What is the formula to determine the change in pressure when there is a change in flow? [Updated to help clarify my question]
I have a current water flow of 9 GPM (gallons per minute) at 50 PSI through a 1/2 inch diameter pipe pouring out at the end.
I understand that if I reduce the flow at the end (exit) of the pipe, it will increase the pressure of the water exiting the pipe. So, if I add a fixture at the end of the pipe which reduces the out-coming water flow to 2 GPM what will my new exiting water PSI be? What is the formula to determine this?
 A: I misunderstood the original question, here is a corrected version of my answer.
First of all I will assume that the source pressure in your plumbing system is 50 PSI. This means at the head end of the pipe the pressure is 50 PSI and at the outlet end of the pipe, the pressure is 0 (i.e., atmospheric).
This means that it requires 50 PSI of pressure to push 9 GPM of water through that piece of pipe.
Now if you put your flow restrictor on the outlet end of that pipe, the source pressure stays the same (50 PSI) but now there are not one but two resistances in series in the pipe: 1) the pipe itself as before plus 2) the restrictor.
Your new flow rate through the pipe is now 2 GPM, a reduction of 2/9ths. The pressure needed to push 2 GPM through that pipe is 2/9 x 50  or ~11 PSI. Since the source pressure is still 50 PSI, the pressure at the end of the pipe just before the flow restrictor will be 50-11 or 39 PSI and at the outlet of the flow restrictor it is still 0 PSI (atmospheric).
This means that the flow restrictor dissipates 39 PSI while passing 2 GPM; meanwhile, the pipe itself dissipates 11 PSI and 39 + 11 = 50 PSI, the source pressure.
Note that you could get your required 2 GPM without the restrictor, simply by turning down the source pressure to 11 PSI. You could do this by partially closing a valve upstream of the pipe inlet, meaning that the flow restriction required to decrease the pipe flow from 9 GPM to 2 GPM can be placed at either end of the pipe.
A: This is an example of incompressible flow.  So first, look at things from a dimensional analysis point of view.  You have a flow rate given, which is a volume per unit time.  You could also think of this as a velocity times an area.  This is useful because you will need velocity later on.
So first, let's get things into better units (my preference).  A flow rate, $dV_{1}/dt$, of 9 gpm is roughly equivalent to ~0.000568 m3 s-1.  A radius, $r_{1}$, of 1/4 inch is equivalent to ~0.00635 m for a cross-sectional area of $A_{1}$ ~ 0.0001267 m2.  Then we can use the following relation:
$$
v_{j} = \frac{ 1 }{ A_{j} } \frac{ d V_{j} }{ d t } \tag{0}
$$
where $v_{j}$ is the velocity in section $j$ of the pipe.  So for the first part, the speed will be $v_{1}$ ~ 4.482 m s-1.
The total pressure in section $j$ of the pipe (assuming horizontal and "short", i.e., not 10s of miles long) is given by:
$$
P_{tot,j} = P_{j} + \frac{ 1 }{ 2 } \rho_{j} v_{j}^{2} = constant \tag{1}
$$
where $\rho_{j}$ is the density of water (which will not change unless the things were under extreme pressures) equivalent to ~1000 kg m-3 (it actually varies by temperature but is typically somewhere near 997 kg m-3 but let's use 1000 for simplicity) and $P_{j}$ is the static pressure in section $j$ of the pipe.
You give a value for $P_{tot,1}$ ~ 50 psi which is equivalent to ~3.44738 x 105 Pa (or ~3.402 atm).  The dynamic pressure (i.e., 2nd term on right-hand side of Equation 1) is equal to ~10043.6 Pa, therefore the static pressure is equivalent to $P_{1}$ ~ 3.34694 x 105 Pa (or ~3.303 atm).  So you can see the static pressure is completely dominating things here.

I understand that if I reduce the flow at the exit of the pipe, it will increase the pressure. So, if I reduce the flow to 2 gpm what will my new psi be? What is the formula to determine this?

