题型分析 — Chi-squared Tests

Question Type 1: GoF for Binomial Distribution

如何识别

题目给出频数分布表，要求检验数据是否服从 Binomial 分布。关键词："test whether a Binomial distribution fits"、"Goodness of fit — Binomial"

标准解题方法

1. 估计 Binomial 参数 $p$（如果未知）：$\hat{p} = \dfrac{\sum x f_x}{n \sum f_x}$
2. 计算 $P(X = x) = \binom{n}{x} p^x (1-p)^{n-x}$ 对于 $x = 0, 1, \dots, n$
3. 计算期望频数 $E = \text{total} \times P(X = x)$
4. 合并尾部使 $E \geq 5$，通常 $P(X \geq r)$ 合并为一项
5. $\chi^2 = \sum \dfrac{(O - E)^2}{E}$
6. $\mathrm{df} = k - 1 - 1$（估计了 $p$）
7. 查表比较

评分标准（MS 模式）

• 估计 $\hat{p}$：M1 正确方法，A1 数值正确
• 计算 $P(X = x)$：每项各 M1 A1（通常前 3 项）
• 合并尾部概率：B1
• 计算期望频数：M1 方法，A1 数值（允许多步误差）
• 计算 $\chi^2$：M1 方法，A1 数值
• 自由度：B1
• 临界值和结论：M1 比较，A1 正确结论

Example 1 — 9231/w20/qp/41 Q3 (9 marks):

The number of defective items, $X$, in a random sample of 8 items is recorded for 200 samples. The results are:

$x$	0	1	2	3	4	5
Frequency	22	79	62	28	7	2

Test, at the 5% significance level, whether a Binomial distribution with $n = 8$ is a suitable model.

📝 展开查看完整解法 + MS

Step 1 — Estimate $p$

Total items checked $= 200 \times 8 = 1600$ Total defectives $= 0 \times 22 + 1 \times 79 + 2 \times 62 + 3 \times 28 + 4 \times 7 + 5 \times 2 = 0 + 79 + 124 + 84 + 28 + 10 = 325$

$\hat{p} = \dfrac{325}{1600} = 0.203125$ M1 A1

Step 2 — Binomial probabilities

$P(X = x) = \binom{8}{x} (0.203125)^x (0.796875)^{8-x}$

$P(0) = 0.796875^8 = 0.1536$ M1 $P(1) = 8 \times 0.203125 \times 0.796875^7 = 0.3131$ $P(2) = \binom{8}{2} \times 0.203125^2 \times 0.796875^6 = 0.2793$ $P(3) = \binom{8}{3} \times 0.203125^3 \times 0.796875^5 = 0.1423$ $P(4) = \binom{8}{4} \times 0.203125^4 \times 0.796875^4 = 0.0454$ $P(5) = \binom{8}{5} \times 0.203125^5 \times 0.796875^3 = 0.0093$ A1 (all probabilities correct)

$P(6) = 1 - \text{sum}(P(0)\text{ to }P(5)) = 0.0013$ (or compute directly)

Step 3 — Expected frequencies

$E = 200 \times P(X = x)$

$x$	$O$	$P$	$E$
0	22	0.1536	30.72
1	79	0.3131	62.62
2	62	0.2793	55.86
3	28	0.1423	28.46
4	7	0.0454	9.08
5	2	0.0093	1.86
$\geq 6$	0	0.0013	0.26

B1 — Combine last three rows: $4+5+6+$ have $E = 9.08 + 1.86 + 0.26 = 11.20$, $O = 7 + 2 + 0 = 9$

Step 4 — Chi-squared

$x$	$O$	$E$	$(O-E)^2/E$
0	22	30.72	2.476
1	79	62.62	4.287
2	62	55.86	0.675
3	28	28.46	0.007
$\geq 4$	9	11.20	0.432

$\chi^2 = 2.476 + 4.287 + 0.675 + 0.007 + 0.432 = 7.877$ M1 A1

Step 5 — Degrees of freedom

$k = 5$ categories after merging, $m = 1$ parameter ($p$) estimated $\mathrm{df} = 5 - 1 - 1 = 3$ B1

Step 6 — Conclusion

$\chi^2_{0.05, 3} = 7.815$ M1 Since 7.877 > 7.815, we reject $H_0$.

