题目
Problem
Chen is treating vines to prevent fungus appearing. One month after the treatment,
Chen monitors the vines to see if fungus is present.
The contingency table shows information about the type of treatment for a sample of
150 vines and whether or not fungus is present.
| Type of treatment | None | Sulphur | Copper sulphate |
|---|
| No fungus present | 20 | 55 | 48 |
| Fungus present | 10 | 8 | 9 |
Test, at the 5% level of significance, whether or not there is any association between the
type of treatment and the presence of fungus.
Show your working clearly, stating your hypotheses, expected frequencies, test statistic
and critical value.
(8)
(Total for Question 1 is 8 marks)
题目中文翻译
Chen 正在对葡萄藤进行处理,以防止真菌出现。处理一个月后,
Chen 观察葡萄藤上是否有真菌。
列联表给出了一个 150 棵葡萄藤样本中处理类型以及是否有真菌的信息。
| 处理类型 | 不处理 | 硫磺 | 硫酸铜 |
|---|
| 无真菌 | 20 | 55 | 48 |
| 有真菌 | 10 | 8 | 9 |
在 5% 显著性水平下,检验处理类型与是否有真菌之间是否存在关联。
请清楚写出你的过程,并写出原假设、期望频数、检验统计量和临界值。
解答
解法一
思路
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这是 2×3 列联表的独立性检验。先求行、列总数,再按“行总数 × 列总数 ÷ 总数”计算全部六格期望频数;随后求卡方统计量、自由度和临界值,最后作情境化结论。
答题过程
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The hypotheses are
H0:H1: The type of treatment and the presence of fungus are independent. The type of treatment and the presence of fungus are associated.
The row totals are 123 and 27, while the column totals are 30, 63 and 57. Under H0,
E=150(row total)(column total).
Therefore, the expected frequencies are
| None | Sulphur | Copper sulphate |
|---|
| No fungus present | 24.6 | 51.66 | 46.74 |
| Fungus present | 5.4 | 11.34 | 10.26 |
For example,
E(no fungus, no treatment)=150(123)(30)=24.6.
The test statistic is
χ2====∑E(O−E)224.6(20−24.6)2+51.66(55−51.66)2+46.74(48−46.74)2+5.4(10−5.4)2+11.34(8−11.34)2+10.26(9−10.26)26.167…6.17(3 s.f.).
The number of degrees of freedom is
(2−1)(3−1)=2.
At the 5% significance level, the critical value is 5.991. Since
6.17>5.991,
H0 is rejected. There is sufficient evidence at the 5% significance level of an association between the type of treatment and the presence of fungus.