For a single mean from a normal distribution with known variance, a two-sided, 100(1 – α)% confidence interval is calculated by
X ˉ − z α / 2 ∗ σ n ≤ μ ≤ X ˉ + z α / 2 ∗ σ n \bar{X}-z_{\alpha/2}*{\sigma \over \sqrt{n}}\le\mu\le\bar{X}+z_{\alpha/2}*{\sigma \over \sqrt{n}} X ˉ − z α /2 ∗ n σ ≤ μ ≤ X ˉ + z α /2 ∗ n σ For a 95% confidence interval for μ
z α / 2 = z 0.025 = 1.96 z_{\alpha/2}=z_{0.025}=1.96 z α /2 = z 0.025 = 1.96 We have that
X ˉ = 60 , σ 2 = 25 , n \bar{X}=60, \sigma^2=25, n X ˉ = 60 , σ 2 = 25 , n Then
60 − 1.96 ∗ 25 n ≤ μ ≤ 60 + 1.96 ∗ 25 n 60-1.96*{\sqrt{25} \over \sqrt{n}}\le\mu\le60+1.96*{\sqrt{25} \over \sqrt{n}} 60 − 1.96 ∗ n 25 ≤ μ ≤ 60 + 1.96 ∗ n 25
60 − 9.8 n ≤ μ ≤ 60 + 9.8 n 60-{9.8 \over \sqrt{n}}\le\mu\le60+{9.8 \over \sqrt{n}} 60 − n 9.8 ≤ μ ≤ 60 + n 9.8
95 % C I [ 60 − 9.8 n , 60 + 9.8 n ] 95\% CI\ [60-{9.8 \over \sqrt{n}}, 60+{9.8 \over \sqrt{n}}] 95% C I [ 60 − n 9.8 , 60 + n 9.8 ]
If n=16
60 − 1.96 ∗ 25 16 ≤ μ ≤ 60 + 1.96 ∗ 25 16 60-1.96*{\sqrt{25} \over \sqrt{16}}\le\mu\le60+1.96*{\sqrt{25} \over \sqrt{16}} 60 − 1.96 ∗ 16 25 ≤ μ ≤ 60 + 1.96 ∗ 16 25
57.55 ≤ μ ≤ 62.45 57.55\le\mu\le62.45 57.55 ≤ μ ≤ 62.45
95 % C I [ 57.55 , 62.45 ] 95\% CI\ [57.55, 62.45] 95% C I [ 57.55 , 62.45 ]
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