An airline was concerned about passengers arriving too late at the airport to allow for the additional security measures. Based on a survey of 1,000 passengers, the mean time from arrival at the airport to reaching the boarding gate was 1 hour, 20 minutes, with a standard deviation of 30 minutes. If the airline wants to make sure at the 95 percent confidence level that passengers have sufficient time to catch their flight, how much time ahead of their flight should passengers be advised to arrive at the airport? A) One hour, fifty minutes. B) Two hours, thirty minutes. C) Two hours, forty-five minutes. D) Two hours, ten minutes. Your answer: A was incorrect. The correct answer was D) Two hours, ten minutes. We can use standard distribution tables because the sample is so large. From a table of area under a normally distributed curve, the Z value corresponding to a 95 percent, one-tail test is: 1.65. (We use a one-tailed test because we are not concerned with passengers arriving too early, only arriving too late.) Here, we do not divide by the standard error, because we are interested in a point estimate of making our flight. The answer is One hour, twenty minutes + 1.65(30 minutes) = 2 hours,10 minutes. -------------------------------------------------------------------------------------------------------------- When I first read this question, I got confused, then I realized it’s one of those use confidence interval to predict value of random variable type of incorrect questions, so I calculated the confidence interval as follows, One hour, twenty minutes +/- 1.65(30 minutes/sqrt(1000)) And since it asked for the latest passengers can arrive before they’ll miss the flight, so I took the upper limit, but then I saw the none of the answers were anywhere close to it (the CI I got was One hour, twenty minutes +/- 1.89 minutes), so I chose the best one… Then when I read their answer, it just seems like a mess to me, my understanding is that the point estimate is supposed to be the sample statistic, and in this case the mean, all of a sudden, they said, “Here, we do not divide by the standard error, because we are interested in a point estimate of making our flight.” Where did they get this from about a point estimate? Does this question make any sense whatsoever?

Oh the humanity! That phrase “point estimate” is used incorrectly in their answer. As you say, a point estimate is a statistic used to estimate some parameter, e.g., X-bar is a point estimate of Mu. Also, this question is kind of messed anyway. They want “to make sure at the 95 percent confidence level that passengers have sufficient time to catch their flight” which I guess means that they are 95% certain that all their passengers will catch their flight. The question they answered was “what amount of time prior to boarding should passengers arrive at the airport so that 5% of them miss their flights”, which would be one dumbass way of running an airline.

maybe they only intended to take 95% of the people to begin with, by advising them of this time limit, they can legitimately take their money and say it’s the passenger’s fault for missing the flight because they were told in advance to arrive early? but they definitely won’t stay in the business too long…

I think you have a great future in hedge fund product development.

But I don’t understand why it is [mean + 1.65 x 30 minutes]; instead of [mean + 1.65 x (30 minutes/sqrt(1,000))]. Anyone please advise.

[X-bar + 1.65 x 30 minutes] => you are 95% sure that a passenger arriving this far ahead of time will make his/her flight. Here they ignore all estimation error by using z-scores and no correction for estimating standard deviation. That’s reasonable because the sample size is so big. [X-bar + 1.65 x (30 minutes/sqrt(1,000))] => this would be a one end of a 90% confidence interval for the mean time to boarding gate. Remember that you still have a sample even if it is huge so you still are trying to estimate the mean. The amount of uncertainty in estimating the mean is the standard error given by s/Sqrt(n).

JoeyDVivre Wrote: ------------------------------------------------------- > => you are 95% sure that a passenger arriving this > far ahead of time will make his/her flight. Here > they ignore all estimation error by using z-scores > and no correction for estimating standard > deviation. That’s reasonable because the sample > size is so big. > I wonder if on the exam, you would see questions similar to this? By question similar to this, I mean using sample mean/standard deviation to make estimates with “CI” or “point estimate”… > => this would be a one end of a 90% confidence > interval for the mean time to boarding gate. > Remember that you still have a sample even if it > is huge so you still are trying to estimate the > mean. The amount of uncertainty in estimating the > mean is the standard error given by s/Sqrt(n).

I don’t know, but I am currently trying to have a discussion about it with the curriculum committee.

Thanks Joey for your explanation. May I ask one more question, why 1.65 in your explanation above is the critical value for 90% confidence interval and at the same time with 95% sure?

Because most C.I.'s and all the ones in CFA are two-sided, e.g., X-bar ± z*s/Sqrt(n) so you need to divide the area in both tails.

ok. It’s clear now. Thanks again.

This is related. What’s your answer? 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 Dreary

For all you math geeks out there: a) Suppose that Y is a continuous r.v. with density f(y) so that f(y) > c > 0. Prove that E(Y) is infinite. b) Suppose that X is a continuous positive r.v. with density g(x) so that g(0) < d < infinity. Prove that E(X/Y) is infinite. c) Use a) and b) to conclude that P/E ratios cannot be normally distributed.

Dreary Wrote: ------------------------------------------------------- > This is related. What’s your answer? > > 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 > > Dreary Of course it’s b)

Joey, your density functions are meaningless, obviously combined probability (integral of the density function) is infinite in your example and it should be equal to 1. Maybe we should start a thread for math geeks. My brother who is in high school gave me a couple of problems last week, I can share those.

Isn’t the answer is A?

maratikus Wrote: ------------------------------------------------------- > Joey, > > your density functions are meaningless, obviously > combined probability (integral of the density > function) is infinite in your example and it > should be equal to 1. > > Maybe we should start a thread for math geeks. My > brother who is in high school gave me a couple of > problems last week, I can share those. C’mon there’s a little typo there - it’s f(0) > c > 0

still don’t get it, Joey. N(0,1) satisfies your condition.

Geez you’re right another typo in that - must have been before my coffee or something - That’s E(1/Y) is infinite.

That’s true because the integral of (1/y)f(y) is going to have special point at 0.