Question

A recent survey of 50 employed male executives showed that it took an average of 26 weeks for them to find another position. The standard deviation of the sample was 6.2 weeks. What was the margin of error and standard error?

Answer 1

Here, confidence interval is not given so, i consider 95%

sample mean, xbar = 26
sample standard deviation, s = 6.2
sample size, n = 50
degrees of freedom, df = n - 1 = 49

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.01

ME = tc * s/sqrt(n)
ME = 2.01 * 6.2/sqrt(50)
ME = 1.7624

MArgin of error = 1.7624

std.error = 6.2/sqrt(50) = 0.8768