Question

Consider the following population: {7, 8, 8, 9, 9}. The value of
*μ* is 8.2, but suppose that this is not known to an
investigator. Three possible statistics for estimating *μ*
are

Statistic 1: the sample mean, *x*

Statistic 2: the sample median

Statistic 3: the average of the largest and the smallest values in the sample

A random sample of size 3 will be selected without replacement. Provided that we disregard the order in which the observations are selected, there are 10 possible samples that might result (writing 8and 8*, 9 and 9* to distinguish the two 8's and the two 9's in the population.)

7, 8, 8* | 7, 8, 9 | 7, 8, 9* | 7, 8*, 9 | 7, 8*, 9* |

7, 9, 9* | 8, 8*, 9 | 8, 8*, 9* | 8, 9, 9* | 8*, 9, 9* |

(a)

For each of these 10 samples, compute Statistics 1, 2, and 3.

Statistic 1 | Statistic 2 | Statistic 3 | |
---|---|---|---|

7, 8, 8* | |||

7, 8, 9 | |||

7, 8, 9* | |||

7, 8*, 9 | |||

7, 8*, 9* | |||

7, 9, 9* | |||

8, 8*, 9 | |||

8, 8*, 9* | |||

8, 9, 9* | |||

8*, 9, 9* |

(b)

Construct the sampling distribution of each of these statistics. (Enter each statistic from smallest to largest.)

Statistic 1 | ||||
---|---|---|---|---|

p(Statistic 1) |

Statistic 2 | ||
---|---|---|

p(Statistic 2) |

Statistic 3 | |||
---|---|---|---|

p(Statistic 3) |

(c)

Which statistic would you recommend for estimating *μ*
and why?

Using the sampling distributions above, the means of the three
statistics are calculated to be *E*(Statistic 1)
= , *E*(Statistic 2) = , and
*E*(Statistic 3) = . Since the mean of
Statistic ? 1 2 3 is closest to *μ* = 8.2, we
pick the ---Select--- sample mean sample median average
of the largest and smallest values in the sample as the best of the
three statistics for estimating *μ*.

Answer #1

a)

Statistic 1 | statistic 2 | Statistic 3 | |

7,8,8* | 7.666667 | 8 | 7.5 |

7,8,9 | 8 | 8 | 8 |

7,8,9* | 8 | 8 | 8 |

7,8*,9 | 8 | 8 | 8 |

7,8*,9* | 8 | 8 | 8 |

7,9,9* | 8.33 | 9 | 8 |

8,8*,9 | 8.33 | 8 | 8.5 |

8,8*,9* | 8.33 | 8 | 8.5 |

8,9,9* | 8.67 | 9 | 8.5 |

8*,9,9* | 8.67 | 9 | 8.5 |

b)

Statistic 1 | 7.67 | 8 | 8.33 | 8.67 |

Probability | 1/10=0.1 | 4/10=0.4 | 3/10=0.3 | 2/10=0.2 |

Statistic 2 | 8 | 9 | ||

probability | 7/10=0.7 | 3/10=0.3 | ||

Statistic 3 | 7.5 | 8 | 8.5 | |

probability | 1/10=0.1 | 5/10=0.5 | 4/10=0.4 |

c) E ( Statistic 1 ) =

= 7.67 * 0.1 + 8 * 0.4 + 8.33 * 0.3 + 8.67 * 0.2

= 8.2

E ( Statistic 2 ) =

= 8 * 0.7 + 9 * 0.3 = 8.3

E ( statistic 3 ) =

= 8.15

We have population mean = μ = 8.2

Since the mean of **Statistic 1** is closest to μ =
8.2, we pick the **sample mean** as the best of the
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