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?

MINITAB

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.