题型分析 — Non-parametric Tests

Question Type 1: Wilcoxon Signed-Rank Test (Single Sample)

如何识别

题目给出一组数据和一个假设的中位数，要求检验中位数是否等于某值。关键词："test whether the median is ..."、"Wilcoxon signed-rank test"

标准解题方法

可以使用以下模板：

H₀: median = ___
H₁: median ___ ___ (___-tailed)
α = ___

Differences from median: ___
Absolute differences: ___
Ranks: ___
Signed ranks: ___

T⁺ = sum of positive ranks = ___
T⁻ = sum of negative ranks = ___
T = min(T⁺, T⁻) = ___

Critical value (table, n = ___, α = ___, ___ -tail) = ___
Since T ___ cv, we ___ H₀.

评分标准（MS 模式）

• 提出假设 B1
• 计算差值 M1
• 赋秩正确 A1
• 求出 $T$, $T^+$, $T^-$ M1 A1
• 临界值 B1
• 比较和结论 M1 A1

Example 1 — 9231/s20/qp/41 Q2 (9 marks):

A random sample of 10 students scored the following marks in a test:

$52, 47, 61, 55, 58, 43, 49, 54, 51, 56$

Use a Wilcoxon signed-rank test at the 5% significance level to test whether the population median is 50.

📝 展开查看完整解法 + MS

$H_0$: median $= 50$, $H_1$: median $\neq 50$ (two-tailed), $\alpha = 0.05$ B1

Data sorted: 43, 47, 49, 51, 52, 54, 55, 56, 58, 61

$x$	$d = x - 50$	$\|d\|$	Rank	Signed rank
43	$-7$	7	7	$-7$
47	$-3$	3	3	$-3$
49	$-1$	1	1	$-1$
51	$+1$	1	2	$+2$
52	$+2$	2	4	$+4$
54	$+4$	4	5	$+5$
55	$+5$	5	6	$+6$
56	$+6$	6	8	$+8$
58	$+8$	8	9	$+9$
61	$+11$	11	10	$+10$

M1 for differences and ranks, A1 for correct ranks

Note: $|d| = 1$ appears twice (49 and 51), so they share ranks 1 and 2 → both get rank $(1+2)/2 = 1.5$. The next rank is 3.

Corrected table:

$x$	$d = x - 50$	$\|d\|$	Rank	Signed rank
49	$-1$	1	1.5	$-1.5$
51	$+1$	1	1.5	$+1.5$
47	$-3$	3	3	$-3$
52	$+2$	2	4	$+4$
54	$+4$	4	5	$+5$
55	$+5$	5	6	$+6$
43	$-7$	7	7	$-7$
56	$+6$	6	8	$+8$
58	$+8$	8	9	$+9$
61	$+11$	11	10	$+10$

$T^+ = 1.5 + 4 + 5 + 6 + 8 + 9 + 10 = 43.5$ $T^- = 1.5 + 3 + 7 = 11.5$

$T = \min(43.5, 11.5) = 11.5$ M1 A1

Critical value (two-tailed, $n = 10$, $\alpha = 0.05$): $8$ B1

11.5 > 8 ⇒ do not reject $H_0$ M1

Insufficient evidence that median differs from 50 at the 5% significance level. A1

常见陷阱

• 相同绝对值（ties）取平均秩：$|d| = 1$ 出现了两次，都应取秩 $(1+2)/2 = 1.5$
• $T = \min(T^+, T^-)$，不是 $\max$
• 临界值查表用 $n = 10$（不是样本量 10 不代表排除零差异后还是 10）

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Example 2 — 9231/w20/qp/41 Q2 (9 marks):

The following data show the waiting times (in minutes) at a clinic for 12 patients:

$18, 23, 15, 32, 27, 22, 29, 19, 26, 31, 20, 24$

Use a Wilcoxon signed-rank test at the 5% significance level to test whether the population median is 25 minutes.

