CHAPTER 6

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CHAPTER 6

SOLUTIONS TO PROBLEMS

John Tukey developed the techniques of EDA to allow researchers to explore the distribution of the variables in our data set. They permit us to visually see and comprehend what our distributions look like, what abnormalities that might be present (e.g. outliers etc.), and what secrets they might hold.
There are five steps involved in the construction of a stem and leaf disply:

Step 1) Rank order your variable distribution in ascending order from lowest to highest values.

Step 2) Decide which digits you will use as your stems and which you will use as your leaves. If each case contains more than two digits, will you truncate or round them?

Step 3) Decide if multiple stems are necessary to properly display your distribution.

Step 4) List your selection of stems and place each leaf value onto its respective stem.

Step 5) Finish the display with a legend containing interpretation information: stem width, leaf width, sample size, and sample interpretation.
The first thing to do, is to rank order the twenty judges' age from low to high:
```
46	58	61	66
49	58	62	67
50	59	63	70
51	60	65	71
57	61	65	71
```
Now we can make our stem and leaf display. Let the stem represent units of ten and the leaves represent units of one. We will let one leaf represent each case as follows:
```
f      Stem       Leaves 
2       4         69
6       5         017889
9       6         011235567
3       7         001

Stem Unit = 10 years
1 leaf represents 1 case
5 | 1 represents 51 years old
N = 20
```
The range in age is 71 - 46 = 25 years.

gifh 20 scores, the position of the median is The median score, then, is the average of the scores in the 10^th and 11^th positions. This is equal to . The median age of th e judges is, therefore, 61 years of age.

The shape of the distribution is approximately symmetrical. There is no negative or positive skew.
A stem and leaf display shows every data point of the distribution. From it you can comprehend the general shape and skewness of the distribution. With a little counting, you can easily determine the median. A box and whisker plot does not display all the data points. It is more graphic and visual, while the stem and leaf display is more numerical. From the box of the box and whisker plot you can determine the shape of the middle 50% of a distribution of scores. From the length of the w hiskers and the location of the median in the box, you can get some sense of the skewness of the distribution.
To construct a box and whisker plot from this data, first rank order all scores from low to high.
```
1224		2803		5713		10081		16532		79868
1303		2959		6174		10232		17881
1497		3184		6520		11435		23463
1562		3596		7326		12090		24947
1663		3619		8335		12306		28653
1769		4101		8564		12896		30476
1811		4438		8632		13665		38541
2434		4594		8941		14041		38583
2756		4782		9355		14136		42384
2777		5497		9556		15728		66261
```
The position of the median with 51 scores is: . The median is the score in the 26^th position. The median, therefore, is equal to 8564 persons.

To find the interquartile range, first find the quartile positions. The quartile positions can be found by taking the truncated median position, adding one, and then dividing by two: . The first quartile, then, can be found by averaging the scores at the 13^th and 14^th positions from the lowest score, and the third quartile can be found by averaging the scores at the 13^th and 14^th positions from the highest score in the distr ibution. The first quartile is equal to:. The third quartile is equal to:. The Interquartile Range, then, is equal to: . Now that we have the interquartile range, we can calculate our fences:
```
Low Inner Fence        3390 - (1.5) 10698.5 = -12,657.75
Lower Outer Fence      3390 - (3.0) 10698.5 = -28,705.50

High Inner Fence       14088.5 + (1.5) 10698.5 = 30,136.25
Higher Outer Fence      14088.5 + (3.0) 10698.5 = 46,184

and our adjacent values:

Low Adjacent Value = 1224
High Adjacent Value = 28,653
```
We can also determine that we have four mild outliers (New Jersey, Illinois, Florida, and Texas) and two extreme outliers (New York and California).

Now we have all the information we need to construct our box and whisker plot. We show this for your below.

Before constructing a double stem and leaf display, you should rank order the data from low to high:

10		14		18		21		25
10		14		18		21		26
11		15		18		22              27
12		17		18		23		28
13		17		19		24		29
13		17		20		24		29

Now we can create double stem and leaf display:

f       Stem        Leaves           
8        1*         00123344
9        1·         577788889
7        2*         0112344
6        2·         567899

Stem Unit = 10 offenses
1 leaf represents 1 case
1|7 represents 17 offenses
N = 30

With 30 cases, the median is in the position. The average of the 15^th and 16^th scores is equal to . The median, therefore, is 18 offenses.

As we have done for other problems, before constructing a box and whisker plot we will rank order the data from low to high.
```
52		74		87		95
60		74		87		95
62		75		88		99
63		75		89		100
66		77		90		101
67		78		92		102
67		78		92		103
68		80		92		104
69		81		93		104
69		84		93		105
```
The position of the median with 40 scores is the score. The average of the 20^th and 21^st score is equal to . The median IQ score, then, is 85.5.

The position of the first and third quartiles is: . The first quartile, then, is the average of the 10^th and 11^th scores from the lowest score: . The third quartile is the average of the 10^th and 11^th scores from the highest score: . The interquartile range is equal to: .

Now we can construct our fences and identify the low and high adjacent values.
```
Low Inner Fence     = 71.5 - (1.5)22 = 38.5
Lower Outer Fence   = 71.5 - (3.0)22 = 5.5

High Inner Fence    = 93.5 + (1.5)22 = 126.5
Higher Outer Fence  = 93.5 + (3.0)22 = 159.5

Low Adjacent Value  = 52
High Adjacent Value = 105
There are no outliers.
```
The completed box and whisker plot is shown below.

Since the bottom whisker is slightly longer than the upper whisker, and the median line in the box is above the half-way point, we would conclude that there is a small negative skewness to the distribution of these IQ scores.

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