I have seen so many videos on YouTube with instructors following each other with descriptions of how to design the stage.
In fact ALL OF THEM ARE WRONG
You design the stage BACKWARDS.
The 3 critical components are the EMITTER RESISTOR, THE LOAD and the OUTPUT CAPACITOR.
These 3 components will decide the output swing.
If the LOAD is directly coupled to the emitter, the circuitry is simple.
But when the output is coupled  via a capacitor (electrolytic) the output will be affected by the surrounding components.
What most instructors fail to take into account is the DISCHARGING of the capacitor and the second cycle will fail to perform.
That's why taking only mathematics into the equation will give you a failed design.  You have to know how the circuit works to create one that works.

An emitter-follower stage is also called a common collector stage and it puts very little load on the stage that is driving it, and it drives into a very low impedance load. 
In other words it has a very high impedance input and a very low impedance output.
The input needs to have an amplitude equal to rail voltage and only requires about 1/100th the current into the load.
We say the transistor is a driver transistor and the stage is an output stage. 

This is the ideal way to drive a LOAD.  It is called DIRECT COUPLING.
When the base is raised, the emitter follows with about 100 to 200 times more strength.
You need a lot of "strength" (current) through the coil of the speaker to move the cone and create a loud sound.
The circuit has NO LOSSES.  There are no other components and thus NO LOSSES.
If the input signal remains HIGH, full current will flow through the coil of the speaker but it will not make any sound and just get hot. That's the disadvantage of this circuit. The input must be oscillating at all times.

If the input has the possibility of remaining HIGH, the solution is to add a capacitor to the input. But it will introduce LOSSES. These losses can be from 1% to 99%With the capacitor, the current flows through the coil of the speaker when the input signal is rising and slightly less flows when the input signal is falling. 
If it stays HIGH, current will stop flowing through the coil of the speaker because the capacitor will charge and the base voltage will drop to almost zero.
This means you have full control of the current through the voice-coil when the input signal is oscillating or when it is stationary, and the speaker will never get hot.
But you will have less control over the current when the signal is falling. because the capacitor is charging and some of the amplitude of the input signal is being lost.
When the next cycle is delivered, the capacitor is partially charged because there is no component to discharge the capacitor. The base emitter junction allows the capacitor to discharge to about 0.7v, but this value remains on the capacitor. And the capacitor may not have time to fully discharge due to the frequency of the input signal and an even higher charge may be present.

To reduce these problems, this is the next improvement to the circuit.
The speaker is now AC coupled to the output of the emitter-follower. This means it is capacitor-coupled.
The load resistor is 8R, the speaker is 8R and a large capacitor is added to transfer as much energy as possible.
The resistance (capacitive Reactance) of 47u at 2kHz is only 2 ohms and it will have little effect on the transfer of energy.
Now we have a surprising advantage. The capacitor becomes a miniature battery and it gets charged when the input is rising and when the input falls, the positive lead of the 47u sees a falling voltage and the other lead of the capacitor falls about the same amount. This pushes a negative voltage through the voice coil and makes the cone move in the opposite direction. All the energy in the 47u is transferred to the speaker to produce a louder effect.
We have actually produced almost twice the amplitude of the signal into the speaker.  
This feature is so important that we will cover it in more detail.
Suppose the emitter is pulled HIGH. During this time the capacitor will charge and the charging current will actually flow from the 0v rail and up through the speaker to the capacitor. We have to describe it this way so you understand what is happening. When the emitter is as high as possible, the capacitor will be fully charged and the current through the speaker will drop to zero.
The base voltage starts to fall and the transistor does not deliver as much current through the 8R resistor. But the 47u is fully charged and it is like a miniature battery. Replace it with a battery symbol to see what is happening.
Suppose the 8R had 10 volts across it and the capacitor (battery) has 6v. The energy in the capacitor (battery) will start to flow through the voice coil and as the voltage across the 8R resistor reduces, the positive of the electrolytic starts to fall and and this makes the negative lead fall and not only will it fall by the voltage in the capacitor but by the 10v across the 8R. This current creates magnetic flux in the opposite direction and pushes the cone in the opposite direction and the speaker sees a waveform that is larger than the input waveform.
Putting this phenomena another way, the charging current through the speaker puts a positive on the top of the speaker. When the capacitor is discharging, the speaker is effective "flipped over" and the top terminal of the speaker is below the 0v rail and becomes negative.

Now we look at fitting different (wrong) value component to the output.
If we fit a high value resistor as the LOAD, This is what happens:
The output becomes a closed-loop with the capacitor as a battery and passing current through 8R and 100R in series. 
You can see the current will be a lot less than the circuit above where the resistance is 16R.
But in the circuit above, the voltage across each component will be about 50%. In this circuit the voltage across the speaker will be about 10%. So the energy will fall from 50% to 10%. This is called a miss-match.

This arrangement creates a miss-match because the circuit is required to deliver a lot of current through the 100R load resistor, whereas the 1k is only going to use 10% of the current.   

The coupling capacitor is too small.  We said the 47u has a capacitive reactance of 2 ohms. The 1u will be 100 ohms.
It will charge very quickly during a cycle and all the rise in voltage from the transistor will be lost in the capacitor. The speaker will get almost nothing.
And it will have very little energy to provide reverse voltage to the speaker. The result will be absolutely zero output from the speaker.