By my calculations*:
70F to 40F you'd go from 13 PSI to 11.43 PSI (27.7 to 26.13 PSI real pressure), or about 1.5 PSI lost.
70F to 50F you'd go from 13 PSI to 11.94 PSI (27.7 to 26.64 PSI real pressure), or about 1 PSI lost.
Note that the nominal pressure of a ball is really shorthand for "X PSI above the normal atmospheric pressure". Normal atmospheric pressure is about 14.7 PSI; a ball that was measured at "13 PSI" was really at 27.7 PSI of total pressure**. If it lost 2 PSI of pressure, then it went from 27.7 PSI to 25.7 PSI of total pressure: about a 7% loss.
For everyone doing the calculation, that difference is crucial. If you throw 13 PSI into a Combined Gas Law Calculator for the pressure and then vary the temperature, you'll underestimate the effect of temperature significantly and come to the conclusion that the ball only loses about 0.4 PSI from the 70 to 50F change rather than something more like 1 full PSI.
*http://www.calculatoredge.com/chemical/combined%20gas%20law.htm double checked at http://www.1728.org/combined.htm
**If it were actually 13 PSI, it'd be lower pressure than the surrounding air--when you opened the valve, it'd suck in air rather than spewing air out.