Given that there is a checkmark-shaped relationship between district size and per pupil spending, we can calculate the theoretically most efficient district size. To do that, we have to know the value of the constant a in the checkmark function f(Size) = ln2(Size) − a * ln(Size). That value can be computed by running the regression with the ln2(Size) and ln(Size) terms being allowed to vary separately, and dividing the ln(Size) coefficient by the ln2(Size) coefficient.[*] Doing so reveals the value of a to be roughly 15.95.

Based on that value of a, the author found that the most efficient school district size in Michigan is 2,911 students. Using the coefficient of 96.2 for the checkmark term (see Graphic 3, Model 4, here Investigating f(Size)), the author calculated that a district of 1,500 students is likely to spend about $40 less per pupil each year than a district of 2,911 students, all other things being equal. Similarly, the spending difference between a district of 500 students and one of 2,911 students is about $300 per pupil.

Those are, of course, only ballpark numbers. A 95 percent confidence interval around the coefficient value for the checkmark term ranges from 46 to 147. As a result, actual differences in spending due to variations in district size could easily be anywhere from less than half the values reported in the previous paragraph, to one and a half times those values. It is also important to realize that a variety of political and geographical[†] considerations might make particular mergers or consolidations difficult or impossible, so this entire confidence interval represents an upper bound on possible savings.

It is also worth considering that districts larger than 2,911 students generally spend more per pupil than optimally sized districts. This is particularly important given that 70 percent of all Michigan’s conventional public school students are currently enrolled in these excessively large districts. The potential savings from consolidating tiny districts are thus modest compared to the potential savings from breaking up overly large districts.

Accentuating the practical advantage of breakups over consolidations is the fact that any excessively large district can be beneficially broken up into smaller districts, but small districts can be beneficially merged together only if they happen to be adjacent to other districts that are also far enough below the optimal size for a consolidation to result in a new district closer to the optimal size.[**]

To put numbers on this discussion, the total cost premium currently being paid due to excessively small districts is on the order of $31 million according to the model presented here, whereas the total cost premium due to excessively large districts is on the order of $363 million.[‡] So, even if every small district in Michigan were located adjacent to other suitably small districts (and hence a viable consolidation candidate), the savings from such consolidations are still predicted to be only about 8.5 percent of those predicted for breaking up excessively large districts. And because many small districts are undoubtedly not located next to other suitably small districts, the difference in potential savings from consolidations versus breakups is almost certainly greater.

Once again, it should be noted that these are only ballpark numbers, since the 95 percent confidence interval on the size term is quite broad.


[*] The reason we don’t simply leave these terms separate in our checkmark model is that they contribute to variance inflation when they’re allowed to vary independently. That variance inflation can be eliminated by grouping the terms together as described in the text.

[†] Optimal consolidations could only happen among adjacent small districts. A small district that is geographically surrounded by large districts could not efficiently be merged with any other.

[**] Noncontiguous school districts are unusual.

[‡]To generate these estimates, the study first calculated the value of the size term for an optimally sized district (i.e., a district of 2,911 pupils). The size term for such a district has the value -6,120. Since that is the minimum possible value for the size term, we can ascertain the cost premium per pupil attached to any other district size by adding 6,120 to the value of its size term. Consequently, we can compute the total possible savings from consolidations and breakups by simply multiplying the per pupil cost premium for each district by that district’s enrollment, and then summing up those net premiums for districts with fewer than 2,911 pupils (to get the maximum theoretical consolidation savings) and for districts of more than 2,911 pupils (to get the maximum theoretical breakup savings).