开发者:上海品职教育科技有限公司 隐私政策详情

应用版本:4.2.11(IOS)|3.2.5(安卓)APP下载

Churning · 2019年10月09日

问一道题:NO.PZ2015120604000145 [ CFA I ]

问题如下图:

选项:

A.

B.

C.

解释:

老师想问下 当选择用哪一个分布时 方差已知和未知是指总体方差吗

1 个答案
已采纳答案

星星_品职助教 · 2019年10月09日

同学你好,

你理解的没问题,方差的已知未知指得都是总体的方差。因为样本的方差其实在抽出样本的同时就可以算出来了,不存在未知的情况。

这其实是中心极限定理的一个思路,如果我们已知总体的方差的话,那么抽出来的样本的样本均值就服从正态分布(N>30)。但绝大多数情况下,我们是不知道总体的方差的,所以只能用抽样抽出来的样本的方差来代替未知的总体方差,这种情况下,样本均值服从t-分布,所以可以发现大部分的题目其实都是用t-分布来解题的。

加油

  • 1

    回答
  • 0

    关注
  • 287

    浏览
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=n​s​(X−μ0​)​=25​0.15​(0.26−0.22)​=1.33t α= 0.01= ±2.797;Because -2.797 1.333 +2.797, therefore, H0 cannot rejecte 这个是双尾T检验,请问B为什么不对呢?

2024-11-18 11:24 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=n​s​(X−μ0​)​=25​0.15​(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

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=n​s​(X−μ0​)​=25​0.15​(0.26−0.22)​=1.33t α= 0.01= ±2.797;Because -2.797 1.333 +2.797, therefore, H0 cannot rejecte 查t表的时候为什么用24呢(25-1))

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=n​s​(X−μ0​)​=25​0.15​(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=n​s​(X−μ0​)​=25​0.15​(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