In: Statistics and Probability
How much experience do supply-chain transportation managers have in their field? Suppose in an effort to estimate this, 41 supply-chain transportation managers are surveyed and asked how many years of managerial experience they have in transportation. Survey results (in years) are shown below. Use these data to construct a 99% confidence interval to estimate the mean number of years of experience in transportation. Assume that years of experience in transportation is normally distributed in the population.
5 | 8 | 10 | 21 | 20 |
25 | 14 | 6 | 19 | 3 |
3 | 9 | 11 | 2 | 3 |
13 | 2 | 4 | 9 | 4 |
5 | 4 | 21 | 7 | 6 |
3 | 28 | 17 | 23 | 2 |
25 | 8 | 13 | 17 | 27 |
7 | 3 | 15 | 4 | 11 |
6 |
Appendix A Statistical Tables
(Round the intermediate values to 3 decimal places. Round your answers to 3 decimal places.)
For the given sample data the mean is calculated as:
Mean = (2 + 2 + 2 + 3 + 3 + 3 + 3 + 3 + 4 + 4 + 4 + 4 + 5 + 5 +
6 + 6 + 6 + 7 + 7 + 8 + 8 + 9 + 9 + 10 + 11 + 11 + 13 + 13 + 14 +
15 + 17 + 17 + 19 + 20 + 21 + 21 + 23 + 25 + 25 + 27 + 28)/41
= 443/41
Mean = 10.8049
and Sample standard deviation as:
Since the population standard deviation is not known hence T-distribution is applicable to calculated the confidence interval as:
μ = ± t(sM)
where:
= sample mean
t = t statistic determined by confidence level
and degree of frredom n-1 , using excel formula for
T-distribution.
sM = standard error =
√(s2/n)
= 10.8049
t = 2.7 Computed using excel formula at 0.01 leve of
significance and degree of freedom n-1 = 41-1= 40 as
=T.INV.2T(0.01,40)
sM = √(7.89692/41) =
1.23
μ =
± t(sM)
μ = 10.8049 ± 2.7*1.23
μ = 10.8049 ± 3.335379
99% CI is [7.470, 14.140].
So, the 99% confidence interval of the mean number of years of experience in transportation is {7.470, 14.140}