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粗眉毛辣椒油 · 2024年02月03日

这里是要通过查表求得吗?

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 [μcσ,μ+cσ]\left[\overset\wedge\mu-c*\overset\wedge\sigma, \overset\wedge\mu+c*\overset\wedge\sigma\right] and so cσc*\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\%

CI corresponds to the point where there is 0.5% in each tail, 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?

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已采纳答案

pzqa39 · 2024年02月03日

嗨,爱思考的PZer你好:


同学你的理解是对的,如果考试的时候给了表格最好,但是像99%、98%、95%、90%这种常见的数字考试的时候也有可能不给的,所以最好的单独记一下。99%对应的是2.58(准确来讲四舍五入之后应该是2.58的),98%对应的是2.33,95%对应的是1.96,90%对应的是1.65。记这几个就够了。

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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

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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% mioint = me这一步还懂, 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

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