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Helen 🎈 · 2023年01月07日

哪个公式

  1. CFA I
  2. Quantitative

NO.PZ2022071202000084

问题如下：

Question

Using the following sample results drawn as 25 paired observations from their underlying distributions, test whether the mean returns of the two portfolios differ from each other at the 1% level of statistical significance. Assume the underlying distributions of returns for each portfolio are normal and that their population variances are not known.

t-statistic for 24 degrees of freedom and at the 1% level of statistical significance = 2.807

Null hypothesis (H0): Mean difference of returns = 0

Based on the paired comparisons test of the two portfolios, the most appropriate conclusion is that H0 should be:

选项：

A.rejected because the computed test statistic exceeds 2.807.

B.accepted because the computed test statistic exceeds 2.807.

C.accepted because the computed test statistic is less than 2.807.

解释：

Solution

A is correct.

In this case, the test statistic equals: (4.25-0)/(6.25/√25)= 3.40. Because 3.40 > 2.807, the null hypothesis that the mean difference is zero is rejected.

B is incorrect because a too large (or too small) test statistic leads to the rejection of the null hypothesis.

C is incorrect because the correctly computed test statistic exceeds the critical value of 2.807. A reasonable way to reach response C is to fail to divide the standard error by the square root of the sample size, erroneously using 4.25/6.25 = 0.68 as the test statistic.

请问是哪个公式，有记忆，但是咋没找到，

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2 个答案

星星_品职助教 · 2023年01月09日

@Helen 🎈

公式中分母为Sd，即“difference”的standard error，所以不能看两个sample各自的standard error，而是要看“difference”这一列的对应值

星星_品职助教 · 2023年01月07日

同学你好，

本题基于paired comparisons test中的t-statistic的计算公式，如下：

分子代入mean return的difference 4.25，分母代入standard error=sample standard deviation/ 根号n=6.25/根号25，得到3.4.


Helen 🎈 · 2023年01月09日

4.25-0/6.25/根号下25是哪个公式a

Helen 🎈 · 2023年01月09日

知道了就是test statistic公式。还想问下standard diviation 不是15.5

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NO.PZ2022071202000084 问题如下 QuestionUsing the following sample results awn 25 paireobservations from their unrlying stributions, test whether the mereturns of the two portfolios ffer from eaother the 1% level of statisticsignificance. Assume the unrlying stributions of returns for eaportfolio are normanththeir population variances are not known.t-statistic for 24 grees of freem anthe 1% level of statisticsignifican= 2.807Null hypothesis (H0): Mefferenof returns = 0Baseon the pairecomparisons test of the two portfolios, the most appropriate conclusion is thH0 shoulbe: A.rejectebecause the computetest statistic excee 2.807. B.acceptebecause the computetest statistic excee 2.807. C.acceptebecause the computetest statistic is less th2.807. SolutionA is correct. In this case, the test statistic equals: (4.25-0)/(6.25/√25)= 3.40. Because 3.40 2.807, the null hypothesis ththe mefferenis zero is rejecteB is incorrebecause a too large (or too small) test statistic lea to the rejection of the null hypothesis.C is incorrebecause the correctly computetest statistic excee the criticvalue of 2.807. A reasonable wto rearesponse C is to fail to vi the stanrerror the square root of the sample size, erroneously using 4.25/6.25 = 0.68 the test statisti 我通过给出的两个标准差算t检验，表格第四行two inpennt populations和pairecomparison有什么区别？

2023-08-05 17:37 1 · 回答