Question

2. Suppose a scheduled airline flight must average at least 60% occupancy in order to be profitable to the airline. An examination of the occupancy rate for 120 10:00 A.M. flights from Atlanta to Dallas showed a mean occupancy per flight of 58% and a standard deviation of 11%. a. If u 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? Explain. c. Do the occupancy data for the 120 flights suggest that this scheduled flight is unprofitable? Test using a = .05. 4990000000000000044999

Answer 1

Solution:

a)

Let us $\mu$ denote the mean occupancy per flight.

The null and alternative hypothesis can be definer as follows:

H₀ : $\mu$ $\geq$ 0.6

H_a : $\mu$ < 0.6

b)

The alternative hypothesis in part (a) is one tailer (left).

c)

under the null hypotheses the test statistics can he defined as follows:

Z =

(1 - )/(s/vn)

= (0.58-0.6 )/0.11/$\sqrt{120}$

= —1.9917

From. normal distribution tables, the critical value at 0 05 level of the significance Mr left tailed test is -1.645
Here, it can he observed that the calculated value Is less than the critical value.

Hence, the null hypothesis should be rejected
Therefore, it can be concluded that there is sufficient evidence to support Mat a scheduled night must average less than 60%.

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