Consider the following time series data.
Quarter  Year 1  Year 2  Year 3 
1  2  5  7 
2  0  2  6 
3  5  8  10 
4  5  8  10 
(a)  Choose the correct time series plot.  


Plot (iii) Select your answer Plot (i)Plot (ii)Plot (iii)Plot (iv)Item 1  
What type of pattern exists in the data?  
Positive trend pattern, no seasonality Select your answer Positive trend pattern, no seasonalityHorizontal pattern, no seasonalityNegative trend pattern, no seasonalityPositive trend pattern, with seasonalityHorizontal pattern, with seasonalityItem 2  
(b)  Use a multiple regression model with dummy variables as follows to develop an equation to account for seasonal effects in the data. Qtr1 = 1 if Quarter 1, 0 otherwise; Qtr2 = 1 if Quarter 2, 0 otherwise; Qtr3 = 1 if Quarter 3, 0 otherwise.  
If required, round your answers to three decimal places. For subtractive or negative numbers use a minus sign even if there is a + sign before the blank. (Example: 300) If the constant is "1" it must be entered in the box. Do not round intermediate calculation.  
ŷ = + Qtr1 + Qtr2 + Qtr3  
(c)  Compute the quarterly forecasts for next year based on the model you developed in part (b).  
If required, round your answers to three decimal places. Do not round intermediate calculation.  


(d)  Use a multiple regression model to develop an equation to account for trend and seasonal effects in the data. Use the dummy variables you developed in part (b) to capture seasonal effects and create a variable t such that t = 1 for Quarter 1 in Year 1, t = 2 for Quarter 2 in Year 1,… t = 12 for Quarter 4 in Year 3.  
If required, round your answers to three decimal places. For subtractive or negative numbers use a minus sign even if there is a + sign before the blank. (Example: 300)  
ŷ = + Qtr1 + Qtr2 + Qtr3 + t  
(e)  Compute the quarterly forecasts for next year based on the model you developed in part (d).  
Do not round your interim computations and round your final answer to three decimal places.  


(f)  Is the model you developed in part (b) or the model you developed in part (d) more effective?  
If required, round your intermediate calculations and final answer to three decimal places.  


 Select your answer  Select your answer Model developed in part (b)Model developed in part (d)Item 22  
Justify your answer.  
The input in the box below will not be graded, but may be reviewed and considered by your instructor.  
I just need help with b c d e f thank you! please show all work
Year  Ft  Q1  Q2  Q3  t (period) 
1  2  1  0  0  1 
1  0  0  1  0  2 
1  5  0  0  1  3 
1  5  0  0  0  4 
2  5  1  0  0  5 
2  2  0  1  0  6 
2  8  0  0  1  7 
2  8  0  0  0  8 
3  7  1  0  0  9 
3  6  0  1  0  10 
3  10  0  0  1  11 
3  10  0  0  0  12 
b) Without t
Excel > Data > Data Analysis > Regression
SUMMARY OUTPUT  
Regression Statistics  
Multiple R  0.698535473  
R Square  0.487951807  
Adjusted R Square  0.295933735  
Standard Error  2.661453237  
Observations  12  
ANOVA  
df  SS  MS  F  Significance F  
Regression  3  54  18  2.541176471  0.129679966  
Residual  8  56.66666667  7.083333333  
Total  11  110.6666667  
Coefficients  Standard Error  t Stat  Pvalue  Lower 95%  Upper 95%  Lower 95.0%  Upper 95.0%  
Intercept  7.666666667  1.536590743  4.98940053  0.001066868  4.123282059  11.21005127  4.123282059  11.21005127 
Q1  3  2.173067468  1.38053698  0.204763892  8.011102568  2.011102568  8.011102568  2.011102568 
Q2  5  2.173067468  2.300894967  0.050400371  10.01110257  0.011102568  10.01110257  0.011102568 
Q3  6.28037E16  2.173067468  2.89009E16  1  5.011102568  5.011102568  5.011102568  5.011102568 
Y = 7.6703.000*Q15.000*Q2+0.000*Q3
c)
Ft = 7.6703.000*Q15.000*Q2+0.000*Q3
Quarter  Year  Ft  Q1  Q2  Q3 
1  4  4.670  1  0  0 
2  4  2.670  0  1  0 
3  4  7.670  0  0  1 
4  4  7.670  0  0  0 
d) with t
SUMMARY OUTPUT  
Regression Statistics  
Multiple R  0.99301021  
R Square  0.986069277  
Adjusted R Square  0.978108864  
Standard Error  0.469295318  
Observations  12  
ANOVA  
df  SS  MS  F  Significance F  
Regression  4  109.125  27.28125  123.8716216  1.42033E06  
Residual  7  1.541666667  0.220238095  
Total  11  110.6666667  
Coefficients  Standard Error  t Stat  Pvalue  Lower 95%  Upper 95%  Lower 95.0%  Upper 95.0%  
Intercept  2.416666667  0.428406053  5.641065646  0.000781715  1.403647325  3.429686009  1.403647325  3.429686009 
Q1  1.03125  0.402878254  2.559706285  0.037567703  1.983905691  0.078594309  1.983905691  0.078594309 
Q2  3.6875  0.392055911  9.405546243  3.19973E05  4.614564915  2.760435085  4.614564915  2.760435085 
Q3  0.65625  0.385416667  1.702702703  0.132407607  0.255115597  1.567615597  0.255115597  1.567615597 
t  0.65625  0.041480238  15.82078687  9.77012E07  0.558164824  0.754335176  0.558164824  0.754335176 
Y = 2.4171.031*Q13.688*Q2+0.656*Q3+0.656*t
e)
Ft = 2.4171.031*Q13.688*Q2+0.656*Q3+0.656*t
Quarter  Year  Ft  Q1  Q2  Q3  t 
1  4  9.918  1  0  0  13 
2  4  7.918  0  1  0  14 
3  4  12.918  0  0  1  15 
4  4  12.918  0  0  0  16 
f)
MSE from above output tables(MS Residuals)
Model developed in part (b)  Model developed in part (d)  
MSE  7.083  0.220 
MSE in part b > MSE part d
So, part d model is effective
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