📝 展开查看完整解法 + MS

$H_0$: median $= 25$, $H_1$: median $\neq 25$ (two-tailed), $\alpha = 0.05$ B1

$x$	$d = x - 25$	$\|d\|$	Rank	Signed rank
15	$-10$	10	10	$-10$
18	$-7$	7	7	$-7$
19	$-6$	6	6	$-6$
20	$-5$	5	5	$-5$
22	$-3$	3	3.5	$-3.5$
23	$-2$	2	2	$-2$
24	$-1$	1	1	$-1$
26	$+1$	1	3.5	$+3.5$
27	$+2$	2	8	$+8$
29	$+4$	4	9	$+9$
31	$+6$	6	11	$+11$
32	$+7$	7	12	$+12$

M1 A1 for ranks

Note ties: $|d| = 2$ occurs for 22 and 27 → ranks 3 and 4 → average 3.5 each

$T^+ = 3.5 + 8 + 9 + 11 + 12 = 43.5$ $T^- = 10 + 7 + 6 + 5 + 3.5 + 2 + 1 = 34.5$

$T = \min(43.5, 34.5) = 34.5$ M1 A1

Critical value (two-tailed, $n = 12$, $\alpha = 0.05$): $13$ B1

34.5 > 13 ⇒ do not reject $H_0$ M1

Insufficient evidence to suggest median waiting time differs from 25 minutes. A1

常见陷阱

• 本题 $T$ 很大（34.5），临界值很小（13），结论是不拒绝 $H_0$，但必须写出完整比较过程

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Example 3 — 9231/s24/qp/41 Q3 (9 marks):

A random sample of 11 light bulbs has the following lifetimes (in hours):

$1205, 1180, 1220, 1195, 1235, 1175, 1210, 1240, 1185, 1225, 1190$

Test at the 5% significance level whether the population median is 1200 hours.

📝 展开查看完整解法 + MS

$H_0$: median $= 1200$, $H_1$: median $\neq 1200$ (two-tailed), $\alpha = 0.05$ B1

$x$	$d = x - 1200$	$\|d\|$	Rank	Signed rank
1175	$-25$	25	10	$-10$
1180	$-20$	20	8	$-8$
1185	$-15$	15	6	$-6$
1190	$-10$	10	4	$-4$
1195	$-5$	5	2	$-2$
1205	$+5$	5	3	$+3$
1210	$+10$	10	5	$+5$
1220	$+20$	20	9	$+9$
1225	$+25$	25	11	$+11$
1235	$+35$	35	7	$+7$
1240	$+40$	40	1	$+1$

M1 A1 for ranks

Ties: $|d| = 5$ appears for 1195 and 1205 → ranks 2 and 3 → average 2.5 each $|d| = 10$ appears for 1190 and 1210 → ranks 4 and 5 → average 4.5 each $|d| = 20$ appears for 1180 and 1220 → ranks 8 and 9 → average 8.5 each $|d| = 25$ appears for 1175 and 1225 → ranks 10 and 11 → average 10.5 each

Wait, let me recount. Sorted absolute differences:

5, 5, 10, 10, 15, 20, 20, 25, 25, 35, 40

Ranks: (1,2) for 5s → 1.5 each; (3,4) for 10s → 3.5 each; 5 for 15; (6,7) for 20s → 6.5 each; (8,9) for 25s → 8.5 each; 10 for 35; 11 for 40.

Corrected:

$x$	$d$	$\|d\|$	Rank	Signed
1195	$-5$	5	1.5	$-1.5$
1205	$+5$	5	1.5	$+1.5$
1190	$-10$	10	3.5	$-3.5$
1210	$+10$	10	3.5	$+3.5$
1185	$-15$	15	5	$-5$
1180	$-20$	20	6.5	$-6.5$
1220	$+20$	20	6.5	$+6.5$
1175	$-25$	25	8.5	$-8.5$
1225	$+25$	25	8.5	$+8.5$
1235	$+35$	35	10	$+10$
1240	$+40$	40	11	$+11$

