Question

The file CO2.txt, found on Blackboard with this assignment, contains 50 numbers, which represent the concentration of atmospheric carbon dioxide (parts per million) recorded at Mauna Loa, HI. The data in the file are the CO2 values on May 15th of each year from 1961 through 2010, (with background level CO2 removed). Fit an exponential model. Use the model to predict the CO2 value for May 15, 2015. Print the result to the screen using fprintf. The actual value was 125.1ppm.

   41.5800
   42.0100
   43.2500
   43.2400
   43.1600
   45.0100
   46.0000
   46.5700
   48.3400
   49.0700
   49.9200
   51.0700
   53.4800
   54.0900
   54.9700
   55.8700
   57.7500
   59.0100
   60.4700
   62.4600
   63.9100
   65.1400
   66.7500
   68.4300
   69.9300
   71.2100
   72.8400
   75.2200
   76.6700
   78.1600
   80.3400
   80.6600
   81.2800
   82.6800
   84.7900
   86.4100
   87.8000
   90.3000
   92.0000
   92.8200
   95.0200
   96.5500
   99.3500
  101.6100
  103.2400
  105.9400
  107.4300
  109.4900
  111.1800
  114.2200

Answer 1

We represent the years as 1 to 50 on x and the Co2 concentration data as y

The data can be represented exponentially as y = a.b^x

Year	x	y	Y = log y	x*Y	X^2
1961	1	41.58	1.6189	1.6189	1
1962	2	42.01	1.6234	3.2467	4
1963	3	43.25	1.6360	4.9080	9
1964	4	43.24	1.6359	6.5435	16
1965	5	43.16	1.6351	8.1754	25
1966	6	45.01	1.6533	9.9199	36
1967	7	46	1.6628	11.6393	49
1968	8	46.57	1.6681	13.3448	64
1969	9	48.34	1.6843	15.1588	81
1970	10	49.07	1.6908	16.9082	100
1971	11	49.92	1.6983	18.6810	121
1972	12	51.07	1.7082	20.4980	144
1973	13	53.48	1.7282	22.4665	169
1974	14	54.09	1.7331	24.2636	196
1975	15	54.97	1.7401	26.1019	225
1976	16	55.87	1.7472	27.9549	256
1977	17	57.75	1.7616	29.9464	289
1978	18	59.01	1.7709	31.8767	324
1979	19	60.47	1.7815	33.8493	361
1980	20	62.46	1.7956	35.9120	400
1981	21	63.91	1.8056	37.9169	441
1982	22	65.14	1.8138	39.9047	484
1983	23	66.75	1.8245	41.9624	529
1984	24	68.43	1.8352	44.0459	576
1985	25	69.93	1.8447	46.1166	625
1986	26	71.21	1.8525	48.1661	676
1987	27	72.84	1.8624	50.2840	729
1988	28	75.22	1.8763	52.5373	784
1989	29	76.67	1.8846	54.6541	841
1990	30	78.16	1.8930	56.7895	900
1991	31	80.34	1.9049	59.0529	961
1992	32	80.66	1.9067	61.0131	1024
1993	33	81.28	1.9100	63.0295	1089
1994	34	82.68	1.9174	65.1916	1156
1995	35	84.79	1.9283	67.4921	1225
1996	36	86.41	1.9366	69.7163	1296
1997	37	87.8	1.9435	71.9093	1369
1998	38	90.3	1.9557	74.3161	1444
1999	39	92	1.9638	76.5877	1521
2000	40	92.82	1.9676	78.7057	1600
2001	41	95.02	1.9778	81.0904	1681
2002	42	96.55	1.9848	83.3596	1764
2003	43	99.35	1.9972	85.8782	1849
2004	44	101.61	2.0069	88.3052	1936
2005	45	103.24	2.0138	90.6232	2025
2006	46	105.94	2.0251	93.1528	2116
2007	47	107.43	2.0311	95.4629	2209
2008	48	109.49	2.0394	97.8900	2304
2009	49	111.18	2.0460	100.2553	2401
2010	50	114.22	2.0577	102.8871	2500
Sum	1275.000		91.980	2441.310	42925.000

?Y = AN+B?X

91.98 = 50A + 1275B       ……(I)

?XY = A?X + B ?X²

2441.31 =1275A + 42925B              …(II)

SOLVING (I) AND (II) WE GET,

A =1.602                               B =0.0092

Now , a = Antilog(1.602) = 39.99,               b = Antilog(0.0092) = 1.021

The exponential regression is thus given as,

Y =40(1.021)^x

To estimate Co2 levels in 2015, we put x = 55

Y =40(1.021)⁵⁵ = 125.4 ppm