Given that the sample mean (x̄) is 20.56, the standard deviation (s) is 2.38, and the sample size (n) is 16, the margin of error (E) can be calculated as:
\[ E = \frac{t_{\alpha/2} \times s}{\sqrt{n}} \]
where \( t_{\alpha/2} \) is the critical value for a 99%
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