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I think that it does make conceptual sense. For means and proportions, we want to compare the amount at which their values are apart. Which is why difference is used. For variances, the factor at which one is different from the other is important since it is the spread. Which is why ratio is used.

Someone told me here in another thread that a difference of variance does exist but I have tried searching and not seen any results. Can you just apply any operation to the multiple population measurements? I guess you could but they would probably not be worth of interest in studying.

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    $\begingroup$ It can make conceptual sense, but there are exceptions. For example, it's climatological folklore that nearby mean temperatures tend to differ by a constant (additively) while nearby mean precipitations (rainfall etc.) tend to differ by a ratio (multiplicatively). Looking at variance ratios is connected to the fact that variances can't be negative. There is a mix of mathematics and of what nature or society does here behind whether additive or multiplicative comparison is the better approximation. $\endgroup$
    – Nick Cox
    Aug 6 '20 at 23:55
  • $\begingroup$ How are exceptions dealt with? How would a researcher deal with the mean precipitations situation then? Do they just ignore that the variable happens to differ by ratios and go on using differences because using ratios would be too complicated? $\endgroup$
    – AndroidV11
    Aug 6 '20 at 23:58
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    $\begingroup$ Ratios really aren't more complicated. We slip between "Joanna is 5 cm (2 inches) taller than her brother" and "that house costs twice as much and is way outside our range". It can also be true that "that house is USD 10000 more than we can afford". All a matter of what is convenient and congenial as well as what is natural quantitatively. Some people use logit scale enough to find it natural, not to mention pH, decibels and octaves. Many economists go straight to log income as their scale of choice and that fits with wage or salary increases or decreases of so many % and 50% off in a sale. $\endgroup$
    – Nick Cox
    Aug 7 '20 at 0:55
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The simple answer is because we have known distributions for those quantities.

We generally believe differences of means and proportions should be asymptotically normal (and probably close to normal in small samples). That means we can use a $t$-test and not be breaking our assumptions too much.

Testing a ratio of means would be very difficult if the denominator were close to 0: the distribution would not be close to normal and the confidence interval for that ratio might not even be a convex set. (Read up on Fieller's Method to see how such a ratio can yield confidence intervals unlike those typically seen.)

A difference of variances is a difference of chi-squared-related variables. For small samples, that is not going to be close to normal. However, we do know that variances are bounded below by zero. Hence we can test the ratio of variances using an $F$-test. Furthermore, we often use $F$-tests on likelihood ratios -- so we aren't getting rid of $F$-tests any time soon.

If the ratio versus difference really bothers you, remember that we could just take logs and test the difference of logged variances. To do that, we would still be using an $F$-test.

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  • $\begingroup$ Does taking the logarithm of the ratio drastically affect the sampling distribution or would it still be similar? I mean, you have to still use the F-test but would it give the exact same result, a close one or something different? $\endgroup$
    – AndroidV11
    Aug 6 '20 at 23:56
  • $\begingroup$ It does affect the distribution. To do the $F$-test, you would exponentiate the difference -- or create a table of logged $F$-test statistics. $\endgroup$
    – kurtosis
    Aug 6 '20 at 23:58

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