Chapter 11 Data Analysis and Statistics - Standardized Test Practice - Short Response - Page 791: 14

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The margin of error formula is $M=\pm \frac{1}{\sqrt n}$ Find $n=\frac{1}{m^2}\\ \rightarrow n=\pm \frac{1}{(\pm 0.02)^2}=2500$ Hence, $M=\pm 0.02$ Since the interval is $[49,51]$, we may not know whether the exact percentage is higher than 50.