问一道题：NO.PZ2015120604000125

问题如下：

The population is 6000 programmer which is supposed to be normally distributed. A sample with 100 size is drawn from the population. Based on z-statistic, 95% confidence interval of sample mean of annual salary is 32.5 (in thousands) dollars ranges from 22 (in thousands) dollars to 43 (in thousands) dollars .Calculate the standard error of mean annual salary:

选项：

1.96.

3.99.

5.36.

解释：

C is correct.

At the 95% level of significance, the critical value is $z_{α /2} = 1.96$ .

So the confidence interval is $32.5 \pm 1.96 σ_{\bar{x}}$ ,

From the equation t 32.5 + 1.96 σ $_{\bar{x}}$ = 43 or 32.5 - 1.96σ $_{\bar{x}}$ = 22, we get $σ_{\bar{x}} = 5.36.$

所以standard error就是样本里面的standard deviation？可以这样理解吗？然后求standard error of the mean是算这个范围？

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星星_品职助教 · 2020年11月26日

同学你好，

standard error本质上就是一个标准差，但是他是样本统计量（例如X bar就是population对应的的样本统计量）的标准差。

所以有些题目中给的是sample standard deviation，这个sample standard deviation是不能直接用的，因为这是sample（样本）的标准差，不是样本统计量的标准差（standard error）。

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the standard error of mean 就是样本均值标准差的意思，在这道题里是通过置信区间的公式解出来的。置信区间是你说的那个“范围”

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NO.PZ2015120604000125问题如下 The population is 6000 programmer whiis supposeto normally stribute A sample with 100 size is awn from the population. Baseon z-statisti95% confinintervof sample meof annusalary is 32.5 (in thousan) llars ranges from 22 (in thousan) llars to 43 (in thousan) llars .Calculate the stanrerror of meannusalary: A.1.96.B.3.99.C.5.36. C is correct.the 95% level of significance, the criticvalue is ±1.96.So the confinintervis 32.5±1.96 σ x ¯ , From the equation t 32.5 + 1.96 stanrerror = 43 or 32.5 - 1.96 stanrerror x ¯ = 22, we get σ x ¯ =5.36.=5.3571. 根据题干的sample mean=32.5，基于z-statistics的95%的置信区间（得到关键值±1.96），和置信区间的公式，可以得到 32.5 + 1.96 stanrerror = 43 或 32.5 - 1.96 stanrerror = 22。得到 stanrerror=5.3571 我可能公式有点搞混了，请问置信区间的计算什么时候使用平均值+z×标准差，什么时候使用平均值+z×标准误

2023-08-09 08:29 1 · 回答

NO.PZ2015120604000125 问题如下 The population is 6000 programmer whiis supposeto normally stribute A sample with 100 size is awn from the population. Baseon z-statisti95% confinintervof sample meof annusalary is 32.5 (in thousan) llars ranges from 22 (in thousan) llars to 43 (in thousan) llars .Calculate the stanrerror of meannusalary: A.1.96. B.3.99. C.5.36. C is correct.the 95% level of significance, the criticvalue is ±1.96.So the confinintervis 32.5±1.96 σ x ¯ , From the equation t 32.5 + 1.96 stanrerror = 43 or 32.5 - 1.96 stanrerror x ¯ = 22, we get σ x ¯ =5.36.=5.3571. 根据题干的sample mean=32.5，基于z-statistics的95%的置信区间（得到关键值±1.96），和置信区间的公式，可以得到 32.5 + 1.96 stanrerror = 43 或 32.5 - 1.96 stanrerror = 22。得到 stanrerror=5.3571 没看懂这道题啥意思，老师可以再讲一下吗

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问一道题：NO.PZ2015120604000125

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