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:

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

At the 0.01 significance level, can the agency conclude that the mean number of complaints per airport is less than 16 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 ≥ 16; H₁ : μ < 16;  (Click to select)  Reject  Accept  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 16.

Answer 1

a)

The population of complaints follows a normal probability distribution.

b)

null hypothesis: HO: μ		=	16
Alternate Hypothesis: Ha: μ		<	16
0.01 level with left tail test and n-1= 19 df, critical t=				-2.539
Decision rule :reject Ho if test statistic t<-2.539
	population mean μ=	16
	sample mean 'x̄=	14.100
	sample size   n=	20.00
	sample std deviation s=	2.573
	std error 'sx=s/√n=	0.575
	test stat t ='(x-μ)*√n/sx=	-3.30
	p value      =	0.0019

from above:

  Reject H₀ if t < -2.539

The value of the test statistic is -3.30

Reject  H₀. There is  enough  evidence to conclude that the mean number of complaints is less than 16