Answer:
(0.01, 3.19)
Explanation:
Given that :
Mean (m) = 1.6 million
Standard deviation (s) = 2.5 million
α = 0.05
Sample size (n) = 12
Confidence interval :
m ± t(α, df = n - 1) * standard error
Standard Error = s/√n = 2.5 / √12 = 0.7216878
t0.05,11 = 2.200985 (t value calculator)
Hence,
1.6 ± 2.200985(0.7216878)
Lower bound /
1.6 - 2.200985(0.7216878) = 0.011575977517
Upper bound:
1.6 + 2.200985(0.7216878) = 3.188424022483
(0.01, 3.19)