Analysis of variance approach to regression
- Page ID
- 229
We divide the total variability in the observe data into two parts - one coming from the errors, the other coming from the predictor.
ANOVA Decomposition
The following decomposition
\[ Y_i - \overline{Y} = (\widehat{Y_i} - \overline{Y}) + (Y_i - \widehat{Y_i} )\]
with \( i=1,2,...,n. \).
represents the deviation of the observed response from the mean response in terms of the sum of the deviation of the fitted value from the mean plus the residual.
Taking the sum of squares, and after some algebra we have:
\[ \sum_{i=1}^n (Y_i - \overline{Y})^2 = \sum_{i=1}^n (\widehat{Y_i} -\overline{Y})^2 + \sum_{i=1}^n (Y_i - \widehat{Y_i})^2. \label{1}\]
or
\[ SSTO = SSR +SSE\]
where
\[SSTO = \sum_{i=1}^n (Y_i - \overline{Y})^2 \]
and
\[SSR = \sum_{i=1}^n (\widehat{Y_i} -\overline{Y})^2. \label{2}\]
is referred to as the ANOVA decomposition to the variation in the response. Note that
\[ SSR = b_1^2 \sum_{i=1}^n (X_i - \overline{X})^2 .\]
Degrees of freedom
The degrees of freedom of different terms in the decomposition Equation \ref{2} are
\[df( SSTO ) = n - 1\]
\[df( SSR ) = 1\]
\[df( SSE ) = n - 2. \]
So,
\[df( SSTO ) = d.f.( SSR ) + d.f.( SSE ). \]
Expected value and distribution
\( E ( SSE ) = ( n - 2) \sigma^2, \) and \( E ( SSR ) = \sigma^2 + \beta_1^2 \sum_{i=1}^n (X_i - \overline{X})^2. \) Also, under the normal regression model, and under \( H_0 : \beta_1 = 0, \)
\[ SSR \sim \sigma^2 \chi_1^2, SSE \sim \sigma^2 \chi_{n-2}^2, \]
and these two are independent.
Mean squares
\[ MSE = \dfrac{SSE}{d.f.(SSE)} = \dfrac{SSE}{n-2}, MSR = \dfrac{SSR}{d.f.(SSR)} = \dfrac{SSR}{1}. \]
Also, \( E ( MSE ) = \sigma^2 , E ( MSR ) = \sigma^2 + \beta_1^2 \sum_{i=1}^n (X_i - \overline{X})^2. \)
F ratio
For testing \( H_0 : \beta_1 = 0 \) versus \( H_1 : \beta_1 \neq 0, \) the following test statistics, called the F ratio, can be used:
\[ F^* = \dfrac{MSR}{MSE}. \]
The reason is that \( \dfrac{MSR}{MSE} \) fluctuates around 1 + \( \dfrac{ \beta_1^2 \sum_{i=1}^n (X_i - \overline{X})^2 }{\sigma^2}. \) So, a significantly large value of \(F^*\) provides evidence against \(H_0\) and for \(H_1.\)
Under \(H_0, F^* \) has the \(F\) distribution with paired degrees of freedom (d.f.( SSR ), d.f.( SSE )) = (1, n - 2 ), (written \(F^* \sim F_{1, n - 2}). \) Thus, the test rejects \(H_0\) at level of significance \(\alpha\) if \(F^* > F( 1 - \alpha; 1, n - 2 ), \) where \(F( 1 - \alpha; 1, n - 2 ) \) is the \( (1 - \alpha ) \) quantile of \(F_{1; n - 2} \) distribution.
Relation between F-test and t-test
Check that \( F^* = ( t^* )^2. \) where \( t^* = \dfrac{b_1}{s ( b_1 )} \) is the test statistic for testing \(H_0 : \beta_1 = 0 \) versus \(H_1 : \beta_1 \neq 0. \) So, the F-test is equivalent to the t-test in this case.
ANOVA table
It is a table that gives the summary of the various objects used in testing \(H_0 : \beta_1 = 0 \) against \(H_1 : \beta_1 \neq 0.\) It is of the form:
Source | df | SS | MS | F* |
---|---|---|---|---|
Regression | d.f.(SSR) = 1 | SSR | MSR | \(\dfrac{MSR}{MSE} \) |
Error | d.f.(SSE) = n - 2 | SSE | MSE | |
Total | d.f.(SSTO) = n - 1 | SSTO |
Example \(\PageIndex{1}\): housing price data
We consider a data set on housing prices. Here Y = selling price of houses (in $1000), and X = size of houses (100 square feet). The summary statistics are given below:
$$ n = 19, \overline{X} = 15.719, \overline{Y} = 75.211, $$
\( \sum_i ( X_i - \overline{X} )^2 = 40.805, \sum_i ( Y_i - \overline{Y} )^2 = 556.078, \sum_i ( X_i - \overline{X} ) ( Y_i - \overline{Y} ) = 120.001. \)
(Example) - Estimates of \(\beta_1 \) and \(\beta_0\)
\( b_1 = \dfrac{\sum_i ( X_i - \overline{X} ) ( Y_i - \overline{Y} ) }{\sum_i ( X_i - \overline{X} )^2} = \dfrac{120.001}{40.805} = 2.941. \)
and
\( b_0 = \overline{Y} - b_1 \overline{X} = 75.211 - (2.941)(15.719) = 28.981. \)
(Example) - MSE
The degrees of freedom (d.f.) = \( n -2 = 17. SSE = \sum_i (Y_i - \overline{Y} )^2 - b_1^2 \sum_i ( X_i - \overline{X} )^2 = 203.17.\) So,
\[ MSE = \dfrac{SSE}{n - 2} = {203.17}{17} = 11.95. \]
Also, SSTO = 556.08 and SSR = SSTO - SSE = 352.91, MSR = SSR/1 = 352.91.
\(F^* = \dfrac{MSR}{MSE} = 29.529 = (t^* )^2,\) where \(t^* = \dfrac{b_1}{s ( b_1 )} = \dfrac{2.941}{0.5412} = 5.434.\) Also, F( 0.95; 1, 17 ) = 4.45, t( 0.975; 17) = 2.11. So, we reject \(H_0 : \beta_1 = 0. \) The ANOVA table is given below.
Source | df | SS | MS | F* |
---|---|---|---|---|
Regression | 1 | 352.91 | 352.91 | 29.529 |
Error | 17 | 203.17 | 11.95 | |
Total | 18 | 556.08 |
Contributors
- Valerie Regalia
- Debashis Paul