题目

Problem

The numbers of green sweets in $200$ randomly chosen packets of Frutos are summarised in the table.

Number of green sweets	$0$	$1$	$2$	$3$	$>3$
Number of packets	$32$	$50$	$97$	$21$	$0$

(a) Calculate an unbiased estimate for the population mean of the number of green sweets in a packet of Frutos, and show that an unbiased estimate of the population variance is $0.783$ correct to $3$ significant figures.

(3)

The manufacturers of Frutos claim that the mean number of green sweets in a packet is $1.65$ .

Anji believes that the true value of the mean, $\mu$ , is less than $1.65$ . She uses the results from the $200$ randomly chosen packets to test the manufacturers’ claim.

(b) State suitable null and alternative hypotheses for the test.

(1)

(c) Show that the result of Anji’s test is significant at the $5\%$ level but not at the $1\%$ level.

(4)

(d) It is given that Anji made a Type I error.

Explain how this shows that the significance level that Anji used in her test was not $1\%$ .

(1)

题目中文翻译

随机抽取 $200$ 袋 Frutos 糖果，每袋绿色糖果的颗数汇总在下表中。

绿色糖果颗数	$0$	$1$	$2$	$3$	$>3$
袋数	$32$	$50$	$97$	$21$	$0$

(a) 计算一袋 Frutos 糖果中绿色糖果颗数的总体均值的一个无偏估计值，并证明总体方差的一个无偏估计值为 $0.783$ ，精确到 $3$ 位有效数字。

(3)

Frutos 的生产商声称，每袋平均有 $1.65$ 颗绿色糖果。

Anji 认为均值 $\mu$ 的真值小于 $1.65$ 。她用随机抽取的 $200$ 袋糖果的结果检验生产商的说法。

(b) 写出适合该检验的原假设和备择假设。

(1)

(c) 证明 Anji 的检验结果在 $5\%$ 水平下显著，在 $1\%$ 水平下不显著。

(4)

(d) 已知 Anji 犯了第一类错误。

解释由此如何看出 Anji 在检验中使用的显著性水平并非 $1\%$ 。

(1)

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

(a)

解法一

思路

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频数表先求总和与平方和，再计算无偏估计。均值保留未舍入值用于方差。

答题过程

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From the frequency table,

\begin{align*} \sum x=&\,0(32)+1(50)+2(97)+3(21)\\[4mm] =&\,307, \end{align*}

and

\begin{align*} \sum x^2=&\,0^2(32)+1^2(50)+2^2(97)+3^2(21)\\[4mm] =&\,627. \end{align*}

The unbiased estimate of the population mean is

\bar x=\frac{307}{200}=\boxed{1.535}.

The unbiased estimate of the population variance is

\begin{align*} s^2=&\,\frac{627-307^2/200}{199}\\[4mm] =&\,0.782688\ldots\\[4mm] \approx&\,\boxed{0.783}\quad\text{(3 s.f.),} \end{align*}

as required.

(b)

解法一

思路

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原假设为厂家声称的总体均数，备择为均数较小。

答题过程

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For the population mean number of green sweets per packet,

\boxed{H_0:\mu=1.65,\qquad H_1:\mu<1.65}.

(c)

解法一

思路

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利用大样本正态均值检验，保留未舍入样本方差，并把同一统计量与5%及1%的左尾临界值比较。

答题过程

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Using the large-sample normal approximation and the unrounded variance from part (a),

z=\frac{1.535-1.65}{\sqrt{0.782688\ldots/200}} =-1.838308\ldots.

The lower-tail critical values are $-1.645$ at 5% and $-2.326$ at 1%. Since

-2.326<-1.838308\ldots<-1.645,

we reject $H_0$ at the 5% level but do not reject it at the 1% level. Thus the result is significant at 5% but not at 1%.

(d)

解法一

思路

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第一类错误必先拒绝原假设；1%水平不会拒绝，因此不可能是Anji实际使用的水平。

答题过程

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A Type I error requires rejection of a true $H_0$ . At the 1% significance level, $H_0$ would not have been rejected, so Anji could not have made a Type I error using that level.

CIE 9709 2024 June Paper 63 Q6

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解答

(a)

解法一

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(b)

解法一

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(c)

解法一

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(d)

解法一

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