Xptpcrewx is correct in that the open circuit test will only measure the losses due to hysteresis and eddy currents. Further, these losses only occur due to the current flow in the primary winding (primary in this sense means the source side of the transformer). The losses measured under the no-load condition are ignored for the most part to simplify calculations based on the very small loss of accuracy of the final result. It appears from your measurements you are attempting determine the %Z using the open circuit type of test. The percent Z represents a measurement of the equivalent resistance and reactance values of the transformer. That is it takes into account the primary and secondary resistance and reactance to give you an equivalent series impedance that you may use to determine the voltage regulation (voltage drop) that will occur based the load current and power factor.
An open circuit test will give you the no-load losses that are used with the short circuit test for the load losses, to calculate the overall efficiency of the transformer. I note here that the maximum efficiency of any transformer occurs when the open circuit test losses and the copper losses are equal, typically around 60% of the full load capacity.
To equate the %Z on the nameplate of the transformer to the actual equivalent %R and %X, you need to perform a short circuit test. The easiest path to this number is to determine the short circuit power (Psc), derive Re (equivalent Resistance) = Psc/(Isc*Isc) and equivalent impedance (Ze) = Esc/Isc, then equivalent reactance, Xe = square root of (Ze squared - Re squared). To get to the %Z on the nameplate, divide Ze by (nameplate voltage squared/nameplate VA).
Last, to your question, “So also a more general question: can we say or make any predictions about the losses given the %Z? I have often heard a phrase like "Utilities are getting more concerned with losses so are using lower impedance transformers". Does the %Z give a better indication of load losses, no load losses, both, or neither?”
I’m note sure what predictions you are referring to here but I can speak to Utilites using lower impedance transformers to reduce losses. Yes, we do that for two reasons, First, it reduces losses. Losses which you as a customer pay for. We take the sum total of our generatEd energy and compare it to the sum total of our measured delivered energy and compute overall losses, the cost of which ultimately is paid for by the customer. So, reducing losses is a good thing, right? Second, reducing %Z raises the amount of short circuit current that a transformer may deliver. At the terminals of the transformer and assuming an infinite supply the available short circuit current = I (full load) / (%Z/100). So you can see the short circuit current gets higher as the equivalent impedance goes down.