The data in the table were collected from randomly selected flights at airports
ID: 3208098 • Letter: T
Question
The data in the table were collected from randomly selected flights at airports in three and indicate the number of minutes that each plane was behind schedule at its departure. Perform a one-way ANOVA using alpha = 0.05 to determine if there is a difference in the average lateness of flights from these three airports. Perform a multiple comparison test to determine which pairs are different using alpha = 0.05. Click the icon to view a studentized range table for a = 0.05 What are the correct hypotheses for the one-way ANOVA test? H_0: Not all the means are equal. H_1: mu _1 = mu_2 = mu _3 H_0:mu _1 notequalto mu _2 notequalto mu _3 H_0: mu _1 = mu_2 = mu_3 H_1:Not all the means are equal. H_0: mu_1 = mu_2 = mu_3 H_1: mu_1 notequalto mu_2 notequalto mu_3Explanation / Answer
Part a
Correct hypotheses:
C
H0: µ1 = µ2 = µ3
H1: Not all the means are equal
The one way ANOVA test by using alpha = 0.05 is given as below:
ANOVA: Single Factor
SUMMARY
Groups
Count
Sum
Average
Variance
City 1
5
48
9.6
56.3000
City 2
5
108
21.6
79.8000
City 3
5
156
31.2
51.7000
ANOVA
Source of Variation
SS
df
MS
F
P-value
F crit
Between Groups
1171.2000
2
585.6000
9.3546
0.0036
3.8853
Within Groups
751.2000
12
62.6000
Total
1922.4000
14
Level of significance
0.05
For this ANOVA test we get p-value < alpha value, so we reject the null hypothesis. This means we conclude that there is sufficient evidence the population averages for the given three cities are not same.
Part b
The multiple comparison by using Tukey Kramer is given as below:
Tukey-Kramer Multiple Comparisons
Sample
Sample
Group
Mean
Size
1: City 1
9.6
5
2: City 2
21.6
5
3: City 3
31.2
5
Other Data
Level of significance
0.05
Numerator d.f.
3
Denominator d.f.
12
MSW
62.6
Q Statistic
3.77
Absolute
Std. Error
Critical
Comparison
Difference
of Difference
Range
Results
Group 1 to Group 2
12
3.538361203
13.34
Means are not different
Group 1 to Group 3
21.6
3.538361203
13.34
Means are different
Group 2 to Group 3
9.6
3.538361203
13.34
Means are not different
(ANOVA table and multiple tests for comparison are calculated by using Excel as manual calculations are not possible due to time constraint.)
ANOVA: Single Factor
SUMMARY
Groups
Count
Sum
Average
Variance
City 1
5
48
9.6
56.3000
City 2
5
108
21.6
79.8000
City 3
5
156
31.2
51.7000
ANOVA
Source of Variation
SS
df
MS
F
P-value
F crit
Between Groups
1171.2000
2
585.6000
9.3546
0.0036
3.8853
Within Groups
751.2000
12
62.6000
Total
1922.4000
14
Level of significance
0.05
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.