There is sufficient evidence at the 5% significance level that the Binomial distribution is not a suitable model. A1

常见陷阱

• 合并时注意总和：合并后 $O$ 和 $E$ 都要相加
• 自由度是 $k - 1 - 1$，不是 $k - 1$
• $\hat{p}$ 是用加权平均，不是简单平均

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Example 2 — 9231/w21/qp/41 Q3 (9 marks):

The number of accidents per day, $X$, at a factory over a period of 200 days is recorded:

$x$	0	1	2	3	4	5
Frequency	78	74	34	11	2	1

Test, at the 5% significance level, whether a Binomial distribution with $n = 5$ is suitable.

📝 展开查看完整解法 + MS

Step 1 — Estimate $p$

Total accidents $= 0 \times 78 + 1 \times 74 + 2 \times 34 + 3 \times 11 + 4 \times 2 + 5 \times 1 = 0 + 74 + 68 + 33 + 8 + 5 = 188$ Total trials $= 200 \times 5 = 1000$

$\hat{p} = \dfrac{188}{1000} = 0.188$ M1 A1

Step 2 — Binomial probabilities ($n = 5$, $p = 0.188$)

$P(0) = (0.812)^5 = 0.3527$ $P(1) = 5 \times 0.188 \times (0.812)^4 = 0.4086$ $P(2) = \binom{5}{2} \times 0.188^2 \times 0.812^3 = 0.1893$ $P(3) = \binom{5}{3} \times 0.188^3 \times 0.812^2 = 0.0438$ $P(4) = \binom{5}{4} \times 0.188^4 \times 0.812 = 0.0051$ $P(5) = 0.188^5 = 0.0002$ A1

Step 3 — Expected frequencies ($E = 200 \times P$)

$x$	$O$	$E$
0	78	70.54
1	74	81.72
2	34	37.86
3	11	8.76
4	2	1.02
5	1	0.04

B1 — Combine $x \geq 3$: $E = 8.76 + 1.02 + 0.04 = 9.82$, $O = 11 + 2 + 1 = 14$

Step 4 — Chi-squared

$x$	$O$	$E$	$(O-E)^2/E$
0	78	70.54	0.789
1	74	81.72	0.729
2	34	37.86	0.393
$\geq 3$	14	9.82	1.779

$\chi^2 = 0.789 + 0.729 + 0.393 + 1.779 = 3.690$ M1 A1

Step 5 — df

$k = 4$, $m = 1$ ⇒ $\mathrm{df} = 4 - 1 - 1 = 2$ B1

Step 6 — Conclusion

$\chi^2_{0.05, 2} = 5.991$ M1 3.690 < 5.991 ⇒ do not reject $H_0$.

Insufficient evidence to suggest Binomial is not suitable. A1

常见陷阱

• 合并后确保总频数匹配：4+5 合并后频数是 $2 + 1 = 3$ 和 $1.02 + 0.04 = 1.06$

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Example 3 — 9231/s25/qp/41 Q3 (9 marks):

A biologist records the number of seeds germinating, $X$, in each of 100 trays containing 6 seeds each:

$x$	0	1	2	3	4	5	6
Frequency	2	8	22	32	24	9	3

Test, at the 5% significance level, whether a Binomial distribution with $n = 6$ is a suitable model.

📝 展开查看完整解法 + MS

Step 1 — Estimate $p$

Total seeds $= 100 \times 6 = 600$ Total germinated $= 0 \times 2 + 1 \times 8 + 2 \times 22 + 3 \times 32 + 4 \times 24 + 5 \times 9 + 6 \times 3$ $= 0 + 8 + 44 + 96 + 96 + 45 + 18 = 307$

