I do not understand why you want to evaluate $y + \Delta y$ as $(z + \Delta z)^{3/2}$? You can evaluate the uncertainty in y(z) where y(z) is a function of the random variable z; for your case $y = z^{3/2}$. If this is the case, the following applies.
I assume 4.480 is the mean for z and $\Delta z$ 0f 0.168 is the standard deviation for z?
You can find discussions of how to evaluate the uncertainty for a function of a random variable in many statistics texts, and use that information to evaluate the uncertainty in y for your function as its standard deviation $\Delta y$. For more complicated functions, you can to do a Taylor series expansion. For example see Dougherty's text on Probability and Statistics or Meyer's text Data Analysis for Scientists and Engineers.
For your function, we have $\Delta y = y_{mean} (m^2 {\Delta z^2/z_m}^2)^{1/2}$ where $m = 3/2$, and $y_{mean}$ = $z_{mean}^{3/2}$. With this I calculate $y_{mean}$ of 9.48 and $\Delta y$ of 0.53; same result as Penguino provides in his answer.
Then you can express your answer for y with uncertainty as $y_{mean} \pm \Delta y$; 9.48 $\pm$ 0.53.