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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.

t at α= 0.01= ±2.797;

请问为什么是2.797 ，我查出来t表是2.368啊

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星星_品职助教 · 2022年12月14日

同学你好，

1）本题自由度为df=25-1=24，处于表的左半部分。

2）根据“whether the mean of bond A is equal to 22%”，可知本题备择假设为μ≠22%，对应的是双尾检验，拒绝域在两侧。由于总的拒绝域面积为significant level=1%，所以单侧拒绝域面积为1%/2=0.005。

根据df=24和单尾面积0.005查得关键值如下：

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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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