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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：

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.

$t=\frac{（X-\mu_0）}{\frac s{\sqrt n}}={\textstyle\frac{(0.26-0.22)}{\textstyle\frac{0.15}{\sqrt{25}}}}=1.33$

t at α= 0.01= ±2.797;

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

不是说方差已知用z吗，为啥还用t

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星星_品职助教 · 2023年09月14日

同学你好，

总体方差已知用z。本题给出的是样本的标准差，总体标准差/方差未知，所以使用t分布。

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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 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=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%？

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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 请问真正考试时候会给卡方分布表让查询吗？不会看表呀关键是？讲义哪里可以下载卡方分布表。

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