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 K dollers ranges from 22 K dollars to 43 K 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 $σ_{\bar{x}} = 5.36.$

没明白为什么不用再除以根号n？

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1 个答案

星星_品职助教 · 2019年11月18日

同学你好，

要理解置信区间的公式和标准误的概念。以本题为例，区间应该是 X bar±1.96*X bar的标准差。这里“X bar的标准差”（或者描述为样本均值的标准差）就已经是标准误的概念了，所以计算出来也不用再除以根号n了。

相反，如果是题干给了“样本标准差”S，要算标准误，就需要用S/根号n了。要注意区分“样本标准差”和“样本均值标准差”，这个辨析很重要，如果不明白可以追问哈。加油

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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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