If you have data on a meaningful scale I would report the difference in geometric means and the CIs using backtransformed (this will make them asymmetrical - but that is appropriate if you have skewed data. Cohen's d doesn't really make sense on a skewed scale because it isn't really interpretable if you don't have constant SD - though it might be OK here because you have very similar SDs indicating that the differences in the skew are probably the result of different tail distributions. In a lognormal distribution a higher mean implies a higher SD. However, probably the most important consideration is to clearly label the quantities reported.

For some analyses I find it useful to report the transformed statistics for the model but plot means and CIs on the original scale to aid interpretation.

Hello Cordelia.  I'm confused.  The t-test results you show for original and transformed data are the same.  Both show Mean Difference = 0.321 [95%CI, 0.251, 0.392], t = 8.925, df = 1514, p < .001. 

Also, given that both sample sizes are quite large, I think that two things follow:

In light of point 2, I think you should be focusing on whether the mean difference is large enough to be practically important. 

HTH.

Thanks so much for getting back to me

So would the following be the correct report provided I stated that I was going to report back transformed values for means and confidence intervals (obviously with the values calculated):

An independent samples t-test revealed that children attending pre-primary school (M=10^2.45, SD = 10^0.684) had significantly higher household assets than children not attending (M=10^2.12, SD = 10^0.676) , M=10^0.321 95%CI [10^0.251, 10^0.392], t(1514)=8.925, p

Hi Brian, I meant that's how I would report the transformed results if they hadn't been transformed based on standard reporting so I could find out which aspects of that need to change due to the data being transformed. Does that make sense?

I think your question is "upside down". You analyse how the assets depend on the primary school visit of children. But I think the relevant question is actually how the primary school visits depend on the assets. The hypothesis may be that higher assets increase the odds of primary school visits of the children.

If this is so, then I would expect to read the odds (and the odds ratio) and their confidence intervals per increase of the household assets. The basis of this analysis is a logistic regression (aka generalized linear model of the binomial family). The response/dependent/outcome variable is binary (preschool visit: yes or no), and the predictor is continuous (assets). You may consider more additional predictors, like the parent's education, religion, urbanity, employment, profession, distance to the next school, family size etc.). This may allow you to see that increasing assets may have a positive impact on the outcome for, say, people living in urban regions, but not so for people in rural regions. 

It might not hold that the (log) odds change by a constant amount for an absolute difference in the assets but rather by a proportional difference in the assets (like: a x% increase in the assets increases the log odds by y). In this case the predictor "houshold assets" is to be used as log(assets).

Hello Cordelia,

I agree with Bruce, given the sample size, thanks to the Central Limit Theorem, you can probably use a t-test on the non-transformed variable (when sample size is high, the mean of the variable is approximately normally distributed, and so the difference of the means also). So, the difference of the 2 means and its 95%CI are appropriate.

Nevertheless, if you want to transform the variable, you can give (in each group) geometric mean GM times / divided by GSD rather than arithmetic mean M +/- SD, or give the 2 GM and their 95%CI. Indeed, I think the difference of 2 GM has little sense, but you can give the ratio: when you log-transform, you move to a multiplicative scale: X'-Y'=log(X)-log(Y)=log(X/Y). 

Another way is to give the median and the interquartile range and to do a non-parametric Mann-Whitney test.

These are my few suggestions... 

Hi Jochen,

Thanks for your reply. I see what you mean and agree that building a logistic regression model would probably be a more interesting way of handling the data.

To be completely honest, I am doing this for an MSc assignment which just has to demonstrate that I understand the basics of how to calculate group differences from a relatively large dataset. 

So whilst I know there are lots of more complex analyses I could do, and which I will definitely explore at a later date, I have already gone "above and beyond" what's expected by doing a data transformation, so I am really just looking to find out about reporting on results of log transformed data (for both this assignment and in the future as I haven't found anything which explains it clearly). 

I see :)

And now I have a different problem...

So you should learn "how to calculate group differences" - but using a transformation changes this question!

You have two groups and you want to estimate the expected difference in the response variable (assets). This estimate is given by the difference of the sample means. The uncertainty associated with this estimate depends on the variance and the shape of the distribution of the values. As you correctly noticed, this distribution is skewed, and this fact can make it difficult to give sensible measures of the uncertainty. Standard methods that calculate you such uncertainty values (like standard errors or confidence intervals) usually assume a symmetric (strictly: normal) distribution. If this assumption is unreasonable in light of your data, one may be sceptical about the usefulness of the uncertainty these methods propose.

Now Bruce already mentioned that for such large sample sizes like yours the central limit theorem tells us that these uncertainty measures will become independent of the actual distribution of your values. So you can really consider the results of the standard analyses as sufficiently good for any practical purpose.

If you are really picky about skewness of your data, you will need a method that will give you a correct estimate of the uncertainty in the expected difference without assuming a symmetric distribution. Here I see two possibilities: either you use a generalized linear model of the gamma family with identity link, or you estimate the uncertainty using a bootstrap technique. I think both is quite beyond the requirements of an MSc assignment.

Using the log transformed values gives you a symmetric distribution, and the standard methods give you correct estimates of the uncertainty of the expected differences - of the transformed values. But this is a difference of logs, not a difference of the original values. A difference of logs is equivalent to a log-ratio of original values, and all the back-transformations will eventually represent your uncertainty in the estimate of a ratio, not of a difference. That's not what you told you wanted.

So your transformation does not give you a correct sultion to your initial question. It gives you a correct solution to a different question. This should be clear to you. Seeing this you may find that it might not make so much sense to report differences in geometric means and that one would be better off reporting [log] ratios instead.

Thanks Jochen, I see what you mean.

In terms of the central limit theorem - what are its "limits"? I've attached images of the histograms, Q-Q plots and box plots for the raw data, does the sample size mean that standard analyses would be acceptable even with this level of skewness and number of outliers?

Difficult to say, as one has to consider three things together:

i) the "degree of incorrectness" can you accept, (ii) the skewness and (iii) the sample size.

A way to get a hint if this might be ok for you is to really bootstrap the expected difference and compare this with the results of the standard procedure (that assumes a normal distribution)*, e.g. by comparing the location and width of the confidence intervals.** The stupid part about that is that the you do have the bootstrap estimates, so there is no need to fall back to mathemaically more elegant procedures at the cost of using possibly unjustified assumptions. However, this comparison may help to get an idea of the adequacy of using standard methods with other datasets with similar skewness and sample sizes.

--

* the standard prodecure means: the expected difference "D" is estimated by the difference of the sample means and the 95% confidence interval of this estimate is given by D ± 2*SED.

** you bootstrap the distribution of D using randomly resampled data. You can use this distribution to get the limits of the 95% confidence interval as the 0.025- and the 0.975-quantile of that distribution.

PS: bootstrap is not assumption-free. It makes the assumptions that the data is representative to the problem and that there is sufficient data to represent all relevant features of the underlying distribution. These assumptions are reasonable for large datasets obtained by "random sampling".

I think the points about inference in this sample that Bruce makes are correct, but you'll get more appropriate CIs from the back-transformed model or from bootstrapping.