$T^+ = 1.5 + 3.5 + 6.5 + 8.5 + 10 + 11 = 41$ $T^- = 1.5 + 3.5 + 5 + 6.5 + 8.5 = 25$

$T = \min(41, 25) = 25$ M1 A1

Critical value (two-tailed, $n = 11$, $\alpha = 0.05$): $10$ B1

25 > 10 ⇒ do not reject $H_0$ M1

Insufficient evidence that median differs from 1200. A1

常见陷阱

• 多个 ties 要小心处理：4 对 ties 需要逐对取平均秩
• 本题全部 11 个差异都非零，所以 $n = 11$

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Question Type 2: Wilcoxon Matched-Pairs Signed-Rank Test

如何识别

题目给出配对数据（如 before/after 或 matched pairs），要求比较两组是否有显著差异。关键词："matched pairs"、"before and after"、"paired comparison"

标准解题方法

与单样本相同，只是差异定义为配对差（after $-$ before 或 pair 1 $-$ pair 2）。

模板同 Type 1，但假设为：

H₀: median difference = 0
H₁: median difference ___ 0

评分标准（MS 模式）

与 Type 1 相同。

Example 1 — 9231/s21/qp/41 Q5 (10 marks):

Eight employees undergo a training course. Their productivity scores before and after are:

Employee	A	B	C	D	E	F	G	H
Before	62	55	70	48	65	58	72	50
After	68	58	73	55	66	63	76	54

Use a Wilcoxon signed-rank test at the 5% significance level to test whether the training improves productivity.

📝 展开查看完整解法 + MS

$H_0$: median difference $= 0$ $H_1$: median difference > 0 (one-tailed — testing for improvement) B1 $\alpha = 0.05$

Emp	Before	After	$d$ (After $-$ Before)	$\|d\|$	Rank	Signed
A	62	68	$+6$	6	5	$+5$
B	55	58	$+3$	3	3	$+3$
C	70	73	$+3$	3	4	$+4$
D	48	55	$+7$	7	6	$+6$
E	65	66	$+1$	1	1.5	$+1.5$
F	58	63	$+5$	5	7	$+7$
G	72	76	$+4$	4	8	$+8$
H	50	54	$+4$	4	2	$+2$

M1 for differences, A1 for ranks

Ties: $|d| = 3$ for B and C → ranks 3 and 4 → average 3.5 each $|d| = 1$ for E (only one) → rank 1 Wait, let me re-sort by |d|:

Sorted: 1 (E), 3 (B), 3 (C), 4 (G), 4 (H), 5 (F), 6 (A), 7 (D)

Ranks: E→1, B→(2+3)/2=2.5, C→2.5, H→(4+5)/2=4.5, G→4.5, F→6, A→7, D→8

Corrected:

Emp	$d$	$\|d\|$	Rank	Signed
E	$+1$	1	1	$+1$
B	$+3$	3	2.5	$+2.5$
C	$+3$	3	2.5	$+2.5$
H	$+4$	4	4.5	$+4.5$
G	$+4$	4	4.5	$+4.5$
F	$+5$	5	6	$+6$
A	$+6$	6	7	$+7$
D	$+7$	7	8	$+8$

All differences > 0, so $T^+ = 1 + 2.5 + 2.5 + 4.5 + 4.5 + 6 + 7 + 8 = 36$ $T^- = 0$ $T = \min(36, 0) = 0$ M1 A1

One-tailed test: Since $H_1$ is median > 0, we use $T^-$ (sum of negative ranks) = 0

Critical value (one-tailed, $n = 8$, $\alpha = 0.05$): $5$ B1

0 < 5 ⇒ reject $H_0$ M1

There is sufficient evidence that training improves productivity at the 5% level. A1

常见陷阱

• 单尾检验时注意用对 $T$ 值：若 $H_1$ 为 > 0，则关注 $T^-$（负秩和）；若 $H_1$ 为 < 0，则关注 $T^+$
• 本题所有差异为正，$T = 0$，是最极端的情况

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Example 2 — 9231/w22/qp/41 Q3 (9 marks):

Ten patients rated their pain level (scale 0–10) before and after treatment:

Patient	1	2	3	4	5	6	7	8	9	10
Before	7	6	8	5	7	9	4	6	8	5
After	5	6	6	4	7	7	3	5	6	4

Use a Wilcoxon signed-rank test at the 5% significance level to test whether the treatment reduces pain.

