Question

A recent survey of 60 randomly sampled unemployed male executives found that it took a mean of 12.4 weeks to find another position. The standard deviation of the sample was 3.1 weeks.

a) Construct a 95% confidence interval for the population mean time for unemployed male executives to find another position.

b) A recent Washington Post article claimed that the mean time for these executives to find another position is less than 10 weeks. Based upon the confidence interval that you computed in part a) above, what do you think of the Post’s claim?

Answer 1

a)
sample mean, xbar = 12.4
sample standard deviation, s = 3.1
sample size, n = 60
degrees of freedom, df = n - 1 = 59

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.001

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (12.4 - 2.001 * 3.1/sqrt(60) , 12.4 + 2.001 * 3.1/sqrt(60))
CI = (11.6 , 13.2)

b)
Based on the above calculated CI, finding a new position in less than 10 weeks is not likely. This is because 10 lies below the lower limit of the calculated CI.

Hence, we reject reject the claim made by Post. This is an incorrect claim.