$\hat{p} = \dfrac{307}{600} = 0.5117$ M1 A1

Step 2 — Binomial probabilities

$P(X = x) = \binom{6}{x} (0.5117)^x (0.4883)^{6-x}$

$P(0) = 0.4883^6 = 0.0136$ $P(1) = 6 \times 0.5117 \times 0.4883^5 = 0.0853$ $P(2) = 15 \times 0.5117^2 \times 0.4883^4 = 0.2228$ $P(3) = 20 \times 0.5117^3 \times 0.4883^3 = 0.3107$ $P(4) = 15 \times 0.5117^4 \times 0.4883^2 = 0.2437$ $P(5) = 6 \times 0.5117^5 \times 0.4883 = 0.1020$ $P(6) = 0.5117^6 = 0.0219$ A1

Step 3 — Expected frequencies ($E = 100 \times P$)

$x$	$O$	$E$
0	2	1.36
1	8	8.53
2	22	22.28
3	32	31.07
4	24	24.37
5	9	10.20
6	3	2.19

B1 — Combine $x = 0$ and $x = 1$: $E = 1.36 + 8.53 = 9.89$, $O = 2 + 8 = 10$ Combine $x = 6$ with $x = 5$: $E = 10.20 + 2.19 = 12.39$, $O = 9 + 3 = 12$

Step 4 — Chi-squared

$x$	$O$	$E$	$(O-E)^2/E$
0–1	10	9.89	0.0012
2	22	22.28	0.0035
3	32	31.07	0.0278
4	24	24.37	0.0056
5–6	12	12.39	0.0123

$\chi^2 = 0.0012 + 0.0035 + 0.0278 + 0.0056 + 0.0123 = 0.0504$ M1 A1

Step 5 — df

$k = 5$, $m = 1$ ⇒ $\mathrm{df} = 5 - 1 - 1 = 3$ B1

Step 6 — Conclusion

$\chi^2_{0.05, 3} = 7.815$ M1 0.0504 < 7.815 ⇒ do not reject $H_0$.

Insufficient evidence to reject Binomial as suitable model. A1

常见陷阱

• E < 5 可能出现在两端，需分别合并
• 对比本题（$p \approx 0.5$，对称）与 Example 1（$p$ 较小，右偏）观察合并方式的不同

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Question Type 2: GoF for Poisson Distribution

如何识别

关键词："Poisson distribution"、"test for randomness"、"Goodness of fit — Poisson"。数据通常是计数数据（事故数、缺陷数等）。

标准解题方法

1. 估计 $\hat{\lambda} = \bar{x}$
2. 计算 $P(X = x) = e^{-\lambda} \lambda^x / x!$
3. 计算期望频数 $E = \text{total} \times P(X = x)$
4. 合并尾部使 $E \geq 5$
5. $\chi^2 = \sum (O-E)^2/E$
6. $\mathrm{df} = k - 1 - 1$（估计了 $\lambda$）
7. 查表比较

评分标准（MS 模式）

• 估计 $\overline{x}$：M1 A1
• 计算 Poisson 概率：每项 M1 A1
• 尾部合并：B1
• $\chi^2$：M1 A1
• 自由度 B1，结论 M1 A1

Example 1 — 9231/s22/qp/41 Q4 (10 marks):

The number of particles emitted per minute by a radioactive source is recorded for 200 minutes:

$x$	0	1	2	3	4	5	6	7
Frequency	10	28	49	52	35	17	7	2

Test, at the 5% significance level, whether a Poisson distribution is a suitable model.

📝 展开查看完整解法 + MS

Step 1 — Estimate $\lambda$

$\bar{x} = \dfrac{0 \times 10 + 1 \times 28 + 2 \times 49 + 3 \times 52 + 4 \times 35 + 5 \times 17 + 6 \times 7 + 7 \times 2}{200}$

$= \dfrac{0 + 28 + 98 + 156 + 140 + 85 + 42 + 14}{200} = \dfrac{563}{200} = 2.815$ M1 A1

