Question

Jacob Lee is a frequent traveler between Los Angeles and San Diego. For the past month,...

Jacob Lee is a frequent traveler between Los Angeles and San Diego. For the past month, he wrote down the flight times in minutes on three different airlines. The results are:

Goust Jet Red Cloudtran
51 50 52
51 53 55
52 52 60
42 62 64
51 53 61
57 49 49
47 50 49
47 49
50 58
60 54
54 51
49 49
48 49
48 50

1. Null hypothesis:

2. Alternate hypothesis:

3.  State the decision rule for 0.05 significance level.

4. Develop a 95% confidence interval for the difference in the means between Goust and Cloudtran.

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Answer #1

using minitab>Stat>ANOVA >one way ANOVA

we have

One-way ANOVA: Goust, Jet Red, Cloudtran

Method

Null hypothesis All means are equal
Alternative hypothesis At least one mean is different
Significance level α = 0.05
Rows unused 6

Equal variances were assumed for the analysis.


Factor Information

Factor Levels Values
Factor 3 Goust, Jet Red, Cloudtran


Analysis of Variance

Source DF Adj SS Adj MS F-Value P-Value
Factor 2 127.1 63.56 3.04 0.062
Error 32 669.9 20.93
Total 34 797.0


Model Summary

S R-sq R-sq(adj) R-sq(pred)
4.57526 15.95% 10.70% 0.00%


Means

Factor N Mean StDev 95% CI
Goust 14 50.50 4.47 (48.01, 52.99)
Jet Red 14 52.07 3.83 (49.58, 54.56)
Cloudtran 7 55.71 6.05 (52.19, 59.24)

Pooled StDev = 4.57526


Tukey Pairwise Comparisons

Grouping Information Using the Tukey Method and 95% Confidence

Factor N Mean Grouping
Cloudtran 7 55.71 A
Jet Red 14 52.07 A B
Goust 14 50.50 B

Means that do not share a letter are significantly different.


Tukey Simultaneous Tests for Differences of Means

Difference SE of Adjusted
Difference of Levels of Means Difference 95% CI T-Value P-Value
Jet Red - Goust 1.57 1.73 (-2.68, 5.83) 0.91 0.639
Cloudtran - Goust 5.21 2.12 ( 0.00, 10.43) 2.46 0.050
Cloudtran - Jet Red 3.64 2.12 (-1.57, 8.85) 1.72 0.213

Individual confidence level = 98.06%

1. the Null hypothesis :

2. Alternate hypothesis: At least one mean is different

3.  since p value of F stat is 0.062 which is greater than 0.05 so we dont reject Ho and conclude   All means are equal

4. a 95% confidence interval for the difference in the means between Goust and Cloudtran. is (-10.43 , 0.00)

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