题目
Problem
5.
The continuous random variable X X X
X ∼ N ( μ , 5 2 ) X \sim N(\mu,5^2) X ∼ N ( μ , 5 2 ) A random sample of 10 observations of X X X X ˉ \bar X X ˉ
(a) Show that a 90% confidence interval for μ \mu μ x ˉ \bar x x ˉ
( x ˉ − 2.60 , x ˉ + 2.60 ) (\bar x - 2.60,\ \bar x + 2.60) ( x ˉ − 2.60 , x ˉ + 2.60 ) (3)
The continuous random variable Y Y Y
Y ∼ N ( μ , 3 2 ) Y \sim N(\mu,3^2) Y ∼ N ( μ , 3 2 ) A random sample of 20 observations of Y Y Y Y ˉ \bar Y Y ˉ
(b) Find a 95% confidence interval for μ \mu μ y ˉ \bar y y ˉ
(3)
(c) Given that X X X Y Y Y X ˉ − Y ˉ \bar X - \bar Y X ˉ − Y ˉ
(7)
(Total for Question 5 is 13 marks)
题目中文翻译
连续随机变量 X X X
X ∼ N ( μ , 5 2 ) X \sim N(\mu,5^2) X ∼ N ( μ , 5 2 ) 抽取了 X X X X ˉ \bar X X ˉ
(a) 证明以 x ˉ \bar x x ˉ μ \mu μ
( x ˉ − 2.60 , x ˉ + 2.60 ) (\bar x - 2.60,\ \bar x + 2.60) ( x ˉ − 2.60 , x ˉ + 2.60 ) (3)
连续随机变量 Y Y Y
Y ∼ N ( μ , 3 2 ) Y \sim N(\mu,3^2) Y ∼ N ( μ , 3 2 ) 抽取了 Y Y Y Y ˉ \bar Y Y ˉ
(b) 求以 y ˉ \bar y y ˉ μ \mu μ
(3)
(c) 已知 X X X Y Y Y X ˉ − Y ˉ \bar X - \bar Y X ˉ − Y ˉ
(7)
解答
(a)
解法一
思路
展开
总体标准差已知，且总体服从正态分布，因此使用标准正态临界值构造置信区间。90 % 90\% 90% 5 % 5\% 5% z 0.95 = 1.6449 z_{0.95}=1.6449 z 0.95 = 1.6449
答题过程
展开
Since the population standard deviation is known, a 90 % 90\% 90% μ \mu μ
x ‾ ± z 0.95 σ X n X . \overline{x}\pm
z_{0.95}\frac{\sigma_X}{\sqrt{n_X}}. x ± z 0.95 n X σ X . Its margin of error is
1.6449 ( 5 10 ) = 2.600815 … = 2.60. \begin{align*}
1.6449\big(\frac{5}{\sqrt{10}}\big)
=&\,2.600815\ldots\\
=&\,2.60.
\end{align*} 1.6449 ( 10 5 ) = = 2.600815 … 2.60. Therefore, the confidence interval is
( x ‾ − 2.60 , x ‾ + 2.60 ) , \boxed{(\overline{x}-2.60,\ \overline{x}+2.60)}, ( x − 2.60 , x + 2.60 ) , as required.
(b)
解法一
思路
展开
同样使用已知总体标准差的正态置信区间。95 % 95\% 95% 1.96 1.96 1.96 3 / 20 3/\sqrt{20} 3/ 20
答题过程
展开
A 95 % 95\% 95% μ \mu μ
y ‾ ± 1.96 ( 3 20 ) . \overline{y}\pm
1.96\big(\frac{3}{\sqrt{20}}\big). y ± 1.96 ( 20 3 ) . The margin of error is
1.96 ( 3 20 ) = 1.3148 … ≈ 1.31. \begin{align*}
1.96\big(\frac{3}{\sqrt{20}}\big)
=&\,1.3148\ldots\\
\approx&\,1.31.
\end{align*} 1.96 ( 20 3 ) = ≈ 1.3148 … 1.31. Therefore, the required confidence interval is
( y ‾ − 1.31 , y ‾ + 1.31 ) . \boxed{(\overline{y}-1.31,\ \overline{y}+1.31)}. ( y − 1.31 , y + 1.31 ) .
(c)(i)
解法一
思路
展开
先分别写出两个样本均值的分布。由于两组样本独立，均值之差仍服从正态分布，均值相减、方差相加。
答题过程
展开
The sample means have distributions
X ‾ ∼ N ( μ , 5 2 10 ) \overline X\sim
\mathrm N\big(\mu,\frac{5^2}{10}\big) X ∼ N ( μ , 10 5 2 ) and
Y ‾ ∼ N ( μ , 3 2 20 ) . \overline Y\sim
\mathrm N\big(\mu,\frac{3^2}{20}\big). Y ∼ N ( μ , 20 3 2 ) . Since X X X Y Y Y D = X ‾ − Y ‾ D=\overline X-\overline Y D = X − Y
E ( D ) = μ − μ = 0 \operatorname E(D)=\mu-\mu=0 E ( D ) = μ − μ = 0 and
Var ( D ) = 5 2 10 + 3 2 20 = 2.95. \begin{align*}
\operatorname{Var}(D)
=&\,\frac{5^2}{10}+\frac{3^2}{20}\\
=&\,2.95.
\end{align*} Var ( D ) = = 10 5 2 + 20 3 2 2.95. Therefore,
X ‾ − Y ‾ ∼ N ( 0 , 2.95 ) . \boxed{\overline X-\overline Y\sim\mathrm N(0,2.95)}. X − Y ∼ N ( 0 , 2.95 ) .
(c)(ii)
解法一
思路
展开
两个区间不重叠有两种对称情形：X X X Y Y Y X ‾ − Y ‾ \overline X-\overline Y X − Y
答题过程
展开
The two confidence intervals do not overlap when either
x ‾ − 2.60 > y ‾ + 1.31 \overline{x}-2.60>\overline{y}+1.31 x − 2.60 > y + 1.31 or
x ‾ + 2.60 < y ‾ − 1.31. \overline{x}+2.60<\overline{y}-1.31. x + 2.60 < y − 1.31. Equivalently,
x ‾ − y ‾ > 3.91 or x ‾ − y ‾ < − 3.91. \overline{x}-\overline{y}>3.91
\quad\text{or}\quad
\overline{x}-\overline{y}<-3.91. x − y > 3.91 or x − y < − 3.91. From part (c)(i),
X ‾ − Y ‾ ∼ N ( 0 , 2.95 ) . \overline X-\overline Y\sim\mathrm N(0,2.95). X − Y ∼ N ( 0 , 2.95 ) . By symmetry,
P ( no overlap ) = 2 P ( X ‾ − Y ‾ > 3.91 ) = 2 P ( Z > 3.91 2.95 ) = 2 P ( Z > 2.276 … ) = 0.0228 . \begin{align*}
\operatorname P(\text{no overlap})
=&\,2\operatorname P(
\overline X-\overline Y>3.91
)\\
=&\,2\operatorname P\big(
Z>\frac{3.91}{\sqrt{2.95}}
\big)\\
=&\,2\operatorname P(Z>2.276\ldots)\\
=&\,\boxed{0.0228}.
\end{align*} P ( no overlap ) = = = = 2 P ( X − Y > 3.91 ) 2 P ( Z > 2.95 3.91 ) 2 P ( Z > 2.276 … ) 0.0228 . 