Question

Random samples of female and male UVA undergraduates are asked to estimate the number of alcoholic drinks that each consumes on a typical weekend. The data is below: Females (Population 1): 2, 4, 4, 5, 4, 3, 5, 4, 2, 2 Males (Population 2): 8, 9, 6, 9, 9, 5, 5, 7, 8, 8 Give a 91.6% confidence interval for the difference between mean female and male drink consumption. (Assume that the population variances are equal.) Confidence Interval =

Answer #1

Ans:

df=10+10-2=18

t*=TINV(0.084,18)=1.829

Females | Males | |

1 | 2 | 8 |

2 | 4 | 9 |

3 | 4 | 6 |

4 | 5 | 9 |

5 | 4 | 9 |

6 | 3 | 5 |

7 | 5 | 5 |

8 | 4 | 7 |

9 | 2 | 8 |

10 | 2 | 8 |

mean | 3.5 | 7.4 |

std. dev. | 1.1785 | 1.5776 |

Point estimate=3.5-7.4=-3.9

standard error for difference=sqrt((1.1785^2/10)+(1.5776^2/10))=0.6227

91.6% confidence interval for the difference between mean female and male drink consumption

=-3.9+/-1.829*0.6227

=-3.9+/-1.139

**=(-5.039, -2.761)**

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