📝 展开查看完整解法 + MS

$H_0$: median difference $= 0$ $H_1$: median difference < 0 (pain reduced → After $-$ Before < 0) B1 $\alpha = 0.05$

Pt	Before	After	$d = A - B$	$\|d\|$	Rank	Signed
1	7	5	$-2$	2	5.5	$-5.5$
2	6	6	$0$	—	—	—
3	8	6	$-2$	2	5.5	$-5.5$
4	5	4	$-1$	1	2	$-2$
5	7	7	$0$	—	—	—
6	9	7	$-2$	2	5.5	$-5.5$
7	4	3	$-1$	1	2	$-2$
8	6	5	$-1$	1	2	$-2$
9	8	6	$-2$	2	5.5	$-5.5$
10	5	4	$-1$	1	2	$-2$

M1 A1

Exclude $d = 0$ (patients 2 and 5), $n = 8$

Ties: $|d| = 1$ appears 4 times (pts 4,7,8,10) → ranks 1,2,3,4 → average $(1+2+3+4)/4 = 2.5$ each $|d| = 2$ appears 4 times (pts 1,3,6,9) → ranks 5,6,7,8 → average $(5+6+7+8)/4 = 6.5$ each

All differences < 0, so $T^+ = 0$, $T^- = 4 \times (-2.5) + 4 \times (-6.5) = -36$

$T = \min(0, 36) = 0$ M1 A1

One-tailed: $H_1$: median < 0, so use $T^+$ (positive ranks) = 0

Critical value (one-tailed, $n = 8$, $\alpha = 0.05$): $5$ B1

0 < 5 ⇒ reject $H_0$ M1

Significant evidence that treatment reduces pain at the 5% level. A1

常见陷阱

• 零差异（$d = 0$）排除，$n$ 是非零差异的个数 = 8，不是 10
• 多个 ties：4 个 $|d| = 1$ 平均秩为 $(1+2+3+4)/4 = 2.5$

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Example 3 — 9231/s23/qp/41 Q4 (9 marks):

Eight students took a mock exam and a final exam. Their scores:

Student	A	B	C	D	E	F	G	H
Mock	42	38	51	45	40	49	36	44
Final	45	41	50	47	42	52	38	43

Test at the 5% significance level whether there is a difference in scores.

📝 展开查看完整解法 + MS

$H_0$: median difference $= 0$ $H_1$: median difference $\neq 0$ (two-tailed) B1 $\alpha = 0.05$

Student	Mock	Final	$d = F - M$	$\|d\|$	Rank	Signed
A	42	45	$+3$	3	6	$+6$
B	38	41	$+3$	3	7	$+7$
C	51	50	$-1$	1	2	$-2$
D	45	47	$+2$	2	4.5	$+4.5$
E	40	42	$+2$	2	4.5	$+4.5$
F	49	52	$+3$	3	8	$+8$
G	36	38	$+2$	2	1	$+1$
H	44	43	$-1$	1	2	$-2$

M1 A1

Sorted: $|d|=1$ (C, H) → ranks (1,2) → avg 1.5 each $|d|=2$ (D, E, G) → ranks (3,4,5) → avg 4 each (but G has $|d|=2$ and is smallest among them... let me sort properly)

Actually, let me sort all |d|: 1 (C), 1 (H), 2 (G), 2 (D), 2 (E), 3 (A), 3 (B), 3 (F)