Step 2 — Poisson probabilities

$P(X = x) = e^{-2.815} \times 2.815^x / x!$

$P(0) = e^{-2.815} = 0.0599$ $P(1) = 0.0599 \times 2.815 = 0.1686$ $P(2) = 0.1686 \times 2.815/2 = 0.2373$ $P(3) = 0.2373 \times 2.815/3 = 0.2227$ $P(4) = 0.2227 \times 2.815/4 = 0.1567$ $P(5) = 0.1567 \times 2.815/5 = 0.0882$ $P(6) = 0.0882 \times 2.815/6 = 0.0414$ $P(7) = 0.0414 \times 2.815/7 = 0.0167$ A1 $P(\geq 8) = 1 - \text{sum} = 0.0085$

Step 3 — Expected frequencies ($E = 200 \times P$)

$x$	$O$	$E$
0	10	11.98
1	28	33.72
2	49	47.46
3	52	44.54
4	35	31.34
5	17	17.64
6	7	8.28
7	2	3.34
$\geq 8$	0	1.70

B1 — Combine $7+$: $O = 2 + 0 = 2$, $E = 3.34 + 1.70 = 5.04$

Step 4 — Chi-squared

$x$	$O$	$E$	$(O-E)^2/E$
0	10	11.98	0.327
1	28	33.72	0.970
2	49	47.46	0.050
3	52	44.54	1.249
4	35	31.34	0.427
5	17	17.64	0.023
6	7	8.28	0.198
$\geq 7$	2	5.04	1.834

$\chi^2 = 0.327 + 0.970 + 0.050 + 1.249 + 0.427 + 0.023 + 0.198 + 1.834 = 5.078$ M1 A1

Step 5 — df

$k = 8$, $m = 1$ ⇒ $\mathrm{df} = 8 - 1 - 1 = 6$ B1

Step 6 — Conclusion

$\chi^2_{0.05, 6} = 12.592$ M1 5.078 < 12.592 ⇒ do not reject $H_0$.

Poisson distribution is a suitable model. A1

常见陷阱

• Poisson 概率用递推公式 $P(X = x) = P(X = x-1) \times \lambda / x$ 可减少计算量
• 合并尾部时注意是否还有 E < 5

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Example 2 — 9231/w24/qp/41 Q3 (9 marks):

The number of goals scored per match in a tournament is recorded for 150 matches:

$x$	0	1	2	3	4	5
Frequency	45	52	30	16	5	2

Test, at the 5% significance level, whether a Poisson distribution is a suitable model.

📝 展开查看完整解法 + MS

Step 1 — Estimate $\lambda$

$\bar{x} = \dfrac{0 \times 45 + 1 \times 52 + 2 \times 30 + 3 \times 16 + 4 \times 5 + 5 \times 2}{150}$ $= \dfrac{0 + 52 + 60 + 48 + 20 + 10}{150} = \dfrac{190}{150} = 1.267$ M1 A1

Step 2 — Poisson probabilities ($\lambda = 1.267$)

$P(0) = e^{-1.267} = 0.2816$ $P(1) = 0.2816 \times 1.267 = 0.3568$ $P(2) = 0.3568 \times 1.267/2 = 0.2260$ $P(3) = 0.2260 \times 1.267/3 = 0.0955$ $P(4) = 0.0955 \times 1.267/4 = 0.0302$ $P(5) = 0.0302 \times 1.267/5 = 0.0077$ A1 $P(\geq 6) = 1 - \text{sum} = 0.0022$

Step 3 — Expected frequencies ($E = 150 \times P$)

$x$	$O$	$E$
0	45	42.24
1	52	53.52
2	30	33.90
3	16	14.33
4	5	4.53
$\geq 5$	2	1.48

B1 — Combine $4+$: $O = 5 + 2 = 7$, $E = 4.53 + 1.48 = 6.01$

Step 4 — Chi-squared

$x$	$O$	$E$	$(O-E)^2/E$
0	45	42.24	0.180
1	52	53.52	0.043
2	30	33.90	0.448
3	16	14.33	0.195
$\geq 4$	7	6.01	0.163

$\chi^2 = 0.180 + 0.043 + 0.448 + 0.195 + 0.163 = 1.029$ M1 A1

Step 5 — df

$k = 5$, $m = 1$ ⇒ $\mathrm{df} = 5 - 1 - 1 = 3$ B1

Step 6 — Conclusion

$\chi^2_{0.05, 3} = 7.815$ M1 1.029 < 7.815 ⇒ do not reject $H_0$. A1

常见陷阱

• 当 $\lambda$ 较小（如 1.267），Poisson 分布偏斜，尾部合并更关键

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Example 3 — 9231/w25/qp/41 Q2 (9 marks):

The number of bacteria colonies per square on a slide is recorded for 120 squares:

$x$	0	1	2	3	4	5	6
Frequency	18	32	31	22	11	4	2

Test, at the 5% significance level, whether a Poisson distribution is a suitable model.

