A sample of 100 college graduates is obtained and the mean
salary is found to be $ 43704 with a standard
deviation of $9879. Find the margin error E associated with a 95%
C. I.
Z for 95% confidence interval = Z0.025 = 1.96
Margin of error, E = Z0.025 * sd / sqrt(n)
= 1.96 * 9879 / sqrt(100)
= 1936.284 (ans)
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