题目

Problem

Dev owns a small company which produces bottles of juice. He uses two machines, $X$ and $Y$ , to fill empty bottles with juice. Dev is investigating the volumes of juice in the bottles. He chooses a random sample of 35 bottles filled by machine $X$ and a random sample of 60 bottles filled by machine $Y$ . The volumes of juice, $x$ and $y$ respectively, measured in suitable units, are summarised by

\begin{gathered} \sum x = 30.8, \qquad \sum x^2 = 29.0, \\ \sum y = 62.4, \qquad \sum y^2 = 76.8. \end{gathered}

Dev claims that the mean volume of juice in bottles filled by machine $Y$ is greater than the mean volume of juice in bottles filled by machine $X$ . A test at the $\alpha$ % significance level suggests that there is sufficient evidence to support Dev’s claim.

Find the set of possible values of $\alpha$ .

[9]

题目中文翻译

Dev 拥有一家生产瓶装果汁的小公司。他使用两台机器 $X$ 和 $Y$ 将果汁灌入空瓶中。Dev 正在研究瓶中果汁的体积。他选取了由机器 $X$ 灌装的 35 瓶随机样本，以及由机器 $Y$ 灌装的 60 瓶随机样本。果汁体积分别记为 $x$ 和 $y$ ，以合适单位测量，其数据汇总为

\begin{gathered} \sum x = 30.8, \qquad \sum x^2 = 29.0, \\ \sum y = 62.4, \qquad \sum y^2 = 76.8. \end{gathered}

Dev 声称由机器 $Y$ 灌装的瓶中果汁平均体积大于由机器 $X$ 灌装的瓶中果汁平均体积。在 $\alpha$ % 显著性水平下进行的检验表明，有足够证据支持 Dev 的说法。

求 $\alpha$ 的可能取值集合。

(9)

原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

解答

解法一

思路

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两组样本量都较大，用均值差的正态近似。把单尾 $p$ 值换成百分数，即得到能拒绝零假设的显著性水平边界。

答题过程

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Test $H_0\colon\mu_Y=\mu_X$ against $H_1\colon\mu_Y>\mu_X$ . The means are $\bar x=30.8/35=0.88$ and $\bar y=62.4/60=1.04$ . The unbiased variance estimates are

\begin{align*} &\,s_X^2=\frac{29.0-30.8^2/35}{34}=0.0557647\ldots,\\ &\,s_Y^2=\frac{76.8-62.4^2/60}{59}=0.2017627\ldots. \end{align*}

For the independent large samples, the estimated variance of $\bar Y-\bar X$ is $s_Y^2/60+s_X^2/35=0.004955989\ldots$ . Thus

\begin{align*} &\,z=\frac{1.04-0.88}{\sqrt{s_Y^2/60+s_X^2/35}}\\ =&\,2.272766\ldots. \end{align*}

The upper-tail $p$ -value is $1-\Phi(2.272766\ldots)=0.0115201\ldots$ . Using rejection when $p<\alpha/100$ , the required levels satisfy $\alpha>100(1-\Phi(2.272766\ldots))$ , with boundary approximately $1.15$ . Thus $\alpha>1.15$ , with the boundary stated to three significant figures.

CIE 9231 2024 Nov Paper 42 Q5

题目

解答

解法一

思路

答题过程