Airline Occupancy Rates High airline occupancy rates on scheduled flights are es

Question

Airline Occupancy Rates High airline occupancy rates on scheduled flights are essential to corporate profitability. Suppose a scheduled flight must average at least 60% occupancy in order to be profitable, and an examination of the occupancy rate for 120 10:00 a.m. flights from Toronto to Calgary showed a mean occupancy per flight of 58% and a standard deviation of 11%. If mu is the mean occupancy per flight and if the company wishes to determine whether or not this scheduled flight is unprofitable, give the alternative and the null hypotheses for the test. Does the alternative hypothesis in part a imply a one- or two-tailed test? Explain. Do the occupancy data for the 120 flights suggest that this scheduled flight is unprofitable? Test using alpha = 0.05.

Explanation / Answer

u be the mean occupancy rate per flight.

it is given that a scheduled flight must average at least 60% occupancy in order to be profitable.

a) hence the null hypothesis is H0: u=0.6    and alternative hypothesis is H1:u<0.6

b) the alternative hypothesis is one tailed test and to be specific a left tailed test

because in the question it is mentioned that the occupancy rate is at least 60%. so a occuapancy rate less than 60% means that the scheduled flight is un profitable.

c) we have a sample of n=120 fligthts with standdard deviation=s=11% and occupancy rate=x=58%

hence to test H0 we have the test statistic T=(x-0.6)/sqrt(s/n) which under H0 follows N(0,1) dsitribution

given that alpha=0.05

so we reject H0 iff t<-tao0.05 where t is the observed value of t and tao0.05 is the upper 0.05 point of a standard normal distribution. [since the alternative hypothesis is left sided rejection region is also left tailed]

so t=(0.58-0.6)/sqrt(0.11/120)=-0.66

and -tao0.05=-1.64 [using MINITAB]

hence t>-tao0.05

hence 5% level of significance we accept H0 and conclude that the scheduled flight is profitable   [answer]

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