1. Cases of F-test Problems with Examples
Lecture 10
Review of Lecture 9
Cases of F-test Problems with Examples
Review of Lecture 9:
Three Basic Concepts:
Full Model: contain ALL coefficients of interest
Reduced Model: contain PART of the coefficients of interest
Nested Model : one model is a SUBMODEL of the other one
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ST3131, Lecture 10

2. Common Cases for Model Testing
Case 1: ALL NON-intercept coefficients are zero
Case 2: SOME of the coefficients are zero
Case 3: SOME of the coefficients are EQUAL to each other
Case 4: Other specified CONSTRAINTS on coefficients
All can be tested using F-test.
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ST3131, Lecture 10

3. Steps for Model Comparison :
RM H0: The RM is adequate vs FM H1: The FM is adequate
Step1: Fit the FM and get SSE (in the ANOVA table)
df (in the ANOVA table)
R_sq (under the Coefficient Table)
Step 2: Fit the RM and get SSE, df, and R_sq.
Step 3: Compute F-statistic:
Step 4: Conclusion: Reject H0 if F>F(r,df(SSE,F),alpha)
Can’t Reject H0 otherwise.
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4. Case 1: ALL NON-intercept coefficients are zero
Statistical Meaning: ALL predictor variables have no explanatory power
the effect of ALL predictor variables are zero.
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ST3131, Lecture 10

5. MSR(F)= Mean Square due to REGRESSION for the Full Model
MSE(F)=Mean Square due to ERROR for the Full Model
=Mean Squared Error for the Full Model
The Test can be conducted using an ANOVA (ANalysis Of VAriance) table:
Source
Sum of Squares df
Mean Square
F-test
P-value
Regression
SSR p
MSR=SSR/p
F=MSR/MSE
Residuals
SSE
n-p-1
MSE=SSE/(n-p-1)
Total
SST
n-1
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ST3131, Lecture 10

6. Thus, Reject H0, i.e., not all coefficients are zero.
Example: the Supervisor Performance Data
Analysis of Variance (ANOVA table)
Source DF SS MS F P
Regression
Residual Error
Total
Thus, Reject H0, i.e., not all coefficients are zero.
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ST3131, Lecture 10

7. The F-test using Multiple Correlation Coefficients R-square
The F-test based on ANOVA table is equivalent to the following test based on R-square:
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ST3131, Lecture 10

8. Example: the Supervisor Performance Data
Results for: P054.txt
Regression Analysis: Y versus X1, X2, X3, X4, X5, X6
The regression equation is
Y = X X X X X X6
Predictor Coef SE Coef T P
Constant X X X X X X
S = R-Sq = 73.3% R-Sq(adj) = 66.3%
F(6,23,.05)=F(6,24,.05)=2.455, F(6,23,.01)=3.80
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9. REMARK: when some of individual coefficients (t-test) are significant, the F-test for
testing if all non-intercept coefficients are zero will usually be significant.
However, it is possible that none of the non-intercept coefficient t-tests are significant but the F-test is still significant. This implies that the combining explanatory power of the predictor variables is larger than that of any one of the individual predictor variables.
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10. Case 2: Some of the Coefficients are zero.
If H0 is not rejected and hence is adequate, we should use the Reduced Model.
The Principle of Parsimony: always use the ADEQUATE SIMPLER model.
Two advantages for using the Reduced Models are
1) Reduced Models are simpler than Full Models
2) The RETAINED predictor variables are emphasized.
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11. SSE(R)=, df(R) =, SSE(F)= df(F)=
Example: the Supervisor Performance Data (Continued)
Regression Analysis: Y versus X1, X3
The regression equation is
Y = X X3
Predictor Coef SE Coef T P
Constant X X
S = R-Sq = 70.8% R-Sq(adj) = 68.6%
Analysis of Variance
Source DF SS MS F P
Regression
Residual Error
Total
SSE(R)=, df(R) =, SSE(F)= df(F)=
F=[(SSE(R )-SSE(F))/(df(R )-df(F))]/[SSE(F)/df(F)]=
F(4,23,.05)=2.8, Can’t reject H0; the RM is adequate!
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12. REMARKS:
(1) F-test can be written in terms of the Multiple Correlation Coefficients of the RM and FM. That is,
Actually
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13. Example: the Supervisor Performance Data (Continued)
R_sq(F)= .733, df(F)=23, R_sq(R)=.708, df(R )=27,
F=[( )/4]/[(1-.733)/23]=.528<2.8=F(4,23,.05), can’t Reject H0
Remark (2): When the RM has only 1 coefficient fewer than the FM, say, beta_j, then r=1,
In this case, F-test is equivalent to t-test.
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14. Two Remarks about the Coefficients Retaining
The estimates of regression coefficients that do not significantly differ from 0 are often replaced by 0 in the equation. The replacement has two advanatges: a simple model and a smaller prediction variance (the Principle of Parsimony).
A variable or a set of variables may particularly be retained in an equation because of their theoretical importance in a given problem, even though the coefficients are statistically insignificant. For example, the intercept is often retained in the equation even it is not significant in statistical meaning.
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15. Case 3: Some Coefficients are equal
(II)
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16. SSE (R ) = , df(R )= SSE(F)= df(F)=
Example: the Supervisor Performance Data
Results for: P054.txt, Regression Analysis: Y versus X1+X3
The regression equation is Y = (X1+X3)
Predictor Coef SE Coef T P
Constant
X1+X
S = R-Sq = 66.8% R-Sq(adj) = 65.7%
Analysis of Variance
Source DF SS MS F P
Regression
Residual Error
Total
SSE (R ) = , df(R )= SSE(F)= df(F)=
F={[(SSE(R )-SSE(F)]/[df(R )-df(F)]}/{SSE(F)/df(F)}=
=1.10, df=(5,23)
F(5,23,.05)=2.49>1.10
So H0 is NOT Rejected.
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ST3131, Lecture 10

17. SSE (R ) = , df(R )= SSE(F)= df(F)=
Example: the Supervisor Performance Data (Continued)
Regression Analysis: Y versus X1, X3
The regression equation is Y = X X3
Predictor Coef SE Coef T P
Constant X X
S = R-Sq = 70.8% R-Sq(adj) = 68.6%
Analysis of Variance
Source DF SS MS F P
Regression
Residual Error
Total
SSE (R ) = , df(R )= SSE(F)= df(F)=
F={[(SSE(R )-SSE(F)]/[df(R )-df(F)]}/{SSE(F)/df(F)}=
=3.65, df=(1,27)
F(1,27,.05)=4.21>3.65
So H0 is NOT Rejected.
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18. Case 4: Other constraints on Coefficients
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19. Example: the Supervisor Performance Data (continued)
Regression Analysis: Y-X3 versus X1-X3
The regression equation is
(Y-X3) = (X1-X3) Y= X X3
Predictor Coef SE Coef T P
Constant
X1-X
S = R-Sq = 57.4% R-Sq(adj) = 55.9%
Analysis of Variance
Source DF SS MS F P
Regression
Residual Error
Total
SSE (R ) = , df(R )= SSE(F)= df(F)=
F={[(SSE(R )-SSE(F)]/[df(R )-df(F)]}/{SSE(F)/df(F)}=
=1.62, df=(1,27)
F(1,27,.05)=4.21>1.62, So H0 is NOT Rejected.
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20. After-Class Questions:
What is the difference between F-test and T-test?
If H0 is rejected, does this show that the full model
is better than the reduced model?
11/16/2018
ST3131, Lecture 10