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Test for Lack of Fit

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    242
  • [ "article:topic", "authorname:pauld" ]

    Lack of Fit 

    When we have repeated measurements for different values of the predictor variables \(X\), it is possible to test whether a linear model fits the data.

    Suppose that we have data that can be expressed in the form:

    \( \{(X_j,Y_{ij}) : i = 1, ..., n_{j}; j = 1, ..., c\}\)where \(c >2\).

    Assume that the data come from the model  :

    \[Y_{ij} = \mu_{j} + \varepsilon_{ij}, i = 1, ..., n_{j}; j = 1, ..., c                                                        (1)\].

    The null hypothesis in which the linear model holds is: \(H_{0}: \mu_{j} = \beta_{0} + \beta_{1}X_{j}\), for all \(j = 1, ..., c\).

    Here (1) is the full model and the model specified by \(H_{0}\) is the reduced model. We follow the usual procedure for the ANOVA, by computing the sum of squares due to errors for the full and reduced models.

    Let \(\bar{Y} = \frac{1}{n_{j}} \sum_{i = 1}^{n_{j}} Y_{ij}\), and \(\bar{Y} = \frac{1}{c}\sum_{j=1}^{c}n_{j}\bar{Y}_{j} = \frac{1}{n}\sum_{j=1}^{n}\sum_{i=1}^{n_j}Y_{ij}\), where \(n = \sum_{j=1}^{c}n_{j}\).

    \(SSTO = \sum_{j=1}^{c}\sum_{i=1}^{n_j}(Y_{ij} - \bar{Y})^2\) and

    \[SSPE = SSE_{full} = \sum_{j=1}^{c}\sum_{i=1}^{n_j}(Y_{ij} - \bar{Y}_{j})^2 = \sum_{j=1}^{c}\sum_{i=1}^{n_j}Y_{ij}^2 - \sum_{j=1}^{c}n_{j}\bar{Y}_{j}^2\]

    \[SSE_{red} = SSE = \sum_{j=1}^{c}\sum_{i=1}^{n_j}(Y_{ij} - \beta_{0} - \beta_{1}X_{j})^2\]

    \[SSLF = SSE_{red} - SSE_{full}\].

    Degrees of freedom

    \[d.f.(SSPE) = n - c; d.f.(SSLF) = d.f.(SSE_{red}) - d.f.(SSPE) = (n - 2) - (n - c) = c - 2.\]

     

    ANOVA Table
    Source d.f. SS MS=SS/d.f. F-statistic
    Regression 1 SSR MSR MSR/MSE
    Error n-2 SSE=SSLF+SSPE MSE  
    Lack of fit c-2 SSLF MSLF MSLF/MSPE
    Pure error n-c SSPE MSPE  
    Total n-1 SSTO=SSR+SSLF+SSPE    

     

    Reject \(H_{0} : (\mu_{j} = \beta_{0} + \beta_{1}X_{j} for  all  j)\) at level \(\alpha\) if \(F^*_{LF} = \frac{MSLF}{MSPE} > F(1 - \alpha; c - 2, n - c)\).

    Example: Growth rate data

    In the following example, data are available on the effect of dietary supplement on the growth rates of rats. Here \(X = \) dose of dietary supplement and \(Y =\) growth rate. The following table presents the data in a form suitable for the analysis.

    \(j = 1\)
    \(X_{1} = 10\)
    \(n_{1} = 2\)

    \(j = 2\)
    \(X_{2} = 15\)
    \(n_{2} = 2\)

    \(j = 3\)
    \(X_{3} = 20\)
    \(n_{3} = 2\)

    \(j = 4\)
    \(X_{4} = 25\)
    \(n_{4} = 3\)

    \(j = 5\)
    \(X_{5} = 30\)
    \(n_{5} = 1\)

    \(j = 6\)
    \(X_{6} = 35\)
    \(n_{6} = 2\)

    \(Y_{11} = 73\)
    \(Y_{21} = 78\)
    \(Y_{12} = 85\)
    \(Y_{22} = 88\)
    \(Y_{13} = 90\)
    \(Y_{23} = 91\)
    \(Y_{14} = 87\)
    \(Y_{24} = 86\)
    \(Y_{34} = 91\)
    \(Y_{15} = 75\) \(Y_{16} = 65\)
    \(Y_{26} = 63\)
    \(\bar{Y}_{1} = 75.5\) \(\bar{Y}_{2} = 86.5\) \(\bar{Y}_{3} = 90.5\) \(\bar{Y}_{4} = 88\) \(\bar{Y}_{5} = 75\) \(\bar{Y}_{6} = 64\)

    So, for this data, \(c = 6, n = \sum_{j = 1}^{c}n_{j} = 12\).

    \(SSTO = \sum_{j}\sum_{i}(Y_{ij} - \bar{Y})^2 = 1096.00\)

    \(SSPE = \sum_{j}\sum_{i}Y_{ij}^2 - \sum_{j}n_{j}\bar{Y}_{j}^2 = 79828 - 79704.5 = 33.50\)

    \(SSE_{red} = \sum_{i}\sum_{j}(Y_{ij} - \beta_{0} - \beta_{1}X_{j})^2 = 891.73 (Note: \beta_{0} = 92.003, \beta_{1} = -0.498)\)

    \(SSLF = SSE_{red} - SSPE = 891.73 - 33.50 = 858.23\)

    \(d.f.(SSPE) = n - c = 6\)

    \(d.f.(SSLF) = c - 2 = 4\)

    \(MSLF = \frac{SSLF}{c - 2} = 214.5575\)

    \(MSPE = \frac{SSPE}{n -c} = 5.5833\)

    ANOVA Table:

    Source d.f. SS MS F*
    Regression
    Error
    1
    10
    204.27
    891.73
    204.27
    89.173
    \(\frac{MSR}{MSE} =\) 2.29
    Lack of fit
    Pure error
    4
    6
    858.23
    33.50
    214.56
    5.583
    \(\frac{MSLF}{MSPE} =\) 38.43
    Total 11 1096.00    

    \(F(0.95;4,6) = 4.534\). Since \(F_{LF}^\star = 38.43 > 4.534\), reject \(H_{0} :\) \( \mu_{j} = \beta_{0} + \beta_{1}X_{j} \) for all j at 5% level of significance.

    Also, if you are testing \(H_{0}' :\)  \(\beta_{1} = 0\) against \(H_{1}' :\)  \(\beta_{1} \neq 0\), assuming that the linear model holds, as in the
    usual ANOVA for linear regression, then the corresponding test statistic \(F^\star = \frac{MSR}{MSE} = 2.29\). Which is less than
    \(F(0.95;1,n-2) = F(0.95;1,10) = 4.964\). So, if one assumes that linear model holds then the test cannot reject \(H_{0}' \) at 5% level of significance.

    Contributors

    • Joy Wei
    • Debashis Paul