Question

: Let ?1, ?2,…. . , ?12 (12 random variables iid) like

a variable ? (?; 1). The following data are observed:

2.5345792; 2.5236928; 0.3303505; 0.1267132; 1.3670369;
0.2349068;

1.5209821; 0.6114980; 2.3096728; 1.6590382; 4.0726550;
4.7865432

(1) Give a point estimator of ?.

(2) Find a 99% confidence interval for ?

Answer #1

1. Point estimate of

2. Create the following table.

data | data-mean | (data - mean)^{2} |

2.5345792 | 0.6947792 | 0.48271813675264 |

2.5236928 | 0.6838928 | 0.46770936189184 |

0.3303505 | -1.5094495 | 2.2784377930503 |

0.1267132 | -1.7130868 | 2.9346663843342 |

1.3670369 | -0.4727631 | 0.22350494872161 |

0.2349068 | -1.6048932 | 2.5756821834062 |

1.5209821 | -0.3188179 | 0.10164485336041 |

0.6114980 | -1.228302 | 1.508725803204 |

2.3096728 | 0.4698728 | 0.22078044817984 |

1.6590382 | -0.1807618 | 0.03267482833924 |

4.0726550 | 2.232855 | 4.985641451025 |

4.7865432 | 2.9467432 | 8.6832954867462 |

Find the sum of numbers in the last column to get.

So standard deviation is

t value for 99% CI is TINV(0.01,11)=3.106

So Margin of Error is

Hence CI is

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