Question

A particular report included the following table classifying 818 fatal bicycle accidents that occurred in a certain year according to the time of day the accident occurred.

Time of Day | Number of Accidents |
---|---|

midnight to 3 a.m. | 47 |

3 a.m. to 6 a.m. | 52 |

6 a.m. to 9 a.m. | 84 |

9 a.m. to noon | 72 |

noon to 3 p.m. | 79 |

3 p.m. to 6 p.m. | 157 |

6 p.m. to 9 p.m. | 193 |

9 p.m. to midnight | 134 |

For purposes of this exercise, assume that these 818 bicycle accidents are a random sample of fatal bicycle accidents.

Suppose a safety office proposes that bicycle fatalities are
twice as likely to occur between noon and midnight as during
midnight to noon and suggests the following hypothesis:
*H*_{0}: *p*_{1} =
1/3, *p*_{2} = 2/3 ,where
*p*_{1} is the proportion of accidents occurring
between midnight and noon and *p*_{2} is the
proportion occurring between noon and midnight. Do the data given
provide evidence against this hypothesis, or are the data
compatible with it? Justify your answer with an appropriate test.
Use a significance level of 0.05.

State the appropriate alternative hypothesis.

*H*_{a}:
*p*_{1} > 1/3,
*p*_{2} < 2/3

*H*_{a}:
*p*_{1} = 1/3,
*p*_{2} = 2/3

*H*_{a}:
*p*_{1} > 1/3,
*p*_{2} > 2/3

*H*_{a}:
*H*_{0} is not true.

*H*_{a}:
*p*_{1} ≠ 1/3,
*p*_{2} = 2/3

Find the test statistic and *P*-value. (Use technology.
Round your test statistic to three decimal places and your
*P*-value to four decimal places.)

*X*^{2}=

*P*-value=

State the conclusion in the problem context.

Reject *H*_{0}. There is convincing evidence to
conclude that bicycle fatalities are not twice as likely to occur
between noon and midnight as during midnight to noon.

Reject *H*_{0}. There is not convincing evidence
to conclude that bicycle fatalities are not twice as likely to
occur between noon and midnight as during midnight to noon.

Fail to reject *H*_{0}. There is convincing
evidence to conclude that bicycle fatalities are not twice as
likely to occur between noon and midnight as during midnight to
noon.

Fail to reject *H*_{0}. There is not convincing
evidence to conclude that bicycle fatalities are not twice as
likely to occur between noon and midnight as during midnight to
noon.

Answer #1

alternative hypothesis
:*H*_{a}:
*H*_{0} is not true.

Applying chi square tesT:

relative | observed | Expected | residual | Chi square | |

category | frequency |
O_{i} |
E_{i}=total*p |
R^{2}_{i}=(O_{i}-E_{i})/√E_{i} |
R^{2}_{i}=(O_{i}-E_{i})^{2}/E_{i} |

midnight to noon | 1/3 | 255 | 272.67 | -1.07 | 1.145 |

noon to midnight | 2/3 | 563 | 545.33 | 0.76 | 0.572 |

total | 1.000 | 818 | 818 | 1.717 |

X^{2} =1.717

p value =0.1901

Fail to reject *H*_{0}. There is not convincing
evidence to conclude that bicycle fatalities are not twice as
likely to occur between noon and midnight as during midnight to
noon.

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38
3
a.m. to 6 a.m.
27
6
a.m. to 9 a.m.
66
9
a.m. to Noon
77
Noon
to 3 p.m.
98
3
p.m. to 6 p.m.
127
6
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164
9
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