统计第3讲：点估计 | Point Estimation · 练习

题目1解答#

(1) $\overline{x} = \frac{1}{20}\sum x_i = 0.89\%$

(2) $s = \sqrt{\frac{\sum(x_i-0.89)^2}{19}} \approx 1.02\%$

(3) $\widehat{SE} = \frac{1.02}{\sqrt{20}} \approx 0.228\%$

题目2证明#

展开期望运算：

$$E(\hat{\sigma}_1^2) = \frac{n-1}{n}\sigma^2 \quad (\text{有偏}) \\ E(\hat{\sigma}_2^2) = \frac{n-1}{n+1}\sigma^2 \quad (\text{仍为有偏}) \\ \text{仅有} S^2 = \frac{1}{n-1}\sum(X_i-\overline{X})^2 \text{是无偏的}$$

题目3解答#

(1) $\hat{p} = 213/300 = 0.71$

(2) $SE \approx \sqrt{\frac{0.71×0.29}{300}} \approx 0.026$

(3) $0.71 \pm 1.96×0.026 = (0.659, 0.761)$

题目4解答#

(1) 修正因子 = $\sqrt{\frac{1000-80}{1000-1}} \approx 0.96$

修正标准误 = $0.96 × \frac{4\%}{\sqrt{80}} \approx 0.43\%$

(2) 说明当抽样比例>5%时，需考虑有限总体对估计精度的影响

题目5解答#

(1) $SE = \frac{0.5}{\sqrt{50}} \approx 0.0707\%$

(2) 在原假设边界 $\mu=1\%$ 下，$Z = \frac{1.2-1}{0.0707} \approx 2.83$

查标准正态分布表得单侧 P 值 $P(Z>2.83) \approx 0.23\%$