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:

18	18	14	15	15	17	17	15	13	13
13	16	18	12	12	14	17	12	13	10

At the 0.01 significance level, can the agency conclude that the mean number of complaints per airport is less than 17 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 ≥ 17; H₁ : μ < 17;  (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 17.

Answer 1

Claim : The mean number of complaints per airport is less than 17 per month.

Assumption :

Population of complaints follows a normal probability distribution and population standard deviation is unknown.

Therefore, we use one-sample t test.

Hypothesis:

H0 : μ ≥ 17; H1 : μ < 17

Left tailed test

Test statistic :

​​​​​​

Therefore, the value of test statistic is (-4.527).

Critical value :

df=20-1=19,

..........(From t-table for one-tailed)

Decision Rule:

i.e. 4.527 > 2.539

Therefore, we reject H0.

Conclusion :

There is enough evidence to conclude that the mean number of complaints is less than 17.