A large sample of 149 units has a mean 61 and a standard deviation σ of 10
Find a 99% confidence interval of the mean μ
Confidence Interval Formula for μ
X - zscoreα/2 * s/√n < μ < X + zscoreα/2 * s/√n where:
X = sample mean, s = sample standard deviation, zscore = Normal distribution Z-score from a probability where α = (1 - Confidence Percentage)/2
Calculate α
α = 1 - Confidence%
α = 1 - 0.99
α = 0.01
Find α spread range:
α = ½(α)
α = ½(0.01)
α = 0.005
Find z-score for α value for 0.005
zscore0.005 = 2.576
<--- Value can be found on Excel using =NORMSINV(0.995)
Calculate the Standard Error of the Mean:
| SEM = | σ |
| √n |
| SEM = | 10 |
| √149 |
| SEM = | 10 |
| 12.206555615734 |
SEM = 0.8192
Calculate high end confidence interval total:
High End = X + zscoreα * s/√n
High End = 61 + 2.576 * 10/√149
High End = 61 + 2.576 * 0.81923192051904
High End = 61 + 2.110341427257
High End = 63.1103
Calculate low end confidence interval total:
Low End = X - zscoreα * s/√n
Low End = 61 - 2.576 * 10/√149
Low End = 61 - 2.576 * 0.81923192051904
Low End = 61 - 2.110341427257
Low End = 58.8897
Final Answer
What this means is if we repeated experiments, the proportion of such intervals that contain μ would be 99%
