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 during the month of March. The information is shown in the table below.14 14 16 12 12 14 13 16 15 1412 15 15 14 13 13 12 13 10 13a. Assuming that the data are approximately normally distributed, is there sufficient evidence for the watchdog agency to conclude that the mean number of complaints per airport is less than 15 per month? Use a .05 level of significance.Complete the following:1. State H0.2. State H1.3. State the value of a.4. State the value of the test statistic.5. State the p-value.6. State the decision in terms of H0 and why.7. State the decision in terms of the problem.b. Is the normality assumption in part a necessary? Explain your answer.c. Using a graphical approach discussed in the course, determine whether or not the assumption of normality appears to be valid. Show your graph and explain your answer.

Answer 1

a) There is not enough evidence to claim that the mean number of complaints is less than 15 per month.

1)
`H_0: \mu\geq15`

2)
`H_1: \mu<15`

3)
`\alpha=0.05`

4)
`z=-1.02`

5)
`P=0.15386`

6) Failed to reject the null hypothesis, because the P-value is greater than the significance level.

7) The sample, although has a mean that is less than 15, is not as different from 15 to be an odd result. It is still the normal sample we can get if the real number of complaints is 15 per month.

b) Yes. We are assuming that the sample means are distributed normally around 15. In other words, we are assuming a sampling distribution that is approximate normal.

c) The histogram shows normality in the distribution of the data. The correct way to prove it is with a normal test plot in which, when the data is graphed, a straight line shows normality.

Explanation:

a) We have to perform a hypothesis test, with level of significance of α=0.05 and this null and alternative hypothesis:

`H_0: \mu\geq15\\\\H_1: \mu<15`

The data has a sample mean of 13.5 and a s.d. of 1.467.

We can calculate the statistic z as

`z=(13.5-15)/(1.467)=(-1.5)/(1.467)= -1.02`

The P-value of z=-1.02 is P(z<-1.02)=0.15.

The P-value (0.15) is greater than the significance level (0.05), so the effect is not significant and it failed to reject the null hypothesis.