题目
Problem
A machine fills bottles with mineral water.
The machine is checked every day to ensure that it is working correctly. On a particular
day a random sample of 100 bottles is taken. The volume of water, x millilitres, for each
bottle is measured and each measurement is coded using
y = x − 1000 y=x-1000 y = x − 1000 The results are summarised below
∑ y = 847 ∑ y 2 = 13 510.09 \sum y=847 \qquad \sum y^2=13\ 510.09 ∑ y = 847 ∑ y 2 = 13 510.09 (a) (i) Show that the value of the unbiased estimate of the mean of x is 1008.47
(ii) Calculate the unbiased estimate of the variance of x
(4)
The machine was initially set so that the volume of water in a bottle had a mean value
of 1010 millilitres.
Later, a test at the 5% significance level is used to determine whether or not the mean
volume of water in a bottle has changed. If it has changed then the machine is stopped
and reset.
(b) Write down suitable null and alternative hypotheses for a 2-tailed test.
(1)
(c) Find the critical region for X ‾ \overline X X
(4)
(d) Using your answer to part (a) and your critical region found in part (c), comment on
whether or not the machine needs to be stopped and reset.
Give a reason for your answer.
(2)
(e) Explain why the use of σ 2 ≈ s 2 \sigma^2 \approx s^2 σ 2 ≈ s 2
(1)
(Total for Question 1 is 12 marks)
题目中文翻译
一台机器正在灌装矿泉水瓶。
机器每天都会检查，以确保它正常工作。某天随机抽取了 100 个瓶子。
记录每个瓶中的水量 x x x
y = x − 1000 y=x-1000 y = x − 1000 进行编码。
结果汇总如下：
∑ y = 847 ∑ y 2 = 13 510.09 \sum y=847 \qquad \sum y^2=13\ 510.09 ∑ y = 847 ∑ y 2 = 13 510.09 (a) (i) 证明 x x x x x x
机器最初被设定为每瓶水的平均容量为 1010 毫升。
后来在 5% 显著性水平下进行检验，以判断每瓶水的平均容量是否已经改变。
如果改变了，机器就会停止并重新设定。
(b) 写出合适的零假设和备择假设，用于双侧检验。
(c) 求上述检验中 X ‾ \overline X X
(d) 利用 (a) 小题的答案和 (c) 小题得到的临界域，说明机器是否需要停止并重新设定。
给出理由。
(e) 解释为什么在这种情况下使用 σ 2 ≈ s 2 \sigma^2 \approx s^2 σ 2 ≈ s 2
解答
(a)(i)
解法一
思路
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先由 ∑ y \sum y ∑ y y ‾ \overline y y x = y + 1000 x=y+1000 x = y + 1000
答题过程
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The sample mean of the coded values is
y ‾ = ∑ y 100 = 847 100 = 8.47. \begin{align*}
\overline y
=&\,\frac{\sum y}{100}\\
=&\,\frac{847}{100}\\
=&\,8.47.
\end{align*} y = = = 100 ∑ y 100 847 8.47. Since x = y + 1000 x=y+1000 x = y + 1000
x ‾ = y ‾ + 1000 = 8.47 + 1000 = 1008.47 , \begin{align*}
\overline x
=&\,\overline y+1000\\
=&\,8.47+1000\\
=&\,1008.47,
\end{align*} x = = = y + 1000 8.47 + 1000 1008.47 , as required.
解法二
思路
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官方评分资料也接受先把编码后的总和还原为原数据总和。100 个观测值各自都要加回 1000，因此 ∑ x = ∑ y + 100 ( 1000 ) \sum x=\sum y+100(1000) ∑ x = ∑ y + 100 ( 1000 )
答题过程
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Using x = y + 1000 x=y+1000 x = y + 1000
∑ x = ∑ y + 100 ( 1000 ) = 847 + 100000 = 100847. \begin{align*}
\sum x
=&\,\sum y+100(1000)\\
=&\,847+100000\\
=&\,100847.
\end{align*} ∑ x = = = ∑ y + 100 ( 1000 ) 847 + 100000 100847. Therefore,
x ‾ = 100847 100 = 1008.47 , \overline x=\frac{100847}{100}=1008.47, x = 100 100847 = 1008.47 , as required.
(a)(ii)
解法一
思路
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所有数据都加上同一个常数不会改变方差，所以 x x x y y y ∑ y \sum y ∑ y ∑ y 2 \sum y^2 ∑ y 2 n − 1 n-1 n − 1
答题过程
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Since x = y + 1000 x=y+1000 x = y + 1000
s x 2 = s y 2 . s_x^2=s_y^2. s x 2 = s y 2 . Therefore,
∑ y 2 − ( ∑ y ) 2 n = 13510.09 − 847 2 100 = 6336. \begin{align*}
\sum y^2-\frac{(\sum y)^2}{n}
=&\,13510.09-\frac{847^2}{100}\\
=&\,6336.
