题目

Problem

A doctor believes that the diet of her patients and their health are not independent.

She takes a random sample of 200 patients and records whether they are in good health or poor health and whether they have a good diet or a poor diet. The results are summarised in the table below.

	Good health	Poor health
Good diet	86	8
Poor diet	91	15

Stating your hypotheses clearly, test the doctor’s belief using a 5% level of significance. Show your working for your test statistic and state your critical value clearly.

(9)

(Total for Question 2 is 9 marks)

题目中文翻译

一位医生认为，她的病人的饮食与健康状况并不是相互独立的。

她随机抽取 200 名病人，记录他们是健康状况良好还是较差，以及他们饮食是良好还是较差。结果汇总如下表。

	健康状况良好	健康状况较差
饮食良好	86	8
饮食较差	91	15

清楚写出假设，在 5% 显著性水平下检验医生的看法。写出你的检验统计量计算过程，并清楚写出临界值。

(9)

解答

解法一

思路

展开

这是一个 $2\times2$ 列联表的卡方独立性检验。先由行总数、列总数与总人数计算四个期望频数，再求卡方统计量。自由度为 $(2-1)(2-1)=1$ ，最后用 $5\%$ 临界值作出与医生观点相关的结论。

答题过程

展开

The hypotheses are

\begin{align*} H_0:&\ \text{diet and health are independent},\\ H_1:&\ \text{diet and health are not independent}. \end{align*}

The row and column totals are

	Good health	Poor health	Total
Good diet	86	8	94
Poor diet	91	15	106
Total	177	23	200

Under $H_0$ , each expected frequency is calculated using

E=\frac{\text{row total}\times\text{column total}}{200}.

Therefore,

Diet	Health	Expected frequency, $E$
Good	Good	$94(177)/200=83.19$
Good	Poor	$94(23)/200=10.81$
Poor	Good	$106(177)/200=93.81$
Poor	Poor	$106(23)/200=12.19$

The test statistic is

\begin{align*} \chi^2 =&\,\sum\frac{(O-E)^2}{E}\\ =&\,\frac{(86-83.19)^2}{83.19}\\ &\,\hspace{2pt}+\frac{(8-10.81)^2}{10.81}\\ &\,\hspace{4pt}+\frac{(91-93.81)^2}{93.81}\\ &\,\hspace{6pt}+\frac{(15-12.19)^2}{12.19}\\ =&\,1.557\ldots\\ \approx&\,1.56. \end{align*}

The number of degrees of freedom is

\nu=(2-1)(2-1)=1.

At the $5\%$ significance level, the critical value is

\chi_1^2(0.05)=3.841.

Since

1.557\ldots<3.841,

the test statistic does not lie in the critical region, so $H_0$ is not rejected. There is insufficient evidence at the $5\%$ significance level of an association between diet and health. Therefore, these data do not support the doctor’s belief.

解法二

思路

展开

用等价公式 $\chi^2=\sum\frac{O^2}{E}-200$ ，沿用解法一的四个期望频数，先求观测频数平方与期望频数之比的总和，再减去样本总人数。

答题过程

展开

We test

\begin{align*} H_0:&\ \text{Diet and health are independent,}\\ H_1:&\ \text{Diet and health are not independent.} \end{align*}

Using the expected frequencies $83.19$ , $10.81$ , $93.81$ and $12.19$ from Method 1,

\begin{align*} \chi^2 =&\,\sum\frac{O^2}{E}-200\\ =&\,\frac{86^2}{83.19}+\frac{8^2}{10.81}\\ &\,\hspace{2pt}+\frac{91^2}{93.81}+\frac{15^2}{12.19}-200\\ =&\,201.557283\ldots-200\\ =&\,1.557283\ldots\\ \approx&\,1.56. \end{align*}

There are $(2-1)(2-1)=1$ degrees of freedom. The $5\%$ critical value is $3.841$ . Since $1.557283\ldots<3.841$ , we do not reject $H_0$ . There is insufficient evidence of an association between diet and health, so the data do not support the doctor’s belief.

S3 2021 June Q2

题目

解答

解法一

思路

答题过程

解法二

思路

答题过程