# Am I using the right statistical tests?

#### galkim

##### New Member
I am conducting a study to check for the distribution and effects of troponin levels in ill children.
I have ~700 patients. I used SPSS version 24. I first wanted to check if there is a correlation between the troponin level (numeric/continuous) and several other lab parameters for each patient - some are numeric/continuous and some are numeric/ordinal (e.g. white blod cell count, hospitalization days, etc..). as the troponin levels did not follow normal distribution, I used spearman's correlation. I then wanted to check whether or not a certain individual is more likely to die if he had a higher troponin result. (so death was is a categorical variable with two levels, - YES/NO right?) . for that I used the Mann-Whitney test, having the Troponin levels as the test variable and death as the grouping variable.

since i got weird results i wanted to make sure that i did the right tests...

#### Karabiner

##### TS Contributor
Looks o.k. What do you mean by weird resultis?

With kind regards

Karabiner

#### galkim

##### New Member
Looks o.k. What do you mean by weird resultis?

With kind regards

Karabiner
well... results that logically dont make sense... especially with the mann-whitney test - in all of the four independent variables that i tested (there were 3 others apart from death) the test result showed exactly the opposite from what i expected... ofcourse that it could be perfectly accurate but i doubt it....

##### Member
galkim,
Relation between troponin and other parameters may be not linear. Firstly see scatterplots, may be the kind of relation will be obvious. Regarding other tests, I would make ANOVA to be sure that the parameters are not related. Another way, may be more correct, is to use log transformation of troponin values and try to make regression models.

#### galkim

##### New Member
galkim,
Relation between troponin and other parameters may be not linear. Firstly see scatterplots, may be the kind of relation will be obvious. Regarding other tests, I would make ANOVA to be sure that the parameters are not related. Another way, may be more correct, is to use log transformation of troponin values and try to make regression models.
thanks for your reply. the spearman correlation made sense completely. My problem is with the Mann-whitney results. regarding you suggestion to use ANOVA and the logarithmic transformation- why would that help me?
thanks for the help again...

##### Member
Log transformed data may have normal distribution. In general, if histogram of your data hasn't heavy tails, you may use ANOVA for analysis.

#### Karabiner

##### TS Contributor
Please, inspect your data graphically first, before you start transforming them or something like that. You could for example use histograms and/or box-and-whisker plots for the troponin in each group.

With kind regards

Karabiner

#### galkim

##### New Member
Please, inspect your data graphically first, before you start transforming them or something like that. You could for example use histograms and/or box-and-whisker plots for the troponin in each group.

With kind regards

Karabiner
i tried that, i'm not sure if did that correctly - what do you think? this is the troponin histogram for all subjects

##### Member
galkim,
this histogram will be more informative, if you provide it with smaller bin, for example 2. May you also make it for both groups you mentioned? Regarding this figure, the data should has normal distribution after log transformation.

#### galkim

##### New Member
galkim,
this histogram will be more informative, if you provide it with smaller bin, for example 2. May you also make it for both groups you mentioned? Regarding this figure, the data should has normal distribution after log transformation.
ok i playes with the histogram a bit -

and this is the histogram for the troponin after a log transformation: (i think)

what other groups are you talking about? because the other values that im testing against the troponin in the Mann-whitney are categorial with two levels

##### Member
galkim,
in the first post you wrote about deaths (yes or no). If you plan to compare these 2 groups, you should check troponin distribution in each group. I think, log transformed data are not met ANOVA assumptions. Sorry, but it was not obvious from the first histogram. So you may finish your analysis with M-W test for measured values. By the way, are extremely high values of troponin correct? What units do you use? I see, more than half subjects have values near 0 and some subjects have >50 and even about 100.

#### galkim

##### New Member
biostat,
isnt comparing the two groups against troponin is what the MW supposed to do?
also - about the values - this is exactly why we are performing this study. the levels of troponin vary a LOT in children and we're trying to find out if the height of their rise above 0.05 means anything

#### Karabiner

##### TS Contributor
It would be useful to compare groups graphically first (boxplots with 2 boxes, one for each group, and 2 histograms, one for each group),
Supposedly, the U-test will be useful here - if it actually covers your research question; as you are probably aware of, the U test is not a
comparison of means. I does not use the original data, but data transformed into ranks.

By the way, you did not explain what you mean by "the test result showed exactly the opposite from what i expected".
A non-significant result where a significant result was expected? O something else?

With kind regards

Karabiner

##### Member
yes, you can use this test without assumptions about distribution.
But usually the first step is to see distribution of analysing parameters and then select data transformation by necessity and select appropriate method of analysis. In more cases in biology log transformation leads to normal distribution.

#### galkim

##### New Member

#Karabiner - hope i got this right:

i'm not sure i know how to present the same data in a boxplot but i tired anyway... so...

regarding "the test result showed exactly the opposite from what i expected" - these are my Mann-Whitney results:

from what I understand, and I could be wrong of course, these result mean that the individuals in group #2 had a higher troponin level, and the two groups are significantly different from one another. that's a bit strange, because troponin is indicative of cardiac damage so i expected exactly the opposite... this obviously could be a correct finding but still, i did not expect that...

##### Member
Sorry, I don't understand exactly your conclusion. The result is that your groups don't differ significantly. In other words, troponin doesn't influent to the death event.

#### galkim

##### New Member
Sorry, I don't understand exactly your conclusion. The result is that your groups don't differ significantly. In other words, troponin doesn't influent to the death event.
but doesn't the mean rank result in MW have meaning?

#### Karabiner

##### TS Contributor
It says that IN YOUR SAMPLE one group has higher ranks than the other.
The problem is, this could be due to chance.

The result of the test tells you that you cannot discard the null hypothesis,
which says (loosely speaking): In the POPULATION from which these two
samples are drawn, the mean ranks are the same in each group. The observed
difference between your two sampled groups cannot be distinguished from
chance.

Since you test group differences with regard to ranked data, maybe important
information contained in the original data was lost; but I don't know that,
it is a matter of knowledge of the subject in question.

With kind regards

Karabiner

#### galkim

##### New Member
wow... i'm confused...
didn't you two just say two different things?
let me get this straight... if the p.value would have been 0.003 instead of 0.2 and the mean ranks would be the same... doesnt that mean that there is no significant difference between the groups = and then troponin has no effect on death (either positive or negative) ?

#### Karabiner

##### TS Contributor
If the mean ranks would have been the same in the sample, then p = 0.003 would not have been possible.

With kind regards

Karabiner