📝 展开查看完整解法 + MS

Step 1 — Estimate $\lambda$

$\bar{x} = \dfrac{0 \times 18 + 1 \times 32 + 2 \times 31 + 3 \times 22 + 4 \times 11 + 5 \times 4 + 6 \times 2}{120}$ $= \dfrac{0 + 32 + 62 + 66 + 44 + 20 + 12}{120} = \dfrac{236}{120} = 1.967$ M1 A1

Step 2 — Poisson probabilities ($\lambda = 1.967$)

$P(0) = e^{-1.967} = 0.1398$ $P(1) = 0.1398 \times 1.967 = 0.2750$ $P(2) = 0.2750 \times 1.967/2 = 0.2704$ $P(3) = 0.2704 \times 1.967/3 = 0.1773$ $P(4) = 0.1773 \times 1.967/4 = 0.0872$ $P(5) = 0.0872 \times 1.967/5 = 0.0343$ $P(6) = 0.0343 \times 1.967/6 = 0.0112$ A1 $P(\geq 7) = 1 - \text{sum} = 0.0048$

Step 3 — Expected frequencies ($E = 120 \times P$)

$x$	$O$	$E$
0	18	16.78
1	32	33.00
2	31	32.45
3	22	21.28
4	11	10.46
5	4	4.12
$\geq 6$	2	1.91

B1 — Combine $5+$: $O = 4 + 2 = 6$, $E = 4.12 + 1.91 = 6.03$

Step 4 — Chi-squared

$x$	$O$	$E$	$(O-E)^2/E$
0	18	16.78	0.089
1	32	33.00	0.030
2	31	32.45	0.065
3	22	21.28	0.024
4	11	10.46	0.028
$\geq 5$	6	6.03	0.000

$\chi^2 = 0.089 + 0.030 + 0.065 + 0.024 + 0.028 + 0.000 = 0.236$ M1 A1

Step 5 — df

$k = 6$, $m = 1$ ⇒ $\mathrm{df} = 6 - 1 - 1 = 4$ B1

Step 6 — Conclusion

$\chi^2_{0.05, 4} = 9.488$ M1 0.236 < 9.488 ⇒ do not reject $H_0$. A1

常见陷阱

• 当 $\chi^2$ 非常小时（如 0.236），模型拟合非常好，但仍需按步骤完整作答

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Question Type 3: Test for Independence (Contingency Tables)

如何识别

题目给出行列分类的频数表，要求检验两个变量是否独立。关键词："test for independence"、"test for association"、"contingency table"

标准解题方法

1. 计算行和 $R_i$、列和 $C_j$、总和 $T$
2. 每格期望频数 $E_{ij} = \dfrac{R_i \times C_j}{T}$
3. 检查 $E_{ij} \geq 5$，否则合并行或列
4. $\chi^2 = \sum \dfrac{(O_{ij} - E_{ij})^2}{E_{ij}}$
5. $\mathrm{df} = (r - 1)(c - 1)$
6. 查表比较

评分标准（MS 模式）

• 行和、列和：B1
• $E_{ij}$ 计算：M1 方法，A1 全部正确
• 合并（若需要）：B1
• $\chi^2$ 计算：M1 A1
• 自由度 B1，结论 M1 A1

Example 1 — 9231/s20/qp/41 Q1 (7 marks):

A survey records favourite ice cream flavour by age group:

	Vanilla	Chocolate	Strawberry	Total
Children	25	30	15	70
Adults	35	20	25	80
Total	60	50	40	150

Test at the 5% significance level whether flavour preference is independent of age group.

