1. Solutions of Tutorial 10 SSE df RMS Cp Radjsq SSE1 F Xs c).
1. a). and b)
SSE df RMS Cp Radjsq SSE1 F Xs
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none X4
X3 X4
X1 X3 X4
*** *** X1 X2 X3 X4
c).
Step 1, X4 is introduced since F=(SSE0-SSE4)/SSE4*38=( )/38863*38= >Fin=1.2
Step 2, X3 is introduced since F=(SSE4-SSE34)/SSE34*37=( )/28804*37=12.92>Fin=1.2
Step 3, X1 is introduced since F=(SSE34-SSE134)/SSE134*36=( )/27554*36=1.633>Fin=1.2
Step 4, X2 should not be introduced since F=(SSE134-SSE1234)/SSE1234*35=.038<Fin=1.2.
The best model is (Y, X1, X3, X4).
d). H0: beta2= vs H1:beta2 is not 0
F=.038 [ see Part c) ], df=(1,35), F-value at alpha=.05 with df=(1,35) is larger than So H0 is not rejected. That is, the Reduced model is adequate.
11/21/2018
ST3131, Solution 10

2. Y X X X X X5 X
0.000 X X X X X
Cell Contents: Pearson correlation
P-Value
2. Step 1. The right-hand side is the correlation coefficient table. According to the forward selection procedure, we should try to first introduce X6 since it has the largest significant correlation with Y, which is See the fitted results in next page.
11/21/2018
ST3131, Solution 10

3. Since the t-test value T=23.60>1, we introduce X6 to the model.
Regression Analysis: Y versus X6
The regression equation is
Y = X6
Predictor Coef SE Coef T P
Constant X
S = R-Sq = 93.6% R-Sq(adj) = 93.4%
Analysis of Variance
Source DF SS MS F P
Regression
Residual Error
Total
Since the t-test value T=23.60>1, we introduce X6 to the model.
11/21/2018
ST3131, Solution 10

4. Step 2. The correlation coefficient table between YoX6 vs X1, X2, X3, X4, and X5 is listed in the right-hand side. It seems that we should introduce X4 into the model since X4 has the largest significant correlation coefficient with YoX6.
Correlations: YoX6, X1, X2, X3, X4, X5
YoX X X X X4 X
0.417 X X X X
Cell Contents: Pearson correlation
P-Value
11/21/2018
ST3131, Solution 10

5. Regression Analysis: YoX6 versus X4
The regression equation is
YoX6 = X4
Predictor Coef SE Coef T P
Constant X
S = R-Sq = 25.9% R-Sq(adj) = 23.9%
Analysis of Variance
Source DF SS MS F P
Regression
Residual Error
Total
It seems we should introduce X4 into the model since the t-test value T=3.64>1.
11/21/2018
ST3131, Solution 10

6. YoX X X X3 X
0.437 X X X
Step 3. It seems we should try to add X1 into the model since it has the largest correlation coefficient with YOX64. However, we can not add X1 into the model since the t-test |T|=.79<1. The p-value is about 43.7% which is very large.
We stop the procedure here and the resulting model is Y vs X4 and X6.
Regression Analysis: YoX64 versus X1
The regression equation is
YoX64 = X1
Predictor Coef SE Coef T P
Constant X
S = R-Sq = 1.6% R-Sq(adj) = 0.0%
11/21/2018
ST3131, Solution 10

7. 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 = 95.6% R-Sq(adj) = 94.8%
3. Step 1. A full model is fitted to the data. The predictor variable with the smallest absolute t-test value |T|=.22 is X6. The |T|<Tout=1. We then first remove X6 from the model.
11/21/2018
ST3131, Solution 10

8. 11/21/2018 ST3131, Solution 10 The regression equation is
Y = X X X X X5
Predictor Coef SE Coef T P
Constant X X X X X
S = R-Sq = 95.6% R-Sq(adj) = 94.9%
Step 2. The data are then fitted Y vs X1, X2, X3, X4, and X5. The predictor variable with smallest absolute t-test value |T|=.75 is X5, which is less than the cutoff value Tout=1. We then remove X5 from the model.
11/21/2018
ST3131, Solution 10

9. Regression Analysis: Y versus X1, X2, X3, X4
The regression equation is
Y = X X X X4
Predictor Coef SE Coef T P
Constant X X X X
Step 3. The model Y vs X1, X2, X3, and X4 is then fitted to the data. All the absolute t-test values |T|>1, the cutoff value. Thus, we stop here and the best model is Y vs X1, X2, x3, X4.
4. The following two models obtained from Problem 2 and 3 are not the same:
Y vs X4 and X6, from the Forward Selection Procedure
Y vs X1, X2, X3, and X4, from the Backward Elimination Procedure.
They share a same predictor X4. It seems X6 plays a similar role as X1, X2, X3. We guess X6 should have a strong linear relationship with X1, X2, X3, and X4, so that its effect on Y can be replaced by the combined effect of X1, X2, X3, and possibly X4 on Y. To verify this, we fit X6 vs X1, x2, X3, and X4. The results are presented in next page.
11/21/2018
ST3131, Solution 10

10. Regression Analysis: X6 versus X1, X2, X3, X4
The regression equation is
X6 = X X X X4
Predictor Coef SE Coef T P
Constant X X X X
S = R-Sq = 100.0% R-Sq(adj) = 100.0%
Regression Analysis: X6 versus X1, X2, X3
X6 = X X X3
Constant X X X
S = R-Sq = 100.0% R-Sq(adj) = 100.0%
We can see that X6 is linearly explained by X1, X2, and X3 with 100%. This verifys our guess.
11/21/2018
ST3131, Solution 10

11. The matrix plot above shows that X6 has strong linear relationship with X1, X2, and X3. This is collinear. We can see that the collinearity often affects the model selection results.
11/21/2018
ST3131, Solution 10