NMR Spectra - Peaks

The CH group in question is adjacent to two very electron withdrawing groups (i.e. 2x CO2H), which remove electron density from the relevant proton and thus deshield it (the local magnetic field is BIGGER than ordinarily expected, since there are on average fewer electrons acting in opposition to the external magnetic field than 'normal' for CH groups).
I'm not sure how familiar you are with shielding constants, but if you write the local magnetic field B as $B=B_0 (1-\sigma)$, where B₀ is the external magnetic field and $\sigma$ is the shielding constant, the principal component of the shielding constant is the diamagnetic shielding constant. If you will, an applied magnetic field causes electron to flow around a nucleus and thus to generate their own magnetic field (as all moving charges do - from elementary electromagnetism), which (it turns out) is in opposition to the external field. So the more electrons you have, the greater this opposing magnetic field, and your nucleus 'feels' as though the external magnet is not as powerful (electrons are said to SHIELD the nucleus) - the diamagnetic shielding constant is big and positive. This means the splitting between the spin-up and spin-down (for protons at least) energy states DECREASES (as B goes down, the difference between $E=E_0 - m_I \hbar \gamma B$ for $m_I = \pm 1/2$ also decreases, where $m_I$ is the magnetic quantum number and $\gamma$ the gyromagnetic ratio).
In contrast, if there are few electrons, the diamagnetic shielding constant will be small (and positive). Thus electron withdrawing groups cause DESHIELDING (compared to 'normal' resonances - there is of course still shielding taking place as compared to the bare nucleus).

Since the proton is deshielded, it means it moves downfield, i.e. to greater chemical shift values (or greater frequency (or energy difference, as per the equation above) of spin-flip transition, if you will). The CH3 group is one carbon further away from the withdrawing groups, and since this sort of withdrawal is an inductive (through-bond) effect, it falls off rapidly with increasing the number of bonds in between the groups in question. So you'd expect to see the CH3 group at its 'ordinary' resonance of ~1.5ppm. So the spectrum assignment looks OK to me.

The reasoning is actually pretty similar for e.g. O-CH2: oxygen is very electronegative and withdraws electron density from the protons, thus deshielding them to higher chemical shift.

I mean, from your spectrum it's pretty obvious which is which proton anyway, from the peak area, as well as from the coupling pattern (doublet vs quartet). I take it you only had problems with rationalising the unusually large chemical shift?

Thanks Sinuhe for your very comprehensive reply, but alas my main problem doesnt lie in the actual NMR theory behind it, but as you rightly pointed out the conflicting chemical shift stated by my datasheet and the other by the mark-scheme;

Its my fault for not explaining myself clearly, sorry , ;

Well the marksheme says that peak at 3.3ppm identifed as due to the CH,

But when i look at the datasheet, it says the following;

type of proton -------Chemical Shift
-O-CH3 or O-CH2-R :_____ 3.3 - 4.3

Either i need to get some glasses and improve my vision, or the markscheme is incorrect,

this is why i hate chemistry -- they have fixed rules with 2139788978213 exceptions for each one.