Percentage of $\hat{g}$

Since $0.1995 < L < 0.2005$ and $0.895 < T < 0.905$ , and the formula gives $g \; = \; \frac{4\pi^2 L}{T^2}$ the experiment tells us that $9.6162244085 < g < 9.88161759$

This experiment uses the values $L = 0.2$ and $T = 0.9$ to calculate $\hat{g} = 9.7477757433$ Thus the measurement tolerances of the experiment allow that the actual value of $g$ could lie between $98.7$ % and $101.4$ % of $\hat{g}$ . This makes the percentage error of the experiment $\boxed{1.4}$ %