Dear Miguel,

thank you for your response. That was an example. In my case, it looks similar. I have two strains of mice and I want to compare the fertility of males. In each strain (A and B) I have a lot of males, each of them was in a cage with one female. For every male (cage) I have 3-4 litters. The easiest way would be to use all the litters to compare in a simple t-test. Then I would have a mean litter size for strain A and B. However, given that I have a few litters for each male I would prefer to have mean litter size for each male and then use this data for the t-test. Actually, I have tried this and mean of means is almost the same as all litters mean and its distribution is normal. But I was wondering if it is a correct way to do this, omitting variances for each male.

Also, I am asking this question because I could not find a way to do statistics for this kind of problem anywhere! All explanations are so complicated that someone who is not very familiar with statistics will have a problem. I can imagine a lot of research with this kind of data. We would like to compare plants kept in light (A) and dark (B). We have 20 plants in every condition. For each plant, we measure the level of chlorophyll in all leaves (so f. ex. in one case we have 5 leaves, in other 7). So we should average the levels of chlorophyll for each plant and then compare two groups - A and B. But how? With a normal t-test? What with the variations of single means?

Dear Przemyslaw,

do you have just 3 people in one group or was this just an example? If you have just 3 persons in one group, the sample size is too small for a valide statistical analysis.

In your example, the measurements are without inacceptable differences - so I think it would be possible to take the mean of the repetitive measurements.

Before starting the analysis, you should perform a test for find out the distribution - in studies with small sample size non-parametric tests are appropriate very often.

Good luck for your study!

Best

Tanja

Dear Tanja,

thank you for your response. It was just an example. Let assume that I have bigger number of people in each group, like 15 or 20? How can I compare these groups?

Hi, what are the multiple measurements? Are you trying to measure the same thing several times? If you are, I'd say you can definitely take the average (or median) per individual and then you'll have one measure per individual in group A and one measure per individual in group B. When you average or take the median of the different measures you've taken for one same individual, you're just trying to obtain a more accurate approximation of what you're actually trying to measure. After you do this, to compare group A with group B you must first (like Tanja said) assess the normality of each group (with normality tests or by graphical analysis). If both groups are normally distributed you can use a t-test, if not you can use a mann-whitney test.

The crucial question here is: what exactly do you mean by "multiple measurements"?

Dear Miguel,

thank you for your response. That was an example. In my case, it looks similar. I have two strains of mice and I want to compare the fertility of males. In each strain (A and B) I have a lot of males, each of them was in a cage with one female. For every male (cage) I have 3-4 litters. The easiest way would be to use all the litters to compare in a simple t-test. Then I would have a mean litter size for strain A and B. However, given that I have a few litters for each male I would prefer to have mean litter size for each male and then use this data for the t-test. Actually, I have tried this and mean of means is almost the same as all litters mean and its distribution is normal. But I was wondering if it is a correct way to do this, omitting variances for each male.

Also, I am asking this question because I could not find a way to do statistics for this kind of problem anywhere! All explanations are so complicated that someone who is not very familiar with statistics will have a problem. I can imagine a lot of research with this kind of data. We would like to compare plants kept in light (A) and dark (B). We have 20 plants in every condition. For each plant, we measure the level of chlorophyll in all leaves (so f. ex. in one case we have 5 leaves, in other 7). So we should average the levels of chlorophyll for each plant and then compare two groups - A and B. But how? With a normal t-test? What with the variations of single means?

Calculate the mean and standard deviation for each group

Subject the results from each group to a test of difference of means, comparing control and experimental groups. Obtain p values for each comparison. Assume P

All the above answers are okay but if time is a factor in your measurements then you need to use what is called repeated measure ANOVA. This then will give you within group differences due to time and between group difference for the groups you are comparing.

If your repeated measures are just replicates of the same thing then you do not need the repeated measure analysis. Again your distribution for normality must be tested. If it is violated then Friedman' test will be most appropriate.

I hope you find this helpful.

There are many statistical analysis methods. Out of them to identify statistically significance in measurements between 2 or more groups, it is best to use ANOVA (Analysis of Variance).

You arrange the data in a two way table for each group.

There will be 2 two way table, one for Group A and the other for or Group B.

The go for analysis of variance.

I would suggest using paired T test

I suggest two way ANOVA with the group on two levels and time with three or four levels after elevation of the ANOVA assumption and if the data is normal.

I find a test called "nested-test" provided by Prism 8 software would be suitable in this situation.

Dear Chuang Liu,

Kindly explain the idea behind the "nested-test" as mentioned by you.

Also, mention the types of hypothesis that can be tested by this method along with the test statistic, with the probability distribution followed by it, used in the method.

I am of the same view with that of Raoof Negaresh that two way ANOVA is applicable here.

In my earlier answer I have already mentioned that two-way ANOVA is suitable here. The broad steps in this method are as follows:

You arrange the data in a two way table for each group.

There will be 2 two way table, one for Group A and the other for or Group B.

The go for analysis of variance.

I think that you could do an ANOVA for each group as suggested, but you do want to compare the groups, right?

I think that Thomas Akkueteh Ndanu indicated the best approach. The problem in this mice case is the non-independence of the data, since the litters are successive in the same females and litter parameters can be assumed to be influenced by the conditions of both male and female, and more seriously, by the successive effect of litters on the females (the litters before the last one, also it was the first litter!). Non-independence is usually the case on observations over time, as in this mice situation.

Repeated measures design of ANOVA is probably the best solution unless the assumption of normality is violated in which case the non-parametric alternative (Friedman test) is in place, as Thomas mentioned.