Most air travelers now use e-tickets. Electronic ticketing allows passengers to

Question

Most air travelers 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 watchdog 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 14 16 12 12 14 13 16 15 14 12 15 15 14 13 13 12 13 10 13 At the .05 significance level, can the watchdog agency conclude the mean number of complaints per airport is less than 15 per month? Conduct a test of hypothesis and interpret the results.

What is the decision rule? (Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.)

Explanation / Answer

Solution:

Here, we have to use one sample t test for the population mean. The null and alternative hypotheses for this test are given as below:

Null hypothesis: H0: Mean number of complaints per airport is 15 per month.

Alternative hypothesis: Ha: Mean number of complaints per airport is less than 15 per month.

H0: H0: µ = 15 versus Ha: µ < 15

This is a one tailed test. This is a lower tailed or left tailed test.

We are given a level of significance or alpha value as 0.05. ( = 0.05)

The test statistic formula is given as below:

Test statistic = t = (Xbar - µ) / [S/sqrt(n)]

From the given data, we have

Sample mean = Xbar = 13.5

Sample standard deviation = S = 1.504379571

Sample size = n = 20

Degrees of freedom = df = n – 1 = 20 – 1 = 19

Level of significance = = 0.05

Lower critical value = -1.7291

(By using t-table or excel)

Decision rule:

Reject H0 if test statistic t < -1.7291

Or

Reject H0 if P-value < = 0.05

Test statistic = t = (13.5 – 15) / [1.504379571/sqrt(20)]

Test statistic = t = -4.4591

P-value = 0.0001

(By using t-table or excel)

P-value < = 0.05

So, we reject the null hypothesis that Mean number of complaints per airport is 15 per month.

There is sufficient evidence to conclude that Mean number of complaints per airport is less than 15 per month.

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