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梦梦 · 2024年11月09日

1.8的条件没用吗?

NO.PZ2023091601000049

问题如下:

A quantitative analyst used a simulation to forecast the S&P 500 index value at the end of the year with an index value of 1.800 at the beginning of the year. He generated 200 scenarios and calculated the average index value at year-end to be 1.980, with a 95% confidence interval of (1.940, 2.020). In order to improve the accuracy of the forecast, the quantitative analyst increased the number of scenarios to attain a new 95% confidence interval of (1.970, 1.9901) with the same sample mean and the same sample standard deviation. How many scenarios were used to generate this result?

选项:

A.

400

B.

800

C.

1600

D.

3200

解释:

年初1.8value这个条件没用?“He generated 200 scenarios and calculated the average index value at year-end to be 1.980, with a 95% confidence interval of (1.940, 2.020). ”

这句话咋翻译?意思难道不是用1.8和95%confidence interval (1.94,2.02)求出1.98?

而是用1.98去求1.94,2.02?

1 个答案
已采纳答案

李坏_品职助教 · 2024年11月09日

嗨,努力学习的PZer你好:


1.8没啥用,1.8只是年初的数值。这道题有用的就是两次区间估计的置信区间范围,至于1.980这个是置信区间的点估计μ,这个是有用的。


第一次置信区间的下限是1.94,所以μ-1.96*σ/根号n1 = 1.94,上限是2.02,所以μ+1.96*σ/根号n1 = 2.02,结合这两个等式,

另外题目说n1 = 200,所以σ=0.2886.

第二次置信区间的下限是1.97,所以μ-1.96*σ/根号n2=1.97,所以1.98-1.96*σ/根号n2 = 1.97。上限是1.9901,所以1.98+1.96*σ/根号n2 = 1.9901.

(σ同样也等于0.2886)。


结合第二次置信区间的这两个等式:

1.98-1.96*0.2886/根号n2 = 1.97,

1.98+1.96*0.2886/根号n2 = 1.9901.

所以n2 = 3136.6,最接近的就是D。

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