John, others,
Depends on the application.

I disagree. Whatever the application, and whatever the precision
required, having to go back and redo work because of careless errors
made is sloppy workmanship.

If I wanted high precision timing,
I would use a crystal and would say that not using this is a
sloppy design to begin with.

And by that logic, not using an OCXO would constitute a sloppy crystal
oscillator design, and so on up the chain...

The lack of use of a crystal by the OP
suggests that the design is permitted to be a little "sloppy".

Lack of crystal? You must be joking; the poor guy doesn't even know
where a 2:1 error in output frequency is coming from, and you've got
him sitting down deciding how much slop he can live with, before the
fact!

If precision timing is not needed, the use of the crystal
would constitute a sloppy thinking process, because it just adds
extra components and cost and maybe even troubleshooting and
additional failure modes to the design.

How on earth did this crystal manage to work its way into the
discussion? It's certainly not relevant to the discussion, which is
about where a 555 timing error is coming from.

At least where I live
there is a drammatic price and time difference in buying precision
components (online usually) and picking the lower precision
components off the shelf at the local surplus shop. It sounds like
the OP will be quite content, and learn much (just like me), with
common lower precision components and a little tuning with a trim
pot.

Yeah, great.

I note that many multimeters have cap testers. No idea how accurate
they are,

Read the multimeter spec's...

but since the timing caps seem to have higher inaccuracy
than the rest of the typical astable 555 circuit, I would think checking
the capacitance would be informative in predicting the frequency. In
worst case, one could get relative values to compare caps, which is
still helpful.

???

BTW... when I write "theoretical", I mean specific equations for
frequency and duty cycle using stated values on components Ra, Rb
and C. These equations are only approximations regardless of how precise
components you add to the 555.

They're _not_ approximations, they're exact.
To determine how closely the device will conform to the equations
however, the device specifications must be read.

A case in point, for instance, a couple
somewhat different constants for the frequency constant in the
numerator for the astable calculation exist (1.49 and 1.44). These values
are mentioned in books about the astable mode and are discussed
independently of vendor, yet they differ by about 3.4%. Until now, I have
never heard anyone claim otherwise and have expected to have easily 5%
deviation from calculated frequency and/or duty cycle. Maybe you know
something I do not?

I suspect that's probably true.

Actually, I have been curious about where that constant comes from to
begin with.

The short answer is, "Because of the voltage divider".
The long answer follows:
If you look at the "front end" of a 555, you'll find a voltage divider
and two voltage comparators hooked up like this:
Vcc

[R]

+\
  >
TH>+/

[R]
__ 
TR>+\
  >
+/

[R]

GND
The first thing to notice is that since the resistors are all equal,
the voltages on the inputs of the comparators connected to the string
will always be 2/3 Vcc and 1/3 Vcc. That is, they will be
ratiometric. On top of that, since the resistors are all made of the
same material and are very nearly isothermal, temperature changes
affecting one resistor will affect all of them, with the result that
the voltages on the inputs of the comparators connected to them will
stay constant for changes in temperature.
Now, looking at a simplified diagram of a 555 hooked up as an astable
multivibrator:
Vcc Vcc
 
[R] [Ra]
 U1A 
+\ +>OUT
  >+  
TH>++/  ++  ++
  +R Q+  
 [R]  _ D 
__   U1B +S QG Q1 
TR>++\  ++ S [Rb]
   >+  
 ++/ GND 
  
++Vth
 
[R] [C]
 
GND GND
Assume that the circuit has just been powered up and that C is at 0V
and is just beginning to charge up toward Vcc through Ra and Rb.
Then, since
T = k (Ra+Rb) C (1)
and, since
Vcc  Vth1
k = ln  (2)
Vcc  Vth2
where Vth1 is Vth at turnon and Vth2 is Vth at 2/3Vcc,
if C needs to charge to 2/3 Vcc to get U1A+ to go more positive than
U1A, we can say:
3  0
k = ln  = ln 3 = 1.1
3  2
and, therefore:
T = k (Ra_Rb) C = 1.1RC
Once U1A+ goes more positive than U1A, the output of the 555 will go
low, Q1 will turn on, and C will begin to discharge through R2.
However, since C only charged to 2/3 Vcc before Q1 turned on and
started discharging C, it will only have to discharge to 1/3 Vcc
before it goes more negative than U1A. Since this a voltage ratio of
2:1 we can write
2/3Vcc
k = ln  = ln 2 = 0.693
1/3Vcc
so the discharge time will be
T = k (Rb) C = 0.693 Rb C
This time, when the cap discharges to 1/3Vcc and starts charging
again, the same situation will prevail. That is, there will be 1/3Vcc
across the cap and it will have to charge to 2/3Vcc before the cycle
will start anew, so it will charge to twice the voltage that was
across it when it started to charge, so the time constant will be the
same as when it was discharging, but now it will be charging through
Ra _and_ Rb, so the time to get to 2/3Vcc will be longer than the time
it took to get from 2/3 Vcc to 1/3 Vcc. That's the reason for the
asymmetrical duty cycle.
Finally, since frequency is the reciprocal of time, we can write
1 1
f =  =  = 1.44 Rc
k RC 0.693 RC
and that's where f = 1.44 RC comes from.
The other values you've seen may have been generated considering
leakage and bias currents.