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Schewser Question on Point Estimate
In Schewser Qbank there is one question on point estimate but I am not able to understand the solution for it, Please help me out
An airline was concerned about passengers arriving too late at the airport to allow for the additional security measures. The airline collected survey data from 1,000 passengers on their time from arrival at the airport to reaching the boarding gate. The sample mean was 1 hour and 20 minutes, with a sample standard deviation of 30 minutes. Based on this sample, how long prior to a flight should a passenger arrive at the airport to
have a 95% probability of making it to the gate on time?
A) Two hours, thirty minutes.
B) Two hours, ten minutes.
C) One hour, fifty minutes.
The correct answer was B) 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%, onetail test is: 1.65. (We use a onetailed 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.
If it is saying that we need to calculate point estimate then for mean point estimate should be same isnt it?
Please explain me……….. |
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