1
$\begingroup$

Suppose there are $N$ types of blocks. The blocks are labelled with their types. The average weights of the types are in the set $\{w_i\,|\,\,i<N\}$. In other words, at least one pair of block types shares an average weight. The variances of each block type are unknown, but samples fall into a normal distribution.

I draw $M_i$ samples of block type $i$. From my samples, I would like to find out which block types have the same weights.

For example, suppose block type $A$ has an average weight of $1$, while types $B$ and $C$ have average weights of $2$. I draw the following weights for the three block types:

Block type A: 1.03,0.89,0.86,0.77,1.36,1.09 (mean=1.00, stddev=0.213)
Block type B: 2.51,2.26,2.83,3.02,1.62,2.34,2.02,2.84  (mean=2.43, stddev=0.439)
Block type C: 1.99,2.14,2.07,2.01,1.75,1.96,2.1  (mean=2.00, stddev=0.119)

I would like to use this data to infer that types $B$ and $C$ likely share an average weight, while type $A$ has a distinct average weight.


One potentially naïve idea I came up with was to run pairwise t-tests between the means of different types of blocks, to quantify the extent that I suspect the means are identical. Then, I could cluster this data into sets of blocks that likely share means. I don't have a rigorous justification for this method, however, and it is possible it is not statistically sound.

Does anyone know an algorithm which attempts to solve this problem? Does anyone have any other suggestions for what I could do?

$\endgroup$
10
  • $\begingroup$ You lost me at "shares an average weight," because that does not appear to follow from what was stated before. Moreover, asserting "normally-distributed weights" contradicts everything that preceded it. Not only that, with a Normal distribution model for weights the chance of any two randomly independently chosen blocks sharing a weight is zero. Please, then, see whether you could explain the actual problem you face, rather than offering these contradictory abstractions of it. Otherwise it's likely you will get contradictory or misleading answers. $\endgroup$
    – whuber
    Aug 28, 2023 at 15:58
  • 1
    $\begingroup$ I agree with @whuber's assessment. Do you mean to say that each block's weight is drawn from a normal distribution with mean $w_i$ (but unknown variance)? If you have fewer mean weights than "types" of blocks, what distinguishes different "types"? Are variances equal within "types"? If so, do different "types" with equal mean differ on variance? $\endgroup$ Aug 28, 2023 at 16:04
  • $\begingroup$ I have updated the question @whuber $\endgroup$
    – Argon
    Aug 28, 2023 at 16:11
  • $\begingroup$ Could you perhaps give us a concrete example of your experiment and the intended result? The problem with the Normality assumption continues to hold. Moreover, the meaning of "likely to have the same weights" is ambiguous and needs clarification. $\endgroup$
    – whuber
    Aug 28, 2023 at 16:21
  • $\begingroup$ The example is helpful, thank you. Are you perhaps trying to describe an experiment in which you obtain independent samples from three Normal distributions with unknown means $\mu_i$ and you are simultaneously testing the hypotheses $\mu_1=\mu_2,$ $\mu_1=\mu_3,$ and $\mu_2=\mu_3$? If so, do you assume all three standard deviations are equal or not? (If you assume they are equal, this is a standard ANOVA setup; otherwise, it requires a little more technique.) $\endgroup$
    – whuber
    Aug 28, 2023 at 16:38

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.