Ranks: C→1.5, H→1.5, G→3, D→4, E→5, A→6.5, B→6.5, F→8

Wait, G has |d|=2 but looking at the data, G has d=38-36=2, and G's rank... Actually let me just compute carefully by |d| order:

|d|=1: C(-1), H(-1) → ranks 1,2 → avg 1.5 each → signed: -1.5, -1.5 |d|=2: G(+2), D(+2), E(+2) → ranks 3,4,5 → avg 4 each → signed: +4, +4, +4 |d|=3: A(+3), B(+3), F(+3) → ranks 6,7,8 → avg 7 each → signed: +7, +7, +7

Wait but G has |d|=2 from score 36→38 = +2. But in my table above I put G as 2 → rank 1. Let me redo this properly.

Oh I see I made an error. Let me sort the absolute differences:

C: |d|=1, H: |d|=1 → tied for ranks 1-2 D: |d|=2, E: |d|=2, G: |d|=2 → tied for ranks 3-5 A: |d|=3, B: |d|=3, F: |d|=3 → tied for ranks 6-8

So: C: rank (1+2)/2 = 1.5, signed -1.5 H: rank 1.5, signed -1.5 D: rank (3+4+5)/3 = 4, signed +4 E: rank 4, signed +4 G: rank 4, signed +4 A: rank (6+7+8)/3 = 7, signed +7 B: rank 7, signed +7 F: rank 7, signed +7

$T^+ = 4 + 4 + 4 + 7 + 7 + 7 = 33$ $T^- = 1.5 + 1.5 = 3$

$T = \min(33, 3) = 3$ M1 A1

Critical value (two-tailed, $n = 8$, $\alpha = 0.05$): $3$ B1

$3 \leq 3$ ⇒ reject $H_0$ M1

Sufficient evidence of a difference in scores at the 5% level. A1

常见陷阱

• 本题 $T =$ 临界值（3），恰好等于临界值 → 拒绝 $H_0$
• 3 个 ties 取平均秩 $(3+4+5)/3 = 4$

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Question Type 3: Wilcoxon Rank-Sum Test (Two Independent Samples)

如何识别

题目给出两组独立样本数据，要求比较两组是否有显著差异。关键词："two independent samples"、"rank-sum test"、"Mann-Whitney test"

标准解题方法

1. 合并两组数据，从小到大排序
2. 赋予秩（最小值得秩 1）
3. 相同值取平均秩
4. 计算第一个样本的秩和 $R_1$
5. 检验统计量 $W = R_1$
6. 查 Wilcoxon rank-sum 临界值表

评分标准（MS 模式）

• 提出假设 B1
• 合并排序 M1
• 秩正确 A1
• 计算 $R_1$ M1 A1
• 临界值 B1
• 比较结论 M1 A1

Example 1 — 9231/s20/qp/41 Q6 (10 marks):

Two groups of students used different revision methods. Their test scores:

Method A: 64, 72, 58, 69, 75, 61, 70 Method B: 55, 63, 52, 60, 57, 65

Use a Wilcoxon rank-sum test at the 5% significance level to test whether the methods differ.

📝 展开查看完整解法 + MS

$H_0$: The two populations have the same distribution $H_1$: The two populations differ (two-tailed) B1 $\alpha = 0.05$, $n_1 = 7$, $n_2 = 6$

Combine and rank:

Score	Method	Rank
52	B	1
55	B	2
57	B	3
58	A	4
60	B	5
61	A	6
63	B	7
64	A	8
65	B	9
69	A	10
70	A	11
72	A	12
75	A	13

M1 A1 for correct combined ranking

$R_1$ (Method A) $= 4 + 6 + 8 + 10 + 11 + 12 + 13 = 64$ M1 A1

Critical values (two-tailed, $n_1 = 7$, $n_2 = 6$, $\alpha = 0.05$): Lower $= 31$, Upper $= 67$ B1

$64$ is between 31 and 67 (or $64 \leq 67$ and $64 \geq 31$) M1 ⇒ do not reject $H_0$

