设有一组来自正态总体N(μ,σ^2)的样本观测值:-|||-0.497, 0.506, 0.518,0.524,0.488, 0.510,0.510,0.515,0.512.-|||-(1)已知 sigma =0.01 ,求μ的置信区间(设置信度为0.95);-|||-(2)σ^2未知,求μ的置信区间(设置信度为0.95).

步骤 1：计算样本均值 $\overline{x}$
样本观测值为：0.497, 0.506, 0.518, 0.524, 0.488, 0.510, 0.510, 0.515, 0.512。
样本均值 $\overline{x} = \frac{1}{n}\sum_{i=1}^{n}x_i = \frac{0.497 + 0.506 + 0.518 + 0.524 + 0.488 + 0.510 + 0.510 + 0.515 + 0.512}{9} = 0.50889$。

步骤 2：计算样本标准差 $s$
样本标准差 $s = \sqrt{\frac{1}{n-1}\sum_{i=1}^{n}(x_i - \overline{x})^2} = \sqrt{\frac{(0.497-0.50889)^2 + (0.506-0.50889)^2 + ... + (0.512-0.50889)^2}{8}} = 0.01088$。

步骤 3：计算置信区间
(1) 已知 $\sigma = 0.01$，置信度为0.95，查表得 $z_{0.025} = 1.96$。
μ的置信区间为 $(\overline{x} - z_{0.025} \cdot \frac{\sigma}{\sqrt{n}}, \overline{x} + z_{0.025} \cdot \frac{\sigma}{\sqrt{n}})$
$= (0.50889 - 1.96 \cdot \frac{0.01}{\sqrt{9}}, 0.50889 + 1.96 \cdot \frac{0.01}{\sqrt{9}})$
$= (0.50889 - 0.00653, 0.50889 + 0.00653)$
$= (0.50236, 0.51542)$。

(2) σ^2未知，置信度为0.95，查表得 $t_{0.025}(n-1) = t_{0.025}(8) = 2.306$。
μ的置信区间为 $(\overline{x} - t_{0.025}(n-1) \cdot \frac{s}{\sqrt{n}}, \overline{x} + t_{0.025}(n-1) \cdot \frac{s}{\sqrt{n}})$
$= (0.50889 - 2.306 \cdot \frac{0.01088}{\sqrt{9}}, 0.50889 + 2.306 \cdot \frac{0.01088}{\sqrt{9}})$
$= (0.50889 - 0.00853, 0.50889 + 0.00853)$
$= (0.50036, 0.51742)$。