sum X^{2}-(\sum X)^{2}}$

(30 × 126)-(22 × 169) (4× 126)-222 = 3.10

$b=\frac{n\sum XY-\sum X\sum Y}{n\sum X^{2}-(\sum X)^{2}}$

(4 × 169)-(22 × 30) 0.80 4 x 126)- 222

So,

Least square regression equation is given by:

у = 3.10+0.802.