Answer:
a. A linear pattern exists in the data.
b. The parameters for the line that minimizes MSE for this time series are as folows:
ß1 = Estimated slope = 1.4567
ß0 = Estimated intercept = 4.7165
Also, we have:
MSE = Mean squared error = 0.4896
c. Forecast for year 10 is 19,280.
Step-by-step explanation:
a. What type of pattern exists in the data?
Note: See Sheet1 of the attached excel file for the line graph.
From the line graph, it can be observed that a linear pattern exists in the data.
b. Use simple linear regression analysis to find the parameters for the line that minimizes MSE for this time series.
Note: See Sheet2 of the attached excel file for all the calculations to obtain the following:
Sample size = 9
Total of X = 45
Total of Y = 108
Mean of X = Total of X / Sample size = 45 / 9 = 5
Mean of Y = Total of X / Sample size = 108 / 9 = 12
SSxx = Total of (X - Mean of X)^2 = 60
SSyy = Total of (Y - Mean of Y)^2 = 130.74
SSxy = Total of (X - Mean of X) * (Y - Mean of Y) = 87.40
Therefore, we have:
ß1 = Estimated slope = SSxy/SSxx = 87.4 / 60 = 1.4567
ß0 = Estimated intercept = Mean of Y – (ß1 * Mean of X) = 12 - (5 * 1.4567) = 4.7165
Therefore, the parameters for the line that minimizes MSE for this time series are as folows:
ß1 = Estimated slope = 1.4567
ß0 = Estimated intercept = 4.7165
Regression equation which also used in the attached excel is as follows:
Y = ß0 + ß1X =
Y = 4.7165 + 1.4567X …………………. (1)
SSE = Sum of squared error = Total of (Y - Y*)^2 = 3.4273
Therefore, we have:
MSE = Mean squared error = (SSE/(n-2)) = (3.4273 / (9 - 2)) = 0.4896
c. What is the forecast for year 10?
This implies that X = 10
Substitute X = 10 into equation (1), we have:
Y = 4.7165 + (1.4567 * 10) = 19.28
Since it is 1,000s, we have:
Y = 19.28 * 1,000 = 19,280
Therefore, forecast for year 10 is 19,280.
