1. Chapter 4: Variability

2. Variability The goal for variability is to obtain a measure of how spread out the scores are in a distribution. A measure of variability usually accompanies a measure of central tendency as basic descriptive statistics for a set of scores.

3. Central Tendency and Variability Central tendency describes the central point of the distribution, and variability describes how the scores are scattered around that central point. Together, central tendency and variability are the two primary values that are used to describe a distribution of scores.

4. Variability Variability serves both as a descriptive measure and as an important component of most inferential statistics. As a descriptive statistic, variability measures the degree to which the scores are spread out or clustered together in a distribution. In the context of inferential statistics, variability provides a measure of how accurately any individual score or sample represents the entire population.

5. Variability (cont'd.) When the population variability is small, all of the scores are clustered close together and any individual score or sample will necessarily provide a good representation of the entire set. On the other hand, when variability is large and scores are widely spread, it is easy for one or two extreme scores to give a distorted picture of the general population.

6. Measuring Variability Variability can be measured with –The range –The standard deviation/variance In both cases, variability is determined by measuring distance. - distance between two scores - distance between a score and the mean

7. The Range (p. 106) The range is the total distance covered by the distribution, from the highest score to the lowest score (using the upper and lower real limits of the range). range = URL for X max – LRL for X min

8. The Range (cont'd.) Alternative definitions of range: –When scores are whole numbers or discrete variables with numerical scores, the range tells us the number of measurement categories. range = X max – X min + 1 –Alternatively, the range can be defined as the difference between the largest score and the smallest score. (commonly used definition, especially for discrete variables) range = X max – X min

9. The Standard Deviation Standard deviation measures the standard (average) distance between a score and the mean. The calculation of standard deviation can be summarized as a four-step process:

10. Dispersion: Deviation (p. 107-) deviation = X – μ  find the mean of the deviation  not a good measure for variability... Why? because..... mean serves as a balance point for the distribution  Σ (X – μ) = 0 (see p. 107 example 4.1) mean deviation is always zero no matter what... So we need to get rid of the +/- sign  sum of squared deviations (but you also squared the “unit of measurement”, e.g. dollar square, age square )  mean square deviation (standard/average measure)  take a square root ( the unit of measurement is normal again!!, e.g. dollar, age )

11. Example 4.1 (p.107) N = 4 X X- μ 85 1-2 30 0-3 sum =120 mean =3

12. The Standard Deviation (cont'd.) 1. Compute the deviation (distance from the mean) for each score. 2. Square each deviation. 3.Compute the mean of the squared deviations. For a population, this involves summing the squared deviations (sum of squares, SS) and then dividing by N. The resulting value is called the variance or mean square and measures the average squared distance from the mean. For samples, variance is computed by dividing the sum of the squared deviations (SS) by n - 1, rather than N. The value, n - 1, is know as degrees of freedom (df) and is used so that the sample variance will provide an unbiased estimate of the population variance. 4. Finally, take the square root of the variance to obtain the standard deviation.

14. 3-14 Measures of Dispersion Range Variance Standard Deviation

15. Computing the Variance Steps in computing the variance: Step 1: Find the mean. Step 2:Find the difference between each observation and the mean, and square that difference. Step 3:Sum all the squared differences found in Step 2. Step 4:Divide the sum of the squared differences by the number of items in the population.

16. Population Variance Sum of Squares: : Variance=mean squared deviation

17. Example 4.2 (p.109) N = 5 X = 1, 9, 5, 8, 7 X X- μ SS 1-525 939 51 824 711 sum =30040 mean = 68  Sum of Squares  Variance

18. Ex 4 (p.110) N = 5, X = 4, 0, 7, 1, 3 Calculate the variance  Sum of Squares  Variance X X- μ SS 411 0-39 7416 1-24 300 sum =15030 mean = 36

19. Example 4.3 (p.111) N = 4 X = 1, 0, 6, 1  Sum of Squares  Variance X X- μ SSX*X 111 0-240 641636 111 sum =802238 mean = 25.5

20. Sample Variance

21. 

