What is the 95 % conference interval for the average distance traveled to work by all employees?

Question

A study of 13 employees found that the average distance they traveled to work was 14.3 miles. The standard deviation of the sample was 2 miles.

Answer 1

If population is normally distributed and n < 30, first find the value of ±t_0.025 so that 2.5% ((100-95)%÷2) of the area is within each tail for n - 1 (or 13 - 1 = 12) degrees of freedom. Enter a t Distribution Table with df = 12 under 0.025 to obtain 2.179.

Then µ = X-bar ± 2.179(s/√n) or 14.3 ± 2.179(2/√13) or 14.3 ± 1.209. This gives 13.09 < µ < 15.51 with 95% confidence.

If population is not normally distributed, then use Chebyshev's Theorem: set 1 - (1/K²) = 0.95 or 0.05 = 1/K² or K² = √(1/0.05) and K ≈ 4.47. Then µ = X-bar ± K(s/√n) or 14.3 ± 4.47(2/√13). This gives 11.82 < µ < 16.78 with a 95% level of confidence.