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

Sayuri999 · 2022年05月09日

检验mean of bond A是否等于22%,为什么零假设是H0=22%?

NO.PZ2015120604000145

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:

A.

The null hypothesis can be rejected.

B.

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

C.

The test statistic value is 1.333.

解释:

A is correct.

The null hypothesis: H0: 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=Xμ0sn=(0.260.22)0.1525=1.33t=\frac{(X-\mu_0)}{\frac s{\sqrt n}}={\textstyle\frac{(0.26-0.22)}{\textstyle\frac{0.15}{\sqrt{25}}}}=1.33

t at α= 0.01= ±2.797;

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

No.PZ2015120604000145 (选择题)

来源: 品职出题

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%) ?


题目中whether the mean of bond A is equal to 22%,为什么零假设是H0=22%而不是 Ha=22%?视频课里老师说相信的那个为备择假设,题目的意思为什么不是相信A的平均收益率为22%呢


1 个答案
已采纳答案

星星_品职助教 · 2022年05月10日

同学你好,

有一个快捷的判断方法:“等号”永远在原假设,所以“=”。“≥”,和“≤”这些带等号的只会在原假设里出现,不会出现在备择假设里。

备择假设里只会有≠,>,和<

从定性角度出发,当检验均值/方差是否等于一个确定的数的时候,原假设都是=,备择假设都是≠。不会有其他的情况出现,也不需要按照“是否相信”来判断。

  • 1

    回答
  • 0

    关注
  • 449

    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