when you draw you line of best fit, can you use the gradient to find the value of 'g' becuase when the formula is re-arranged, you get T^2 =((4*pi)l)/g
in terms of x and y, T^2 =y and l=x therefore, from y=mx, the gradiaent will equal 4*pi on g. when i do this i get g to be 11.25ms^-2??? or...