Example of fitting a quadratic response surface model
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In the example for a linear response surface model, you determined that the linear model did not adequately represent the response surface. The next step is to fit the quadratic model. The quadratic model allows detection of curvature in the response surface.

To reduce the impact of non-orthogonal terms, Minitab fits the model in coded units.

1    Open the worksheet CCD_EX1.MTW. (The design and response data have been saved for you.)

2    Choose Stat > DOE > Response Surface > Analyze Response Surface Design.

3    In Responses, enter BeanYield.

4    Click Terms.

5    From Include the following terms, choose Full quadratic. Click OK.

6    Click Graphs.

7    Under Residual Plots, choose Four in one. Click OK in each dialog box.

Session window output

Response Surface Regression: BeanYield versus Nitrogen, PhosAcid, Potash

 

 

Analysis of Variance

 

Source                 DF   Adj SS   Adj MS  F-Value  P-Value

Model                   9  36.4653   4.0517     4.08    0.019

  Linear                3   7.7886   2.5962     2.62    0.109

    Nitrogen            1   4.4960   4.4960     4.53    0.059

    PhosAcid            1   0.4593   0.4593     0.46    0.512

    Potash              1   2.8332   2.8332     2.86    0.122

  Square                3  13.3857   4.4619     4.50    0.030

    Nitrogen*Nitrogen   1   6.5943   6.5943     6.65    0.027

    PhosAcid*PhosAcid   1   4.5640   4.5640     4.60    0.058

    Potash*Potash       1   1.0772   1.0772     1.09    0.322

  2-Way Interaction     3  15.2909   5.0970     5.14    0.021

    Nitrogen*PhosAcid   1   3.6721   3.6721     3.70    0.083

    Nitrogen*Potash     1  11.1865  11.1865    11.28    0.007

    PhosAcid*Potash     1   0.4325   0.4325     0.44    0.524

Error                  10   9.9198   0.9920

  Lack-of-Fit           5   7.3802   1.4760     2.91    0.133

  Pure Error            5   2.5396   0.5079

Total                  19  46.3851

 

 

Model Summary

 

       S    R-sq  R-sq(adj)  R-sq(pred)

0.995984  78.61%     59.37%       0.00%

 

 

Coded Coefficients

 

Term               Effect    Coef  SE Coef  T-Value  P-Value   VIF

Constant                   10.462    0.406    25.76    0.000

Nitrogen           -1.148  -0.574    0.270    -2.13    0.059  1.00

PhosAcid            0.367   0.183    0.270     0.68    0.512  1.00

Potash              0.911   0.455    0.270     1.69    0.122  1.00

Nitrogen*Nitrogen  -1.353  -0.676    0.262    -2.58    0.027  1.02

PhosAcid*PhosAcid   1.126   0.563    0.262     2.14    0.058  1.02

Potash*Potash      -0.547  -0.273    0.262    -1.04    0.322  1.02

Nitrogen*PhosAcid  -1.355  -0.678    0.352    -1.92    0.083  1.00

Nitrogen*Potash     2.365   1.182    0.352     3.36    0.007  1.00

PhosAcid*Potash     0.465   0.233    0.352     0.66    0.524  1.00

 

 

Regression Equation in Uncoded Units

 

BeanYield = 12.45 + 0.96 Nitrogen - 2.28 PhosAcid - 1.48 Potash - 0.268 Nitrogen*Nitrogen

            + 1.116 PhosAcid*PhosAcid - 0.239 Potash*Potash - 0.600 Nitrogen*PhosAcid

            + 0.695 Nitrogen*Potash + 0.306 PhosAcid*Potash

 

 

Fits and Diagnostics for Unusual Observations

 

Obs  BeanYield     Fit   Resid  Std Resid

  2     11.060  12.362  -1.302      -2.09  R

  7      8.260   9.514  -1.254      -2.01  R

 16     13.190  12.004   1.186       2.07  R

 

R  Large residual

Graph window output

Interpreting the results

Because the linear model suggested that a higher-order model is needed to adequately model the response surface, you fit the full quadratic model. For the full quadratic model, the p-value for lack of fit is 0.133 suggesting that this model adequately fits the data.

The Analysis of Variance table summarizes the linear terms, the squared terms, and the interactions. The small p-values for the interactions (p = 0.021) and the squared terms (p = 0.030) suggest there is curvature in the response surface. Small p-values for the Nitrogen by Potash interaction (p = 0.007), Nitrogen squared (p = 0.027), and Phosphoric acid squared (p = 0.058) indicate that these effects are statistically significant.

The coefficient table gives the coefficients for all the terms in the model. Because you used an orthogonal design, each effect is estimated independently. Therefore, the coefficients for the linear terms are the same as when you fit just the linear model. The error term, s = 0.995984, is smaller because you reduced the variability accounted for by error.

The VIFs are all close to 1, which indicates that the predictors are not correlated. VIF values greater than 5-10 suggest that the regression coefficients are poorly estimated due to severe multicollinearity.

In addition, Minitab draws four residual plots. The residual plots do not indicate any problems with the model. For assistance in interpreting residual plots, see Residual plot choices.

For contour and surface plots of this response surface, see Example of a Contour Plot and a Surface Plot.