Spss political analysts

Variation, Dispersion, Homogeneity

Introduction to standardized error terms

STATISTICS as MEASUREMENTS

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HOW WELL (GOOD) DOES THE MODE “FIT” THE DATA? How much do the data vary from the mode?

CA UT Rep 4,483,810 515,231 Dem 8,753,788 310,676 Libert. 478,500 243,690 Green 278,657 9,438 McMullin 39,596 243,690 Other 203,533 12,787

Total 14,237,884 1,131,430

CA UT Rep 31.5% 45.5% Dem 61.5% 27.5% Libert. 3.4% 3.5% Green 2.0% 0.8% McMullin 0.3% 21.5% Other 1.4% 1.1%

Total 100.0% 100.0%

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How “varied” is a population?

 Start with nominal data—2 measures

 Simplest: Variation Ratio—by how much does the population “vary from” the modal category? How useful is the mode as summary?

 More complicated, but relevant—How diverse is the population? IQV (not covered in this class due to time constraints—in text)

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The mode “fits” the data better in California than Utah

 Why?

 Lower Probability of making a wrong guess

 Measure of “goodness of fit”: mode is more “accurate” /”gooder” in summarizing CA

 —we lose less information

 What do we lose?

 Variation Ratio:  =proportion not in the modal category

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Variation ratio

N

f VR

mk 1

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Variation Ratio =proportion not in modal category

=1-pmk

 VRc = 1-.615 = .385  Mode does not “fit” 38.5% of the cases; 38.5% vary from the mode;  If I guessed the mode, I would have a .385 probability of making a wrong

guess with any randomly drawn individual.

 VRU = 1-.455 = .545 (higher=worse fit)  Methodological concern—perhaps use an election (HR) with less 3rd party activity

 Lowest VR ?  = 0 (all in mode—perfect guess)

 Highest VR ?  ~1 but never 1 (why not?)

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VR is a standardized error measurement

 Variation Ratio=proportion not in modal category

mpVR 1

N

f VR

mk 1

N

f

N

N VR

mk 

N

fN VR

mk 

•Number of errors

•standardized by total number

•(why: more errors possible in California)

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New Concept –actually not (observations and expectations)

(hypotheses=expected/tests=observed)

 If fe = frequency expected if our guess was always correct = N

 If fo = actual frequency observed in modal category

 Then: proportionately, how much are the data off when I use the mode?

 Those who took statistics

—chi-square?

N

fN VR

mk 

e

oe

f

ff VR

 

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Variation, Dispersion, Homogeneity

Interval Data

STATISTICS as MEASUREMENTS

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Interval Distribution (Quantitative)

 Range: Difference between highest (xh) and lowest (xl) value

 If you randomly pulled 2 individuals, what would be the maximum difference? Always correct if you state range or less.

 In our example: $850,000-$180,000 =$670,000 (but $4 million?)

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But what about those other 2?

Less dispersion / better fit

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So, let’s use the precision of all of the data, not just 2 points:

 Concept of Deviation Score

 Deviation score: how much does each case vary from the mean?

 Difference between what we would expect (e) if our best guess were the mean, and what we actually observe (o) for each individual xi.

 It’s an individualized error term.

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Deviation Score

 Where is the mean of the distribution

 … and is the value for each individual value

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List of deviation scores

xi di

x=1 $180,000 $-145,000

x=2 $180,000 $-145,000

x=3 $200,000 $-125,000

x=4 $210,000 $-115,000

x=5 $230,000 $-95,000

x=6 $250,000 $-75,000

x=7 $500,000 $175,000

x=8 $850,000 $525,000

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Mean of deviation scores always equals 0 WHY?

POINT OF REDISTRIBUTION—

AMOUNT BELOW EQUALS

AMOUNT ABOVE

Need to take care of ‘0’

problem/negatives canceling

out positives

1. Take absolute value of

each and calculate the mean

= MAD (mean absolute deviation)

2. Take square of each value

and calculate the mean = Variance

Variance will usually be:

1. A very large number

2. Expressed in squared units of measurement

xi di

x=1 $180,000 $-145,000

x=2 $180,000 $-145,000

x=3 $200,000 $-125,000

x=4 $210,000 $-115,000

x=5 $230,000 $-95,000

x=6 $250,000 $-75,000

x=7 $500,000 $175,000

x=8 $850,000 $525,000

Mean = $325,000 $0

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xi di | di | di 2

x=1 $180,000 $-145,000 $145,000 $$21,025,000,000

x=2 $180,000 $-145,000 $145,000 $$21,025,000,000

x=3 $200,000 $-125,000 $125,000 $$15,625,000,000

x=4 $210,000 $-115,000 $115,000 $$13,225,000,000

x=5 $230,000 $-95,000 $95,000 $$9,025,000,000

x=6 $250,000 $-75,000 $75,000 $$5,625,000,000

x=7 $500,000 $175,000 $175,000 $$30,625,000,000

x=8 $850,000 $525,000 $525,000 $$275,625,000,000

Mean = $325,000 $0 MAD=

$175,000 VARIANCE=

$$48,975,000,000

Standard Deviation = SQRT (VARIANCE) = $221,303

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Mean of Individual Values =$325,000

N

N

i ix

1

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Mean of Individual Deviation Scores

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Mean of Absolute Deviations of Individual Deviation Scores = MAD

Mean Absolute Deviation (MAD) —often listed as Absolute Deviation:

If I guessed the mean PAC contribution for each candidate, I would be off, on mean average, by $175,000

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Mean of Squared Individual Deviation Scores = Variance

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Square Root of Variance = standard deviation

Standard deviation= $221,303 =square root of the variance=gets us back in

line, but, will usually be greater than the MAD, with the greatest outliers creating a greater influence.

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