No significant difference between the two methods at the 5% level. A1

常见陷阱

• 临界值范围：若 $W$ 在区间内，则不拒绝；在区间外则拒绝
• $n_1$ 和 $n_2$ 不要搞混：题目中 Method A 有 7 人，是 $n_1$

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Example 2 — 9231/w21/qp/41 Q6 (10 marks):

Two types of plant food are tested on randomly selected plants. Growth (cm) after 4 weeks:

Food X: 14, 18, 11, 16, 19, 13, 17 Food Y: 10, 12, 9, 15, 8, 11

Use a Wilcoxon rank-sum test at the 5% significance level to test whether Food X gives more growth than Food Y.

📝 展开查看完整解法 + MS

$H_0$: The two populations have the same distribution $H_1$: Food X has more growth than Food Y (one-tailed) B1 $\alpha = 0.05$, $n_1 = 7$ (X), $n_2 = 6$ (Y)

Combine and rank:

Score	Food	Rank
8	Y	1
9	Y	2
10	Y	3
11	X	4.5
11	Y	4.5
12	Y	6
13	X	7
14	X	8
15	Y	9
16	X	10
17	X	11
18	X	12
19	X	13

M1 A1 — tie: score 11 appears for both X and Y → ranks 4 and 5 → average 4.5

$R_1$ (Food X) $= 4.5 + 7 + 8 + 10 + 11 + 12 + 13 = 65.5$ M1 A1

One-tailed critical value ($n_1 = 7$, $n_2 = 6$, $\alpha = 0.05$): Upper $= 64$ (for one-tailed test) B1

65.5 > 64 ⇒ reject $H_0$ M1

Food X gives significantly more growth than Food Y at the 5% significance level. A1

常见陷阱

• 单尾检验的临界值与双尾不同。双尾用 $\alpha/2$ 的临界值
• 本题是"greater than"的单尾检验，只需看上临界值

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Question Type 4: Sign Test

如何识别

题目要求使用符号检验，只需计数正负号而不考虑大小。关键词："sign test"、"ignore magnitude"

标准解题方法

1. 记录每项符号（+、-、0）
2. 排除 0，$n$ = 非零符号数
3. $S$ = 较少符号的个数
4. 在 $H_0$ 下，$S \sim B(n, 0.5)$
5. 计算 p-value（双尾 $\times 2$）

评分标准（MS 模式）

• 提出假设 B1
• 符号计数 M1
• $S$ 和 $n$ 正确 A1
• 二项概率计算 M1
• p-value 正确 A1
• 比较结论 M1 A1

Example 1 — 9231/s24/qp/41 Q2 (7 marks):

A company claims that its new packaging increases customer satisfaction. 12 customers rate their satisfaction (1-10) before and after:

Customer	1	2	3	4	5	6	7	8	9	10	11	12
Before	6	5	7	4	8	5	6	7	3	6	5	7
After	7	5	8	6	7	6	8	8	5	7	6	7

Use a sign test at the 5% significance level to test whether satisfaction has increased.

📝 展开查看完整解法 + MS

$H_0$: P(\text{After} > \text{Before}) = P(\text{After} < \text{Before}) = 0.5 (no difference) $H_1$: P(\text{After} > \text{Before}) > 0.5 (one-tailed) B1 $\alpha = 0.05$

Customer	Before	After	Sign
1	6	7	+
2	5	5	0
3	7	8	+
4	4	6	+
5	8	7	−
6	5	6	+
7	6	8	+
8	7	8	+
9	3	5	+
10	6	7	+
11	5	6	+
12	7	7	0

M1

$+$ : 9, $-$ : 1, $0$ : 2 $n = 12 - 2 = 10$ (non-zero signs) A1 $S = 1$ (number of $-$ signs, the less frequent)

Under $H_0$, $S \sim B(10, 0.5)$

$P(S \leq 1) = P(S = 0) + P(S = 1)$ $= \binom{10}{0}(0.5)^{10} + \binom{10}{1}(0.5)^{10}$ $= (1 + 10) \times 0.5^{10} = 11 \times 0.0009766 = 0.01074$

