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

In order to test whether there is difference in the mean hours wasted by the car drivers, we have to perform as

H0:

That is there is no significance difference among the mean hours wasted by the drivers in the three cities.

H1: Not all population means are equal.

That is there is significance difference among the mean hours wasted by the drivers in the three cities.

The Analysis of Variance is carried out using

- Select ‘
**Data**’ from the main menu. Then click ‘**Data Analysis’**. - Select ‘
**ANOVA – one factor’**from the analysis tools. - Input the given data range in ‘
**input range’**. - Then click ‘
**OK**’ to get the result.

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**.

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