This exercise illustrates a reason for the exceptions to

Chapter 8, Problem 10SE

(choose chapter or problem)

This exercise illustrates a reason for the exceptions to the rule of parsimony (see page 623).

a. A scientist fits the model \(Y=\beta_1C+\varepsilon\), where C represents temperature in \(^\circ \mathrm C\) and Y can represent any outcome. Note that the model has no intercept. Now convert \(^\circ \mathrm C\) to \(^\circ \mathrm F\) \((C=0.556F-17.78)\). Does the model have an intercept now?

b. Another scientist fits the model \(Y=\beta_0+\beta_2C^2\), where C and Y are as in part (a). Note the model has a quadratic term, but no linear term. Now convert \(^\circ \mathrm C\) to \(^\circ \mathrm F\) \((C=0.556F-17.78)\). Does the model have a linear term now?

c. Assume that x and z are two different units that can be used to measure the same quantity, and that \(z=a+bx\), where \(a \ne 0\). (\(^\circ \mathrm C\) and \(^\circ \mathrm F\) are an example.) Show that the no-intercept models \(y=\beta x\) and \(y=\beta z\) cannot both be correct, so that the validity of a no-intercept model depends on the zero point of the units for the independent variable.

d. Let x and z be as in part (c). Show that the models \(y=\beta_0+\beta_2x^2\) and \(y=\beta_0+\beta_2z^2\) cannot both be correct, and, thus, that the validity of such a model depends on the zero point of the units for the independent variable.

Equation Transcription:

Text Transcription:

Y=beta_1C+varepsilon

^oC

^oC

^oF

(C=0.556F-17.78)

Y=beta_0+beta_2C^2

 oC

 oF

(C=0.556F-17.78)

z=a+bx

a{not=}0

^oC

^oF

y=beta{x}

y=beta{z}

y=beta_0+beta_2x^2

y=beta_0+beta_2z^2