题目

Problem

Claire grows strawberries on her farm. She wants to compare two brands of fertiliser, brand A and brand B.

She grows two sets of plants of the same variety of strawberries under the same conditions, fertilising one set with brand A and the other with brand B. The yields per plant, in grams, from each set of plants are summarised below.

Fertiliser	Mean	Standard deviation	Number of plants
A	1377	17.8	50
B	1368	18.4	40

(a) Stating your hypotheses clearly, carry out a suitable test to assess whether the mean yield from plants using fertiliser A is greater than the mean yield from plants using fertiliser B. Use a 1% level of significance and state your test statistic and critical value.

(7)

The total cost of fertiliser A for Claire’s 50 plants was £75 The total cost of fertiliser B for Claire’s 40 plants was £50 Claire sells all her strawberries at £3 per kilogram.

(b) Use this information, together with your answer in part (a), to advise Claire on which of the two brands of fertiliser she should use next year in order to maximise her expected profit per plant, giving a reason for your answer.

(3)

(Total for Question 5 is 10 marks)

题目中文翻译

Claire 在她的农场种草莓。她想比较两种肥料品牌，A 品牌和 B 品牌。

她在相同条件下种植了两组相同品种的草莓植株，一组施 A 品牌肥料，另一组施 B 品牌肥料。每株的产量（克）汇总如下。

肥料	平均值	标准差	植株数
A	1377	17.8	50
B	1368	18.4	40

(a) 清楚写出原假设，进行适当检验以判断使用 A 品牌肥料的植株平均产量是否大于使用 B 品牌肥料的植株平均产量。使用 1% 显著性水平，并写出检验统计量和临界值。

A 品牌肥料用于 50 株植株的总成本为 £75。 B 品牌肥料用于 40 株植株的总成本为 £50。 Claire 以每千克 £3 的价格出售所有草莓。

(b) 利用这些信息，并结合 (a) 小题的答案，建议 Claire 明年应使用哪一种肥料品牌以最大化每株的期望利润，并给出理由。

解答

(a)

解法一

思路

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要检验 A 品牌是否带来更高的总体平均产量，因此对 $\mu_A-\mu_B$ 进行右尾检验。两个样本量都较大，以样本标准差估计总体标准差，先求均值差的标准误，再计算 $z$ 统计量并与 $1\%$ 单尾临界值比较。

答题过程

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Let $\mu_A$ and $\mu_B$ be the population mean yields per plant using fertilisers A and B respectively. The hypotheses are

\begin{aligned} H_0:&\ \mu_A=\mu_B,\\ H_1:&\ \mu_A>\mu_B. \end{aligned}

The estimated variance of $\overline X_A-\overline X_B$ is

\begin{align*} \operatorname{SE}^2 =&\,\frac{17.8^2}{50}\\ &\,\hspace{2pt}+\frac{18.4^2}{40}\\ =&\,14.8008. \end{align*}

Therefore,

\operatorname{SE} =\sqrt{14.8008} =3.84718\ldots.

The test statistic is

\begin{align*} z =&\,\frac{1377-1368}{3.84718\ldots}\\ =&\,2.3393\ldots\\ =&\,2.34. \end{align*}

For a one-tailed test at the $1\%$ significance level, the critical value is $2.3263$ , giving the critical region

z\geqslant2.3263.

Since $2.34>2.3263$ , the test statistic lies in the critical region, so $H_0$ is rejected. There is sufficient evidence at the $1\%$ significance level that the mean yield from plants using fertiliser A is greater than that from plants using fertiliser B.

解法二

思路

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官方评分资料也接受用置信区间判断。右尾 $1\%$ 检验可对应到 $\mu_A-\mu_B$ 的双侧 $98\%$ 置信区间；若整个区间都大于 0，便说明 A 的总体平均产量显著较高。

答题过程

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Using the same hypotheses and estimated standard error,

\operatorname{SE}(\overline X_A-\overline X_B) =3.84718\ldots.

A $98\%$ confidence interval for $\mu_A-\mu_B$ is

\begin{align*} I =&\,(1377-1368)\\ &\,\hspace{2pt}\pm2.3263(3.84718\ldots)\\ =&\,9\pm8.9497\ldots\\ =&\,(0.0503\ldots,\ 17.9497\ldots). \end{align*}

The whole interval is above zero. Therefore, $H_0$ is rejected and there is sufficient evidence at the $1\%$ significance level that the mean yield using fertiliser A is greater than the mean yield using fertiliser B.

(b)

解法一

思路

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(a) 只比较了产量，但本题要求最大化每株期望利润。分别把每株平均产量从克换算为千克并乘售价，再减去每株肥料成本，最后比较两种肥料的期望利润。

答题过程

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For fertiliser A, the expected profit per plant is

\begin{align*} 3(1.377)-\frac{75}{50} =&\,4.131-1.50\\ =&\,\pounds2.631. \end{align*}

For fertiliser B, the expected profit per plant is

\begin{align*} 3(1.368)-\frac{50}{40} =&\,4.104-1.25\\ =&\,\pounds2.854. \end{align*}

Although part (a) provides evidence that fertiliser A gives a greater mean yield, fertiliser B gives the greater expected profit per plant. Therefore, Claire should use $\boxed{\text{fertiliser B}}$ next year.

IAL 2023 Jan S3 Q5

题目

解答

(a)

解法一

思路

答题过程

解法二

思路

答题过程

(b)

解法一

思路

答题过程