Question

Suppose a scheduled airline flight must average at least 63% occupancy in order to be profitable to the airline. Occupancy rates were recorded daily for a regularly scheduled flight on each of 120 days, showing a mean occupancy per flight of 60% and a standard deviation of 10%.

(a). If μ 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.

(b). Does the alternative hypothesis in part (a) imply a one- or two-tailed test?

(c). Do the occupancy data for the 120 flights suggest that this scheduled flight is unprofitable? Test using α = 0.05. (Round your answers to two decimal places. If the test is one-tailed, enter NONE for the unused region.)

test statistic: z=  

rejection region:

z>

z<

Answer 1

Given that :

n = 120

$\hat{p}$ = 0.60

s = 0.10

Ho : p = 0.63

H1 : p < 0.63

p is scheduled flight occupancy

Z =
n
* ($\hat{p}$-P) / s = $\sqrt{120}$ ​​​​​​​ * (0.60-0.63) / 0.10 = -3.28

P ( Z < -3.28 ) = 0.0005

standred normal distibution under the null hyppothesis

P value is 0.0005 is less than 0.1. Reject Ho is 0.05 level of significance