Required Coefficients for Econometrics - Assignment Example

Question 1
Required coefficients in the estimate table 1 become evident through the writing of the needed estimates using the fitted model, μ12 will be
= (ΣkY12k)/n12 
 = (Σkμ)/n12 + (Σkα1)/n12 + (Σkβ2)/n12 + (Σkαβ12)/n12 + (Σkε12k)/n12
We then assume that the averages of the errors in this context are zero (εijk): 
 = μ + α1 + β2 + αβ12 
Question 2
In the same way, the mean in Table 1 will be:
μ11 = μ + α1 + β1 + αβ11
Getting the estimates of Table 1 mean, add the estimates at μ, α1, β2, and αβ12. 
Achieving this goes by multiplying the solution vector with a vector coefficient (e.g. Sum-product)
Solution for Coefficient estimates 
Effect a b Estimate Standard Error DF t Value Pr > |t|
Intercept 21.6100 0.2958 72 83.21 <.0001
a 1 -11.5879 72 -31.53 <.0001
a 2 -9.4008 0.4243 72 -21.31 <.0001
a 3 -5.7874 0.3735 72 -16.02 <.0001
A 4 -2.2722 0.3675 72 -6.18 <.0001
Question 3
In this structure, several elements are in the parameter estimates within the solution vector. First, interpreting μ (Long, 2007, 21). Subsequently, five other parameters run from A, α1 through α5. They follow two sets of elements through B, β1, and β2. 
Question 4
Estimating syntax within the table:
Intercept 1
 a 1 0 0 0 0
 b 0 1
 a*b 0 1 0 0 0 0 0 0 0 0;
Coefficients for a*b Means (Least Squares)
Effect a b Row1 Row2 Row3 Row4 Row5 Row6 Row7 Row8 Row9 Row10
Intercept 1 1 1 1 1 1 1 1 1 1
a 1 1 1  
a 2 1 1  
b 2 1 1 1 1 1
a*b 1 1 1  
a*b 1 2 1  
a*b 2 1 1  
a*b 2 2 1  
a*b 3 1 1  
a*b 3 2 1  
a*b 4 1 1  
a*b 4 2 1  
a*b 5 1 1  
a*b52
Question 5
The interaction effects of estimates A*B have a second estimate of αβ12. It is because of the changes in level B taking place before those in level A. Allowing B to precede A would result in modifications in A before B and therefore, giving positive feedback as opposed to the current negative. 
Question 6
Understanding the changes in variables in the context of the set parameter estimates for an influence is essential in grasping the negativity in this case. 
Question 7
The coefficients of the same estimates were found after rearranging them in a new model. 
μ11 = μ + α1 + β1 + αβ11
μ12 = μ + α1 + β2 + αβ12
Question 8
It entails the process of yielding the right linear combination of parameter estimates. The result in the table (R1) will lead to a table of LS-means coefficients (Long, 2007, 37). 
Question 10
Proc mixed data=test;
 class a b;
 model y=a b a*b;
 means a*b;
 estimate 'I' intercept 1 
 a 1 0 0 0 0 
 b 1 0
 a*b 1 0 0 0 0 0 0 0 0 0;
 estimate 'R2' intercept 1 
 a 1 0 0 0 0 
 b 0 1
 a*b 0 1 0 0 0 0 0 0 0 0;
run;
Estimates
Label Estimate Standard Error DF t Value Pr > |t|
R2 7.5193 0.2905 74 25.88 <.0001
R1 10.0341 0.2598 74 38.62 <.0001
Question 11
  Testing the average of I and R2 in the table requires:
 H0: ½(μ11 + μ12) = ½(μ21 + μ22)
The mean for the two are: 
 ½(μ11+μ12) = ½[(μ + α1 + β1 + αβ11) + (μ + α1 + β2 + αβ12)] 
 = μ + α1 + ½β1 + ½β2 + ½αβ11 + ½αβ12 
Question 12
Proc mixed data=test;
 class a b;
 model y= a b a*b;
 estimate 'avg I, R2’ intercept 1 
 a 1 
 b .5 .5
 a*b .5 .5;
Question 13
 Estimating 'avg AB21, AB22’ intercept 1 
 a 0 1 
 b .5 .5
 a*b 0 0 .5 .5;
 Contrast 'avg I, R2 - avg AB21+AB22' a 1 -1
 a*b .5 .5 -.5 -.5;
estimate a*b 'avg I, R2 - avg AB21+AB22' 1 1 -1 -1 / divisor=2;
run;
Estimates
Label Estimate Standard Error DF t Value Pr > |t|
avg I,R2 8.7767 0.1949 74 45.04 <.0001
avg AB21,AB22 11.5001 0.2165 74 53.11 <.0001
Contrasts
Label Num DF Den DF F Value Pr > F
avg I,R2 - avg AB21+AB22 1 74 87.39 <.0001
Question 14
Similarly, the other estimates follow the same pattern. However, before subtracting the vector yields of the two coefficients, other considerations appear within the coefficient vector applied in testing the differences between the two average elements. The resultant coefficients appear in the CONTRAST and ESTIMATE. It is essential to note that, analysts can leave out trailing zeros in the set of group coefficients.  Read More