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In Exercises 1鈥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

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 鈥淗emispheric Dominance and Cell Phone Use,鈥 by Seidman, Siegel, Shah, and Bowyer, JAMA Otolaryngology鈥擧ead & 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

Police Calls The police department in Madison, Connecticut, released the following numbers of calls for the different days of the week during February that had 28 days: Monday (114); Tuesday (152); Wednesday (160); Thursday (164); Friday (179); Saturday (196); Sunday (130). Use a 0.01 significance level to test the claim that the different days of the week have the same frequencies of police calls. Is there anything notable about the observed frequencies?

Flat Tire and Missed Class A classic story involves four carpooling students who missed a test and gave as an excuse a flat tire. On the makeup test, the instructor asked the students to identify the particular tire that went flat. If they really didn鈥檛 have a flat tire, would they be able to identify the same tire? The author asked 41 other students to identify the tire they would select. The results are listed in the following table (except for one student who selected the spare). Use a 0.05 significance level to test the author鈥檚 claim that the results fit a uniform distribution. What does the result suggest about the likelihood of four students identifying the same tire when they really didn鈥檛 have a flat tire?

Tire

Left Front

Right Front

Left Rear

Right Rear

Number Selected

11

15

8

16

Questions 6鈥10 refer to the sample data in the following table, which describes the fate of the passengers and crew aboard the Titanic when it sank on April 15, 1912. Assume that the data are a sample from a large population and we want to use a 0.05 significance level to test the claim that surviving is independent of whether the person is a man, woman, boy, or girl.


Men

Women

Boys

Girls

Survived

332

318

29

27

Died

1360

104

35

18

Identify the null and alternative hypotheses corresponding to the stated claim.

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