\end{align*} ∑ y 2 − n ( ∑ y ) 2 = = 13510.09 − 100 84 7 2 6336. Hence
s x 2 = s y 2 = 6336 99 = 64 . s_x^2=s_y^2=\frac{6336}{99}=\boxed{64}. s x 2 = s y 2 = 99 6336 = 64 .
解法二
思路
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也可以把 ∑ x 2 \sum x^2 ∑ x 2 x = y + 1000 x=y+1000 x = y + 1000 x 2 x^2 x 2
答题过程
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Since
x 2 = y 2 + 2000 y + 1000000 , x^2=y^2+2000y+1000000, x 2 = y 2 + 2000 y + 1000000 , we have
∑ x 2 = ∑ y 2 + 2000 ∑ y + 100 ( 1000000 ) = 13510.09 + 2000 ( 847 ) + 100000000 = 101707510.09. \begin{align*}
\sum x^2
=&\,\sum y^2+2000\sum y\\
&\,\hspace{2pt}+100(1000000)\\
=&\,13510.09+2000(847)\\
&\,\hspace{2pt}+100000000\\
=&\,101707510.09.
\end{align*} ∑ x 2 = = = ∑ y 2 + 2000 ∑ y + 100 ( 1000000 ) 13510.09 + 2000 ( 847 ) + 100000000 101707510.09. From part (a)(i), x ‾ = 1008.47 \overline x=1008.47 x = 1008.47
∑ x 2 − n x ‾ 2 = 101707510.09 − 100 ( 1008.47 ) 2 = 6336. \begin{align*}
\sum x^2-n\overline x^{\,2}
=&\,101707510.09\\
&\,\hspace{2pt}-100(1008.47)^2\\
=&\,6336.
\end{align*} ∑ x 2 − n x 2 = = 101707510.09 − 100 ( 1008.47 ) 2 6336. Hence
s x 2 = 6336 99 = 64 . s_x^2=\frac{6336}{99}=\boxed{64}. s x 2 = 99 6336 = 64 .
(b)
解法一
思路
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题目要判断平均灌装量是否“发生改变”，可能变大也可能变小，因此使用双尾检验，并以机器原先设定的总体均值 1010 ml 作为原假设。
答题过程
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Let μ \mu μ
H 0 : μ = 1010 , H 1 : μ ≠ 1010. \begin{aligned}
H_0:&\ \mu=1010,\\
H_1:&\ \mu\ne1010.
\end{aligned} H 0 : H 1 : μ = 1010 , μ = 1010.
(c)
解法一
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由 (a) 得 s 2 = 64 s^2=64 s 2 = 64 s = 8 s=8 s = 8 8 / 100 = 0.8 8/\sqrt{100}=0.8 8/ 100 = 0.8 5 % 5\% 5% ± 1.96 \pm1.96 ± 1.96 X ‾ \overline X X
答题过程
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Under H 0 H_0 H 0 s = 8 s=8 s = 8 σ \sigma σ
X ‾ ≈ N ( 1010 , 8 2 100 ) . \overline X\approx
\mathrm N\left(1010,\frac{8^2}{100}\right). X ≈ N ( 1010 , 100 8 2 ) . For a two-tailed test at the 5 % 5\% 5% z z z ± 1.96 \pm1.96 ± 1.96
X ‾ = 1010 − 1.96 ( 8 100 ) = 1008.432 , \begin{align*}
\overline X
=&\,1010-1.96\left(\frac8{\sqrt{100}}\right)\\
=&\,1008.432,
\end{align*} X = = 1010 − 1.96 ( 100 8 ) 1008.432 , and the upper boundary is
X ‾ = 1010 + 1.96 ( 8 100 ) = 1011.568. \begin{align*}
\overline X
=&\,1010+1.96\left(\frac8{\sqrt{100}}\right)\\
=&\,1011.568.
\end{align*} X = = 1010 + 1.96 ( 100 8 ) 1011.568. Therefore, the critical region is
X ‾ ⩽ 1008.432 or X ‾ ⩾ 1011.568 . \boxed{
\overline X\leqslant1008.432
\quad\text{or}\quad
\overline X\geqslant1011.568
}. X ⩽ 1008.432 or X ⩾ 1011.568 .
(d)
解法一
思路
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将 (a) 得到的样本均值 1008.47 1008.47 1008.47
答题过程
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The observed sample mean is
x ‾ = 1008.47. \overline x=1008.47. x = 1008.47. Since
1008.432 < 1008.47 < 1011.568 , 1008.432<1008.47<1011.568, 1008.432 < 1008.47 < 1011.568 , the sample mean does not lie in the critical region. Therefore, H 0 H_0 H 0
(e)
解法一
思路
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样本量 100 较大，因此样本方差通常能较稳定地估计总体方差，以 s 2 s^2 s 2 σ 2 \sigma^2 σ 2
答题过程
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It is reasonable to use σ 2 ≈ s 2 \sigma^2\approx s^2 σ 2 ≈ s 2
