# standard error problem

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, ten minutes. D) Two hours, forty-five minutes. Your answer: A was incorrect. The correct answer was C) 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. ------------------------------------------ I don’t understand why we don’t divide by the standard error. Can someone tell me when to not divide by the standard error? Why does being interested in a point estimate of making our flight matter? What else would we be interested in?

hi, you are not interested in the standard deviation of the mean, but of the population. you want to be sure that all passengers make it, NOT that the mean passenger is right on time.

hmmm…I think I might be getting it…thanks, but is there another way of looking at it? Any other takers on this?

i understand why you don’t use the standard error, but why do you use 1.65 - isn’t that for a 90% CI?

it’s very easy if you think of it. what’s the mean age in this forum? what’s the standard deviation? If you want to be 95% sure that a randomly sampled person, is younger than x, which number would you choose? mu + 1.65*sd mu + 1.65*sd/sqrt(number of persons writing here)

perfect, thanks!

ridgefield Wrote: ------------------------------------------------------- > i understand why you don’t use the standard error, > but why do you use 1.65 - isn’t that for a 90% CI? 1.65 ~ 95%-Quantile of standard normal. meaning 5% prob mass is to the right of 1.65. and to the left of -1.65 meaning 90% is in between. (so you get your 90%)

Am I missing something here? Why are we using 1.65 when the interval is 95%? I am assuming there is a type somewhere…

5+5=10

got it - it’s one tailed

I’m still confused by the logic of using or not using the SE. In the example above, is SE avoided because the mean is a fact and not an estimation of the mean?

Sorry for bringing it back up again. I was searching for threads on standard error when I chanced upon here. Not wanting to start a new thread and felt that this was a great question, I decided to ask my question here. 1) Does someone elaborate the concept of when do we divide by SE and when by SD? 2) Isn’t SD of sample = SE? Thanks.