how to choose resistors' value for common emitter amplifier?
Its not really practical to design any  circuit without calculations and Alfred's answer gives a comprehensive approach (so +1 from me). However, there are other 'rule of thumb' type" approaches that might suit a 'practical' designer to get started. The thinking behind this approach is very much based in the theory developed by scientists and engineers over decades and if you are to advance in any design work I would recommend that you try to develop a solid mathematical approach. (Note: I still need to use Ohm's law  V= IxR  and an understanding of percentages/fractions  with this approach!)

The circuit is pretty much standard, well understood and lends itself to a simple analysis.

Let's start with the value of emitter resistor (R4). The voltage across it should be between 10 and 20% of the supply (rule of thumb) to give an 80 - 90% voltage swing at the collector.

Decision 1. Let's take Ve as 10% of the supply (= 0.5V for a 5V supply). 

Now choose the current you want through the transistor. 100uA, 1mA, 10mA? This will depend on your application. Do you need to minimise current used or do you need to drive a high current into the next stage? 

Decision 2. Let's choose a typical value, say 1mA 

Now we need a little bit of (simple) maths (using V=IR). 

For a 0.5V drop @ 1mA we need a 500R resistor - BUT this is not a 'standard' E24 or 5% value so choose a nearest preferred value (n.p.v.) - either 470R or 510R 

Decision 3. I choose the 510R 

This value represents 10% supply drop leaving 90% to be divided between the transistor (c-e) and the collector resistor. Allowing for a minimum saturation voltage (say 0.2V) we need about 40% of the supply dropped across R2. (to give maximum output swing at collector). As the emitter resistor represents 10% supply drop we can simply calculate the collector resistor at about 4 X emitter resistor. (= 4 X 510 = 2040R). Once again not a preferred value so we can choose from 2k0 or 2k2. ( the collector voltage will be set around 3V and have a +/- swing of 2V)  

Decision 4. 2k0 

The size of the decoupling capacitor is next. This basically sets the low frequency response of the stage. The bigger the value the lower the cut off frequency. Let's assume you are designing an audio stage. A 100uF capacitor will have a reactance of about 80 ohms @ 20Hz. A 10uF will have about 800 ohms. Doubling the capacitance will half the reactance. A 22uF will be about 400 ohms which is about the size of the emitter resistor. Choosing a higher or lower value will decrease or increase the cut off frequency of the amplifier stage.

Decision 5. I choose 22uF (a preferred value capacitor)

The final parts of the design is choosing suitable values for R1,R3 and Cin.

R1 and R3 form a voltage divider. We 'know' that the voltage at the emitter is about 0.5V and that there will be a 0.6V drop from the base to the emitter (assuming its a silicon transistor). So we know the voltage at the base will be 0.5 + 0.6 = 1.1V.

For a small signal transistor the current gain will be at least 100. This means that the current going into the base  will be a maximum of 1/100 th of the emitter current. This current will be taken from the voltage divider circuit. The rule of thumb is to have at least 10 x this base current going through the bias resistors so that we can 'ignore' this loading effect in the calculation. This means that we need about 1/10th of the emitter current through R1 and R3. this is 1mA/10 = 0.1mA.

R3 can now be calculated because we know it needs 1.1V across it and a current of about 0.1mA through it. (V=IR). This gives R3 = (1.1/0.1) * 1000 = 11k (a preferred value).

R1 has a voltage drop of 5V - 1.1V = 3.9V for the same current. This gives a value of 39k (a preferred value). If the calculations don't give preferred values then adjust to the n.p.v. You can always tweak the values later by measuring the voltage at the base but generally they will be of the correct magnitude.

Decision 6. R1 = 39k and R3 = 11k

These values set the input impedance of the stage. Very roughly R3 will be reduced by about 10% due to current taken through the base of the transistor making it about 10k. This will effectively be in parallel with R1. It unlikely you require a precise answer, so rounding the 39k to 40k we have approx. 10k // 40k. Without calculating anything we can say the input impedance just under the 10k. (lowest resistance will dominate).

This value is about 20x the size of R4 and as we chose 22uF for the by-pass capacitor we need something about 20x smaller at the input to give the same cut off frequency - say 1u0 (preferred value) - which by coincidence matches the value shown.

Decision 7.  Cin = 1u0

What if you want a different current say Ic = 100uA or 10mA? In the words of the great Alexandre Orloff - SIMPLES!

Just scale values by ratio. 

for Ic = 100uA  (= 1/10th), Resistors = 10 x values, capacitors = 1/10th values
for Ic = 10mA (= x 10) , Resistors = 1/10 x values, capacitors = 10 x values