(e) An ezymatic method for determining alcohol in wine is evaluated by comparison with gas chromatography method. The same sample was analyzed by two methods giving following results. Determine if the means of the two methods differ significantly at 95\% confidence level. \( (t=2.776 \) for \( \mathrm{V}=4 \) at \( 95 \% \) confidence level). \begin{tabular}{|l|l|l|l|l|l|} \hline Enzymatic method (\%) & 13.1 & 12.7 & 12.6 & 13.3 & 13.3 \\ \hline GC method (\%) & 13.5 & 13.3 & 13.0 & 12.9 & 13.0 \\ \hline \end{tabular}
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For the enzymatic method: Mean = (13.1 + 12.7 + 12.6 + 13.3 + 13.3) / 5 = 12.8 For the GC method: Mean = (13.5 + 13.3 + 13.0 + 12.9 + 13.0) / 5 = 13.14 Show more…
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