The margin of error in a sample of 100 people with a standard deviation of 12.5 is 2.45.
Let us assume a confidence of 95%.
Given that the confidence level (C) = 95% = 0.95
α = 1 - C = 0.05
α/2 = 0.05/2 = 0.025
The z score of α/2 corresponds with the z score of 0.475 (0.5 - 0.025) which is equal to 1.96
The margin of error (E) is:
[tex]E=z_\frac{\alpha}{2} *\frac{\sigma}{\sqrt{n} } \\\\E=1.96*\frac{12.5}{\sqrt{100} } =2.45[/tex]
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