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Clement Clement
wrote...
Posts: 510
4 years ago
One common method of evaluating the performance of a mutual fund is to compare its returns to those of a recognized benchmark such as an index of the returns on all securities of the type that  the fund accumulates. The Janus Worldwide Fund considers its benchmark to be the MSCI World IndexSM. The table below depicts the annual returns (percent) for a recent ten-year period. Is this fund a good investment? That is, does this fund significantly outperform its benchmark?

Source: https://ww3.janus.com/advisor/Documents/Advisor%20Lit%20System/Fact%20Sheets/4Q12%20Fact%20Sheet%20(Janus%20Worldwide%20Fund -Class%20A)_exp%2004-15-13.pdf

Carry out the appropriate test and state your conclusion in context.
Textbook 
Stats: Modeling the World

Stats: Modeling the World


Edition: 4th
Authors:
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Replies
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4 years ago
Let d = Janus Worldwide Fund return – MSCI World Index return.
H0: μd = 0 The mean difference in the returns is zero.
HA: μd > 0 The mean difference in the returns is greater than zero.
* Paired data: The data are paired because they are measurements of annual returns for the same years for the fund as well as the index.
* Independence: We need to assume that the differences in these returns in each year are independent. This may not be entirely true since market conditions may impact consecutive years similarly, so we will proceed with some caution.
* Randomization: We must consider the differences in these years as an SRS of the differences of all possible differences of Janus Worldwide Fund – MSCI World Index.
* 10% condition: We must assume that our sample of 10 years is less than 10% of the years in which differences could be calculated.
* Nearly Normal condition: The dotplot of the differences does not indicate any extreme skewness or obvious outliers.

Under these conditions the sampling distribution of the differences can be modeled by a Student's t-model with (n – 1) = 9 degrees of freedom, and we will use a paired t-test.

We find from the data: n = 10 , = -2.01, and sd = 6.05
We estimate the standard deviation of using: SE() = = = 1.9.
t9 = = = -1.05
P = P(t9 > -1.05) < 0.84

With a P-value this large, we fail to reject the null hypothesis that the difference between the fund returns and the index returns is zero. The evidence does not support the claim that the fund outperforms the general market as measured by the index returns.
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