为什么是双尾？

NO.PZ2015120604000117

问题如下：

A random sample is 100 CFA candidate's exam scoring. The mean of the scoring is 64. The standard deviation of the population scoring is 15. The distribution is supposed to be normal. The 95% confidence interval for the population mean should be:

选项：

61.06 to 66.94.

61.06 to 69.94.

65.06 to 66.94.

解释：

A is correct.

Because the variance of the population is know and n ≥ 30,so The confidence interval is:

$\bar{x} \pm z_{α /2} (σ / \sqrt{n})$

$z_{α /2} = z_{0.025} = 1.96$

64 ± 1.96(15/10) = 64 ± 2.94 = 61.06 to 66.94.

因为样本方差已知且n>30，我们直接使用置信区间计算：

64 ± 1.96(15/10) = 64 ± 2.94 = 61.06 至 66.94.

根据题目分数只可能是正数，拒绝域应该是在右边，95%的单尾应该对应的是1.65呀？

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1 个答案

星星_品职助教 · 2022年10月20日

同学你好，

本题为置信区间考点，不涉及假设检验和拒绝域。

数量科目中所有置信区间都是双尾情况。

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NO.PZ2015120604000117 问题如下 A ranm sample is 100 CFA cante's exscoring. The meof the scoring is 64. The stanrviation of the population scoring is 15. The stribution is supposeto normal. The 95% confinintervfor the population meshoulbe: A.61.06 to 66.94. B.61.06 to 69.94. C.65.06 to 66.94. A is correct.Because the varianof the population is know ann ≥ 30,so The confinintervis: x¯ ± z α/2 (σ/ n ) z α/2 = z 0.025 =1.96 64 ± 1.96(15/10) = 64 ± 2.94 = 61.06 to 66.94.因为样本方差已知且n 30，我们直接使用置信区间计算64 ± 1.96(15/10) = 64 ± 2.94 = 61.06 至 66.94.

2022-11-19 10:58 1 · 回答

NO.PZ2015120604000117问题如下 A ranm sample is 100 CFA cante's exscoring. The meof the scoring is 64. The stanrviation of the population scoring is 15. The stribution is supposeto normal. The 95% confinintervfor the population meshoulbe: A.61.06 to 66.94. B.61.06 to 69.94. C.65.06 to 66.94. A is correct.Because the varianof the population is know ann ≥ 30,so The confinintervis: x¯ ± z α/2 (σ/ n ) z α/2 = z 0.025 =1.96 64 ± 1.96(15/10) = 64 ± 2.94 = 61.06 to 66.94.因为样本方差已知且n 30，我们直接使用置信区间计算64 ± 1.96(15/10) = 64 ± 2.94 = 61.06 至 66.94. 老师您好，PPT上的公司用的是sigma，所以我的理解是直接用population S好，为什么用SE呢？

2022-04-03 04:05 1 · 回答

NO.PZ2015120604000117 61.06 to 69.94. 65.06 to 66.94. A is correct. Because the varianof the population is know ann ≥ 30,so The confinintervis: x¯ ± z α/2 (σ/ n ) z α/2 = z 0.025 =1.96 64 ± 1.96(15/10) = 64 ± 2.94 = 61.06 to 66.94. 因为样本方差已知且n>30，我们直接使用置信区间计算 64 ± 1.96(15/10) = 64 ± 2.94 = 61.06 至 66.94. 如果只抽一次，一次数量大于等于30，那么，这一次数据的标准差等于总体标准差除以根号n么？如果抽100次，每一次数量都大于等于30，每一次都会有一个均值，这些均值又形成一个新的分布，这个新的分布的方差叫做标准误，也等于总体的标准差除以根号n吗？

2022-02-04 12:33 1 · 回答

NO.PZ2015120604000117 老师请问样本标准差和样本均值标准分别是怎么定义的，区别是什么？

2021-06-06 14:57 2 · 回答

为什么是双尾？

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