Question

Are Flights Cheaper When Scheduled Earlier? Listed below are the costs (in dollars) of flights from New York (JFK) to Los Angeles (LAX). Use a 0.01 significance level to test the claim that flights scheduled one day in advance cost more than flights scheduled 30 days in advance. What strategy appears to be effective in saving money when flying?

	Delta	Jet Blue	American	Virgin	Alaska	United
1 day in advance	501	634	633	646	633	642
30 days in advance	148	149	156	156	252	313

Short answer

There is enough evidence to conclude that flights booked one day in advance are costlier than flights booked 30 days in advance.

The passengers must book their flights well in advance and plan their trips accordingly (a month before).

Step-by-step solution

Given information

The cost of 6 flights is tabulated corresponding to the two types of booking: one day in advance and 30 days in advance.

Hypotheses

It is claimed that the flights booked one day in advance are more expensive than flights booked 30 days in advance.

Null Hypothesis: The mean cost of flights booked one day in advance is more than the cost of flights booked 30 days in advance.

\({H_0}:{\mu _d} = 0\)

Alternative Hypothesis: The mean cost of flights booked one day in advance is more than the cost of flights booked 30 days in advance.

\({H_1}:{\mu _d} > 0\)

Here,\({\mu _d}\)is the population difference in the costs of the flights obtained by subtracting the cost of the flights booked 30 days in advance from the cost of the flights booked 1 day in advance.

The test is right-tailed.

Differences in the values of each matched pair

The following table shows the differences in the systolic blood pressure levels for each matched pair:

	Delta	Jet Blue	American	Virgin	Alaska	United
1 day in advance	501	634	633	646	633	642
30 days in advance	148	149	156	156	252	313
Differences \(\left( {{d_i}} \right)\)	353	485	477	490	381	329

The number of pairs is equal to\(n = 6\).

The mean value of the differences is computed below:

\(\begin{aligned} \bar d &= \frac{{\sum\limits_{i = 1}^n {{d_i}} }}{n}\\ &= \frac{{353 + 485 + ...... + 329}}{6}\\ &= 419.17\end{aligned}\)

The standard deviation of the differences is computed below:

\(\begin{aligned}{c}{s_d} &= \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({d_i} - \bar d)}^2}} }}{{n - 1}}} \\ &= \sqrt {\frac{{{{\left( {353 - 419.17} \right)}^2} + {{\left( {485 - 419.17} \right)}^2} + ....... + {{\left( {329 - 419.17} \right)}^2}}}{{6 - 1}}} \\ &= 73.02\end{aligned}\)

The mean value of the differences for the population of matched pairs \(\left( {{\mu _d}} \right)\) is considered to be equal to 0.

Compute the test statistic, critical value and the p-value

The value of the test statistic is computed as shown:

\(\begin{aligned} t &= \frac{{\bar d - {\mu _d}}}{{\frac{{{s_d}}}{{\sqrt n }}}}\\ &= \frac{{419.17 - 0}}{{\frac{{73.02}}{{\sqrt 6 }}}}\\ &= 14.06\end{aligned}\)

The degrees of freedom are computed below:

\(\begin{aligned} df &= n - 1\\ &= 6 - 1\\ &= 5\end{aligned}\)

Refer to t-table:

The critical value of t at\(\alpha = 0.01\)and degrees of freedom equal to 5 for a right-tailed test is equal to 3.3649.

The p-value obtained using the test statistic value is equal to 0.00002.

Since the absolute value of the test statistic (14.06) is greater than the critical value and the p-value is less than 0.01, the null hypothesis is rejected.

Conclusion

There is enough evidence to conclude that flights booked one day in advance are costlier than flights booked 30 days in advance.

In order to save money, the passengers must plan their trips a month in advance and book their flights well in advance rather than going for flights just a day before the trip