Since you were not sure what test to use, we evaluate our sample size of n = 1007
Since our sample size was greater than 30, we use the Large Sample Normal Distribution Confidence Interval Test
A large sample of 1007 units has a mean 11.3 and a standard deviation σ of 16.6
Find a 90% 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 - 0.9
α = 0.1
Find α spread range:
α = ½(α)
α = ½(0.1)
α = 0.05
Find z-score for α value for 0.05
zscore0.05 = 1.645
<--- Value can be found on Excel using =NORMSINV(0.95)
Calculate the Standard Error of the Mean:
| SEM = | σ |
| √n |
| SEM = | 16.6 |
| √1007 |
| SEM = | 16.6 |
| 31.733263305245 |
SEM = 0.5231
Calculate high end confidence interval total:
High End = X + zscoreα * s/√n
High End = 11.3 + 1.645 * 16.6/√1007
High End = 11.3 + 1.645 * 0.5231103980805
High End = 11.3 + 0.86051660484242
High End = 12.1605
Calculate low end confidence interval total:
Low End = X - zscoreα * s/√n
Low End = 11.3 - 1.645 * 16.6/√1007
Low End = 11.3 - 1.645 * 0.5231103980805
Low End = 11.3 - 0.86051660484242
Low End = 10.4395
Final Answer
What this means is if we repeated experiments, the proportion of such intervals that contain μ would be 90%
