## Answers

a)

degree of freedom(df) =(rows-1)*(columns-1)= | 2 |

b)

Applying chi square test of independence: |

Expected | E_{i}=row total*column total/grand total | c | d | e | Total |

a | 61.431 | 28.038 | 19.532 | 109 | |

b | 133.569 | 60.962 | 42.468 | 237 | |

total | 195 | 89 | 62 | 346 | |

chi square χ^{2} | =(O_{i}-E_{i})^{2}/E_{i} | c | d | e | Total |

a | 0.505 | 0.148 | 0.639 | 1.2916 | |

b | 0.232 | 0.068 | 0.294 | 0.5940 | |

total | 0.7371 | 0.2162 | 0.9323 | 1.8857 | |

test statistic X^{2}= | 1.886 |

c)

p value = | 0.390 | from excel: chidist(1.8857,2) |

d)

since p value >0.05

option C is correct

Do not reject that they are independent at 5% significance level and reserve judgement. We may accept independence, but with some unknown probability.

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