The article “The Ball-on-Three-Ball Test for Tensile Strength: Refined Methodology and Results for Three Hohokam Ceramic Types” (M. Beck, American Antiquity, 2002:558-569) discusses the strength of ancient ceramics. Several specimens of each of three types of ceramic were tested. The loads (in kg) required to crack the specimens are as follows:

Ceramic Type |
Loads (kg) |

Sacaton |
15.30,51,20. 17, 19, 20, |

32. 17,15,23, 19. 15. 18, | |

16, 22,29, 15, 13, 15 | |

Gila Plain |
27. 18, 28,25, 55,21, 18, |

34,23, 30,20, 30,31,25, | |

28, 26,17,19, 16, 24,19, | |

9,31, 19, 27, 20, 43.15 | |

Casa Grande |
20,16,20, 36, 27, 35, |

66,15,18,24,21,30, | |

20,24, 23,21, 13,21 |

a. Construct comparative boxplots for the three samples.

b. How many outliers does each sample contain?

c. Comment on the features of the three samples.

Step 1 of 4:

Here the data regarding loads(in kg) required to crack specimens of three types of ceramic were given.

The data is as follows.

Loads( in kg)

15,30,51,20,17,19,20,32,17,15,23,19,15,18,16,22,29,15,13,15. |
27,18,28,25,55,21,18,34,23,30,20,30,31,25,28,26,17,19,16,24,9,19,31,19,27,20,43,15 |
20,16,20,36,27,35,66,15,18,24,21,30,20,24,23,21,13,21. |

Step 2 of 4:

(a)

For the above data we have to construct the comparative box plots.

To get the box plot the steps to be followed are

>Sort the data in increasing order.

>Find the median,minimum,maximum,1st quartile and 3rd quartile.

Thus,the first sample of data sorted in increasing order as,

13,15,15,15,15,15,16,17,17,18,19,19,20,20,22,23,29,30,32,51.

Second sample of data sorted in increasing order as,

9 15 16 17 18 18 19 19 19 20 20 21 23 24 25 25 26 27 27 28 28 30 30 31 31

34 43 55

Third sample of data sorted in increasing order as,

13 15 16 18 20 20 20 20 21 21 21 23 24 24 27 30 35 36 66

Now we have to find the median,minimum,maximum,first and third quartiles.

For the first sample of data,

Median=(n+1)/2 th value in the arranged series where n is the sample size.

=th value

=10.5th value

=

=18.5

Minimum=13

Maximum=51

First quartile,=median of the lower part.

=Median of the values below the median value.

=median of 13,15,15,15,15,15,16,17,17,18

=15

Third quartile,=median of the upper part

=median of the values above the median

=median of 19,19,20,20,22,23,29,30,32,51

=22.5

In the same the median,minimum,maximum,, for the second sample is

Median=24

Minimum=9

Maximum=55

=19

=28.5

For the third sample

Median=21

Minimum=13

Maximum=66

=20

=25.5

Using these,the boxplot obtained is,