Question

In a random sample of 18 people, the mean commute time to work was 32.4 minutes and the standard deviation was 7.2 minutes. Assume the population is normally distributed and use a t-distribution to construct a 99% confidence interval for the population mean mu. What is the margin of error of mu? Interpret the results.

Answer #1

Sample mean = x̅ = 32.4 |

Sample size = n = 18 |

Sample S.D = s = 7.2 |

Standard Error = s/√n = 7.2/√18 = 1.69706 |

Confidence Level = 99 |

Significance Level = α = (100-99)% = 0.01 |

Degrees of freedom = n-1 = 18 -1 = 17 |

Critical value = t* = 2.898 [ using Excel =TINV(0.01,17) ] |

Margin of Error = t*(s/√n )= 2.898*1.69706 = 4.91808 |

Lower Limit = x̅ - Margin of Error = 32.4 - 4.91808 = 27.4819 |

Upper Limit = x̅ + Margin of Error = 32.4 + 4.91808 = 37.3181 |

27.4819 < ? < 37.3181

Margin of Error = 4.91808

Interpretation : |

We are 99% confident that the true population mean is lies between 27.4819 and 37.3181 . |

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