In: Statistics and Probability
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) |
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 |
(a) | Choose the correct time series plot. | ||||||||||||
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- Select your answer -Plot (i)Plot (ii)Plot (iii)Plot (iv)Item 1 | |||||||||||||
What type of pattern "significantly" exists in the data? (Use 1% level of significance when needed) | |||||||||||||
- Select your answer -Only randomnessRandomness & Linear trendRandomness & SeasonalityRandomness, Linear trend & SeasonalityItem 2 | |||||||||||||
(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 year 10? | ||||||||||||
Do not round your interim computations and round your final answer to two decimal places. | |||||||||||||
(d) | Use the Holt's method with smoothing constants of 0.3 for alpha and 0.6 for gamma. Find the equation of the forecast line and the MSE for this method. | ||||||||||||
If required, round your answers to two decimal places. | |||||||||||||
y-intercept, b0 = | |||||||||||||
Slope, b1 = | |||||||||||||
MSE = | |||||||||||||
(e) | What is the forecast for year 10? | ||||||||||||
Do not round your interim computations and round your final answer to two decimal places. | |||||||||||||
(f) | Which of the following methods perform better with respect to MSE? - Select your answer -RegressionHolt's with alpha=0.3, gamma=0.6Holt's with alpha=0.2, gamma=0.2 |
a)
Linear Trend
b)
ΣX | ΣY | Σ(x-x̅)² | Σ(y-ȳ)² | Σ(x-x̅)(y-ȳ) | |
total sum | 45 | 108 | 60 | 130.7 | 87.40 |
mean | 5.00 | 12.00 | SSxx | SSyy | SSxy |
sample size , n = 9
here, x̅ = Σx / n= 5.00 ,
ȳ = Σy/n = 12.00
SSxx = Σ(x-x̅)² = 60.0000
SSxy= Σ(x-x̅)(y-ȳ) = 87.4
estimated slope , ß1 = SSxy/SSxx = 87.4
/ 60.000 = 1.4567
intercept, ß0 = y̅-ß1* x̄ =
4.7167
so, regression line is Ŷ =
4.7167 + 1.4567 *x
period | demand | forcast | forecast error=demand value-forecast value | absolute forecast error | squared forcast error |
t | Dt | Ft | et=Dt-Ft | | et | | (et)² |
1 | 6.5 | 6.173333 | 0.33 | 0.33 | 0.11 |
2 | 8.1 | 7.630 | 0.47 | 0.47 | 0.22 |
3 | 8.4 | 9.087 | -0.69 | 0.69 | 0.47 |
4 | 10.2 | 10.543 | -0.34 | 0.34 | 0.12 |
5 | 12.5 | 12.000 | 0.50 | 0.50 | 0.25 |
6 | 13.3 | 13.457 | -0.16 | 0.16 | 0.02 |
7 | 13.7 | 14.913 | -1.21 | 1.21 | 1.47 |
8 | 17.2 | 16.370 | 0.83 | 0.83 | 0.69 |
9 | 18.1 | 17.827 | 0.27 | 0.27 | 0.07 |
MSE= Σ(et)²/n =
0.38
c)
Predicted Y at X= 10 is
Ŷ = 4.71667 +
1.456667 * 10 =
19.28
THANKS
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