M1 A1

One-tailed test: p-value $= 0.01074$ A1

0.01074 < 0.05 ⇒ reject $H_0$ M1

There is sufficient evidence that satisfaction has increased. A1

常见陷阱

• 零差异排除后 $n = 10$（不是 12）
• 单尾检验，p-value 不需 $\times 2$
• 本题用较少符号数（$S = 1$）而非较多符号数

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Example 2 — 9231/s25/qp/41 Q5 variation (8 marks):

A taste test compares two cola brands. 15 tasters each pick their preferred brand:

Prefer A: 11, Prefer B: 4

Use a sign test at the 5% significance level to test whether there is a preference.

📝 展开查看完整解法 + MS

$H_0$: $P(\text{prefer A}) = P(\text{prefer B}) = 0.5$ (no preference) $H_1$: There is a preference (two-tailed) B1 $\alpha = 0.05$

$n = 15$ (all tasters expressed a preference, no ties) $S = 4$ (the smaller count) M1 A1

Under $H_0$, $S \sim B(15, 0.5)$

$P(S \leq 4) = \sum_{r=0}^4 \binom{15}{r}(0.5)^{15}$ $= (1 + 15 + 105 + 455 + 1365) \times 0.5^{15}$ $= 1941 \times 0.00003052 = 0.0592$ M1 A1

Two-tailed: p-value $= 2 \times 0.0592 = 0.1184$ A1

0.1184 > 0.05 ⇒ do not reject $H_0$ M1

No significant preference for either brand at the 5% level. A1

常见陷阱

• 双尾检验 p-value 需要 $\times 2$
• 本题没有零差异（所有 taster 都选了），所以 $n = 15$

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Question Type 5: Normal Approximation for Wilcoxon Tests

如何识别

样本量较大（signed-rank: n > 20，rank-sum: n_1, n_2 > 10），题目要求使用正态近似。关键词："normal approximation"、"large sample"、"use a suitable approximation"

标准解题方法

Signed-rank: $\mu_T = \frac{n(n+1)}{4}, \quad \sigma_T = \sqrt{\frac{n(n+1)(2n+1)}{24}}$ $z = \frac{T \pm 0.5 - \mu_T}{\sigma_T}$

Rank-sum: $\mu_W = \frac{n_1(n_1+n_2+1)}{2}, \quad \sigma_W = \sqrt{\frac{n_1 n_2 (n_1+n_2+1)}{12}}$ $z = \frac{W \pm 0.5 - \mu_W}{\sigma_W}$

评分标准（MS 模式）

• 计算 $\mu$：B1
• 计算 $\sigma$：M1 A1
• 计算 $z$ 含连续性校正：M1 A1
• 比较 $|z|$ 与 $z_{\text{crit}}$：M1
• 结论：A1

Example 1 — 9231/s22/qp/41 Q6 (10 marks):

A sample of 30 households recorded their monthly electricity bills ($). The differences from the national average of$120 were calculated for a Wilcoxon signed-rank test. The sum of positive ranks $T^+ = 312$.

Use a normal approximation at the 5% significance level to test whether the population median differs from $120.

📝 展开查看完整解法 + MS

$H_0$: median $= 120$ $H_1$: median $\neq 120$ (two-tailed) B1 $\alpha = 0.05$, $n = 30$ (assume no zero differences)

$T^+ = 312$, total sum of ranks $= \frac{30 \times 31}{2} = 465$ $T^- = 465 - 312 = 153$ $T = \min(312, 153) = 153$

$\mu_T = \frac{n(n+1)}{4} = \frac{30 \times 31}{4} = 232.5$ B1

$\sigma_T = \sqrt{\frac{n(n+1)(2n+1)}{24}} = \sqrt{\frac{30 \times 31 \times 61}{24}} = \sqrt{\frac{56730}{24}} = \sqrt{2363.75} = 48.62$ M1 A1

