Ron Meder, CFA, has examined the price/earnings ratios of 100 companies listed on the NASDAQ. Based on this analysis, he has determined that the average price/earnings ratio of these companies is 35 with a standard deviation of 20. Assuming that P/E ratios are normally distributed, Meder can be reasonably assured that 95% of the price/earnings ratios of all stocks listed on the NASDAQ have price/earnings ratios that lie within which of the following ranges? a. 31 to 39 b. -5 to 75 c. 15 to 55 d. 25 to 45
Use T-table … since sample size is high (n > 30) and population variance in unknown S.E = sample STD/ SQRT(sample size) S.E. = 20/SQRT(100) = 2.0 t-statistic(99,0.05) = 1.9842 C.I = (point statistic) ± (rel factor)*(S.E.) C.I = 35 ± (1.9842 * 2) = 35 ± 3.9684 31.0316 to 38.9684 Answer is A?? - Dinesh S
I got A just by eyeballing it…haha…hope its right
the answer is b. 35-1.96*20 = -4.2 35+1.96*20 = 74.2
any reason, maratikus u r not using the s/sqrt(n) above? anything special that 100 cos. in nasdaq means population that we should “know”?
sqrt(n) should be used for mean. for example, if the question was to come up with a confidence interval for the mean of the population, I would’ve used sqrt(n). The question was: Assuming that P/E ratios are normally distributed, Meder can be reasonably assured that 95% of the price/earnings ratios of all stocks listed on the NASDAQ have price/earnings ratios that lie within which of the following ranges?
It cannot be A based on reasoning. I get B as well. If the standard deviation of his small sample of 100 stocks is 20, and he wants a range to include 95% of P/E’s, then the standard deviation of the population should be roughly similar to that of the sample. I think the times that you want to use sigma_m=s/sqrt(n) is when using a confidence interval to determine the MEAN of the population given a sample. In that case as your sample size gets larger, your error of the MEAN drops. But in this case we want the STANDARD DEVIATION of the population, because we want to include 95% of all P/E ratios. Different kind of question.
maratikus Wrote: ------------------------------------------------------- > sqrt(n) should be used for mean. for example, if > the question was to come up with a confidence > interval for the mean of the population, I > would’ve used sqrt(n). > > The question was: > > Assuming that P/E ratios are normally distributed, > Meder can be reasonably assured that 95% of the > price/earnings ratios of all stocks listed on the > NASDAQ have price/earnings ratios that lie within > which of the following ranges? maratikus, I still don’t understand why it should not be divided by SQRT(n) ?? Sorry for being such a dumbo!! - Dinesh S
dinesh, you estimate mean and standard deviation of population or we can assume that you know those parameters apriori. Then you are asked to define an interval that contains 95% of the population. how would you approach that problem?
Doesn’t 95% entail 2 Stnd Dev? If 1 Stnd deviation is 20, double it. 35-40=-5 35+40=75 Probably not the right answer, but Quant is my weakest section by far.
maratikus - is it correct understanding: If population mean and population standard deviation is given then we don’t need to divide by sqrt(n) but if it is for sample mean and sample st. dev then we have to divide the st. dev by sqrt(n) ??
Hari, When we are estimating the population parameter using a sample of a given size, there is room for error that the popln mean is not exactly same as the sample mean. In this case, we build a CI around the sample mean to determine the range for popn mean at a given confidence level. In this question, however he is not estimating the popln parameter. Rather he is looking for a range around popn parameter where the values in original population would fall at 95% confidence. Since we don’t know the popln parameter exactly and not even from sample[as there is room for error determined by s2/sqrt(n)], we are taking a point estimate that sample mean is equal to popn mean and constructing CI for popln data set. Hope this clarifies. -Ravi
that’s perfect explanation Ravi. Thanks.
So, what’s the answer by the way? A or B? Cheers.
apnesapne Wrote: ------------------------------------------------------- > Hari, > > When we are estimating the population parameter > using a sample of a given size, there is room for > error that the popln mean is not exactly same as > the sample mean. In this case, we build a CI > around the sample mean to determine the range for > popn mean at a given confidence level. > > In this question, however he is not estimating the > popln parameter. Rather he is looking for a range > around popn parameter where the values in original > population would fall at 95% confidence. > > Since we don’t know the popln parameter exactly > and not even from sample, we are taking a point > estimate that sample mean is equal to popn mean > and constructing CI for popln data set. > > Hope this clarifies. > > -Ravi Wow apnesapne, Thanks for such a lucid explanation. I think that’s what Central Limit Theorem is all about… And also the question asks us "95% of the price/earnings ratios of all stocks listed on the NASDAQ” the "all” is the keyword here, so they needed a interval around the population parameter, which as per the central limit theorem should be equal to sample statistic [population mean = sample mean] Thanks apnesapne and maratikus, the answer is undoubtedly B - Dinesh S