📝 展开查看完整解法 + MS

Step 1 — Expected frequencies

$E_{11} = \dfrac{70 \times 60}{150} = 28.0$ $E_{12} = \dfrac{70 \times 50}{150} = 23.33$ $E_{13} = \dfrac{70 \times 40}{150} = 18.67$ $E_{21} = \dfrac{80 \times 60}{150} = 32.0$ $E_{22} = \dfrac{80 \times 50}{150} = 26.67$ $E_{23} = \dfrac{80 \times 40}{150} = 21.33$ M1 A1

All $E \geq 5$ ✓

Step 2 — Chi-squared

Cell	$O$	$E$	$(O-E)^2/E$
C–V	25	28.0	0.321
C–Ch	30	23.33	1.907
C–S	15	18.67	0.721
A–V	35	32.0	0.281
A–Ch	20	26.67	1.668
A–S	25	21.33	0.631

$\chi^2 = 0.321 + 1.907 + 0.721 + 0.281 + 1.668 + 0.631 = 5.529$ M1 A1

Step 3 — df

$r = 2$, $c = 3$ ⇒ $\mathrm{df} = (2-1)(3-1) = 2$ B1

Step 4 — Conclusion

$\chi^2_{0.05, 2} = 5.991$ M1 5.529 < 5.991 ⇒ do not reject $H_0$.

No evidence of association between age and flavour preference. A1

常见陷阱

• $E$ 用分数或小数，不要四舍五入到整数
• $(r-1)(c-1)$ 不是 $r \times c$

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Example 2 — 9231/s21/qp/41 Q2 (8 marks):

Students are classified by grade (A, B, C) and study method (Self-study, Tutored, Mixed):

	A	B	C	Total
Self-study	12	20	8	40
Tutored	18	15	17	50
Mixed	15	25	10	50
Total	45	60	35	140

Test at the 5% significance level whether grade is independent of study method.

📝 展开查看完整解法 + MS

Step 1 — Expected frequencies

$E_{11} = \dfrac{40 \times 45}{140} = 12.86$ $E_{12} = \dfrac{40 \times 60}{140} = 17.14$ $E_{13} = \dfrac{40 \times 35}{140} = 10.00$ $E_{21} = \dfrac{50 \times 45}{140} = 16.07$ $E_{22} = \dfrac{50 \times 60}{140} = 21.43$ $E_{23} = \dfrac{50 \times 35}{140} = 12.50$ $E_{31} = \dfrac{50 \times 45}{140} = 16.07$ $E_{32} = \dfrac{50 \times 60}{140} = 21.43$ $E_{33} = \dfrac{50 \times 35}{140} = 12.50$ M1 A1

All $E \geq 5$ ✓

Step 2 — Chi-squared

$\chi^2 = \dfrac{(12-12.86)^2}{12.86} + \dfrac{(20-17.14)^2}{17.14} + \dfrac{(8-10)^2}{10} + \dfrac{(18-16.07)^2}{16.07} + \dfrac{(15-21.43)^2}{21.43} + \dfrac{(17-12.50)^2}{12.50} + \dfrac{(15-16.07)^2}{16.07} + \dfrac{(25-21.43)^2}{21.43} + \dfrac{(10-12.50)^2}{12.50}$

$= 0.057 + 0.477 + 0.400 + 0.232 + 1.929 + 1.620 + 0.071 + 0.595 + 0.500 = 5.881$

$\chi^2 = 5.881$ M1 A1

Step 3 — df

$r = 3$, $c = 3$ ⇒ $\mathrm{df} = (3-1)(3-1) = 4$ B1

Step 4 — Conclusion

$\chi^2_{0.05, 4} = 9.488$ M1 5.881 < 9.488 ⇒ do not reject $H_0$. A1

常见陷阱

• $3 \times 3$ 表自由度是 4，不是 9
• 每个 $E$ 都要分别计算，不要以为对称就一定相等

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Example 3 — 9231/w23/qp/41 Q2 (8 marks):

A survey records preferred mobile OS by gender:

	iOS	Android	Other	Total
Male	28	35	7	70
Female	32	25	13	70
Total	60	60	20	140

Test at the 5% significance level whether OS preference is independent of gender.

