Question L1-00353 from STALLA A manager claims that his portfolios will on average outperform the S&P by 2% per year after fees. An analyst collects a sample of 25 of the manager’s portfolios and finds on average they outperform by 2.1% with a sample standard deviation of 1%. The confidence interval approach to test the manager’s claim at the 1% level of significance (t at .005 probability and 24 degrees of freedom is 2.797) sets confidence limits closest to: a) -0.1% to 4.1% b) 0.0% to 4.0% c) 1.5% to 2.7% d) 1.4% to 2.6%
C .021 +/- [2.797 * (.01/(25^.5))] = 1.5% - 2.7%. H0 = 2.0%, cannot be rejected.
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D. Using the confidence interval approach, you would use the population mean to calculate instead of the sample mean. 2.0 +/- (2.797)(1/5) =.5594 1.4 to 2.6
so D or C ???
That would be C. If you work the problem with at least 2 decimals you can see that the interval spreads between 1.541% and 2.659%. While the inferior limit is covered by D, the superior is not. If you select C, both limits are covered.
I got C, and my calculation is below: 1) x bar +/- t alpha/2 (stand. dev./sq. root of n) 2) 2.1 +/- 2.797(1/sq.root 25) 3) 2.1 +/- .5594 = 1.5406 to 2.6559
Confidence intervals are build around the mean of the sample, while acceptance range is build around the mean of the population, so you are calculating the aceptance range. jimc28 Wrote: ------------------------------------------------------- > D. > > Using the confidence interval approach, you would > use the population mean to calculate instead of > the sample mean. > > 2.0 +/- (2.797)(1/5) > =.5594 > > 1.4 to 2.6
^ aksa’s Looks good. Another inappropriate hypothesis testing question. Imagine the poor manager who, in fact, does beat the S&P by an average of 2.1% and you suggest to him that you don’t have enough evidence to believe his claim…
I would suggest a larger sample:)
map1 Wrote: ------------------------------------------------------- > Confidence intervals are build around the mean of > the sample, while acceptance range is build around > the mean of the population, so you are calculating > the aceptance range. > > jimc28 Wrote: > -------------------------------------------------- > ----- > > D. > > > > Using the confidence interval approach, you > would > > use the population mean to calculate instead of > > the sample mean. > > > > 2.0 +/- (2.797)(1/5) > > =.5594 > > > > 1.4 to 2.6 I don’t know exactly what this means, but if you are doing this confidence interval thing for hypothesis testing (really ugly, btw) you can see if the sample mean falls in an interval around the hypothesized mean or whether the hypothesized mean falls in an interval around the sample mean. They are both the same (thus equally gross).
I think the acceptance range says: I will accept your claim if your sample mean is between this and this, and the confidence interval says: I am X% sure that your claim is true and your sample mean will be between this and this, if your t-statistic is between such and such. hmmm …
Actually this is a testing of the claim/null hypothesis. To decide whether to “not reject” the claim, one should calculate the range around the claim (in this case, the manager is claiming outperformance by 2.0%). The analyst collects a sample and results in an outperformance by 2.1%, the analyst’s opinion is the alternative hypothesis. H0: μ=2.0 Ha: μ≠2.0 In the calculation, you are building the non-acceptance range around the null hypothesis: 2.0 +/- (2.797)(1/5) =.5594 1.4 to 2.6 Does the analyst’s sample 2.1% fall within the non-rejection range? Yes, therefore conclude that we do not reject the manager’s claim of outperformance by 2.0% Does anyone else think differently?
I do. A hypothesis test does not build a range around anything. Its purpose is simply to state whether or not there is significant evidence to accept or reject a claim. In this case, the claim is that the manager does as he says (2%). The question asked you to make an estimation of the actual population mean, not to make an inference as to whether the mean is a hypothesized value or not. C is the correct answer.
Isn’t the calculation for (1) estimating the population mean using the sample, a bit different from (2) testing the hypothesized value? (1) I agree with wyantjs, that if you are attempting to estimate the population mean, you calculate the range around the sample mean with a given “confidence” that the actual population mean may fall within. 2.1 +/- (t/z-value)(std error) (2) However, isnt the question about hypothesis testing? I just recall that the calculation calls for the hypothesized value into the equation. I think Joey’s right, in the real world it can go either way. Ronyk5, what does the book say the answer is?
The correct answer in Stalla PassMaster for this question is C.