Question

The table below represents the values of x (fork-lift driving experiences in years) and y (monthly insurance in australian dollar).

x	y
5	64
2	87
12	50
9	71
15	44
6	56
25	42
16	60

1. Find the least square regression line.
2. Calculate and critically comment on the value of r and r².

Answer 1

(a)

Sample	x	y	x2	y2	x.y
1	5	64	25	4,096	320
2	2	87	4	7,569	174
3	12	50	144	2,500	600
4	9	71	81	5,041	639
5	15	44	225	1,936	660
6	6	56	36	3,136	336
7	25	42	625	1,764	1,050
8	16	60	256	3,600	960
Totals	90	474	1,396	29,642	4,739
	Σx	Σy	Σx2	Σy2	Σx.y
Sample size (n) =					8
x̅ = Σx/n =					11.3
ȳ = Σy/n =					59.3
α = Σx.y - n.x̅ .ȳ =					-593.5
β = Σx2 - n.x̅ 2 =					383.5
Slope (b) = α/β =					-1.548
Intercept (a) = ȳ - b.x̅ =					76.660

The regression equation will be written as ŷ = 76.66 - 1.548 * x

(b)

Sample	x	y	x2	y2	x.y
1	5	64	25	4,096	320
2	2	87	4	7,569	174
3	12	50	144	2,500	600
4	9	71	81	5,041	639
5	15	44	225	1,936	660
6	6	56	36	3,136	336
7	25	42	625	1,764	1,050
8	16	60	256	3,600	960
Totals	90	474	1,396	29,642	4,739
	Σx	Σy	Σx2	Σy2	Σx.y
Sample size (n) =					8
Numerator = n.Σxy - Σx.Σy =					-4748.00
Denominator = [n.Σx2 - (Σx)2]1/2.[n.Σy2 - (Σy)2]1/2 =					6182.82
Correlation coefficient (r) = Numerator / Denominator					-0.7679

So,

The correlation coefficient, r = -0.768

The correlation coefficient suggests that there is a strong (since > 0.5) and indirect (since negative) correlation between 'x' and 'y'.

The coefficient of determination, r² = (-0.7679)^2 = 0.5897

This value of r2 says that about 59% of the total variation of 'y' can be explained by the independent variable 'x'. This is an indication that the explanatory power of the model is satisfactory.