Solutions to Practice Problems for Unit 3

1. Is the regression line a good fit? If 'yes', explain why. If 'no', suggest how the fit might be improved.

No. The fit is lousy. Values at the extreme ends are over-predicted, and values in the middle are under-predicted. This could be fixed, perhaps, by using a quadratic model or by taking a transformation of the Depth variable.

Comments: You should have guessed that a linear model would not fit. When you open a spigot, the lemonade flows faster when the urn is full, and gradually slows as the urn is emptied, right? So this means the depth inside is changing at a different rate, and so this cannot be a linear phenomenon. In fact, you'll get a pretty good fit if you use a quadratic model: depth = 0.0000254*time^2 - 0.03556*time + 12.46. Fathom doesn't do quadratic models, but you can get the same model by taking a square-root transformation of depth: sqrt(Depth) = 3.5295-0.0050*time.

2. A student says that the high r-squared value means this is a good fit. What would you say to this student?

R-squared measures how well a linear model fits. It does not tell us whether or not the model is linear. In this case, the model is not linear, and so we can't interpret R-squared reliably.

3. What temperature does the linear model predict we will see after 3 hours (180 minutes). Assuming the temperature in the room stays at 81 degrees, what temperature should the model predict we will see?

The model predicts that the temperature will be -.722*180 + 137 = 7.04 degrees. This is probably colder than the coffee will ever get. The temperature of the coffee should level off at 81 degrees.

4. Comment on the fit of the model. How would you improve the fit?

Again, the residuals show that the fit is poor. A transformation of some kind is called for. From what we know of cooling, we might try an exponential. You need to be careful, though, to make sure the asymptote goes through 81 degrees. Here's the fit that works best:

log(Temp-81) = -0.02740*time + 4.296