📝 展开查看完整解法 + MS

Step 1 — Expected frequencies

$E_{11} = \dfrac{70 \times 60}{140} = 30.0$ $E_{12} = \dfrac{70 \times 60}{140} = 30.0$ $E_{13} = \dfrac{70 \times 20}{140} = 10.0$ $E_{21} = \dfrac{70 \times 60}{140} = 30.0$ $E_{22} = \dfrac{70 \times 60}{140} = 30.0$ $E_{23} = \dfrac{70 \times 20}{140} = 10.0$ M1 A1

All $E \geq 5$ ✓

Step 2 — Chi-squared

Cell	$O$	$E$	$(O-E)^2/E$
M–iOS	28	30.0	0.133
M–And	35	30.0	0.833
M–Oth	7	10.0	0.900
F–iOS	32	30.0	0.133
F–And	25	30.0	0.833
F–Oth	13	10.0	0.900

$\chi^2 = 0.133 + 0.833 + 0.900 + 0.133 + 0.833 + 0.900 = 3.732$ M1 A1

Step 3 — df

$r = 2$, $c = 3$ ⇒ $\mathrm{df} = (2-1)(3-1) = 2$ B1

Step 4 — Conclusion

$\chi^2_{0.05, 2} = 5.991$ M1 3.732 < 5.991 ⇒ do not reject $H_0$. A1

常见陷阱

• 本题行列对称（行等和、列等和），可直接用公式：$E = 70 \times 60 / 140 = 30$ 等，注意 Others 列的期望是 10

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Example 4 — 9231/s24/qp/41 Q5 (9 marks):

Data on smoking habits and exercise frequency:

	Never smoke	Occasional	Regular	Total
Exercises	45	20	10	75
Does not exercise	25	25	25	75
Total	70	45	35	150

Test at the 5% significance level whether smoking habit is independent of exercise.

📝 展开查看完整解法 + MS

Step 1 — Expected frequencies

$E_{11} = \dfrac{75 \times 70}{150} = 35.0$ $E_{12} = \dfrac{75 \times 45}{150} = 22.5$ $E_{13} = \dfrac{75 \times 35}{150} = 17.5$ $E_{21} = \dfrac{75 \times 70}{150} = 35.0$ $E_{22} = \dfrac{75 \times 45}{150} = 22.5$ $E_{23} = \dfrac{75 \times 35}{150} = 17.5$ M1 A1

All $E \geq 5$ ✓

Step 2 — Chi-squared

Cell	$O$	$E$	$(O-E)^2/E$
E–N	45	35.0	2.857
E–O	20	22.5	0.278
E–R	10	17.5	3.214
NE–N	25	35.0	2.857
NE–O	25	22.5	0.278
NE–R	25	17.5	3.214

$\chi^2 = 2.857 + 0.278 + 3.214 + 2.857 + 0.278 + 3.214 = 12.698$ M1 A1

Step 3 — df

$r = 2$, $c = 3$ ⇒ $\mathrm{df} = (2-1)(3-1) = 2$ B1

Step 4 — Conclusion

$\chi^2_{0.05, 2} = 5.991$ M1 12.698 > 5.991 ⇒ reject $H_0$.

Evidence of association between smoking and exercise. A1

常见陷阱

• 本题 $\chi^2$ 很大（12.698），明显拒绝 $H_0$，仍需写出完整比较过程

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Question Type 4: GoF for a Given Probability Function

如何识别

题目给出一个概率质量函数（PMF）或概率密度函数（PDF），要求检验数据是否服从该分布。

标准解题方法

1. 使用给定分布公式计算 $P(X = x)$
2. 计算期望频数 $E = n \times P(X = x)$
3. 合并 E < 5 的类别
4. $\chi^2 = \sum (O-E)^2/E$
5. 自由度注意：若分布参数固定（无需估计），则 $m = 0$

评分标准（MS 模式）

• 使用给定分布计算概率：M1 A1
• 期望频数：M1 A1
• 合并：B1
• $\chi^2$：M1 A1
• 自由度：B1
• 结论：M1 A1

Example 1 — 9231/s23/qp/41 Q3 (10 marks):

A discrete random variable $X$ has probability function $P(X = x) = \dfrac{k}{x + 1}$ for $x = 0, 1, 2, 3, 4$. The following observed frequencies are obtained from 200 trials:

$x$	0	1	2	3	4
Frequency	68	56	38	24	14

Test at the 5% significance level whether the data fit the given distribution.

