By bootstrapping the distribution of the difference. If the (1-alpha) percentile range of this distribution includes 0, the difference is not significant at alpha.

EDIT:

The above answer is the simple answer to the questiona as it was asked. However, as Timothy stated below, it is likely that comparing mean and median values does not really make sense in your context. You can surely bootstrap the difference between means, or the difference between medians, or the ratio between means etc. - whatever is the interesting statistic in the context of your research question.

If your variables are some biological/physiological measures, it is very likely that they are approximately log-normal distributed, and that the relative change (i.e. the ratio of values) is actually more interesting than the absolute difference (because biological responses/changes are typically proportional to the amount/level that is changing). The ratio of the original measures will map to the difference of the logarithms.

Also, a log-normal distribution looks very similar to a normal distribution when the mean is considerably larger than the variance. I suspect that this was just the case for some of your variables.

It's a wild guess, but I'd say that you should analyze the mean difference in the log values, and that it will be ok doing that with standard tools that assume normal distributed variables.

If understand correctly, you want to test the difference between mean A and median B? This is not possible because they are already different.

The contrasts A vs B and mean vs median are both different.

Consider a list of numbers:

2, 4, 5, 8, 15; mean=6.8 median=5

Now, if you change the last number to 500 to give

2, 4, 5, 8, 500; mean=103.8 median=5.

Sometimes the mean and median will have the same value, but this is a random behaviour. Comparing them across two groups will only propagate this randomness.

Treat the distribution of measurements for both groups as one population. Depending on that distribution use a parametric test for (mean A, mean B) and a non-parametric test for (median A , median B).

For a symmetrical distribution such as the normal distribution, the mean is (almost) equal to the median. It is therefore not necessary to use any statistical test to compare them. When the distribution is not symmetrical (not normal), these two parameters can be very different. In such a situation parametric methods cannot be applicable since they are sensitive to the nature of the data distribution. When non-parametric methods are used, they are no longer compared means but rather medians. Then you understand with me that it makes no sense to compare a median to itself. So I don't think there is a test to compare a median and a mean.

By bootstrapping the distribution of the difference. If the (1-alpha) percentile range of this distribution includes 0, the difference is not significant at alpha.

EDIT:

The above answer is the simple answer to the questiona as it was asked. However, as Timothy stated below, it is likely that comparing mean and median values does not really make sense in your context. You can surely bootstrap the difference between means, or the difference between medians, or the ratio between means etc. - whatever is the interesting statistic in the context of your research question.

If your variables are some biological/physiological measures, it is very likely that they are approximately log-normal distributed, and that the relative change (i.e. the ratio of values) is actually more interesting than the absolute difference (because biological responses/changes are typically proportional to the amount/level that is changing). The ratio of the original measures will map to the difference of the logarithms.

Also, a log-normal distribution looks very similar to a normal distribution when the mean is considerably larger than the variance. I suspect that this was just the case for some of your variables.

It's a wild guess, but I'd say that you should analyze the mean difference in the log values, and that it will be ok doing that with standard tools that assume normal distributed variables.

With two independent samples, il one or both are not normally distributed, the

Mann Whitney U Test (Wilcoxon Rank Sum Test) is statistical test to be used.

Have a look at this : http://sphweb.bumc.bu.edu/otlt/mph-modules/bs/bs704_nonparametric/BS704_Nonparametric4.html

Thank you very much to you all for the wonderful answers. Let me clarify my study design more. We conducted an unmatched case-control study that included continuous variables. We found for certain variables that distribution was not normally distributed for "cases" but was normally distributed for "controls", and vice versa. This situation perplexed me as I do not know what is the statistical approach to its analysis. Thank you very much for your help.

Rashad Abdul-Ghani

You say certain variables.How many? All of them? Which were you trying to control?

Were cases treated?

Rashad Abdul-Ghani: The difference of the mean and the median of a distribution is defined by the statistic called skewness. If -2 < sk < 2, then it is regarded is close to a normal distribution.

Kurtosis estimates the likelihood of outlier values. For a Gaussian distribution, Kurtosis is 3. If kutosis is betweem 2 and 4, it can be regarded as a normal distribution.

If each of the two groups have 20 or more values then use a randomization test. You could also use a bootstrap as Jochen suggested.

