题目

从一批钢管中抽取16根测其内直径,计算得样本均值 overline (x)=100.21 ,样本标准-|||-差s=0.42 设这批钢管内直径服从正态分布N(μ,σ^2),试分别在下列条件下求-|||-参数μ的区间估计-|||-1.已知 sigma =0.5 2.σ未知.( u0.975=1.96 _(0.975)(15)=2.13)

$\begin{aligned} &从一批钢管中抽取16根测其内直径,计算得样本均值\overline x=100.21,样本标准\\ &\text{差}s=0.42.\text{ 设这批钢管内直径服从正态分布 }N(\mu,\sigma^{2}),\text{试分别在下列条件下求} \\ &\text{参数μ的区间估计} \\ &1.\text{已知}\sigma=0.5;\quad2.\sigma\text{未知}.(u_{0.975}=1.96,t_{0.975}(15)=2.13) \\ \end{aligned}$

题目解答

答案

$\begin{aligned} &\text{1. 已知$\sigma = 0.5$，样本均值$\overline{x} = 100.21$，样本量$n = 16$，$u_{0.975} = 1.96$}\\ &\mu\text{的置信区间为:} \\ &\overline{x}-u_{\alpha/2}\frac\sigma{\sqrt{n}}<\mu<\overline{x}+u_{\alpha/2}\frac\sigma{\sqrt{n}} \\ &100.21-1.96\times\frac{0.5}{\sqrt{16}}<\mu<100.21+1.96\times\frac{0.5}{\sqrt{16}} \\ &100.21-1.96\times\frac{0.5}4<\mu<100.21+1.96\times\frac{0.5}4 \\ &100.21-0.245<\mu<100.21+0.245 \\ &99.965<\mu<100.455 \\ &\mu\text{的置信区间为:} \\ &\overline{x}-t_{\alpha/2}(n-1)\frac s{\sqrt{n}}<\mu<\overline{x}+t_{\alpha/2}(n-1)\frac s{\sqrt{n}} \\ &100.21-2.13\times\frac{0.42}{\sqrt{16}}<\mu<100.21+2.13\times\frac{0.42}{\sqrt{16}} \\ &100.21-2.13\times\frac{0.42}4<\mu<100.21+2.13\times\frac{0.42}4 \\ &100.21-0.22365<\mu<100.21+0.22365 \\ &99.98635<\mu<100.43365 \end{aligned}$