Say you have data on the average food consumed per type of animal for a 30 day period. Not the total amount of food itself, but the averages per day per that type of animal.

Also assume the number of dogs, cats, and snakes would vary and are not consistent between groups or across days within groups, but what that variation and samples are exactly are not known from the data.

For instance the data could look like this:

                     dogs, cats, snakes
Avg Food lbs. Day 1    2    1.5    .3
Avg Food lbs. Day 2    3.1  1.2    .43
Avg Food lbs. Day 3    1.9  1.3    .2
Avg Food lbs. Day 4    2.2  1.6    .25

etc etc

How can I calculate the central tendency for each group, and how can I compare groups' respective central tendencies? Would taking the median of the averages accomplish the first part of the question?

If these questions can't be answered given the above constraints, what more information would be necessary to find each respective list of averages' central tendencies?

  • $\begingroup$ Because the number of animals can vary by day and species, you obviously can't use an ordinary ANOVA which requires equal variances. But you may get some useful tentative results from oneway.test in R, which does not require equal variances for the three species. // Fictitious data: set.seed(804); y1 = rnorm(30, 2.3, .5); y2 = rnorm(30, 1.2, .2); y3 = rnorm(30, .3, .1); format: y=c(y1,y2,y2); gp = rep(1:3, each=30) then analysis of means uses; oneway.test(y~gp), which returns significant P-value $\approx 0.$ // If your data significant, then follow up with properly designed expt. $\endgroup$
    – BruceET
    Aug 5 at 3:15

Since the group sizes are NOT assumed to all be equal here, you need to know the sample sizes in each individual group (or the proportion of data in each group) to compute the averages you want.

If you had that info, then you could compute the averages using the law of total expectation.

  • $\begingroup$ Applying the "the proportion of data in each group" is in essence a weighted average, yes? $\endgroup$
    – JLuu
    Aug 4 at 23:12
  • $\begingroup$ @JLuu Yes, absolutely. The proportions serve as the weights in a weighted average. That is what the law of total expectation says. $\endgroup$ Aug 5 at 0:57

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