Test for Lack of Fit
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- 242
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
\[d.f.(SSPE) = n - c; d.f.(SSLF) = d.f.(SSE_{red}) - d.f.(SSPE) = (n - 2) - (n - c) = c - 2.\]
| 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.
| \(j = 1\) | \(j = 2\) | \(j = 3\) | \(j = 4\) | \(j = 5\) | \(j = 6\) |
| \(Y_{11} = 73\) \(Y_{21} = 78\) | \(Y_{12} = 85\) \(Y_{22} = 88\) | \(Y_{13} = 90\) \(Y_{23} = 91\) | \(Y_{14} = 87\) \(Y_{24} = 86\) \(Y_{34} = 91\) | \(Y_{15} = 75\) | \(Y_{16} = 65\) \(Y_{26} = 63\) |
| \(\bar{Y}_{1} = 75.5\) | \(\bar{Y}_{2} = 86.5\) | \(\bar{Y}_{3} = 90.5\) | \(\bar{Y}_{4} = 88\) | \(\bar{Y}_{5} = 75\) | \(\bar{Y}_{6} = 64\) |
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