问一道题：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.

$t_{24} = \frac{x - μ_{0}}{\frac{s}{\sqrt{n}}} = \frac{0.26 - 0.22}{\frac{0.15}{\sqrt{25}}} = 1.333$

±t₂₄ at α= 0.01= ±2.797;

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

a不是说方差已知用Z分布，方差未知用T分布吗？

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星星_品职助教 · 2020年01月05日

同学你好，

（总体）方差已知用Z，（总体）方差未知用t哈。这个需要理解的，不要硬背。之所以是总体的方差，根源在于中心极限定理。X bar服从的正态分布的方差为总体方差/n。

这道题题干给的是样本的方差，所以总体方差是未知的。样本方差在这里的用处是算标准误的。加油

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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 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？

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2024-04-09 12:01 1 · 回答

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问一道题：NO.PZ2015120604000145

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