Question

Delta Air Lines reports that the average delay time of all arriving domestic flights in 2018 was 60 minutes. A flight is considered delayed if it arrives more than 15 minutes later than the scheduled arrival time. Average delayed minutes are calculated from delayed flights only. A random sample of 20 Delta flights in 2018 had an average delay of 65 minutes with a standard deviation of 5 minutes.

a) Categorical or Quantitative Variable?

b) Is 60 minutes a Sample statistic or Population parameter?

c) Is 65 minutes and 5 minutes a Sample statistics or Population parameters?

d) Should we use a z-distribution or t-distribution?

e) Calculate a hypothesis test to test if there is an increase in average delay from the 60 minutes reported using STAT TESTS #2.

f) What do you conclude in the context of the problem.

g) Which could occur based on your previous conclusion a Type I or Type II error?

Answer 1

a) Quantitative Variable

Variable has been numerically expressed.

b) Population parameter

It is provided for the entire population of domestic flights arriving in 2018

c) Sample statistics

As mentioned it is provided for a random sample of 20 Delta flights.

d) t distribution

The sample size is very small and t distribution would better model the situation under consideration.

e) and f)

The provided sample mean is and the sample standard deviation is , and the sample size is .

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ = 60

Ha: μ > 60

This corresponds to a right-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the critical value for a right-tailed test is tc​=1.729.

The rejection region for this right-tailed test is R=t:t>1.729

(3) Test Statistics

The t-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that t=4.472>tc​=1.729, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.0001, and since p=0.0001<0.05, it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ is greater than 60, at the 0.05 significance level.

Graphically

g) A type 1 error can occur since we have rejected the null hypothesis. Type 1 error is when we reject the null hypothesis when it is infact true.