$z = \frac{T + 0.5 - \mu_T}{\sigma_T} = \frac{153 + 0.5 - 232.5}{48.62} = \frac{-79}{48.62} = -1.625$ M1 A1

Two-tailed critical value: $z_{0.025} = 1.960$ M1

|z| = 1.625 < 1.960 ⇒ do not reject $H_0$ M1

No evidence that median bill differs from $120 at the 5% level. A1

常见陷阱

• $T = \min(T^+, T^-)$，因为双尾检验
• 连续性校正：$T + 0.5$（因为 T < \mu，下尾校正）
• $\sigma_T$ 公式中有 $2n+1$，不要写成 $2n$ 或 $2n-1$

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Example 2 — 9231/w23/qp/41 Q6 (10 marks):

A Wilcoxon signed-rank test on 25 pairs of data (matched pairs) gave $T^+ = 198$ and $T^- = 127$.

Use a normal approximation at the 5% significance level to test whether there is a difference.

📝 展开查看完整解法 + MS

$H_0$: median difference $= 0$ $H_1$: median difference $\neq 0$ (two-tailed) B1 $\alpha = 0.05$, $n = 25$

$T = \min(198, 127) = 127$

$\mu_T = \frac{n(n+1)}{4} = \frac{25 \times 26}{4} = 162.5$ B1

$\sigma_T = \sqrt{\frac{n(n+1)(2n+1)}{24}} = \sqrt{\frac{25 \times 26 \times 51}{24}} = \sqrt{\frac{33150}{24}} = \sqrt{1381.25} = 37.17$ M1 A1

$z = \frac{T + 0.5 - \mu_T}{\sigma_T} = \frac{127 + 0.5 - 162.5}{37.17} = \frac{-35}{37.17} = -0.942$ M1 A1

$z_{0.025} = 1.960$ M1

|z| = 0.942 < 1.960 ⇒ do not reject $H_0$ M1

No evidence of a significant difference at the 5% level. A1

常见陷阱

• $n = 25$ 是排除零差异后的样本量，如果原数据有零差异需先排除
• 连续性校正符号：T < \mu 时用 $T + 0.5$

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Example 3 — 9231/w25/qp/41 Q4 (9 marks):

A Wilcoxon rank-sum test comparing two groups with $n_1 = 14$ and $n_2 = 16$ gave $R_1 = 289$ as the rank sum for the first group.

Use a normal approximation at the 5% significance level to test whether the two groups differ.

📝 展开查看完整解法 + MS

$H_0$: The two populations are identical $H_1$: The two populations differ (two-tailed) B1 $n_1 = 14$, $n_2 = 16$, $W = R_1 = 289$ $\alpha = 0.05$

$\mu_W = \frac{n_1(n_1 + n_2 + 1)}{2} = \frac{14 \times (14 + 16 + 1)}{2} = \frac{14 \times 31}{2} = 217$ B1

$\sigma_W = \sqrt{\frac{n_1 n_2 (n_1 + n_2 + 1)}{12}} = \sqrt{\frac{14 \times 16 \times 31}{12}} = \sqrt{\frac{6944}{12}} = \sqrt{578.67} = 24.06$ M1 A1

$W = 289$ is greater than $\mu_W = 217$, so upper tail: $z = \frac{W - 0.5 - \mu_W}{\sigma_W} = \frac{289 - 0.5 - 217}{24.06} = \frac{71.5}{24.06} = 2.972$ M1 A1

$z_{0.025} = 1.960$ M1

2.972 > 1.960 ⇒ reject $H_0$ M1

Significant evidence of a difference between the two groups at the 5% level. A1

常见陷阱

• Rank-sum 的正态近似 $\mu$ 和 $\sigma$ 公式与 signed-rank 不同，不要混淆
• 上尾检验用 $W - 0.5$（因为 W > \mu），下尾检验用 $W + 0.5$（因为 W < \mu）