Question

Supose you have a table with the next values that represent the change of temperature in 10 hours, based on the data of the table respond the next questions

Hour	Temperature (K)
9:00	287.5
10:00	289.3
11:00	290.75
12:00	292.8
13:00	294.5
14:00	296.37
15:00	298.15
16:00	300.00
17:00	301.8
18:00	303.5

Considering that the measurements can only be taken at those hours and there can be small errors in them, if you were asked to know what was the temperature at 14:30 it would need to use linear regression.
1- Describe as better you can the algorithm of that method
2- Explain and express clearly the outcome of that hour (14:30) using linear regression

3- Respecting the linear regression, Is the given temperature exactly what it was at that hour?
4- How could you know exactly the answer to this question?

Answer 1

The given data is:

Hour (H)	Temperature (K)
9	287.5
10	289.3
11	290.75
12	292.8
13	294.5
14	296.37
15	298.15
16	300
17	301.8
18	303.5

We can fit this data using the Linear Regression Algorithm. Hence, it would follow an equation of the form K = mH + c, where m is the slope of the line and c is the y-intercept. To solve for the values of these parameters, we solve the equation:

where N is the number of data points. Hence, N = 10.

First, we shall evaluate the sums that are required in the formulae. Doing so, we get:

	Hour (H)	Temperature (K)	HK	H^2
	9	287.5	2587.5	81
	10	289.3	2893	100
	11	290.75	3198.25	121
	12	292.8	3513.6	144
	13	294.5	3828.5	169
	14	296.37	4149.18	196
	15	298.15	4472.25	225
	16	300	4800	256
	17	301.8	5130.6	289
	18	303.5	5463	324
TOTAL	135	2954.67	40035.88	1905

Substituting the values in the formulae, we get:

Hence, the equation of the linear regression model becomes:

Using this model, we can calculate the temperature at 14:30. This can be done by substituting H=14.5 in the model. Doing so, we get:

Hence, the temperature as predicted by the model at 14:30 is 297.26 K.

To check for how accurate the model is predicting the values, we can determine the coefficient of determination.

To do so, we first normalize each of the values of H and K. This is done by subtracting each value from the mean of that column and then dividing it by the standard deviation. Let Z_H be the normalized H and Z_K be the normalized K. Thus, we get:

Hour (H)	Temperature (K)	Z_H	Z_K
9	287.5	-1.4863	-1.4682
10	289.3	-1.156	-1.1365
11	290.75	-0.8257	-0.8693
12	292.8	-0.4954	-0.4915
13	294.5	-0.1651	-0.1782
14	296.37	0.1651	0.1664
15	298.15	0.4954	0.4944
16	300	0.8257	0.8354
17	301.8	1.156	1.1671
18	303.5	1.4863	1.4804

Next, we determine the product of Z_H and Z_K, we get:

Hour (H)	Temperature (K)	Z_H	Z_K	Z_H * Z_K
9	287.5	-1.4863	-1.4682	2.1822
10	289.3	-1.156	-1.1365	1.3138
11	290.75	-0.8257	-0.8693	0.7178
12	292.8	-0.4954	-0.4915	0.2435
13	294.5	-0.1651	-0.1782	0.0294
14	296.37	0.1651	0.1664	0.0275
15	298.15	0.4954	0.4944	0.2449
16	300	0.8257	0.8354	0.6898
17	301.8	1.156	1.1671	1.3492
18	303.5	1.4863	1.4804	2.2003

The coefficient of determination si given using the formula:

Summing the last column, we get:

Substituting this in the above formula, we get:

Since, R2 = 1, the model is giving exact values for each of the intermediate points. Thus, the value obtained using the model at time 14:30 is exact and reliable.