Buy

Books
Click images for more details

Twitter
Support

 

Recent comments
Recent posts
Links

A few sites I've stumbled across recently....

Powered by Squarespace

Discussion >

For reference, an ARGO float generates a 200 data point temperature profile every ten days. Annually that is 3500*200*36.5=2555000 data points. Sample size for an annual average ocean temperature, n=25 million.
Feb 28, 2016 at 8:00 PM | Unregistered CommenterEntropic man


Precision of the mean=precision of measurement/√sample size

If you assume that, very conservatively, each float measures to the nearest 1C the precision becomes 1/√25*10^7=0.0002C

Thus a change in mean temperature of 0.06C is well within the measurement resolution of the ARGO system.

Feb 28, 2016 at 8:25 PM | Unregistered CommenterEntropic man

Sorry EM, that's wrong. You are blindly applying what you memorised in the "statistics 101" class that you boasted about not so long ago. Mickey H Corbett is right, but it is worse than what he correctly points out.

Your √n formula depends on assumptions such as:

- the errors are solely due to the instrument
- the errors are random, independent, and identically distributed


You are trying to estimate m, the mean ocean temperature. An individual measurement i will give you a value
xi = ( m + delta i + epsilon i) where

m = the mean ocean temperature you hope to estimate.

delta i = the difference between the mean temperature m and the local temperature in the vicinity of the float at the ith measurement.

epsilon i = the instrument error at the ith measurement

If the float is somewhere warrm - eg in the Gulf of Mexico - delta will be positive and will remain so from one measurement to the next. If the float is somewhere cold - eg within the arctic circle - delta will be negative and remain so. So the delta values are not independent from measurement to measurement. Nor is there any reason to think they will be identically distributed. Moreover, they will generally be very much bigger than the error in the float's temperature sensor.

Your √n formula does not apply and the assumption that (x₁ + x₂+ ... + xₙ)/n will give you an estimate of m to a precision ± epsilon / √n is just plain wrong.

Another for the collection.

Feb 29, 2016 at 9:48 AM | Registered CommenterMartin A