题目

Problem

At a local talent show 2 judges were asked to rank 10 performances, in order of preference. The results are shown in the following table.

Performance	A	B	C	D	E	F	G	H	I	J
Judge 1	1	3	5	7	9	10	2	6	8	4
Judge 2	2	1	3	7	9	4	5	8	10	6

(a) Calculate Spearman’s rank correlation coefficient for these data. Show your working clearly.

(3)

(b) Stating your hypotheses clearly, test at the 1% level of significance, whether or not there is evidence of a positive correlation between the ranks of Judge 1 and the ranks of Judge 2.

(4)

(Total for Question 2 is 7 marks)

题目中文翻译

在一场地方才艺表演中，两位评委被要求按喜好顺序对 10 个表演进行排名。结果如下表所示。

表演	A	B	C	D	E	F	G	H	I	J
评委 1	1	3	5	7	9	10	2	6	8	4
评委 2	2	1	3	7	9	4	5	8	10	6

(a) 计算这些数据的 Spearman 秩相关系数，并清楚写出过程。

(b) 清楚写出假设，在 1% 的显著性水平下检验两位评委的排名之间是否有正相关证据。

解答

(a)

解法一

思路

展开

两行数据已经是排名，因此逐项计算两位评委的排名差 $d$ ，求出 $\sum d^2$ 后代入无并列排名时的 Spearman 公式。

答题过程

展开

Let $d$ be the rank given by Judge 1 minus the rank given by Judge 2.

Performance	A	B	C	D	E	F	G	H	I	J
$d$	$-1$	2	2	0	0	6	$-3$	$-2$	$-2$	$-2$
$d^2$	1	4	4	0	0	36	9	4	4	4

Therefore,

\sum d^2=1+4+4+0+0+36+9+4+4+4=66.

Hence

\begin{align*} r_s =&\,1-\frac{6\sum d^2}{n(n^2-1)}\\ =&\,1-\frac{6(66)}{10(10^2-1)}\\ =&\,\boxed{0.6}. \end{align*}

(b)

解法一

思路

展开

题目询问是否存在正相关，所以使用单尾检验。写出总体 Spearman 秩相关系数的假设后，将 (a) 的 $r_s=0.6$ 与 $n=10$ 、 $1\%$ 单尾检验的临界值比较。

答题过程

展开

Let $\rho_s$ be the population Spearman rank correlation coefficient. The hypotheses are

\begin{aligned} H_0:&\ \rho_s=0,\\ H_1:&\ \rho_s>0. \end{aligned}

For $n=10$ , the critical value for a one-tailed test at the $1\%$ significance level is $0.7455$ . The rejection region is

r_s\geqslant0.7455.

Since

0.6<0.7455,

$H_0$ is not rejected. There is insufficient evidence at the $1\%$ significance level of a positive correlation between the ranks of Judge 1 and Judge 2.

IAL 2025 Jun S3 Q2

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程