The Texas Transportation Institute at Texas A and M University conducted a survey to determine the number of hours per year drivers waste sitting in traffic of 75 urban areas studied; the most jammed urban area was Los Angeles where drivers wasted an average of 90 hours per year. Other jammed urban areas included Denver, Miami and San Francisco. Assume sample data for six drivers in each of these cities show the following number of hour’s waster per year sitting in traffic.
Denver | Miami | San Francisco |
---|---|---|
70 | 66 | 65 |
62 | 70 | 62 |
71 | 55 | 74 |
58 | 65 | 69 |
57 | 56 | 63 |
66 | 66 | 75 |
Mean of Denver = =
= = 64
Similarly the other for the other cities can be calculated. Hence we have
Ø Mean of Denver = 64 hours
Ø Mean of Miami = 63 hours
Ø Mean of San Francisco = 68 hours
The above mentioned steps are followed using Microsoft Excel and the output is given below:
Anova: Single Factor | ||||
SUMMARY | ||||
Groups | Count | Sum | Average | Variance |
Denvver |
6 | 384 | 64 | 35.6 |
Miami | 6 | 378 | 63 | 36.8 |
San Francisco | 6 | 408 | 68 | 31.2 |
ANOVA | ||||||
Source of Variation | SS | df | MS | F | P-value | F crit |
Between Groups | 84 | 2 | 42 | 1.216216 | 0.323966 | 3.68232 |
Within Groups | 518 | 15 | 34.53333333 | |||
Total | 602 | 17 |
Hence the P-value corresponding to F test statistic is 0.323966.
Statistical decision:
From the Anova table, we see that that P-value (0.323966) corresponding to the F test statistic is greater than 0.05 (level of significance). Hence we do not reject the null hypothesis at 0.05 level of significance.
i.e:, there is no significance difference among the mean hours wasted by the drivers in the three cities.
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