rel="nofollow" href="#fb3_img_img_aa2bb8d3-312c-5b4b-9202-67ae4fc35ada.gif" alt="in50-3"/> is computed using uncorrected sums of squares.
There are alternative ways to define R2 for the no-intercept model. One possibility is
However, in cases where
Example 2.8 The Shelf-Stocking Data
The time required for a merchandiser to stock a grocery store shelf with a soft drink product as well as the number of cases of product stocked is shown in Table 2.10. The scatter diagram shown in Figure 2.14 suggests that a straight line passing through the origin could be used to express the relationship between time and the number of cases stocked. Furthermore, since if the number of cases x = 0, then shelf stocking time y = 0, this model seems intuitively reasonable. Note also that the range of x is close to the origin.
The slope in the no-intercept model is computed from Eq. (2.50) as
Therefore, the fitted equation is
This regression line is shown in Figure 2.15. The residual mean square for this model is MSRes = 0.0893 and
We may also fit the intercept model to the data for comparative purposes. This results in
The t statistic for testing H0: β0 = 0 is t0 = −0.65, which is not significant, implying that the no-intercept model may provide a superior fit. The residual mean square for the intercept model is MSRes = 0.0931 and R2 = 0.9997. Since MSRes for the no-intercept model is smaller than MSRes for the intercept model, we conclude that the no-intercept model is superior. As noted previously, the R2 statistics are not directly comparable.
TABLE 2.10 Shelf-Stocking Data for Example 2.8
| Times, y (minutes) | Cases Stocked, x |
| 10.15 | 25 |
| 2.96 | 6 |
| 3.00 | 8 |
| 6.88 | 17 |
| 0.28 | 2 |
| 5.06 | 13 |
| 9.14 | 23 |
| 11.86 | 30 |
| 11.69 | 28 |
| 6.04 | 14 |
| 7.57 | 19 |
| 1.74 | 4 |
| 9.38 | 24 |
| 0.16 | 1 |
| 1.84 | 5 |
Figure 2.14 Scatter diagram of shelf-stocking data.
Figure 2.15 The confidence and prediction bands for the shelf-stocing data.
Figure 2.15 also shows the 95% confidence interval or E(y|x0) computed from Eq. (2.54) and the 95% prediction interval on a single future observation y0 at x = x0 computed from Eq. (2.55). Notice that the length of the confidence interval at x0 = 0 is zero.
SAS handles the no-intercept case. For this situation, the model statement follows:
model time = cases/noint
2.12 ESTIMATION BY MAXIMUM LIKELIHOOD
The method of least squares can be used to estimate the parameters in a linear regression model regardless of the form of the distribution of the errors ε. Least squares produces best linear unbiased estimators of β0 and β1. Other statistical procedures, such as hypothesis testing and CI construction, assume that the errors are normally distributed. If the form of the distribution of the errors is known, an alternative method of parameter estimation, the method of maximum likelihood, can be used.
Consider the data (yi, xi), i = 1, 2, …, n. If we assume that the errors in the regression model are NID(0, σ2), then the observations yi in this sample are normally and independently distributed random variables with mean β0 + β1xi and variance σ2. The likelihood function is found from the joint distribution of the observations. If we consider this joint distribution with the observations given and the parameters β0, β1, and σ2 unknown constants, we have the likelihood function. For the simple linear regression model with normal errors, the likelihood function is
(2.56)
The maximum-likelihood estimators are the parameter values, say
