开发者:上海品职教育科技有限公司 隐私政策详情

应用版本:4.2.11(IOS)｜3.2.5(安卓)APP下载

返回

ditto · 2024年04月20日

看了其他解答，还是没有看懂c∗ σ ∧ = 12.78% - 7.55% = 5.23%.是怎么算出来的

  1. FRM I
  2. Quantitative

NO.PZ2020010304000050

问题如下：

If you are given a 99% confidence interval for the mean return on the Nasdaq 100 of [2.32%, 12.78%], what is the sample mean and standard error? If this confidence interval is based on 37 years of data, assumed to be iid, what is the sample standard deviation?

选项：

解释：

The mean is the midpoint of a symmetric confidence interval (the usual type), and so is 7.55%.

The 99% CI is constructed as [μ∧−c∗σ∧,μ∧+c∗σ∧]\left[\overset\wedge\mu-c*\overset\wedge\sigma, \overset\wedge\mu+c*\overset\wedge\sigma\right] and so c∗σ∧c*\overset\wedge\sigma = 12.78% - 7.55% = 5.23%. The critical value for a 99% CI corresponds to the point where there is 0.5% in each tail, or 2.57, and so σ∧\overset\wedge\sigma =5.23% /2.57=2.03% needs to be surveyed.

If this confidence interval is based on 37 years of data, assumed to be iid, the sample standard deviation is 2.03%×37=12.34%2.03\%\times\sqrt{37}=12.34\%

midpoint = mean 这一步还懂， c∗ σ ∧ 为什么等于一半的 置信区间。

还有一个解答里面5.23不是减法，是用除法算的，更不理解了，求解答

添加评论
2 个答案
已采纳答案

品职答疑小助手雍 · 2024年04月20日

同学你好，中点本来就是用除法算的啊。

区间是 [2.32%, 12.78%]，那区间的中点就是（2.32%+12.78%）/2=7.55%。

然后用上限12.78%减一下均值7.55%就得到c∗ σ =5.23%

ditto · 2024年04月21日

为什么上限减去7.55c乘以西塔等于

品职答疑小助手雍 · 2024年04月21日

因为上限是μ+c∗ σ，题目给了这个值等于12.78%。

现在求出来μ=7.55%了

减一下就是c∗ σ了。

这种题不是靠答疑一步一步的重新描述一遍解析就对考试有用的，而是理解整个计算的过程原理，建议先听一下基础班的课程再刷题库。

  • 2

    回答

  • 0

    关注

  • 179

    浏览
相关问题

NO.PZ2020010304000050问题如下If you are given a 99% confinintervfor the mereturn on the Nasq 100 of [2.32%, 12.78%], whis the sample meanstanrerror? If this confinintervis baseon 37 years of tassumeto ii whis the sample stanrviation? The meis the mioint of a symmetric confininterv(the usutype), anso is 7.55%.The 99% is constructe[μ∧−c∗σ∧,μ∧+c∗σ∧]\left[\overset\wee\mu-c*\overset\wee\sigm\overset\wee\mu+c*\overset\wee\sigma\right][μ∧​−c∗σ∧,μ∧​+c∗σ∧] anso c∗σ∧c*\overset\wee\sigmac∗σ∧ = 12.78% - 7.55% = 5.23%. The criticvalue for a 99% correspon to the point where there is 0.5% in eatail, or 2.57, anso σ∧\overset\wee\sigmaσ∧ =5.23% /2.57=2.03% nee to surveye If this confinintervis baseon 37 years of tassumeto ii the sample stanrviation is 2.03%×37=12.34%2.03\%\times\sqrt{37}=12.34\%2.03%×37​=12.34%我的问题在直接用2.58倍的viation算吗 为什么都没有用到u

2024-04-24 07:51 1 · 回答

NO.PZ2020010304000050 问题如下 If you are given a 99% confinintervfor the mereturn on the Nasq 100 of [2.32%, 12.78%], whis the sample meanstanrerror? If this confinintervis baseon 37 years of tassumeto ii whis the sample stanrviation? The meis the mioint of a symmetric confininterv(the usutype), anso is 7.55%.The 99% is constructe[μ∧−c∗σ∧,μ∧+c∗σ∧]\left[\overset\wee\mu-c*\overset\wee\sigm\overset\wee\mu+c*\overset\wee\sigma\right][μ∧​−c∗σ∧,μ∧​+c∗σ∧] anso c∗σ∧c*\overset\wee\sigmac∗σ∧ = 12.78% - 7.55% = 5.23%. The criticvalue for a 99% correspon to the point where there is 0.5% in eatail, or 2.57, anso σ∧\overset\wee\sigmaσ∧ =5.23% /2.57=2.03% nee to surveye If this confinintervis baseon 37 years of tassumeto ii the sample stanrviation is 2.03%×37=12.34%2.03\%\times\sqrt{37}=12.34\%2.03%×37​=12.34% correspon to the point where there is 0.5% in eatail, or 2.57。样本置信区间求总体均值，首先根据对称性求得样本均值，由于总体方差和样本容量未知，所以用t分布，由于置信区间是99%，所以α=0.5。这样理解对吗？但是之后怎么求2.57呢？

2024-02-03 16:31 1 · 回答

NO.PZ2020010304000050问题如下If you are given a 99% confinintervfor the mereturn on the Nasq 100 of [2.32%, 12.78%], whis the sample meanstanrerror? If this confinintervis baseon 37 years of tassumeto ii whis the sample stanrviation? The meis the mioint of a symmetric confininterv(the usutype), anso is 7.55%.The 99% is constructe[μ∧−c∗σ∧,μ∧+c∗σ∧]\left[\overset\wee\mu-c*\overset\wee\sigm\overset\wee\mu+c*\overset\wee\sigma\right][μ∧​−c∗σ∧,μ∧​+c∗σ∧] anso c∗σ∧c*\overset\wee\sigmac∗σ∧ = 12.78% - 7.55% = 5.23%. The criticvalue for a 99% correspon to the point where there is 0.5% in eatail, or 2.57, anso σ∧\overset\wee\sigmaσ∧ =5.23% /2.57=2.03% nee to surveye If this confinintervis baseon 37 years of tassumeto ii the sample stanrviation is 2.03%×37=12.34%2.03\%\times\sqrt{37}=12.34\%2.03%×37​=12.34%此题已经给定均值方差，解题思路不明。

2023-03-26 04:42 1 · 回答

NO.PZ2020010304000050 老师好，想请问一下这个题的StanrError和Stanrviation是怎么计算的

2022-02-13 09:49 1 · 回答