Lecture 4
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Standard Observations
How far is an item from the mean?
Are items in two different data-sets at comparable distance from their means?
Standardization is extremely useful for comparing individual items in a data-set
What is the relative position of the item?
Standardized Value, Yi = (Xi - X bar) / s
Example - Exam Scores
70 72 75 80 85 90 100
n = 7 X bar = 81.71 s = 10.75
X2 = 70 Y2 = (72 - 81.71) / 10.75 = -0.90
What does this number tell you?
-1.09 -0.90 -0.62 -0.16 +0.31 +0.77 +1.70
What should these numbers add up to?
Percentiles
A second method for giving relative position of items in data-setMedian is the 50th percentile
P-th percentile is the value with at most p% below and at most (100-p)% above
Example: 25th Percentile for US-Open
108 112 106 114 109 114 105 115121 109 102 108 98 115 109 110
Arrayed: 98 102 105 106 108 108 109 109
109 110 112 114 114 115 115 121 Item Rank Cummulative Percentage98 1 6.25
102 2 12.50
105 3 18.75
106 4 25.00
108 5 31.25
108 6 37.50
109 7 43.75
109 8 50.00
Box-and-Whisker Plot
Very effective visual summary of data
Elements needed for constructing box plot
1st Quartile
Median
3rd Quartile
Inner Fences
Outer Fences
The Ist Q and 3rd Q values are referred to as the hinges
Interquartile range (IQR) = 3rd Q - 1st Q
Inner Fences are 1.5 times the IQR from the hinges
Outer Fences are 3.0 times the IQR from the hinges
The whiskers extend to the most extreme values in the inner fences
Example : US-Open
Arrayed: 98 102 105 106 108 108 109 109
109 110 112 114 114 115 115 1213rdQ = 114.0…..median = 109.0
1stQ = 107.0…..IQR = 114.0 - 107.0 = 7.0
Lower inner fence = 107 - 1.5 * 7 = 96.5
Upper inner fence = 114 + 1.5*7 = 124.5
Lower whisker extend to 98
Upper whisker extend to 121
Lower outer fence = 107 - 3*7 = 86
Upper outer fence = 114 + 3*7 = 135
values between inner and outer fences appear as * values beyond outer fences appear as o
95-------100-------105-------110-------115-------120-------125
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