SOLUTION: 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. assuming that this data is us

Algebra ->  Probability-and-statistics -> SOLUTION: 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. assuming that this data is us      Log On


   

Question 1197909: 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. assuming that this data is used to test the airline's claim, find the p-value for the test.
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
test mean is assumed to be .06.
that's the value of p.
the value of q is 1 - p = .94
sample size is 380.
no shows are 18 from thqt sample.
sample p = 18 / 380 = .047368411 = .047368 rounded to 6 decimal digits.
standard error for the test = sqrt(p * q / n)
p is the population proportion.
q is equal to 1 minus the population proportion.
n is equal to the sample size.
standard error is therefore equal to sqrt(.06 * .94 / 380) = .01218 rounded to 5 decimal digits.
z = (x-m)/s = (.047368 - .06) / .01218 = -1.037 rounded to 3 decimal digits.
z is the z-score
x is the sample proportion.
m is the assumed population proportion.
s is the standard error
probability of z smaller than -1.037 = .1499 rounded to 4 decimal digits.
that's the p-value for the test.