Given you have these values from publications, then these publications should also provide a measure of precision/uncertainty of these estimates. If this is the case, then you can use statistical tools to estimate the precision/uncertainty that should be assigned to the difference between these two values. Then you can judge what "significance" you are willing to attribute to this difference.

If not, well, then the papers are examples of "bad practice in science" and I would not take these values very serious anyway.

Hi Kamran,

do the values you want to compare represent just two measurement points or mean values (from a larger sample)? You typically use inferential statistics to make inferences from a sample to an unknown population. In the first case, with just two values to compare (i..e. one per group), there is not much basis for generalization and no statistical test you can use. In the second case case (mean value from  a sample) you can try to get the standard deviation (the spread of the distribution) and use an independent students  t-test.  Hope this helps. 

You can always compare two values. It seems to be too simple to be recognized.

2 < 4, so given these data you should expect that spcies A is more sensitive than B.

The key is that you can not assign any confidence on this expectation. These two givem values are already expectations, and it is not communicated what the confidence in these expextations is, but this would be required to assign a confidence in the difference.

In practice, such confidences are obtained from the combination of statistical model and variability of the data. The contributions are not always similar and depend on the model. For instance, in a Poisson model the variance is completely defined by the expected value, so the confidence does not depend on the actually observed variability of the data but only on the amount of the observed data (=the sample size). In a Normal model both factors (variability anf amount of the observed data) are equally influential on the confidence.

So, depending on the statistical model, at least the sample sizes must be given, and in many cases the variance of the data is required, too, to assign a confidence on such expectations.

There is finally also a (rather crude) way to (roughly) assign a confidence on the expected difference from the given confidences of two expectations. This is obtained by error-propagation. It is based on large-sample theory and normal-approximation and often gives some upper bound for the confidence that should be assigned to the expected difference (or any other statistic one likes to calculate from the given values).

Comparison of a single value or statistical comparison of mean values from a population,,whatever  it be,, but LC50 is to be compared between two different species.. Just see if you need to body take surface area of that species into consideration,, because one may have larger body surface than another,, so some factor may have to be considered while calculations, to normalize the data for species difference.. !!! 

With simple elementary mathematics, you can compare which specie is more sensitive to Cu if Cu has LC50 of 2 mg/L in species A and 4 mg/L in species B. Application of statistics would come in if you have a series of data for Species A and a series of data for Species B, then you can compare using t-test or you can use a non-parametric approach.

i also suggest t-test

Compare significance between two groups or more need to have many observations in each group , if your data has normal distribution you can use T test for two independent samples , in case of not normal distribution you can perform the test with non parametric tests .

Good Luck

Thanks for the answers everyone. You all are right in saying that I can apply statistics if I have series of data for species A as well as B. If I were to compare species A and B in my laboratory, I would conduct series of experiments to generate LC50 values for A and B and then apply t-test to assign a level of confidence. The problem is that I do not have to conduct test in my lab and just have to rely on published data. As 2 < 4, we can simply say that species A is more sensitive than species B. Now the questions is: As published material do not have raw data, will it be appropriate scientifically to say that A is more sensitive than B?

I am worried because I know that even if I take a single species and expose it to metal (e.g CU) several times in exactly similar experimental conditions, the LC50 value will differ each time (at least a little). This issue was raised once in 1970's by Sprague (1977). A method known as standard error of the difference (Sprague and Fogels 1977) was suggested to deal with such issue. This method detects significant differences between a pair of toxicity values but does not provide a p value. However, I am not sure about the credibility of this method.

To summarize: Could you please let me know if it will be acceptable scientifically to say that species A is more sensitive than species B if we do not have raw data and simply looking at the the two published LC50 values from separate publications (of course different experimental conditions)?

Given you have these values from publications, then these publications should also provide a measure of precision/uncertainty of these estimates. If this is the case, then you can use statistical tools to estimate the precision/uncertainty that should be assigned to the difference between these two values. Then you can judge what "significance" you are willing to attribute to this difference.

If not, well, then the papers are examples of "bad practice in science" and I would not take these values very serious anyway.

