This exercise illustrates a reason for the exceptions to the rule of parsimony.

a. A scientist fits the model Y = β₁C + ε, where C represents temperature in °C and Y can represent any outcome. Note that the model has no intercept. Now convert °C to °F (C = 0.556F − 17.78). Does the model have an intercept now?

b. Another scientist fits the model Y = β₀ +β₂C², where C and Y are as in part (a). Note the model has a quadratic term, but no linear term. Now convert °C to °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 = 0. (°C and °F are an example.) Show that the no-intercept models y = βx and y = βz 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 = β₀ + β₂x² and y = β₀ + β₂z² 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.