问一道题：NO.PZ2015120604000145 [ CFA I ]

问题如下：

Here is a table discribing sample statistics from two bonds' rate of return which are both normally distributed over the past decades. If an investor is considering whether the mean of bond A is equal to 22%,

which of the following conclusion is least appropriate (significant level=1%) ?

选项：

The null hypothesis can be rejected.

It is appropriate to use a two-tailed t-test.

The test statistic value is 1.333.

解释：

A is correct.

The null hypothesis: $H_{0} : μ = 22 %.$

Because the sample size is 25, which is less than 30, so it is appropriate to use the two-tailed t-test.

$t_{24} = \frac{x - μ_{0}}{\frac{s}{\sqrt{n}}} = \frac{0.26 - 0.22}{\frac{0.15}{\sqrt{25}}} = 1.333$

±t₂₄ at α= 0.01= ±2.797;

Because -2.797 <1.333<+2.797, therefore, H₀ cannot be rejected.

为什么不是2.58？题目不是说符合正态分布？

添加评论

1 个答案

星星_品职助教 · 2021年03月14日

同学你好，

根据题干，这道题符合正态分布的是“two bonds rate of return”，这是“总体”服从正态分布的概念。

而假设检验看的是检验统计量的分布，在这道题里就是bond A return的样本均值的分布。并不是bond A return自身（也就是总体）的分布。

由于本题总体方差未知，所以检验统计量服从的是t分布。

如果sample size n＞30，可以选择用z-test来近似替代t-test，但本题bond A也就是sample 1的sample size=25。所以不能用z-test。

t分布的关键值是需要根据自由度（df=25-1=24）和significance level（1%）查表求得的。即本题中的±2.797。

添加评论

1
回答
0
关注
624
浏览

我要回答关注问题

相关问题

NO.PZ2015120604000145 问题如下 Here is a table scribing sample statistifrom two bon' rate of return whiare both normally stributeover the past cas. If investor is consiring whether the meof bonA is equto 22%,whiof the following conclusion is least appropriate (significant level=1%) ? A.The null hypothesis crejecte B.It is appropriate to use a two-tailet-test. C.The test statistic value is 1.333. A is correct.The null hypothesis: H0: μ=22%.Because the sample size is 25, whiis less th30, so it is appropriate to use the two-tailet-test.t=（X−μ0）sn=(0.26−0.22)0.1525=1.33t=\frac{（X-\mu_0）}{\frs{\sqrt n}}={\textstyle\frac{(0.26-0.22)}{\textstyle\frac{0.15}{\sqrt{25}}}}=1.33t=ns（X−μ0）=250.15(0.26−0.22)=1.33t α= 0.01= ±2.797;Because -2.797 1.333 +2.797, therefore, H0 cannot rejecte 请问: 考试这题怎么做？significant leval=1%, t检验是n-1. 这个考试中有表可以查吗？

2024-08-09 07:08 1 · 回答

2024-04-09 12:01 1 · 回答

NO.PZ2015120604000145问题如下Here is a table scribing sample statistifrom two bon' rate of return whiare both normally stributeover the past cas. If investor is consiring whether the meof bonA is equto 22%,whiof the following conclusion is least appropriate (significant level=1%) ?A.The null hypothesis crejecteB.It is appropriate to use a two-tailet-test.C.The test statistic value is 1.333. A is correct.The null hypothesis: H0: μ=22%.Because the sample size is 25, whiis less th30, so it is appropriate to use the two-tailet-test.t=（X−μ0）sn=(0.26−0.22)0.1525=1.33t=\frac{（X-\mu_0）}{\frs{\sqrt n}}={\textstyle\frac{(0.26-0.22)}{\textstyle\frac{0.15}{\sqrt{25}}}}=1.33t=ns（X−μ0）=250.15(0.26−0.22)=1.33t α= 0.01= ±2.797;Because -2.797 1.333 +2.797, therefore, H0 cannot rejecte 总体方差不是15%？

2024-04-03 23:48 1 · 回答

NO.PZ2015120604000145 问题如下 Here is a table scribing sample statistifrom two bon' rate of return whiare both normally stributeover the past cas. If investor is consiring whether the meof bonA is equto 22%,whiof the following conclusion is least appropriate (significant level=1%) ? A.The null hypothesis crejecte B.It is appropriate to use a two-tailet-test. C.The test statistic value is 1.333. A is correct.The null hypothesis: H0: μ=22%.Because the sample size is 25, whiis less th30, so it is appropriate to use the two-tailet-test.t=（X−μ0）sn=(0.26−0.22)0.1525=1.33t=\frac{（X-\mu_0）}{\frs{\sqrt n}}={\textstyle\frac{(0.26-0.22)}{\textstyle\frac{0.15}{\sqrt{25}}}}=1.33t=ns（X−μ0）=250.15(0.26−0.22)=1.33t α= 0.01= ±2.797;Because -2.797 1.333 +2.797, therefore, H0 cannot rejecte 请解答者详细解析下这个题目，如果遇到百分数是不是直接去单位进行计算，这个funb是不是不予理会。我觉得品职出题是很细但也很奇怪。

2023-10-31 16:02 1 · 回答

NO.PZ2015120604000145问题如下 Here is a table scribing sample statistifrom two bon' rate of return whiare both normally stributeover the past cas. If investor is consiring whether the meof bonA is equto 22%,whiof the following conclusion is least appropriate (significant level=1%) ?A.The null hypothesis crejecteB.It is appropriate to use a two-tailet-test.C.The test statistic value is 1.333. A is correct.The null hypothesis: H0: μ=22%.Because the sample size is 25, whiis less th30, so it is appropriate to use the two-tailet-test.t=（X−μ0）sn=(0.26−0.22)0.1525=1.33t=\frac{（X-\mu_0）}{\frs{\sqrt n}}={\textstyle\frac{(0.26-0.22)}{\textstyle\frac{0.15}{\sqrt{25}}}}=1.33t=ns（X−μ0）=250.15(0.26−0.22)=1.33t α= 0.01= ±2.797;Because -2.797 1.333 +2.797, therefore, H0 cannot rejecte 不是说方差已知用z吗，为啥还用t

2023-09-12 15:14 1 · 回答

问一道题：NO.PZ2015120604000145 [ CFA I ]

1 个答案

1

0

624

相关问题