Hello,
I have more than two groups each containing the unequal number of samples.
(group 1= 5 samples; Group 2, 3, 6=4 samples; Group 4,5=3 samples).
I ran western blot of each sample.
Now, I am willing to do the statistical analysis of their means.
Kindly, suggest an appropriate statistical test!!!!
I am not specialist in statistical, but I suggest that each two groups be tested together, and then compare the results. If there is a similar between them, the two groups are combined, and then compared the combined results with the third group.
Or, test each two groups together and then find the arithmetic mean of the results.
Completely Randomized Design (CRD) for ANOVA can be used in this situation.
You may check these for examples:
http://www.fao.org/docrep/003/X6831E/X6831E07.htm
and
http://pba.ucdavis.edu/files/45004.pdf
One way analysis of variance, paying attention to possible unequal within group variances and the multiple comparison problem.
ANOVA does not compare means. It compared models, and only if these models differ in exactly one coefficient for a group mean, ANOVA reduces to a t-test.
Only the t-test tests the hypothesis that two group means are equal (rejecting this hypothesis means that you are sufficiently confident to decide in which group you expect higher values).
If you have more than two groups, it is advantgous to use the total information on the variance from all the groups in each t-test (each one comparing two of the groups). This is obtained by using the pooled variance estimate. This statistic is an interim result of an ANOVA, and therefore most software offer the t-tests based on such a pooled estimate in the context of ANOVAs. However, there is no logical link between ANOVA and the t-tests. The actual ANOVA is irrelevant to your problem. It is only about the t-tests.
If there are several groups you can perform a whole series of tests, each one giving a p-value. Usually, a p-value is interpreted in the context of a single, specific comparison. But here your context is a whole series of comparisons, what makes it neccesary to interpret the p-value a bit differently, because the more tests you do, the more likely you will get at least one quite small p-value even under the hypothesis that all groups have the same means. So getting one p < 0.01 might not be sufficient to confidently reject the null hypothesis when this is one out of many p-values. There are many different approaches to tackle this problem, and they are vastly misused and abused because the scientists don't understand the (statistical) context, don't consider the experimental context, and what these procedures actually do or are supposed to do. (I don't blame it on the scientists, I blame it on the bad education system)
Finally you must understand that the t-test is based on some assumptions. One of these assumptions is that the test statistic is t-distributed, and this follows exactly when the data is normal distributed. However, in real life there is nothing like a (perfect) normal distribution. It is only important that the data is approximately normal distributed, in fact it is often absoultely sufficient when the distribution is symmetric. The approximation should be better when the sample size is small, and the larger the sample size the less important is a good approximation. If the sample size is alrger than 30-50 (or so) even an evident assymmetry of the distribution is practically irrelevant. Further, as the variance is estimated from different groups together, the variance should be expected to be the same in all groups. If this expectation is clearly unjustified, then the usual interpretation of the test result will also be unjustified. Thus, if your sample size is small you should give the distribution a thought:
The densitiy values (-> concentration) you measure cannot be negative. Thus, low concentrated samples that have a low average band density can show considerably higher values but not much lower vlaues for individual bands. Therefore can the distribution not be symmetric (it must be right-skewed). Also, band densities of samples with high average concentartions can (and do) vary over a larger range - they have a larger variance. Both "problems" can usually be solved by using log(band density) or log(conc) as the value that are compared. But take care: this changes your question. Instead of asking what the difference conc is, you ask what the difference in log(conc) is. The former test is about the hypothesis "the mean conc are equal", the latter is about "the mean log(conc) are equal", what translates to "the ratio of conc is 1". The two hypotheses are equivalent in so far as: if one of them is true, then the other is also true, and if one is false, the other is also false - but they differ by the amount and interpretation of uncertainty associated with the data.
So I suggest using log(band densities) for your analysis (t-tests using a pooled variance estimate). Whether or not and how you should care about multiple testing depends on the context.
You can use ANOVA, then use Tukey for inter-group comparison. Tukey is powerful for all-possible pairwise comparisons when sample sizes are either equal or unequal.
Stevens (1999): Post Hoc Tests in ANOVA: pages.uoregon.edu/stevensj/posthoc.pdf
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