Since water is incompressible, for all intents and purposes here, the change in flow rate at the end of the pipe will result in a similar change in flow rate in the larger section of pipe according to the continuity equation given by:
$$
A_{1} v_{1} = A_{2} v_{2} \tag{2}
$$
Recall from Equation 0 that the flow rate is just a velocity times an area, so for incompressible flow, the flow rate is constant.  Therefore, we can redo all of the above calculations (and still use the cross-sectional area of the first part of the pipe) but replace the flow rate with 2 gpm ~ 0.000126 m3 s-1.  So the new velocity of water (in the 1/4 inch section of pipe) will be $v_{1}$ ~ 0.996 m s-1.  The new dynamic pressure will be ~495.98 Pa thus the new static pressure will be, assuming the same 50 psi total pressure, ~3.44241 x 105 Pa (or ~3.397 atm).
The difference in static pressures between the two cases is ~9547.6 Pa (or ~0.094 atm or ~1.38 psi).  These values are before the constriction that reduced the flow rate.  In the constriction, the flow speed (and dynamic pressure) will increase.
Update
So there is a slight misunderstanding or mistatement in the question which requires assumptions on the readers part.  That is, is the 50 psi the total pressure or the static pressure?
If we assume it's the total pressure and we assume the pipe is short (i.e., we can ignore frictional losses) then the above answer is correct and the change in apparent pressrue is due to the loss of dynamic pressure.  That is, when you restrict the flow, the dynamic pressure is converted to static pressure.  As I show above, the static pressure increases by ~9500 Pa or ~1.4 psi.
Now let's assume the 50 psi is the upstream static pressure for both cases.  In a realistic scenario the pipe is long enough that much of the pressure at the start is lost by the point one measures it to frictional losses etc.  When restricted, the frictional losses greatly reduce and so the static pressure on the object restricting the flow will increase.  This can be seen by looking at the Hagen–Poiseuille equation given by:
$$
\Delta P = \frac{ 8 \ \pi \ \mu \ L \ Q }{ A^{2} } \tag{3}
$$
where $\Delta P$ is the loss of pressure over length $L$ of pipe, $\mu$ is the dynamic viscosity, $L$ is the length of the pipe, $Q$ is the volumetric flow rate (e.g., see Equation 2 above), and $A$ is the cross-sectional area of the pipe.
So if we divide by length (since we don't know or care about it) and look at each situation, for the 9 gpm flow rate the water would lose ~791.4 Pa m-1 while it would only lose ~175.9 Pa m-1 in the 2 gpm flow rate.  The ratio of these two is, as you might expect, 9/2 = 4.5.  So assuming the upstream static pressure stays the same at 50 psi in both cases, the increase in pressure on the constriction would be close to 4.5.
A: At a flow rate of 9 GPM translates into 1.2 CFM = 1.25 lb_m/sec.  Water viscosity of 0.01 Poise = 0.00067 $lb_m/(ft-sec)$.  So the Reynolds number is $$Re=\frac{4m}{\pi D \mu}=\frac{(4)(1.25)}{(3.14159)(1/24)(0.00067)}=57000$$That would give a Fanning friction factor of about f=0.005.  Dividing the volume flow rate by the cross sectional area of the pipe gives and water velocity of v = 14.7 ft/sec.
So the shear stress at the wall would be $$\tau=\rho \frac{v^2}{2g_c}f=(62.4)\frac{14.7^2}{(2)(32.2)}(0.005)=1.04\ psf=0.0072 psi$$That would make the pressure gradient in the pipe $$\frac{\Delta p}{L}=\frac{4}{D}\tau=(4)(24)(0.0072)=0.70\ \frac{psi}{ft}$$
A: According to Bernoulli’s theorem
$$P_1+\frac{1}{2}\rho{v_1}^2+\rho gh_1 = P_2+\frac{1}{2}\rho{v_2}^2+\rho gh_2$$
Since both ends of the pipe is in the same height
$$h_1=h_2$$
Now the equation can be written as
$$P_1+\frac{1}{2}\rho{v_1}^2= P_2+\frac{1}{2}\rho{v_2}^2$$
$\rho$ is the density of liquid
You can use this equation to get the relation between pressure and flow velocity.
Additionally, you can use equation of continuity to get the relation between flow velocity and area of cross section of pipe
$$A_1V_1=A_2V_2$$
