Question

Problem 8-19 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. Click on the datafile logo to reference the data. Year Period (t) Enrollment (1,000s) 1 1 6.5 2 2 8.1 3 3 8.4 4 4 10.2 5 5 12.5 6 6 13.3 7 7 13.7 8 8 17.2 9 9 18.1 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 = 4.7.17
Slope, b1 = 1.46
MSE = ???????? NEED THIS
What is the forecast for year 10? 19.283
Round your interim computations and final answer to two decimal places.

Answer 1

Answer:

a) find the attached graph

b) find the attachment no 4 and 5

c)$T_{10}= 4.72+1.46(10) = 19.28$

Step-by-step explanation:

a) A trend pattern exist if the time series plot gradually shifts to higher or lower values over a long period of time

find the attached graph

b) Liner Trend Equation

$T_{1} =b_{0} +b_{1}t$

Where $T_{1}$ is the linear trend forecast in period t , $b_{0}$ is the intercept of the linear trend time, $b_{1}$ is the slope of the linear trend line, t is the time period

now computing the slope and intercept

formula is attached ( 3 no attachment)

$Y_{t}$ is the value of the time series in period t, n is the number of time periods

Y(bar) is the average value and t(bar) is the average value of t

due to unavailability of equation in math-script i attached the calculation part of this question( 4th and 5th no attachment)

thus the linear trend equation is $T_{t}= 4.72+1.46t$                         (1)

$T_{10}= 4.72+1.46(10) = 19.28$