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 Refer to the same data from the preceding exercise. Use the Wilcoxon signed ranks test to test the claim that differences between fares for flights scheduled 1 day in advance and those scheduled 30 days in advance have a median equal to 0. What do the results suggest?

Short answer

There is enough evidence to reject the claim that the differences between the costs of flights scheduled oneday in advance and those scheduled 30 days in advance have a median equal to 0.

Step-by-step solution

Given information

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

It is claimed that the differences between the fares of flights scheduled oneday in advance and those scheduled 30 days in advance have a median equal to 0.

Identify the hypothesis of the test

The Wilcoxon signed-ranks test is used to test the claim that there is no difference in the cost between the flights scheduled oneday prior and the flights scheduled 30 days prior.

The null hypothesis is as follows:

The difference between the costs of flights scheduled oneday in advance and those scheduled 30 days in advance have a median equal to 0.

The alternative hypothesis is as follows:

The difference between the costs of flights scheduled oneday in advance and those scheduled 30 days in advance does not have a median equal to 0.

If the value of the test statistic is less than the critical value, the null hypothesis is rejected, else not.

Assign signed-ranks

Compute the differences of the values by subtracting the cost of the flights scheduled 30 days prior from the cost of thosescheduled oneday prior.

Compute the ranks of the absolute differences by assigning rank 1 to the smallest difference, rank 2 to the next smallest difference, and so on.

Mark the sign of the ranks corresponding to the sign of the difference.

The following table shows the signedranks:

Flights scheduled oneday prior	584	490	584	584	584	606	628	717
Flights scheduled 30 days prior	254	308	244	229	284	509	394	258
Differences (d)	330	182	340	355	300	97	234	459
Rank of |d|	5	2	6	7	4	1	3	8
Signed-Ranks	+5	+2	+6	+7	+4	+1	+3	+8

Calculate the test statistic

The total number of observations (n) is 8.

Since \(n < 30\), the value of the test statistic (T) needs to be determined.

Compute the sum of the positive ranks as shown:

\(\begin{array}{c}Su{m_{positive}} = 5 + 2 + .... + 8\\ = 36\end{array}\)

Since there are no negative ranks, \(Su{m_{negative}} = 0\).

T assumes a value corresponding to the smaller of the two sums.

Thus, \(T = 0\).

Determine the critical value and the conclusion of the test

The critical value of T for n=8 and \(\alpha = 0.05\) for a two-tailed test is equal to 4.

Since the value of the test statistic is less than the critical value, the null hypothesis is rejected.

There is enough evidence to conclude that the difference between the costs of flights scheduled one day in advance and those scheduled 30 days in advance does not have a median equal to 0.