If you have included a plate calibrator (what you call an "amplification control", I assume), you can use it to adjust the Ct values to make them comparable between the plates.

If there is no plate-to-plate variation, then this adjustment would simply do nothing, and in any other case it will correct for this variation.

There is no need to demonstrate that the plate calibrators show that there is no between-plate variation.

If you still request some "test": there is no statistical test in that sense that would show equality. Statistical tests can only answer the question if your sample size is large enough to have sufficient confidence in concluding that there is some difference. Failing to reject the hypothesis of "no difference" is not evidence for the absence of a difference.

If you want to compare the amplification controls on different qPCR plates, you can use a statistical test called a two-sample t-test. This test is appropriate when you have two independent samples, and you want to compare their means to determine if they are significantly different.

Assuming that your amplification controls are normally distributed and have equal variances, you can use the standard two-sample t-test. However, if the normality assumption is violated, you may need to use a nonparametric test, such as the Mann-Whitney U test.

Before applying the two-sample t-test, you should first check the assumptions of the test. These assumptions include the normality of the data, equal variances, and independence of the samples. You can check the normality assumption using a normal probability plot or a Shapiro-Wilk test. You can check the equal variance assumption using Levene's test.

If the assumptions of the two-sample t-test are met, you can proceed with the test to determine if there is a statistically significant difference between the means of the two groups. If the p-value is less than the significance level (usually set at 0.05), then you can reject the null hypothesis and conclude that there is a significant difference between the means of the two groups. Otherwise, you fail to reject the null hypothesis and conclude that there is no significant difference between the means of the two groups.

It is important to note that statistical significance does not necessarily imply practical significance. Therefore, you should also consider the magnitude of the difference between the means and the context of your experiment when interpreting the results of the test.

If I understand your question, I think you could compare the mean amplification of controls (if you have used same controls for every primer) between the plates either by a t-test or an ANOVA depending upon the number of plates.

Jochen Wilhelm Thank you for your reply. I don't know what you call "plate calibrator" but my "amplification control" is actually one of my samples, which I placed in duplicate on each of my 9 plates. I therefore have no starting value or calibration curve since I work in relative quantification. Ok, if I understand correctly: accepting the null hypothesis does not imply an absence of difference between my plates.

Ma'Mon Abu Hammad Abid Wani Thank you for your reply. At the beginning I used an independent non-parametric ANOVA test (Kruskal-Wallis) but after reflection, the sample measured on the different (9) plates is always the same. So the proper test would be a non-parametric dependent ANOVA?

Any sample (or a pool) put on each plate can serve as a plate calibrator, because you can use the differences in Ct-values between plates of this sample (or pool) to correct for plate-to-plate variation.

And yes: "accepting the null hypothesis does not imply an absence of difference between my plates.". However, I wouldn't talk about "accepting the null hypothesis". You never accept the null hypothesis. You try to reject it, and this may fail. Tests are NOT to accept hypotheses. Tests are to find out if the information of your data is already sufficient to interpret a sample estimate with respect to the null hypothesis. Failing to reject usually only means that your sample size is too small for such an interpretation. "Accepting" the null would be an interpretation (and quite a strong one!), and this is not at all warranted by the failure to reject it. It would be easy to accept any null hypothesis just by designing experiments sample sizes of n=1. They will never allow to reject any hypothesis, so every hypothesis has to be accepted - what is nonsense.

Another point-of-view: You typically try to reject the hypothesis that the mean difference is zero. Such a rejection would allow you to interpret the sign of the sample estimate of the mean difference. But you can test other null hypotheses about the mean difference, e.g. that the mean difference is 2.5, for instance. Now, depending on the (unknown) variance, the (unknown) true difference, and the sample size you may fail to reject both null hypotheses. What would be your conclusion? Accept that the true difference is zero? Or accept that the true difference is 2.5? (or some other value?). It does not make any sense.

Actually, the set of all null hypotheses that can not be rejected (at a significance level of alpha) is known as the (1-alpha) confidence interval, which is a set of values from which the sample estimate. is not "statistically discernible" (your data is not sufficiently "statistically incompatible" with all the values in this set).

Jochen Wilhelm is corret saying that we try to reject null hypothesis based on our collected data. But a hypothesis test tests mean difference between the groups based on the parameters of any underlying distribution, accounting for variances/errors as well (not only difference in means).

Léa Dupont Why did you use non-parametric ANOVA? Did you check assumptions of the ANOVA before going for non-parametric ANOVA?

Furthemore, I think your controls are not dependent samples. Isn't it so?

To wade in from a more practical standpoint, you assume that within a plate, all data is comparable. That's an assumption, yes, but without that assumption, qPCR becomes an entirely meaningless exercise.

We have to assume we can measure things.

As a corollary, you assume that the same samples measured on two different plates will be comparable WITHIN each plate, again because if they're not, all qPCR is meaningless.

So if samples A, B and C are being measured for a GOI and give Cq values of 22, 23 and 24, we determine that sample A has twice as much as sample B, which in turn has twice as much as sample C.

If we were to run these samples again, we might see 23, 24 and 25. Note these are not the same numbers!

Perhaps the machine had a bad day. Perhaps the humidity was different. Perhaps you waiting slightly longer between loading the plate and running it. Perhaps a hundred different reasons. It's an exponential amplification, so even tiny differences can add up to substantial variations.

However: the relationship between the samples is identical. Sample A still has twice as much as sample B, which in turn has twice as much as sample C. Variations might occur, but they occur en bloc, to every sample in the plate. Samples within a single plate remain comparable.

These are assumptions you HAVE to make to even employ qPCR meaningfully.

So, if you had 180 samples, and needed to run these in triplicate, you'd have to split your samples. You would thus run the first 90 on one plate, and then the second 90 on another plate, but also including 3 samples from the first run. We shall call these calibrator samples "A, B and C", for reasons which should now be obvious.

You compare your data for A, B and C from each plate and conclude "huh, so all the Cq values for plate two are 1 cycle higher".

So then you just subtract 1 from the Cq values of every sample on plate 2.

Easy. No stats really needed, because...what are the stats going to tell you? If there's no difference, it's irrelevant. If there is a difference...well, you're still going to use the data, aren't you? In which case...why bother running stats?

Obviously use some common sense (if A, B and C give you values 10 cycles later in one run, that run clearly didn't work correctly), but otherwise just do it as described above.

Practical advice: I would recommend using at least three calibrator samples, because qPCR is still quite variable from run to run. The consensus Cq shift of three samples is a much more accurate read out for plate-to-plate variation than the Cq shift of a single sample (and if that one sample just amplifies weirdly one time, it invalidates all the data from that plate).

I would also always recommend you run ALL SAMPLES for ONE GENE on any given plate, rather than SOME SAMPLES for MULTIPLE GENES. You cannot directly compare PCRs for different genes, whereas you absolutely CAN compare different samples for the same PCR (this is the point, after all), so maximising the number of samples for which you can make direct, within plate comparisons...is the better strategy.

I.e. if I had 180 samples and 5 genes, I would run ten plates, each plate being for a single gene, using either the first 90 samples, or the second 90 samples + the 3 calibrators from the first set.

Hope this helps.

In general, when you have several means to compare, it’s not valid just to compare all possible pairs with t tests. Instead, you follow a two-stage process:

Are all the means equal? Then, a computation called ANOVA (analysis of variance) answers this question.

If ANOVA shows that the means aren’t all equal, then which means are unequal, and by how much? There are many ways to answer this question (and they give different answers), but we’ll use a process called Tukey’s HSD (Honestly Significant Difference).