Question

Consider the following time series.

t	1	2	3	4	5
yt	5	11	8	14	15

(b)	Use simple linear regression analysis to find the parameters for the line that minimizes MSE for this time series.
	If required, round your answers to two decimal places.
	y-intercept, b0 =
	Slope, b1 =
	MSE =
(c)	What is the forecast for t = 6?
	If required, round your answer to one decimal place.

Answer 1

Let the Regression line be y = b_o + b₁t

where,

b_o = ( Σy Σt² - Σt Σty ) / ( nΣt² - (Σt)² )

b₁ = ( nΣty - ΣtΣy ) / ( nΣt² - (Σt)² )

	t	y	t2	ty
	1	5	1	5
	2	11	4	22
	3	8	9	24
	4	14	16	56
	5	15	25	75
Total	15	53	55	182

b_o = ( 53*55 - 15*182 ) / ( 5*55 - 15² ) = 3.7

b₁ = ( 5*182 - 15*53 ) / ( 5*55 - 15² ) = 2.3

Hence, y = 3.7 + 2.3t

Forecasting the value using the above trend line.

MSE is the average of squared error

t	y	forecast	error	squared error
1	5	6	-1	1
2	11	8.3	2.7	7.29
3	8	10.6	-2.6	6.76
4	14	12.9	1.1	1.21
5	15	15.2	-0.2	0.04
			MSE	3.26

Hence, MSE = 3.26

(c) when t = 6, forecast y = 3.7 + 2.3*6 = 17.5