Equivalence ratio explained
You are correct.  There is a theoretical ratio of air-to-fuel that provides exactly enough air to completely burn the fuel, converting it to $\ce{CO2}$ and $\ce{H2O}$.  This theoretical ratio can be calculated knowing (a) the exact composition of the fuel and (b) how much oxygen is in the air (generally ~21%), and (c) the formula weight(s) of the molecules that are in the fuel.  For fuel that are pure octane, the combustion reaction is:

$$\ce{\frac{25}{2}O2 + C8H18 -&gt; 8CO2 + 9H2O}$$

Which means 12.5 moles of oxygen are required for every mole of octane.  For 12.5 moles of oxygen you need about 60 moles of air because air is only 21% oxygen.  That works out to 1740 grams of air (~29 grams per mole).  One mole of octane is 114 grams.  So the theoretical, or stoichiometric ratio of air to fuel is 1740/114 or about 15.

Most motor fuels are not pure octane.  The additives complicate the math a bit, but the stoichiometric ratio is around 14 for many fuels.  

The equivalence ratio is just the actual air-to-fuel ratio divided by the theoretical.  


  My understanding of the equivalence ratio is that if it is below 1, then there is too little air (compared to the stoichometric mix). If it is above 1, that means that there is too much fuel, and not all of it can be used leading to what is called incomplete combustion.


You said the second part backwards.  If the ratio is below one, there is too little air (i.e. too much fuel), and if it above one, there is too much air (not enough fuel).  So incomplete combustion should happen more at lower equivalence ratios.

Also, the exhaust temperature will be maximized for equivalence ratios close to 1.  Otherwise, there are extra reactants, either fuel or air, which remain after combustion and absorb some of the combustion energy, resulting in a lower temperature increase.  Since temperature is important for NOx formation, I figured it was worth mentioning.