Question

x	1	2	3	4	5	6
y	608	619	674	672	676	721

Use exponential regression to find an exponential function that best fits this data.

f(x) =     

Use linear regression to find an linear function that best fits this data.

g(x) =    

Answer 1

a) Let the equation of the curve is : y = ax^b.

Taking log₁₀ on both sides, we get, log₁₀y = log₁₀a + b*log₁₀x

Let us set v = log₁₀y and u = log₁₀x and let the curve is transferred to v = A+Bu.

Therefore, S =

Thus the normal equations are :

6A+B =

A+B = ........................(i)

x	y	u	v	u2	uv
1	608	0	2.783903579	0	0
2	619	0.301029995	2.791690649	0.090619058	0.840382624
3	674	0.477121254	2.828659897	0.227644691	1.349613759
4	672	0.602059991	2.827369273	0.362476233	1.70224592
5	676	0.698970004	2.829946696	0.488559067	1.978047854
6	721	0.77815125	2.857935265	0.605519368	2.2239059
		2.857332496	16.91950536	1.774818419	8.094196057

Thus, by equation (i), we have,

6A+(2.857332496)B = (16.91950536)

(2.857332496)A+(1.774818419)B = (8.094196057)

Solving we get, A = 2.777649203 and B = 0.088757657

Therefore, the curve is : v = (2.777649203) + (0.088757657)u

i.e., y = (599.3067967)*x^{(1.226754492)}

i.e., f(x) = (599.3067967)*x^{(1.226754492)}

b) Let the equation of the curve is : y = A+Bx

Then, S =

Thus the normal equations are :

6A+B =

A+B = ........................(i)

x	y	x2	xy
1	608	1	608
2	619	4	1238
3	674	9	2022
4	672	16	2688
5	676	25	3380
6	721	36	4326
21	3970	91	14262

Thus, by equation (i), we have,

6A+21B = 3970

21A+91B= 14262

Solving we get, A = 588.2666667 and B = 20.97142857

Therefore, the curve is : y = (588.2666667)+(20.97142857)x

i.e., g(x) = (588.2666667)+(20.97142857)x