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

mushroom1008 · 2018年10月27日

问一道题:NO.PZ2015120604000145 [ CFA I ]

问题如下图:

    

选项:

A.

B.

C.

解释:


原假设为什么不是不等于22%呢?不是想要拒绝的放在原假设吗?这个想要拒绝的到底该怎么理解呢?

1 个答案
已采纳答案

菲菲_品职助教 · 2018年10月27日

同学你好,这里还有另外一个关于假设设立的原则是,“=”一般都在原假设。遇到这种比如题目说他们想要检验bondA的均值是否等于22%时,我们只要把等于号放在原假设即可,不用去考虑把想要拒绝的放在原假设这一点,不然也容易让你自己产生混乱。

  • 1

    回答
  • 0

    关注
  • 320

    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