Econometrics Assignment
. Estimate the values of β_{0 }andβ_{1}
Y_{1}= β_{0}+ β_{1}X_{1}+u_{1}
Where Y_{1 }is the dependent variable
X is the independent variable
β_{0 }is the y intercept (the value of Y when X is zero)
β_{1 }is the slope of the line
u_{1 }is the residual or error term
β_{1}= N (∑XY)-(∑X) (∑Y)/N (∑X^{2})- (∑X)^{2}
β_{1}= 10(4485) -(155) (278)/10(2453) -(155)^{2}
β_{1 }= 44850-43090/ 24530-24025
β_{1 }= 1760/505
β_{1 }= 3.49
β_{0 }= M_{y }– β_{1}M_{x}
β_{0 }= 27.8- (15.5×3.49)
β_{0 }= 27.8-54.095
β_{0 }= -26.30
2. Calculate the error u_{1 }for all the observations made
Y_{1 }= -26.30+ 3.49X_{1}+u_{1}
1st observation
Replace the y and x values in the equation
22 = -26.30+ (3.49×12) + u_{1}
22 = 15.58+ u_{1}
U_{1 }= 6.42
2^{nd} Observation
38 = -26.30+(3.49×17) + u_{1}
38 = 33.03+u_{1}
U_{1 }= 4.97
3^{rd} Observation
31 = -26.30+ (3.49×16) + u_{1}
31 = 29.54+ u_{1}
U_{1}= 1.46
4^{th} Observation
15 = -26.30 + (3.49x 13) + u_{1}
U_{1 }= -4.07
5^{th} Observation
20 = -26.30+ (3.49×16) + u_{1}
U_{1 }= -9.54
6^{th} Observation
27 = -26.30+ (3.49×17) + u_{1}
U_{1 }= -6.03
7^{th} Observation
41 = -26.30+ (3.49×19) + u_{1}
U_{1 }= 0.99
8^{th} Observation
22 = -26.30+ (3.49×14) + u_{1}
U_{1 }= -0.56
9^{th} Observation
43 = -26.30+ (3.49×18) + u_{1}
U_{1 }= 6.48
10^{th} Observation
19 = -26.30+ (3.49×13) + u_{1}
U_{1 }= -0.07
Summary of the error values_{}
Observations | Error (u_{1}) | (u_{1})^{2} | Deviation from mean | (Deviation)^{2} |
1 | 6.42 | 41.22 | 3.69 | 13.62 |
2 | 4.97 | 24.70 | 2.24 | 5.02 |
3 | 1.46 | 2.13 | -1.27 | 1.61 |
4 | 4.07 | 16.56 | 1.34 | 1.80 |
5 | 9.54 | 91.01 | 6.81 | 46.38 |
6 | -6.03 | 36.36 | -8.76 | 76.74 |
7 | 0.99 | 0.98 | -1.74 | 3.03 |
8 | -0.56 | 0.31 | -3.29 | 10.82 |
9 | 6.48 | 41.99 | 3.75 | 14.06 |
10 | -0.07 | 0.00 | -2.8 | 7.84 |
Mean | 2.73 | |||
Sum | 255.26 | 180.92 |
3. Calculate and interpret the standard error of regression
The formula for calculating the regression error is Standard Error = √Sum of error squared/number of observations
The error values give the distance between the actual data points and the regression line. The regression error evaluates the precise nature of the model in predictions based on the dependent variable units. A smaller value of the standard error signifies a lower distance between the regression line and the data points and is therefore desired. A smaller value indicates that the model is sufficiently precise.
The standard error of regression for the sample is calculated below
Standard Error = 5.05
The value obtained is relatively high, thus it can be concluded that the model is not highly sufficient in making the predictions.
4. Calculate and interpret the standard error for B_{1}
Standard Error = 1.10
5. Test the null hypothesis H_{0}:B_{1}=0 (p=0.05)
Test statistics = x-µ/SD/√n
SD = 9.43
Test Statistics = (27.8-0)/(9.43/3.16)
Test Statistics = 9.32
The z score falls outside the range, justifying the rejection of the null hypothesis.
Therefore, there is no sufficient evidence to accept the hull hypothesis
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