Another option is to use non-parametric statistics. While parametric statistics assumes a specific distribution (usually Gaussian), the non-parametric statistics does not assume a distribution. That means the distribution could be Gaussian, or anything else.

The test that showed that one sample was normally distributed did not actually show this. The mistake is using a failure to reject the null hypothesis as proof that the null hypothesis is true. At best all you can say is that my sample size was not sufficient to reject the null hypothesis. It also works to state something like: because I was unable to reject the null hypothesis I will behave as if it were true. This might all seem like playing with words. However, if you prove that X is normally distributed then when I redo your experiment I must reconcile a difference in outcome. If you simply behave as if the data were normally distributed and then if I get a different outcome our results are not necessarily inconsistent.

I am not sure why one would test each sample. The assumption of normality is on the residuals, not the independent or dependent variables.

Thank you all for your contributions!

Joseph L Alvarez It is not a controlled trial. Data of most variables were conflicting regarding the normality of distribution.

Ranga Migara Weerakkody Can I depend on skewness and kurtosis? I used Shapiro-Wilk test for testing the normality of distribution.

Timothy A Ebert Not variables but data which I tested for normality.

The data must consist of something. There would seem little point to compare the median score of the New York Giants (American) football team and the mean number of tuna caught per day in the Indian Ocean. However, even here you have two variables (football score, and tuna harvest).

Even if the numbers are generated using a computer random number generator, you have to store them if you are then going to calculate a mean or median. The storage location has a name, and the computer programming term for that name is variable. Thus, one could state that the variable foot1 contains 27 random numbers generated from a Gaussian distribution with mean=5 and standard deviation = 2.8. True, these are just data, but I still have a variable (foot1).

One could also think of it this way. I calculated a mean. The formula is the sum of all Xi divided by the number of observations. Xi is a variable.

.

maybe you should state precisely why you want to perform a statistical test ; what is your ultimate research goal here ?

for instance your research question could be "are the treatment and control drawn from different distributions", you seem to already have the answer (obviously normal vs non normal if i understood correctly)

your research question could be "is the control mean larger than the treatment mean" ; then you could just bootstrap on the control/test differences

and so on ...

.

Dear All,

Thank you very much for your kind support and cooperation. I may not present my problem well and still cannot understand how to deal with this issue. To clarify my question further, I have to compare certain continuous variables between two groups (malaria-positive patients and malaria-negative patients) as in the attached snippet of the table. My concern that the data of malaria-negative individuals were normally distributed (expressed as means and SD -- denoted with the symbol "a"), while that of malaria-positive patients were not (expressed as median and IQR -- denoted with the symbol "b", or vice versa. In some instances, parameters were/ were not normally distributed in both (I know that the sample size is small). I am very perplexed by this problem and did not know how to present data appropriately in tables or what type of statistical tests I should use for comparison. Even if I assume that data were not normally distributed, I did not know what to present (medians for both or median and mean). If I have to present the means of normally distributed data in the table, should I use the median corresponding to it in the analysis when comparing to the median of the other groups? You can see a snippet of one of the tables for clarification. THANK YOU VERY MUCH FOR YOUR HELP!

Best regards,

Rashad

Rashad Abdul-Ghani

Can you get the original data?

I am concerned about comparing the mean of one group to the median of another group.

Mean and median are different, that is why they are defined separately.

Rashad Abdul-Ghani : You cannot compare mean and median as numbers, unless they are in the same population or representative samples of the same population.

you have to look at the raw numbers and do a transform the variables to a normal distribution. Then you could run the comparison. Raising to logarithm usually works very well.

Conversely, you can use non parametric tests on both distributions if you know the raw numbers.

Thank you Ranga Migara Weerakkody !

Thank you Ranga Migara Weerakkody ! Using non-parametric tests for comparison seems good. I will read about this and adopt it if logical. My concern will be how to present them in the table because the median and mean can be found in the same raw as in the table above. Highly appreciate your input. Kind regards!

@Rashad Absul Ghani: Once you know the raw numbers, you can calculate and present both samples' medians with IQRs.

I personally prefer transformation, because parametruc tests are more accurate than non parametric tests, but tge difference is slight.