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

安在 · 2022年10月13日

怎么知道是T的双尾分布呢

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.

(1)根据 被择假设 判断双尾,根据样本量<30 且不知道总体方差,所以用T ----------对吗?

(2)1% 的T分位点又没有给出表 怎么知道具体等于2.797呢?

(3)表格中 sample B 的信息是不是完全没有使用在本题?

1 个答案

星星_品职助教 · 2022年10月14日

同学你好,

(1)根据 被择假设 判断双尾,根据样本量<30 且不知道总体方差,所以用T ,对吗?---用t的原因是总体方差未知。在正态总体的前提下,小样本应该选择用t分布。

(2)1% 的T分位点又没有给出表 怎么知道具体等于2.797呢?---可查原版书对应附表,该表可在学员资料下载处下载。

(3)表格中 sample B 的信息是不是完全没有使用在本题?---对

  • 1

    回答
  • 0

    关注
  • 282

    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 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

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 不是说方差已知用z吗,为啥还用t

2023-09-12 15:14 1 · 回答