Empirical Modeling: Day 12

One thing I noticed about your addenda is that several groups didn't produce estimates for the half inches. Those were the clearest markings for the 10 foot rule. So I'm still wondering if some of you know exactly how the ruler operates....

I'd rather that you NOT email me with this info, as emails are public records. I've told you that these are anonymous, but -- officially, at least -- emails may be requested through Freedom of Information requests.

It's harder to obtain paper records!

That's why I asked you to hand them to me.

Please read it over, and compare it to your own reports. I intend for these "final reports" to be models which you might follow in your own reports (with a few snarky remarks thrown in sometimes, because these two, in particular, were written for family/friends!:).

If you read the report prior to Sunday, then you missed my structural model for log rules (which we'll discuss soon). It's in the appendix.

So we'll have finished three projects by then:

1. Cross-Country
2. Log Ruler
3. Keeling

Upshot: the variation in the dependent variable, $y$, is split into two parts:

\[ SS_{Total}=SSE+SS_{Reg} \]

and \[ R^2 = \frac{SS_{Reg}}{SS_{Total}} = \frac{SS_{Total}-SSE}{SS_{Total}} = 1-\frac{SSE}{SS_{Total}} \]

which we think of as the proportion of the variance explained by the model (versus the proportion of "error").

• Often these are expressed in one of two formats: with 95% confidence,
  $a \in [338.07, 340.07]$

  or

  $a = 339.07 \pm 0.997$

  (these were the values of $a$ from the quadratic model to Stewart's Keeling data).

• Confidence intervals are related to p-values, but perhaps more informative).

• The standard errors of the parameters pop out of the inverse matrix we compute, multiplied by the mean SSE. Once we have those, we have everything we need for confidence intervals.

  We're often interested to know whether we can exclude a certain value from a confidence interval -- e.g., can we conclude that the slope parameter $b$ in a linear regression $y(x)=a+bx$ is not 0? If so, we conclude that there is a non-zero slope, and the model suggests that $x$ drives values up or down, depending on the sign of $b$.

  check of confidence intervals for Keeling

• Let's check out my discussion of the regressions in the report for Dave.

  $R^2$ is ridiculously high in this case, because we know the function generating the marks on the stick. The errors -- the residuals -- are simply rounding errors, to the nearest board foot. Check out the graphs of data and model....

1. if one does a log-log transformation of both $x$ and $y$, and the data look linear, then you'll be expecting a power model: $y=\alpha x^\beta$.

   (Riegel's model)

2. If you log-transform only the $y$, and get a linear looking cloud, then you'll be looking for an exponential model: $y=\alpha e^{b x}$.

   (I've asked you for an exponential model for CO₂. You log-transform the trend data, and plot. Then linear regress, if appropriate.)

3. What about the oscillations in Keeling's model? We typically model those with sinusoidal functions. Let's write out the model and see what we need to do.... Hint: \[ \sin[2\pi\frac{t-t_0}{T}] \]

[ael: by the way -- ecosystems worldwide are already suffering serious damage.... The world won't "suddenly heat up" if we reach 450 ppm -- it's heating up right now, and we're seeing bumblebees struggle with the heat, for example. They pollinate our food -- think about it....]