Problem 1RE

Auto Fatalities The table below lists auto fatalities by day of the week for a recent year (based on data from the Federal Highway Administration). Minitab results are also shown. Use a 0.01 significance level to test the claim that auto fatalities occur on the different days of the week with the same frequency. Can you provide an explanation for the results?

Day |
Mon |
Tues |
Wed |
Thurs |
Fri |
Sat |
Sun |
|||

Frequency |
3797 |
3615 |
3724 |
4004 |
4867 |
5786 |
5004 |

Answer:

Step 1 of 2

The expected values for each category, Ei , could be determined. With these observed and expected numbers of cases, the hypotheses can be written as

H0 : the claim that auto fatalities occur on the different days of the week with the same frequency

H1 : the claim that auto fatalities occur on the different days of the week with the different frequency.

Use a 0.05 significance level

SL. No. |
O |
p = 1/7 |
E = np |
(O - E )2 |
(O - E)2/E |

1 |
3797 |
0.142857 |
4399.571 |
363092.3 |
82.52902 |

2 |
3615 |
0.142857 |
4399.571 |
615552.3 |
139.9119 |

3 |
3724 |
0.142857 |
4399.571 |
456396.8 |
103.7366 |

4 |
4004 |
0.142857 |
4399.571 |
156476.8 |
35.56636 |

5 |
4867 |
0.142857 |
4399.571 |
218489.5 |
49.66153 |

6 |
5786 |
0.142857 |
4399.571 |
1922184 |
436.9026 |

7 |
5004 |
0.142857 |
4399.571 |
365333.9 |
83.03852 |

Sum |
30797 |
1 |
30797 |
4097526 |
931.3466 |

The first step in conducting the significance test is to compute the expected frequency for each outcome given that the null hypothesis is true. For example, the expected frequency of a "1" is 7 since the probability of a "1" coming up is 1/7 and there were a total of 30797.

Expected frequency (E) = np

= 30797(1/7)

= 4399.571

The calculation continues as follows. Letting E be the expected frequency of an outcome and O be the observed frequency of that outcome.