22. Sample Standard Deviation

23. degrees of freedom: df Remember population mean won’t change no matter what, but sample mean will change whenever you collect a new set of sample! Example 4.6 (p.117) n = 3, M = 5, X = 2, 9, ? the 3 rd score is determined by M=ΣX/n  only n-1 scores are free to vary or independent of each other and can have any values  degrees of freedom = n-1

24. degrees of freedom: df=1, n=2

25. degrees of freedom: df=1, n=2 (cont.)

26. degrees of freedom: df=2, n=3

27. degrees of freedom: df=2, n=3 (cont.) Actually, SS can be reformulated into 2 distinct “distances”:

28. degrees of freedom: df=n-1 with n scores in the sample SS can be reformulated into n-1 distinct “distances”, so “mean” variation has n-1 degrees of freedom:

29. Example 4.5 (p.116) n= 7 X = 1, 6, 4, 3, 8, 7, 6  Sum of Squares  Variance X X- μ SSX*X 1-4161 61136 4116 3-249 83964 72449 61136 sum =35036211 mean = 56

30. Ex 2 (p.118) N = 5, X = 1, 5, 7, 3, 4 if instead this is a sample of n=5  Sum of Squares X X- μ SSX*X 1-391 51125 73949 319 40016 sum =200 100

31. Unbiased, Biased If the average value of all possible sample statistics = population parameter  unbiased e.g. we select 100 samples (n=5)  calculate 100 sample means and variances  calculate the average mean (ΣM/100) and average variance (Σs 2 /100), if  unbiased Otherwise, the sample statistic is biased

32. Example 4.7 (p.120) Population: N=6, X = 0, 0, 3, 3, 9, 9  μ = 4, σ 2 = 14 Select 9 samples with n=2 (sample statistics in Table 4.1) SampleMSS/nSS/(n-1) 1000 21.52.254.5 3 20.2540.5 41.52.254.5 5300 66918 74.520.2540.5 86918 9900 sum =3663126 mean = 4 7 14 BiasedUnbiased

33. Visualize your data set by μ, σ or M, s using histogram or other graphs If your score = 85  high enough to get the scholarship?

34. Properties of the Standard Deviation If a constant is added to every score in a distribution, the standard deviation will not be changed. If you visualize the scores in a frequency distribution histogram, then adding a constant will move each score so that the entire distribution is shifted to a new location. The center of the distribution (the mean) changes, but the standard deviation remains the same.

35. Properties of the Standard Deviation (cont'd.) If each score is multiplied by a constant, the standard deviation will be multiplied by the same constant. Multiplying by a constant will multiply the distance between scores, and because the standard deviation is a measure of distance, it will also be multiplied.

36. The Mean and Standard Deviation as Descriptive Statistics If you are given numerical values for the mean and the standard deviation, you should be able to construct a visual image (or a sketch) of the distribution of scores. As a general rule, about 70% of the scores will be within one standard deviation of the mean, and about 95% of the scores will be within a distance of two standard deviations of the mean.

37. 3-* The Empirical Rule

38. Example 4.8 (p.124-125) any consistent difference between two treatment? Experiment A: 2 sets of data, sd is quite small  so ∆M = 5 is distinct and easy to see Experiment B: 2 sets of data, sd is quite large  ∆M = 5: it is difficult to discern/recognize the differences error variance: the result of unsystematic differences (unexplained and uncontrolled differences between scores), e.g. static noise (radio)

40. Demo 4.1 (p. 129) Sample: n=6, X = 10, 7, 6, 10, 6, 15 compute the variance and standard deviation. ΣX = 54, ΣX 2 = 546, SS = 546-(54 2 /6) = 546 – 486 = 60 s 2 = SS/(n-1) = 60/5 = 12 s = 3.4641