题目
4.设(X1,X2,···,Xn)是来自 approx N(mu ,(sigma )^2) 的一个简单随机样本,σ^2-|||-未知 overline (X)=dfrac (1)(n)sum _(i=1)^n(X)_(i),(s)^2=dfrac (1)(n-1)sum _(i=1)^n(({X)_(i)-overline (X))}^2 ta(n)为t分布的上侧α分位-|||-点,则e^n的置信度为 https:/img.zuoyebang.cc/zyb_c8dd42467158d2d3dffe73a5a550deb6.jpg-a 的置信区间为 () .-|||-(A) ((e)^dfrac (pi {x)}-dfrac ({s)_(t)}(sqrt {n)}sum _(i=i)_(i)(n-1),(e)^overline (x)+dfrac ({s)_(i)}(sqrt {n)}i(n-1))-|||-(B) ((e)^dfrac (pi {x)}-dfrac ({s)_(n)}(sqrt {n)}(t)_(i-i)(n-1),(e)^dfrac (pi {x)}+dfrac ({s)_(i)}(sqrt {n)}(t)_(i-i)(n-1))-|||-(C) (exp overline {X)-dfrac ({s)_(n)}(sqrt {n)}(t)_(dfrac {n)(2)}(n-1)} , overrightarrow {x)+dfrac ({S)_(n)}(sqrt {n)}(t)_(dfrac {n)(2)}(n-1)} )-|||-D) (exp overrightarrow {x)-dfrac ({S)_(n)}(sqrt {n)}(t)_(1-dfrac {a)(2)}(n-1)} overrightarrow {x)+dfrac ({S)_(n)}(sqrt {n)}(t)_(1-dfrac {n)(2)}(n-1)} )-|||-其中 {t)^3=(e)^t.
题目解答
答案
解析
步骤 1：确定μ的置信区间
由于μ的置信度为 $1-a$ 的置信区间为 $(\overline {X}-\dfrac {{S}_{n}}{\sqrt {n}}{t}_{\dfrac {n}{2}}(n-1),\overline {X}+\dfrac {{S}_{n}}{\sqrt {n}}{t}_{\dfrac {n}{2}}(n-1))$ , 即P $|\overrightarrow {x}-\dfrac {{S}_{n}}{\sqrt {n}}(n-1)\lt \mu \lt \overline {x}+\dfrac {{S}_{n}}{\sqrt {n}}{(n-1)}\} =1-a$ ,
步骤 2：转换为e^μ的置信区间
从而 $||x|-\dfrac {{s}_{2}}{\sqrt {{n}_{1}}(n-1)}|\lt e\lt ex||\overrightarrow {x}+\dfrac {{s}_{x}}{\sqrt {n}}(n-1)||=1-$ α,
步骤 3：选择正确的选项
故选(C).
由于μ的置信度为 $1-a$ 的置信区间为 $(\overline {X}-\dfrac {{S}_{n}}{\sqrt {n}}{t}_{\dfrac {n}{2}}(n-1),\overline {X}+\dfrac {{S}_{n}}{\sqrt {n}}{t}_{\dfrac {n}{2}}(n-1))$ , 即P $|\overrightarrow {x}-\dfrac {{S}_{n}}{\sqrt {n}}(n-1)\lt \mu \lt \overline {x}+\dfrac {{S}_{n}}{\sqrt {n}}{(n-1)}\} =1-a$ ,
步骤 2：转换为e^μ的置信区间
从而 $||x|-\dfrac {{s}_{2}}{\sqrt {{n}_{1}}(n-1)}|\lt e\lt ex||\overrightarrow {x}+\dfrac {{s}_{x}}{\sqrt {n}}(n-1)||=1-$ α,
步骤 3：选择正确的选项
故选(C).