Statistically it does not work out well to compare two single samples that are independent which represent two different populations. It is advisable that you obtain more observations for both samples which represent two separate populations. In other-words, unpair t-test can therefore be use to test for significant difference between the two samples. However, if the sample size is large the Z test statistic would be appropriate.

Hi Kamran,

I’m agree with Hochen about that the variability of the response must to be reported. In pharmacology and toxicology an LC50 value is not enough, you must to report a Confidence Interval and that could be the point of comparison.

Good luck

Hi Kamran,

I agree with Jochen's explaination to your question. Of course, if we have no any other information,  we can say that spcies A is more sensitive than B. But there are some uncertainity..

Short answer: No.

Long Answer: It is not possible to check the statistical significance of the difference between the two values (which should be two averages) published unprofessional, without the two variability of measurements (wich can be obtained only by replicates) .

Justification: Statistics starts at two numbers up. Only for a minimum number of two numbers make sense an average, standard deviation, etc. In the example given is invocated one number: 2 or 4. When comparing two with four and observe that 2 < 4 we do algebra, not do statistics, much less we not make assessment of statistical significance.

Obviously, 2 or 4 of the two publications can be only average values of concentrations and their publication without the SD has no scientific value. It's like saying that a certain man has a height of 3 m, without identifying the error of measurement (SD). If SD = ± 1.5m the estimation could be true. But if SD = ± 0.01m the estimation couldn't be true. So without SD we said nothing.

Dear colleagues, in respect of the measures that Kamran shows in his question, they are LC50 or lethal concentration 50, it means that are doses of a compound that reduces the population (Artemia salina or rodents) in 50%. In pharmacological and toxicological assays, the LC50 is not an average or mean, it is a prediction value on basis of a regression equation (log linear or probit), for that reason LC50 of a compound can’t be compared with LC50 of other compound strictly like a two populations hypothesis test. It is necessary to calculate the 95% CI of each LC50 and then this intervals can be compared, considering if there are differences or similarities between the range and precision.

Best regards.

I agree with Federico's explanation

@Liviu:

"Justification: Statistics starts at two numbers up." - I do not agree here. Statistics starts where we must consider uncertainty, and this has noting to do with the amount of values or sample sizes (n) one has. You are correct that some methods or calculations do not work when n=1, but this is not the whole of statistics. Meaningful and constructive statistics is possible even for n=1. Comparing 2 and 4 (to pick up your example) while assuming stochastic processes resulting in these observations is at the very heart of statistics and not just algebra. It is algebra to calculate the difference, but statistics provides tools to estimate the uncertainties associated with these observations and with the difference.

To make the example with your values: Given these are counts from a Poisson process, then the log-likelihood contour for the model E(x) = exp(b0 + b1*x) with E(x)~Pois(lamda) is attached below. You can see the profile and select likelihood or confidence ellipses. Don't tell me that this is "just algebra". There is a way to infere reasonable (and unreasonable) model parameters based on the data, and this involves all the core of inferential statistics, what is, to my opinion, way beyond "just algebra".

This example may or may not be sensible (it may simply be not reasonable to assume an non-overdisperdes Poisson process, who knows?) - but at least it shows that even for n=1 there can a lot of statistics and inference be done!

I would even be more provocative and say that statistics does not at all need data. Statistics can be based only on sensible assumptions from which yet sensible concusions can be derived. Statistics is about thinking and knowledge. Data just adjusts this, so to say, to empirical findings, so that we can use empirical data to modify our knowledge.

It can be considered as observations only.

You can compare the two values simply using Z-test. In most scientific papers the authors have to show SD (standard deviation)or SE (standard error) of mean values. So, at first see the published papers to know the SD or SE.

If the SD were provided, then calculate variance for each values through power of SD (variance= SD ^2).

If the SE were provided, you need to know about the number of replications (n). then calculate SD and finally calculate the variance for each values.

SE= SD/sqrt(n) -----> SD=SE*sqrt(n)

SD ^2= variance

Now you know the Variances of LC50 2 and 4.

At last step do the Z-test in excel:

https://www.dummies.com/software/microsoft-office/excel/how-to-perform-z-test-calculations-in-excel/