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IAL 2021 June S3 Q4

A Level / Edexcel / S3

IAL 2021 June Paper · Question 4

题目

Problem

A college runs academic and vocational courses. The college has 1680 academic students and 2520 vocational students.

(a) Describe how a stratified sample of 70 students at the college could be taken.

(3)

All students at the college take a basic skills test. A random sample of 50 academic students has a mean score of 57 and a variance of 60. An independent random sample of 80 vocational students has a mean score of 62 with a variance of 70

(b) Stating your hypotheses clearly, test at the 5% level of significance, whether or not the mean basic skills score for vocational students is greater than the mean basic skills score for academic students.

(7)

(c) Explain the importance of the Central Limit Theorem to the test in part (b).

(2)

(d) State an assumption that is required to carry out the test in part (b).

(1)

All the academic students at the college take a basic skills course. Another random sample of 50 academic students and another independent random sample of 80 vocational students retake the basic skills test. The hypotheses used in part (b) are then tested again at the same level of significance.

The value of the test statistic z is now 1.54

(e) Comment on the mean basic skills scores of academic and vocational students after taking this course.

(2)

(f) Considering the outcomes of the tests in part (b) and part (e), comment on the effectiveness of the basic skills course.

(1)
(Total for Question 4 is 16 marks)
题目中文翻译

一所学院开设学术课程和职业课程。该学院有 1680 名学术课程学生和 2520 名职业课程学生。

(a) 说明如何从学院中抽取 70 名学生的分层样本。

学院所有学生都参加基础技能测试。随机抽取的 50 名学术课程学生的平均分为 57,方差为 60。另一个相互独立的 80 名职业课程学生样本的平均分为 62,方差为 70。

(b) 清楚写出假设,在 5% 显著性水平下检验:职业课程学生的基础技能平均分是否大于学术课程学生的基础技能平均分。

(c) 解释中心极限定理在 (b) 中检验中的重要性。

(d) 写出进行 (b) 中检验所需的一条假设。

所有学术课程学生都参加了一门基础技能课程。另取一组 50 名学术课程学生的随机样本,以及一组 80 名职业课程学生的相互独立随机样本,再次参加基础技能测试。随后在同样的显著性水平下检验 (b) 中的假设。

此时检验统计量 zz 的值为 1.54。

(e) 评论接受这门课程后学术课程学生和职业课程学生的平均基础技能分数情况。

(f) 结合 (b) 与 (e) 的检验结果,评论这门基础技能课程的有效性。

解答

(a)

解法一

思路

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以学术课程和职业课程学生作为两个层。先按各层人数占全校人数的比例分配 7070 个样本名额,再在每一层内用随机数独立抽取相应人数。

答题过程

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The total number of students is

1680+2520=4200.1680+2520=4200.

The required numbers from the two strata are

academic:  16804200(70)=28,vocational:  25204200(70)=42.\begin{align*} \text{academic: }&\ \frac{1680}{4200}(70)=28,\\ \text{vocational: }&\ \frac{2520}{4200}(70)=42. \end{align*}

Label the academic students from 11 to 16801680 and the vocational students from 11 to 25202520. Use random numbers to select, without replacement, 2828 academic students and 4242 vocational students.

(b)

解法一

思路

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设职业课程与学术课程学生的总体平均分分别为 μV\mu_VμA\mu_A。题目检验 μV>μA\mu_V>\mu_A,因此使用右尾两样本均值检验。两个样本较大,用样本方差估计总体方差,计算均值差的标准误与 zz 统计量。

答题过程

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Let μV\mu_V and μA\mu_A be the population mean basic skills scores for vocational and academic students, respectively.

H0: μVμA=0,H1: μVμA>0.\begin{aligned} H_0:&\ \mu_V-\mu_A=0,\\ H_1:&\ \mu_V-\mu_A>0. \end{aligned}

Under H0H_0, the estimated standard error is

sV2nV=7080=0.875,sA2nA=6050=1.2.\begin{align*} \frac{s_V^2}{n_V} =&\,\frac{70}{80}=0.875,\\ \frac{s_A^2}{n_A} =&\,\frac{60}{50}=1.2. \end{align*}

Hence

SE=0.875+1.2=1.4404\begin{align*} \operatorname{SE} =&\,\sqrt{0.875+1.2}\\ =&\,1.4404\ldots \end{align*}

The test statistic is

z=62571.4404=3.4713.47.\begin{align*} z =&\,\frac{62-57}{1.4404\ldots}\\ =&\,3.471\ldots\\ \approx&\,3.47. \end{align*}

For a one-tailed test at the 5%5\% significance level, the critical value is 1.64491.6449.

Since

3.471>1.6449,3.471\ldots>1.6449,

the test statistic lies in the critical region, so H0H_0 is rejected. There is sufficient evidence at the 5%5\% significance level that the mean basic skills score for vocational students is greater than the mean basic skills score for academic students.

(c)

解法一

思路

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两个样本量分别为 80805050,都足够大。中心极限定理使职业课程和学术课程两个样本均值的抽样分布都近似正态,因此二者之差也近似正态,无须假设个别学生的分数本身服从正态分布。

答题过程

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Both sample sizes are large. By the Central Limit Theorem, the sampling distributions of the mean basic skills scores for both academic and vocational students are approximately normal. Hence their difference is also approximately normally distributed, which justifies the normal test in part (b).

(d)

解法一

思路

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(b) 使用样本方差 60607070 代替了未知的总体方差,因此需要假设两个样本都足够大,使样本方差可视为相应总体方差。

答题过程

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It is assumed that both samples are large enough for the sample variances to be treated as the corresponding population variances:

sA2=σA2,sV2=σV2.\begin{aligned} s_A^2=&\,\sigma_A^2,\\ s_V^2=&\,\sigma_V^2. \end{aligned}

(e)

解法一

思路

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再次使用 (b) 的右尾检验与同一个 5%5\% 临界值。新的统计量 1.541.54 小于 1.64491.6449,所以检验不再显著;结论必须表述为证据不足,而不能断言两个总体均值完全相等。

答题过程

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For the same one-tailed test at the 5%5\% significance level, the critical value remains 1.64491.6449.

Since

1.54<1.6449,1.54<1.6449,

the new test statistic does not lie in the critical region, so H0H_0 is not rejected. After the academic students have taken the course, there is insufficient evidence that the mean basic skills score for vocational students is greater than that for academic students.

(f)

解法一

思路

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(b) 显示课程前职业课程学生的总体平均分显著较高,而 (e) 显示学术课程学生上课后,这一差异不再显著。由此前后变化判断,该基础技能课程有效。

答题过程

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Before the course, there was significant evidence that vocational students had a higher mean basic skills score than academic students. After the academic students took the course, this difference was no longer significant. Therefore, there is evidence that the basic skills course was effective.