### I’ve been pondering since this morning, what’s up with this heat?

Near-gratuitous One Piece reference (watch every episode here!). One Piece is awesome. Make a note. Awesome.

So. Today’s weather:

Walking around with the missus (we went to Grant’s tomb – boring, but, in their defence, closed for the holiday. Still), she commented on the heat, its unseasonality, how freezing cold it had been in her youth, etc. Honestly, she should have known better – a statistician and a contrarian for a husband? Crazy idea.

My counter-arguments: (1) that the temperature on the day of Thanksgiving is a variable, like any other. The week of Thanksgiving, maybe. But the day? Temperature and time both are continuous variables. A single day is just way too precise to pin something like that on; (2) my wife was probably remembering particularly cold days from her youth, which was affecting her memory of the true average temperatures for this period.

So, being the manner of econometrician that I am. After dinner, I jumped on the web and started looking. I found my way to Almanac.com, and started pulling out the temperature for all of the November 22nds since the birth of my wife (1981; a fine year for pretty girls).

I had to jump from Central Park to JFK in 1994 (no Central Park data past then, but I checked a handful of the dates since, and there’s no apparent measurement problem). Descriptive statistics:

So, what do I do? I start looking for 60 degrees (today’s) in the 95% Confidence Interval for the mean temperature on Thanksgiving Day.

$latex begin{eqnarray*} 95\%CI_{\mu } &=&\overline{x}\pm t_{.025,26}\times \sigma _{\overline{x}} \\ &=&\overline{x}\pm t_{.025,26}\times \frac{\sigma }{\sqrt{n}} \\ &=&45.4\pm 2.056\times 1.76 \\ &=&41.78\text{ to }49 \end{eqnarray*}$

WordPress’ latex plug-in is ass. It was supposed to look like this:

So today’s 60 degrees just (just) misses out. Like fun it does.

The conclusion: Confidence Interval conclusions can vary, but we can say that, for example, 95% of Thanksgivings will not have a mean temperature of 60 degrees (97.5% of them will be less than 60 degrees – by a long way). We are 95% certain that the population mean (the true mean for Thanksgiving Day) is not 60 degrees.

Is today’s temperature therefore extreme? We can visit the 99% confidence interval:

$latex begin{eqnarray*} 99\%CI_{\mu } &=&\overline{x}\pm t_{.005,26}\times \sigma _{\overline{x}} \\ &=&\overline{x}\pm t_{.005,26}\times \frac{\sigma }{\sqrt{n}} \\ &=&45.4\pm 2.779\times 1.76 \\ &=&40.51\text{ to }50.29 \end{eqnarray*}$

Or, again,

Again, um, just? Barely. There is less than a 1% chance that, on any given Thanksgiving Day, the mean temperature will be this far from the average (and less than half of one percent that it will be this high). Statistically, it was an extreme event.

There is far less than a 1% chance: the p-value for an average temperature of 60 degrees is, in fact, basically 0 (the t-statistic being 8.3)

I managed a partial victory. The two lowest days – and the only two below 30 degrees – occurred in my wife’s youth, and the lowest occurred the year she was freezing cold out in the parade itself. So I scored a minor point.

A caveat is the mean. I used the entire series. This is to say, a caveat to the numbers – that result isn’t going anywhere.

Suppose I used only the Thanksgivings up until this one? Today’s was the maximum for the series: the next highest was 59 degrees. Cutting out today’s 60 degrees lowers the average to 44.8, and the standard deviation to 8.84. It also lowers the sample size, increasing the “critical value” t-statistic slightly (due to the lower degrees of freedom) as well as the standard error. The new 99% Confidence interval narrows, slightly, but also shifts downwards: 40 degrees to 49.6. Either way, 60 degrees (new t-statistic 8.77) is still nowhere near likely to be another average temperature in a hurry.