题目

Problem

Luka wants to carry out a survey of students at his school.

He obtains a list of all 280 students.

(a) Explain how he can use this list to select a systematic sample of 40 students.

(2)

Luka is trying to make his own random number table. He generates 400 digits to put in his table. Figure 1 shows the frequency of each digit in his table.

Digit generated	0	1	2	3	4	5	6	7	8	9
Frequency	36	42	33	41	44	43	48	38	32	43

A test is carried out at the 10% level of significance to see if the digits Luka generates follow a uniform distribution.

For this test

\sum\frac{(O-E)^2}{E}=5.9.

(b) Determine the conclusion.

(3)

The digits generated by Luka are taken two at a time to form two-digit numbers.

Figure 2 shows the frequency of two-digit numbers in his table.

Two-digit numbers generated	00–19	20–39	40–59	60–79	80–99
Frequency	31	49	30	42	48

(c) Test, at the 10% level of significance, whether the two-digit numbers generated by Luka follow a uniform distribution. You should state the hypotheses, the degrees of freedom and the critical value used for this test.

(8)

There are 70 students in Year 12 at his school.

(d) State, giving a reason, the advice you would give to Luka regarding the use of his table of numbers for generating a simple random sample of 10 of the Year 12 students.

(2)

(Total 15 marks)

题目中文翻译

Luka 想对他所在学校的学生进行调查。

他取得了一份全校 280 名学生的名单。

(a) 说明他怎样利用这份名单抽取 40 名学生的系统样本。

Luka 正尝试制作自己的随机数表。他生成了 400 个数字放入表中。图 1 给出了表中每个数字的频数。

生成的数字	0	1	2	3	4	5	6	7	8	9
频数	36	42	33	41	44	43	48	38	32	43

在 10% 显著性水平下进行检验，判断 Luka 生成的数字是否服从均匀分布。

对于这项检验，

\sum\frac{(O-E)^2}{E}=5.9.

(b) 写出结论。

把 Luka 生成的数字每两个一组，组成两位数。

图 2 给出了他的表中两位数的频数。

生成的两位数	00–19	20–39	40–59	60–79	80–99
频数	31	49	30	42	48

他的学校有 70 名 Year 12 学生。

(d) 对 Luka 使用这张数表生成 10 名 Year 12 学生的简单随机样本提出建议，并说明理由。

解答

(a)

解法一

思路

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系统抽样的抽样间隔是总体人数除以样本人数，即 $280/40=7$ 。先在名单的前七人中随机选一个起点，之后每隔七个编号抽取一人，便会得到 40 人。

答题过程

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The sampling interval is

\frac{280}{40}=7.

Randomly select a starting position from 1 to 7 on the list. Starting from that student, select every seventh student until 40 students have been selected.

(b)

解法一

思路

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共有十种数字，且均匀分布模型没有由样本估计参数，所以自由度为 $10-1=9$ 。题目已经给出检验统计量 $5.9$ ，只需与 $10\%$ 上尾临界值比较，并针对 Luka 生成的个位数字写结论。

答题过程

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The number of degrees of freedom is

\nu=10-1=9.

At the $10\%$ significance level, the critical value is

\chi^2_9(0.10)=14.684.

Since

5.9<14.684,

the result is not significant. There is insufficient evidence that the individual digits generated by Luka do not follow a uniform distribution.

(c)

解法一

思路

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五个区间在 $00$ 至 $99$ 中各包含二十个两位数，所以在均匀分布下期望频数相同，都是 $200/5=40$ 。求各组的卡方贡献并相加，再用 $4$ 个自由度的 $10\%$ 临界值完成检验。

答题过程

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The hypotheses are:

$H_0$ : a uniform distribution is a suitable model for the two-digit numbers generated.
$H_1$ : a uniform distribution is not a suitable model for the two-digit numbers generated.

The expected frequency in each of the five groups is

E=\frac{200}{5}=40.

The contributions to the test statistic are

Range	$00$ – $19$	$20$ – $39$	$40$ – $59$	$60$ – $79$	$80$ – $99$
$O_i$	31	49	30	42	48
$E_i$	40	40	40	40	40
$\dfrac{(O_i-E_i)^2}{E_i}$	2.025	2.025	2.5	0.1	1.6

Therefore,

\begin{align*} \chi^2=&\,\sum\frac{(O_i-E_i)^2}{E_i}\\ =&\,2.025+2.025+2.5+0.1+1.6\\ =&\,8.25. \end{align*}

Equivalently,

\chi^2=\sum\frac{O_i^2}{E_i}-200=8.25.

The number of degrees of freedom is

\nu=5-1=4.

At the $10\%$ significance level, the critical value is

\chi^2_4(0.10)=7.779.

Since

8.25>7.779,

the result is significant, so $H_0$ is rejected. There is sufficient evidence that the two-digit numbers generated by Luka do not follow a uniform distribution.

(d)

解法一

思路

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若要从 70 名学生中抽取简单随机样本，可把他们编号为 $00$ 至 $69$ ，再用两位随机数选人，并舍去超出范围或重复的编号。但 (c) 已表明 Luka 生成的两位数不服从均匀分布，因此每名学生未必有相同的入选机会，这张表不适合使用。

答题过程

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Luka would need to label the 70 Year 12 students from 00 to 69 and use two-digit numbers to select 10 different students, ignoring numbers from 70 to 99 and any repeats.

However, part (c) provides evidence that Luka’s two-digit numbers do not follow a uniform distribution. His table is therefore not suitable for generating this simple random sample, because the students would not all have an equal chance of selection.

IAL 2020 Oct S3 Q4

题目

解答

(a)

解法一

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(b)

解法一

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(c)

解法一

思路

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(d)

解法一

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