📝 展开查看完整解法 + MS

Step 1 — Find $k$

$\sum_{x=0}^4 \dfrac{k}{x+1} = k\left(1 + \dfrac12 + \dfrac13 + \dfrac14 + \dfrac15\right) = k\left(\dfrac{60 + 30 + 20 + 15 + 12}{60}\right) = k \times \dfrac{137}{60} = 1$

$k = \dfrac{60}{137}$ M1 A1

Step 2 — Theoretical probabilities

$P(0) = \dfrac{60}{137} \times 1 = 0.4380$ $P(1) = \dfrac{60}{137} \times \dfrac12 = 0.2190$ $P(2) = \dfrac{60}{137} \times \dfrac13 = 0.1460$ $P(3) = \dfrac{60}{137} \times \dfrac14 = 0.1095$ $P(4) = \dfrac{60}{137} \times \dfrac15 = 0.0876$ A1

Step 3 — Expected frequencies ($E = 200 \times P$)

$x$	$O$	$E$
0	68	87.60
1	56	43.80
2	38	29.20
3	24	21.90
4	14	17.52

All $E \geq 5$ ✓ B1

Step 4 — Chi-squared

$\chi^2 = \dfrac{(68-87.60)^2}{87.60} + \dfrac{(56-43.80)^2}{43.80} + \dfrac{(38-29.20)^2}{29.20} + \dfrac{(24-21.90)^2}{21.90} + \dfrac{(14-17.52)^2}{17.52}$

$= 4.388 + 3.397 + 2.652 + 0.201 + 0.707 = 11.345$ M1 A1

Step 5 — df

$k = 5$ categories, $m = 0$ (no parameters estimated — $k$ was derived from the distribution, not estimated from data) $\mathrm{df} = 5 - 1 - 0 = 4$ B1

Step 6 — Conclusion

$\chi^2_{0.05, 4} = 9.488$ M1 11.345 > 9.488 ⇒ reject $H_0$.

Data does not fit the given distribution. A1

常见陷阱

• 注意 $m = 0$ 还是 $m = 1$：本题中分布参数 $k$ 由概率总和为 1 确定，不是从数据估计的，所以 $m = 0$
• 如果分布有未知参数（如 Binomial 的 $p$）且从数据估计，则 $m = 1$

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Example 2 — 9231/s20/qp/41 Q1 variation (6 marks):

Given $P(X = x) = \dfrac{1}{5}$ for $x = 1, 2, 3, 4, 5$ (uniform distribution), observed frequencies from 100 trials:

$x$	1	2	3	4	5
Frequency	15	25	22	18	20

Test at the 5% level.

📝 展开查看完整解法 + MS

Step 1 — Expected frequencies

$P(X = x) = 0.2$ for all $x$ $E = 100 \times 0.2 = 20$ for all $x$ B1

Step 2 — Chi-squared

$\chi^2 = \dfrac{(15-20)^2}{20} + \dfrac{(25-20)^2}{20} + \dfrac{(22-20)^2}{20} + \dfrac{(18-20)^2}{20} + \dfrac{(20-20)^2}{20}$

$= 1.25 + 1.25 + 0.20 + 0.20 + 0.00 = 2.90$ M1 A1

Step 3 — df

$k = 5$, $m = 0$ ⇒ $\mathrm{df} = 5 - 1 = 4$ B1

Step 4 — Conclusion

$\chi^2_{0.05, 4} = 9.488$ M1 2.90 < 9.488 ⇒ do not reject $H_0$. A1

常见陷阱

• 均匀分布没有估计参数，$m = 0$