Question

In: Statistics and Probability

A public administrator for the Federal Aviation Administration, in light of terrorist bombings and hijacking of...

A public administrator for the Federal Aviation Administration, in light of terrorist
bombings and hijacking of airplanes, was concerned about how four different
methods of screening passengers affect traveler satisfaction. The four methods were
(1) no screen, (2) walk-through metal detector, (3) frisk search, and (4) handheld
metal detector. Random samples of traveler satisfaction were taken when the
travelers were screened by one of the four methods. The sample sizes were n1=2, n2=3,
n3=3, and n4=2 for the four methods. The results are below. Test the hypothesis
that the mean satisfaction values are the same for all four methods. Use the .05
level of significance and assume normal distributions and equal variances for
each method.
No Screen Walk-Through Frisk Hand-Held
8 10 11 16
14 6 9 12
18 8 7 18
• State H0 and H1.
• Should you reject or not reject H0?   Defend your answer. There
    are 2 possible ways to do this.
• The three means (are / are not) the same.

Solutions

Expert Solution

Result:

A public administrator for the Federal Aviation Administration, in light of terrorist bombings and hijacking of airplanes, was concerned about how four different     methods of screening passengers affect traveler satisfaction. The four methods were            (1) no screen, (2) walk-through metal detector, (3) frisk search, and (4) handheld             metal detector. Random samples of traveler satisfaction were taken when the              travelers were screened by one of the four methods. The sample sizes were n1=2, n2=3, n3=3, and n4=2 for the four methods. The results are below. Test the hypothesis that the mean satisfaction values are the same for all four methods. Use the .05 level of significance and assume normal distributions and equal variances for each method.

No Screen

Walk-Through

Frisk

Hand-Held

8

10

11

16

14

6

9

12

18

8

7

Ho: µ1= µ2= µ3= µ4

H1: At least one of the mean is different from the others


Calculated F=2.2803

Table value of F with (DF1=3, DF2=7 ) =4.35

Calculated F=2.2803 < 4.35 the table value

The null hypothesis is not rejected. ( we do not reject H0)

The data indicate there is no difference among the four groups

• The three means (are ) the same.

ANOVA: Single Factor

SUMMARY

Groups

Count

Sum

Average

Variance

No Screen

3

40

13.33333333

25.3333

Walk-Through

3

24

8

4.0000

Frisk

3

27

9

4.0000

Hand-Held

2

28

14

8.0000

ANOVA

Source of Variation

SS

df

MS

F

P-value

F crit

Between Groups

72.9697

3

24.3232

2.2803

0.1664

4.3468

Within Groups

74.6667

7

10.6667

Total

147.6364

10

Level of significance

0.05


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