## Answers

a)

Null hypothesis: Babies are born uniformly in all 12 months

ALternate hypothesis Ha: Babies are not born uniformly in all 12 months

b)

expected value =np=162*(1/12) =13.5

c)

applying chi square goodness of fit test: |

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

category | frequency(p) | 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} |

Jan | 1/12 | 14.000 | 13.500 | 0.14 | 0.019 |

Feb | 1/12 | 12.000 | 13.500 | -0.41 | 0.167 |

Mar | 1/12 | 12.000 | 13.500 | -0.41 | 0.167 |

Apr | 1/12 | 10.000 | 13.500 | -0.95 | 0.907 |

May | 1/12 | 16.000 | 13.500 | 0.68 | 0.463 |

Jun | 1/12 | 9.000 | 13.500 | -1.22 | 1.500 |

Jul | 1/12 | 16.000 | 13.500 | 0.68 | 0.463 |

Aug | 1/12 | 11.000 | 13.500 | -0.68 | 0.463 |

Sep | 1/12 | 21.000 | 13.500 | 2.04 | 4.167 |

Oct | 1/12 | 8.000 | 13.500 | -1.50 | 2.241 |

Nov | 1/12 | 24.000 | 13.500 | 2.86 | 8.167 |

Dec | 1/12 | 9.000 | 13.500 | -1.22 | 1.500 |

total | 1.000 | 162 | 162 | 20.222 |

test statistic X^{2} = | 20.222 |

d)

as test statistic is higher than critical value we reject null and conclude that babies were born at different rates in different months.

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