Question

Most air travellers now use e-tickets. Electronic ticketing allows passengers to not worry about a paper ticket, and it costs the airline companies less to handle than paper ticketing. However, in recent times, the airlines have received complaints from passengers regarding their e-tickets, particularly when connecting flights and a change of airlines were involved. To investigate the problem, an independent agency contacted a random sample of 20 airports and collected information on the number of complaints the airport had with e-tickets for the month of March. The information is reported below:

15	17	16	10	11	10	14	15	12	16
13	11	10	17	12	17	14	13	10	12

At the 0.02 significance level, can the agency conclude that the mean number of complaints per airport is less than 15 per month?

a. What assumption is necessary before conducting a test of hypothesis?

(Click to select)  The population of complaints follows a normal probability distribution.  The population of complaints not follows a normal probability distribution.  The population of complaints follows a uniform probability distribution.

b. Not available in Connect.

c. Conduct a test of hypothesis and interpret the results.

H₀ : μ ≥ 15; H₁ : μ < 15;  (Click to select)  Accept  Reject  H₀ if t <  . (Round the final answer to 3 decimal places.)

The value of the test statistic is  . (Negative answer should be indicated by a minus sign. Round the final answer to 2 decimal places.)

(Click to select)  Do not reject  Reject  H₀. There is  (Click to select)  not enough  enough  evidence to conclude that the mean number of complaints is less than 15.

Answer 1

a)
The population of complaints follows a normal probability distribution.

c)
Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 15
Alternative Hypothesis, Ha: μ < 15

Rejection Region
This is left tailed test, for α = 0.02 and df = 19
Critical value of t is -2.205.
Hence reject H0 if t < -2.205

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (13.25 - 15)/(2.5314/sqrt(20))
t = -3.092

reject the null hypothesis.
There is enough evidence to conclude that the mean number of complaints is less than 15.