In Exercises 1–4, use the following listed arrival delay times (minutes) for American Airline flights from New York to Los Angeles. Negative values correspond to flights that arrived early. Also shown are the SPSS results for analysis of variance. Assume that we plan to use a 0.05 significance level to test the claim that the different flights have the same mean arrival delay time.

Flight 1

-32

-25

-26

-6

5

-15

-17

-36

Flight 19

-5

-32

-13

-9

-19

49

-30

-23

Flight 21

-23

28

103

-19

-5

-46

13

-3

P-Value If we use a 0.05 significance level in analysis of variance with the sample data given in Exercise 1, what is the P-value? What should we conclude? If a passenger abhors late flight arrivals, can that passenger be helped by selecting one of the flights?

Short Answer

Expert verified

The p-value for the test is 0.285.

The null hypothesis is failed to be rejected at 0.05 level of significance.

The passenger cannot be helped as all flights have statistically the same mean arrival delay time.

Step by step solution

01

Given information

The SPSS output for the arrival delay times, along with the observations of three flights,are given.

02

Identify the p-value

P-value is the probability of obtaininga value as extreme as the test statistic. If the value is large, the chances of rejecting the null hypothesis lower.

From the output table, the value computed in the significance column against the test statistic is the p-value,which is 0.285.

03

Decision rule using the p-value

The decision rule states two criteria:

  • If the p-value is greater than the significance level, the null hypothesis fails to be rejected.
  • If the p-value is lower than the significance level, the null hypothesis is rejected.

The significance level is 0.05. The p-value is larger than 0.05, implying that the null hypothesis would fail to be rejected at a 0.05 level of significance.

The statistical hypothesis for the test is as follows:

Null hypothesis: The mean of arrival delays for all flights is the same.

Alternative hypothesis: The mean of arrival delays for at least one flight differs.

As a result, it can be stated that there is sufficient evidence to conclude that the mean arrival delay for all flights is equal.

04

Explain if one of the flights can be selected

As per the conclusion, the mean arrival delay for all three flights is equal. Thus, no flight is different from the rest based on the arrival delay time.

Thus, it is difficult to help the passenger select one of these flights.

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Most popular questions from this chapter

A randomized controlled trial was designed to compare the effectiveness of splinting versus surgery in the treatment of carpal tunnel syndrome. Results are given in the table below (based on data from “Splinting vs. Surgery in the Treatment of Carpal Tunnel Syndrome,” by Gerritsen et al., Journal of the American Medical Association,Vol. 288,

No. 10). The results are based on evaluations made one year after the treatment. Using a 0.01 significance level, test the claim that success is independent of the type of treatment. What do the results suggest about treating carpal tunnel syndrome?

Successful Treatment

Unsuccessful Treatment

Splint Treatment

60

23

Surgery Treatment

67

6

Equivalent Tests A\({\chi ^2}\)test involving a 2\( \times \)2 table is equivalent to the test for the differencebetween two proportions, as described in Section 9-1. Using the claim and table inExercise 9 “Four Quarters the Same as $1?” verify that the\({\chi ^2}\)test statistic and the zteststatistic (found from the test of equality of two proportions) are related as follows:\({z^2}\)=\({\chi ^2}\).

Also show that the critical values have that same relationship.

Cybersecurity What do the results from the preceding exercises suggest about the possibility that the computer has been hacked? Is there any corrective action that should be taken?

The accompanying table is from a study conducted

with the stated objective of addressing cell phone safety by understanding why we use a particular ear for cell phone use. (See “Hemispheric Dominance and Cell Phone Use,” by Seidman, Siegel, Shah, and Bowyer, JAMA Otolaryngology—Head & Neck Surgery,Vol. 139, No. 5.)

The goal was to determine whether the ear choice is associated with auditory or language brain hemispheric dominance. Assume that we want to test the claim that handedness and cell phone ear preference are independent of each other.

a. Use the data in the table to find the expected value for the cell that has an observed frequency of 3. Round the result to three decimal places.

b. What does the expected value indicate about the requirements for the hypothesis test?

Right Ear

Left Ear

No Preference

Right-Handed

436

166

40

Left-Handed

16

50

3

Cybersecurity The accompanying Statdisk results shown in the margin are obtained from the data given in Exercise 1. What should be concluded when testing the claim that the leading digits have a distribution that fits well with Benford’s law?

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