题样本均值不就是2，32吗？12，78开根号就是样本均值的标准误已经给定了？把均值和方差加起来除以2？37年的数据为啥还开根号？

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

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

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？

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