3
$\begingroup$

I have obtained some mean summer temperature records (June 1st- September 30th) for all years within the 1930s (1930-1939) and I have obtained summer temperature records for all years within the 2000s (2000-2009), for all regions in Great Britain (i.e. South West Scotland, South East Scotland, Midlands..).

I want to know whether the mean temperature values recorded in the 1930s significantly differ from the mean temperatures values for the same period in the 2000s.

The data is skewed because naturally I have greater temperature values for the South East and Southern England compared to temperature values in the North (Scotland). Therefore, my data does not meet the t-tests normality assumption.

However, because I am literally comparing one variable (1930s mean temperatures) against another variable (2000s mean temperatures), I am not sure whether the alternative non-parametric Mann-Whitney U-test, is appropriate? Because the data is continuous and not ordinal?

I am not sure what other tests to conduct, and would really appreciate some guidance!

Improve this question
$\endgroup$
6
  • 1
    $\begingroup$ You don't have independent data; they're matched by region (if you have the same regions for both you can look at the temperature differences for each region leading to some paired test; failing that it may still be possible to do something) $\endgroup$
    – Glen_b
    Commented Mar 13, 2016 at 13:00
  • $\begingroup$ Oh so not a Mann-Whitney U-test then! I think the Wilcoxon-signed ranked test is a better shout then. $\endgroup$
    – Pixie
    Commented Mar 13, 2016 at 13:03
  • 1
    $\begingroup$ Its a possibility -- but you specifically ask about change in means, which the signed rank test doesn't test for (at least not without additional assumptions). I'll attempt to come up with an answer. $\endgroup$
    – Glen_b
    Commented Mar 13, 2016 at 13:04
  • 1
    $\begingroup$ This appears to need a model with spatial correlations, and be sure to treat time as a continuous variable, using a flexible modeling approach without assuming linearity. $\endgroup$
    – Frank Harrell
    Commented Mar 13, 2016 at 13:05
  • 1
    $\begingroup$ I have edited the title and tags to make it clear that the data set is spatial in nature, I think that's a key point in this question. Feel free to revert or improve my edits. $\endgroup$
    – Silverfish
    Commented Mar 13, 2016 at 13:40

1 Answer 1

Reset to default
3
$\begingroup$

It would be better to go back to the design of your study. You have taken two time periods as examples, but much more data are available than that. I suggest obtaining a sample that encompasses at least 100 years, using calendar time as a continuous variable, and that incorporates highly localized geographical location in the model, plus allows for spatial correlations between locations. Once you have the best sample and a well-specified model you can look at the model assumptions. Here normality of residuals is at issue, not normality or symmetry of raw temperatures.

The time trend can be modeled using a flexible approach such as regression splines with lots of knots to allow for complexity. A primary focus would be estimation of the time effect, with confidence bands. You can extend this analysis by adding seasonal trends into the mean model, e.g., you can have one spline function for long-term trends and another for repeated yearly trends.

Once you are comfortable with the type of main effects model described above, you can allow time to interact to location to estimate and test for different temperature trends by region.

Note that nonparametric tests work very well on continuous data. All continuous variables are ordinal. The limitation of Wilcoxon and other tests is their inability to incorporate covariates. That's why their generalizations (e.g., proportional odds ordinal logistic model) are gaining in popularity.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.