Average Percent Error?
I'm not sure if I should find the average of 1.112 and 1.069 and then solve. If so, I think it would be ((1.0905 g/L - 1.429 g/L) / (1.429)) X 100%. Can you tell me if I'm doing it right? I got an average percent error of 23.69%, but I'm just not sure. Thanks for your help!
Answer:
You want the average of the two errors... so ((1.112-1.429) + (1.069-1.429)) / 2 = -0.3385. Expressed as a % of 1.429, it is -23.7%.
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