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千人民调的误差范围

千人调查

专题
Statistics / 统计
难度
L4

题目详情

统计题：千人民调的误差范围。

英文原题

In a survey of 1,000 people, 60%60\% said they would vote for Candidate A for president (and 40%40\% said they would vote for someone else). How can you calculate a margin of error on the 60%60\% estimate?

解析

把“支持 A”记为 1，否则记为 0，则样本均值 p^\hat p 的方差近似为

Var⁡(p^)=p(1−p)N.\operatorname{Var}(\hat p)=\frac{p(1-p)}{N}.

用 p^=0.60,N=1000\hat p=0.60, N=1000 代入得到标准误

SE(p^)=0.6⋅0.41000≈0.0155.\mathrm{SE}(\hat p)=\sqrt{\frac{0.6\cdot 0.4}{1000}}\approx 0.0155.

95% 置信区间（正态近似）为

p^±1.96 SE≈0.60±0.030.\hat p\pm 1.96\,\mathrm{SE}\approx 0.60\pm 0.030.

因此误差范围约为

±3%.\boxed{\pm 3\%}.

英文解析

Var⁡(p^)=p(1−p)N.\operatorname{Var}(\hat p)=\frac{p(1-p)}{N}. SE(p^)=0.6⋅0.41000≈0.0155.\mathrm{SE}(\hat p)=\sqrt{\frac{0.6\cdot 0.4}{1000}}\approx 0.0155. p^±1.96 SE≈0.60±0.030.\hat p\pm 1.96\,\mathrm{SE}\approx 0.60\pm 0.030. ±3%.\boxed{\pm 3\%}.