Question

Using Nonparametric Tests. In Exercises 1–10, use a 0.05 significance level with the indicated test. If no particular test is specified, use the appropriate nonparametric test from this chapter.

Airline Fares Listed below are the costs (in dollars) of eight different flights from New York (JFK) to San Francisco for Virgin America, US Airways, United Airlines, JetBlue, Delta, American Airlines, Alaska Airlines, and Sun Country Airlines. (Each pair of costs is for the same flight.) Use the sign test to test the claim that there is no difference in cost between flights scheduled 1 day in advance and those scheduled 30 days in advance. What appears to be a wise scheduling strategy?

Flight scheduled one day in advance	584	490	584	584	584	606	628	717
Flight scheduled 30 days in advance	254	308	244	229	284	509	394	258

Short answer

There is enough evidence to reject the claim that there is no difference in the costs of flights between those scheduled one day in advance and those scheduled 30 days in advance.

A wise scheduling strategy would be to buy the flight tickets 30 days in advance.

Step-by-step solution

Given information

Two samples show the airfares of eight different flights from New York to San Francisco.

Identify the hypothesis of the test

The sign test examinesthe claim that there is no difference in the cost between flights scheduled oneday prior and flights scheduled 30 days before.

The null hypothesis is as follows:

There is no difference in the cost between flights scheduled oneday in advance and those scheduled 30 days in advance.

The alternative hypothesis is as follows:

There is no difference in the cost between flights scheduled oneday in advance and those scheduled 30 days in advance.

Assign the sign of difference

Assign a negative sign to the difference if the cost corresponding to sample 1 (flights scheduled oneday in advance) is less than the cost corresponding to sample 2 (flights scheduled 30 days in advance).

Assign a positive sign to the difference if the cost corresponding to sample 1 (flights scheduled one day in advance) is greater than the cost corresponding to sample 2 (flights scheduled 30 days in advance).

The following table shows the sign of differences:

Flight scheduled one day in advance	584	490	584	584	584	606	628	717
Flight scheduled 30 days in advance	254	308	244	229	284	509	394	258
Sign of Difference	+	+	+	+	+	+	+	+

Calculate the test statistic

The total number of observations (n) is 8.

Since \(n \le 25\), the value of the test statistic (x) needs to be determined.

The number of times the positive sign occurs is 8.

The number of times the negative sign occurs is 0.

Here, the less frequent sign is the negative sign.

Thus, the test statistic (x) is the number of times the less frequent sign occurs, which is equal to 0.

Determine the conclusion of the test

The critical value of x for n=8 and \(\alpha = 0.05\) for a two-tailed test is 0.

Since the value of the test statistic is equal to the critical value, the null hypothesis is rejected.

There is enough evidence to reject the claim that there is nodifference in the costs of flights between those scheduled one day in advance and those scheduled 30 days in advance.

Appropriate scheduling strategy

As all of the eight air fares are higher for a flight scheduled oneday in advance than the flights scheduled 30 days in advance, it would be wise to schedule the flight 30 days in advance to save the extra cost.