In: Statistics and Probability
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. |
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 |