You don't explain what you are doing in the experiment, but I would guess it is supposed to demonstrate conservation of angular momentum. If we calculate the angular momentum $L$ using:
$$ L = I\omega $$
then unless some external force is applied $L$ will be a constant.
When you start the experiment the disk has moment of inertia $I_\text{Disk}$ and is rotating at a speed $\omega_i$. So the initial angular momentum is:
$$ L_i = I_\text{Disk}\omega_i \tag{1} $$
When you put the hoop onto the the disk the combined angular momentum of the disk and hoop is:
$$ I_{\text{total}} = I_\text{Disk} + I_\text{Hoop} $$
where $I_\text{Hoop}$ is the moment of inertia of the hoop. If the final angular velocity is $\omega_f$ then the final angular momentum is:
$$ L_f = (I_\text{Disk} + I_\text{Hoop})\omega_f \tag{2}$$
If angular momentum is conserved that means $L_i = L_f $, and that means the right hand sides of equations (1) and (2) must be equal:
$$ I_\text{Disk}\omega_i = (I_\text{Disk} + I_\text{Hoop})\omega_f $$
and rearranging this equation gives:
$$ \frac{\omega_f}{\omega_i} = \frac{I_\text{Disk}}{I_\text{Disk} + I_\text{Hoop}} \tag{3} $$
Now, in your experiment you are changing $\omega_i$, and therefore $\omega_f$, but you are keeping $I_\text{Disk}$ and $I_\text{Hoop}$ constant. That means the ratio $\omega_f/\omega_i$ given by equation (3) should be constant i.e. it will have the same value for each of your six experiments and it should be the same as the value you calculated in the Measured ratio column. That means the values in the Measured ratio column should all be the same as well, and actually they're all pretty close - the differences are presumably due to experimental error.
The column title Theoretical ratio is possibly a bit misleading. I would have called it Calculated ratio meaning it's what you calculate the ratio should be.
Incidentally, on your sheet you have $I_\text{Hoop} = 2204.900958$ (without any units!). I don't know where you got this from. The moment of inertia of an annulus is:
$$ I_\text{hoop} = \tfrac{1}{2}M\left(r_\text{Outer}^2 + r_\text{Inner}^2\right) $$
And for your hoop I make this $ I_\text{hoop} \approx 0.00508$ kgm$^2$.