Basic Statistics Measures of Variability

The Range Deviation Score The Standard Deviation The Variance

STRUCTURE OF STATISTICS STATISTICS DESCRIPTIVE INFERENTIAL TABULAR GRAPHICAL NUMERICAL CONFIDENCE INTERVALS TESTS OF HYPOTHESIS Continuing with numerical approaches. NUMERICAL

STRUCTURE OF STATISTICS NUMERICAL DESCRIPTIVE MEASURES DESCRIPTIVE TABULAR GRAPHICAL NUMERICAL CENTRAL TENDENCY VARIABILITY SYMMETRY

STRUCTURE OF STATISTICS NUMERICAL DESCRIPTIVE MEASURES NUMERICAL CENTRAL TENDENCY VARIABILITY SYMMETRY RANGE VARIANCE STANDARD DEVIATION

You are an elementary school teacher who has been assigned a class of fifth graders whose mean IQ is 115. Because children with IQ of 115 can handle more complex, abstract material, you plan many sophisticated projects for the year. Do you think your project will succeed ? General population 85% We need the variability of IQs in the class!

Having graduated from college, you are considering two offers of employment. One in sales and the other in management. The pay is about same for both. After checking out the statistics for salespersons and managers at the library, you find that those who have been working for 5 years in each type of job also have similar averages. Can you conclude that the pay for two occupations is equal? Is the average salary enough? We need the variability! management Sales Much more Much less $20,000

Group of scores Single score Central Tendency measures IQ of 100 students Mean IQ=118

Central Tendency Measures ? ? ? Measures of Central Tendency do not tell you the differences that exist among the scores More homogeneous

Central Tendency Same Mean---Different Variability So What? 60 How many are out here ?

1. The Range = The difference between the largest (X max ) and the smallest (X min ) , 21, 22, 23, 28, 26, 24, Range = 29 –21 = A large range means there is a lot of variability in data.

10, 28, 26, 27, Range = 29 –10 = ? Drawbacks of The Range Range = 29 –26=3 The Range depends on only the two extreme scores

Because the range is determined by just two scores in the group, it ignores the spread of all scores except the largest and smallest. One aberrant score or outlier can be greatly increase the range Range and Extreme Observations R R R

Range and Measurement Scales 1, 2, 3, Country Code SESFAge 3-1=2 1=American 2=Asian 3=Mexican 1=Upper 2=Middle 3=Lower Before you determine the Range, all scores must be arranged in order

3. The Variance ? Differences among Scores

Total Variability = Sum of Individual Variability How can you determine the variability of each individual in group? The amount of Individual difference entirely depends on comparison criteria.

Can you figure out total amount of differences among scores ? Can you figure out how much each score is different from other scores ?

48 Reference score ? Mean Score You need a Common Criteria for computing Total Variability

Reference score ? 49 You need a Common Criteria for computing Total Variability Deviation Scores A Deviation score tells you that a particular score deviate, or differs from the mean

DEVIATION SCORE = (X i - Mean) A score a great distance from the mean will have large deviation score. Mean ABCDEF 1 2 3

Total amount of variability?! Sum of all distance values! Sum of Deviation Scores No way! conceptually mathematically

The idea makes sense…but If you compute the sum of the deviation scores, the sum of the deviation scores equals zero! Sum of Deviation scores = (-4) + (-3) + (-2) (1) + (2) + (3) + (4) = 0

The Sum of Absolute Deviation Scores Sum of absolute deviation scores ( ) = 20 The sum of absolute deviations is rarely used as a measure of variability because the process of taking absolute values does not provide meaningful information for inferential statistics.

Sum of Squares of deviation scores “SS” Conceptually And Mathematically

Sum of Squares of Deviation Scores, SS Instead of working with the absolute values of deviation scores, it is preferable to (1) square each deviation score and (2) sum them to obtain a quantity know as the Sum of Squares. SS=(-4)+(-3)+(-2)+(-1)+0+(1)+(2)+(3)+(4) = = SS= i

Group of scores “A” Group of scores “B” SS(A)=30 SS(B)=40 Can you say that the variability of the data in Group B is greater than the data in Group A? So !

3, 4 What happens to SS when we look at some data? Group A Group B Mean = 3.5 SS = ( ) + ( ) + ( ) + ( ) = SS = ( ) + ( ) =.50

i=1 N i SS tends to increase as number of data(N) increase. SS is not appropriate for comparing variability among groups having unequal sample size. How can you overcome the limitation of SS Mean

If SS is divided by N The resulting value will be Mean of the Deviation Scores (Mean Square) VARIANCE

3, 4 Group A Group B Mean = 3.5 V = ( ) + ( ) =.50/2 =.25 V = ( ) + ( ) + ( ) + ( ) = 1.00/4 =

Variance Population Variance Sample Variance

POPULATION VARIANCE Sigma Square Population size Population meanIndividual value

SAMPLE VARIANCE Sample variance The sample variance (S 2 ) is used to estimate the population variance (  2 ) Individual value Sample Mean Sample size-1 Degree of freedom

Why n-1 instead n ?

Sampling n Sampling error = ?<100<? 100 population sample

The value of the squared deviations is less from X than from any other score. Hence, in a sample, the value of (X-X) n would be less than n. > n n Ideally, a sample variance would be based on  (x -  )2. This is impossible since  is not known if one has only a sample of n cases.  is substituted by.

> Ideal sample variance One could correct for this bias by dividing by a factor somewhat less than n n-1

sample n=5 If we know that the mean is equal to 5, and the first 4 scores add to 18, then the last score MUST equal 7. n-1 are free to change Degree of freedom 7 We know that ? must equal 25.

4. Standard Deviation SD Positive square root of the variance PopulationSample

The Standard Deviation and the Mean with Normal Distribution 

Normal Distribution   -1  -2  -3  +2  +3  +1  Relationship between  and 

Normal Distribution Relationship between and S -1S -2S -3S +1S +2S +3S 68% 95% 99.9%

EMPIRICAL RULE For any symmetrical, bell-shaped distribution, approximately 68% of the observations will lie within  1 standard deviation of the mean; approximately 98% within  2 standard deviations of the mean; and approximately 99.9% within  3 standard deviations of the mean.

You can approximately reproduce your data! If a set of data has a Mean=50 and SD=10, then… % 95% 99%