12th Mar 2010
An airline claims that the no-show rate for passengers booked on its flights is less than 6%. Of 380 randomly selected reservations, 18 were no-shows. Calculate the P-value for a test of the airline's claim.Lets see, I really do not know what P-value is, but thinking it through to the likely logical conclusion, first I would compute 6% of 380 = 22.8, less 18 no shows, = 4.8 or 5 for the value of P, or is it that the P stands for probability of the airline's claims being correct, 4.8 divided by 18 = 26%, (1.00 - .26 = .74), therefore the value of P is 74%.
Another, although less likely conclusion: the P-value may be the amount of P in the toilets after the flight, which for 362 passengers, assuming average of 4 out of 10 Pee, and assuming 24 average ounces per pee is 24 x 362 x 40% =3,475 ounces divided by 128 ounces in a gallon = P value of 27.2 gallons.I used a GDC with 1-PropZTest. I ended up with p=0.149906276 and p(cap)=0.0473684211.18 as a percentage of 380 is
(18 x 100)/ 380 = 1800/380 = 4.74%
Therefore the test works in this case.Well you need to this.
18/380 x 100 = 4.74%.
So with this test, their claim is correct as it is below 6%.#If you have any other info about this subject , Please add it free.# |
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Posted by wktd under enart.88083.com |
