Because of high tuition costs at state and private universities, enrollments at community colleges have increased dramatically in recent years. The following data show the enrollment (in thousands) for Jefferson Community College for the nine most recent years. Year Period (t) Enrollment (1,000s)
2001 1 6.5
2002 2 8.1
2003 3 8.4
2004 4 10.2
2005 5 12.5
2006 6 13.3
2007 7 13.7
2008 8 17.2
2009 9 18.1

Required:
a. What type of 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.
c. What is the forecast for year 10?

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.

nope

step-by-step explanation:

answer: 12

step-by-step explanation:

because yes