在查表的时候n要-1？

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.

查t表的时候为什么用24呢（25-1））

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1 个答案

品职助教_七七 · 2024年04月10日

嗨，从没放弃的小努力你好：

本题的检验为检验均值是否为特定的一个数字（whether the mean of bond A is equal to 22%），且总体方差未知。这种情况对应的检验是t检验，自由度为n-1，如下：

或者可以看教材表格，更清晰一些

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就算太阳没有迎着我们而来，我们正在朝着它而去，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？

2024-08-09 07:08 1 · 回答

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在查表的时候n要-1？

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