n=11
x
ˉ
=17.3
σ
2
=18
σ=4.24
The point estimate of the population parameter
\mu = 17.3 \\ α = 1 -0.95 = 0.05 \\ α/2 = 0.025 \\ Z_{0.025} = ±1.96μ=17.3
α=1−0.95=0.05
α/2=0.025
Z
0.025
=±1.96
Confidence interval for the population parameter:
\mu ± Z_{0.025} \times \frac{4.24}{\sqrt{11}} \\ 17.3 ± 1.96 \times 1.278 \\ 17.3 ± 2.50μ±Z
0.025
×
11
4.24
17.3±1.96×1.278
17.3±2.50
The 95% confidence interval for the population parameter: (14.8, 19.8)
α = 1 -0.99 = 0.01 \\ α/2 = 0.005 \\ Z_{0.005} = ±2.58α=1−0.99=0.01
α/2=0.005
Z
0.005
=±2.58
Confidence interval for the population parameter:
\mu ± Z_{0.005} \times \frac{4.24}{\sqrt{11}} \\ 17.3 ± 2.58 \times 1.278 \\ 17.3 ± 3.29μ±Z
0.005
×
11
4.24
17.3±2.58×1.278
17.3
초아?
