Question

We revisit Example 3.6 in Chapter 3’s lecture note, which shows the relationship between the price of a diamond and its carat.

Carat (C)	Price (P)	Carat (C)	Price (P)
0.25	3110	0.7	16896
0.3	3786	0.8	22602
0.35	4685	0.9	29775
0.4	5966	1	40022
0.45	7718	1.5	71031
0.5	10478	2	121734
0.55	13156	3	391770
0.6	13877	4	562342

Now we assume that the relationship is of the form P = a 0 e a 1 C .

a) (10pt) Discuss how to estimate a 0 and a 1 using linear regression technique.

b) (10pt) Using the method discussed in a), by Excel, estimate a 0 and a 1 , determine a CER equation to fit these data, and plot the data points and the associated CER curve.

c) (5pt) Using this method, what will be the price for a 1.2 carat diamond?

d) (10pt) Compute the standard error of this method.

e) (10pt) Using the coefficient obtained on page 31 of Chapter 3’s lecture note, compute the standard error of the method assuming the power form P = a 0 C a 1 , and compare it with the result in d). (Note that “SE = 0.1136" on the page is not an answer for this question.)

Answer 1

b)

using excel

d)

standerd error

C	P	Forecast	Error2
0.25	3110	7086.17	15,809,912.95
0.3	3786	7595.46	14,511,978.87
0.35	4685	8141.35	11,946,379.28
0.4	5966	8726.48	7,620,260.05
0.45	7718	9353.66	2,675,396.99
0.5	10478	10025.92	204,373.96
0.55	13156	10746.50	5,805,704.02
0.6	13877	11518.86	5,560,823.40
0.7	16896	13234.11	13,409,455.08
0.8	22602	15204.77	54,719,027.87
0.9	29775	17468.88	151,440,667.91
1	40022	20070.13	398,077,199.08
1.5	71031	40176.85	951,978,288.69
2	121734	80426.97	1,706,270,446.82
3	391770	322294.81	4,826,802,152.08
4	562342	1291531.18	531,716,862,245.78
		Total	539,883,694,312.81