Introduction to Probability and Statistics for Engineers and Scientists

366 Chapter 9: Regression

REMARK

The result that

B−β

√

σ^2 /Sxx

∼N(0, 1)

cannot be immediately applied to make inferences aboutβsince it involves the unknown
parameterσ^2. Instead, what we do is use the preceding statistic withσ^2 replaced by its
estimatorSSR/(n−2), which has the effect of changing the distribution of the statistic
from the standard normal to thet-distribution withn−2 degrees of freedom.

EXAMPLE 9.4b Derive a 95 percent confidence interval estimate ofβin Example 9.4a.

SOLUTION Sincet.025,5=2.571, it follows from the computations of this example that
the 95 percent confidence interval is

−.170±2.571

√

1.527

3500

=−.170±.054

That is, we can be 95 percent confident thatβlies between−.224 and−.116. ■

9.4.1.1Regression to the Mean

The termregressionwas originally employed by Francis Galton while describing the laws
of inheritance. Galton believed that these laws caused population extremes to “regress
toward the mean.” By this he meant that children of individuals having extreme values
of a certain characteristic would tend to have less extreme values of this characteristic
than their parent.
If we assume a linear regression relationship between the characteristic of the off-
spring (Y), and that of the parent (x), then a regression to the mean will occur when
the regression parameterβis between 0 and 1. That is, if

E[Y]=α+βx

and 0<β<1, thenE[Y]will be smaller thanxwhenxis large and greater thanx
whenxis small. That this statement is true can be easily checked either algebraically or
by plotting the two straight lines

y=α+βx

and

y=x

A plot indicates that, when 0<β<1, the liney=α+βxis above the liney=xfor
small values ofxand is below it for large values ofx.