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Chapter 5 Individual measures Raw scores (Xi) Differential scores (xi) Standard scores (Zi) Fundamental statistical characteristics Group indexes Central tendency Variability (Dispersion) Bias (Asymmetry) Skewness (Kurtosis) Individual indexes Position Centiles (Ci) Percentiles (Pi) Quartiles (Qi) Raw scores (Xi) Differentials scores (xi) Standard scores (Zi) 2 1) X: it’s the score given to each subject to be subjected to any test. Xi 2) x: it’s the raw score minus the mean. xi X i X 3) Z: it’s the differential score divided by the standard deviation. Xi X Zi S 3 Scores comparison 1) The comparison of raw scores (Xi) can lead us to misleading conclusions. 2) The solution based on the distances to the mean (xi) is not an entirely satisfactory solution. 3) The solution is to use the Typical scores, Typed or standardized. 4 Raw scores interpretation: Xi 5 CASE 1: a subject obtains a score on an IQ test: 1 subject = 1 variable How can I interpret the score of X = 25? A raw score such as 25 does not give us more than a number. But, 25 is too much? Is it enough? The answer is: IT DEPENDS. What is the factor on which this score depend? * group mean and * their variability So, raw scores are not ENOUGH. X=25 6 CASE 2: 1 subject 2 variables (not comparable) John weighs 75 kg and is 1.80 m. tall His weight, is it more or less than his height? The answer is that they are not directly comparable. 7 CASO 3: 1 subject 2 variables (supposedly comparable) A student X has recently been examined in Motivation and Methods. If the scores have been respectively 30 and 15: Can we say that the student has done better in Motivation than in Methods? 8 Differential scores interpretation: xi 9 Scores of Deviation Diversions Errors or bias Symbolically xi = X i - X 10 Differentials scores: xi Tell us if a raw score is HIGHER, SMALLER than the mean or THE SAME as the mean. This information is inferred from the sign of the differential score. There exist differential scores positive and negative. 11 Raw and differential scores: Xi = 7 – 9 – 10 – 11 - 13 Xi 7 Xi2 49 (Xi - X ) 7-10 = -3 (Xi - X )2 9 9 81 9-10 = -1 1 10 100 10-10 = 0 0 11 121 11-10 = 1 1 13 169 13-10 = 3 9 50 520 0 20 12 3 3 1 1 X 7 9 10 11 13 x -3 -1 3 0 1 1 1 3 3 13 Conclusions • RAW X 10 S 4 2 X SX 2 DIFFERENTIALS x 0 S 4 2 x Sx 2 14 Example Two groups (A and B) were measured in their intellectual level: A: 97 102 B: 92 97 107 102 112 117 107 112 117 122 15 X A 107 X B 107 Supposing that 2 students, one belonging to group A and another to the group B, received the same intellectual level of 117 points. x X - X 117 107 10 16 This equality in differential scores may be masking very different realities. • Group A 97 • Group B 102 107 112 117 92 97 102 107 112 117 122 17 Conclusions As we can see, group A’s scores are much more homogeneous (less scattered, less variable) than group B’s scores. This means that, while the score 117 in the group A represents an extreme value – because it’s the maximum score-, the same score in group B is not as extreme, there is another score above it (122). 18 Standards scores: Zi One solution to this situation is: Xi X Zi SX 19 Typing or Standardization The process of obtaining the typical scores is known as: Typing or Standardization. Typical scores are also known as scores: Typified or Estandardized. 20 DEFINITION “The standard score, typified or estandardized indicates the number of standard deviations that a particular raw score is separated from its mean." 21 Standard scores calculation: Zi Xi Zi Zi2 7 -3/2 = -1.5 2.25 9 -1/2 = -0.5 0.25 10 0 0 11 1/2 = 0.5 0.25 13 3/2 = 1.5 2.25 Tot. 50 0 n=5 22 Graphically there is a shift to the left side of the x-axis when we convert raw scores in differential: xi [-3 -1 0 Xi 1 3] [7 9 10 11 13] And there is a concentration of scores when they become standards. Zi -1.5 –0.5 0 0.5 1.5 23 Zs properties 1 The sum of standard scores is 0: Zi = 0 24 Demonstration Xi X 0 Zi 0 S 1 ( X i X) 0 S 25 2 The mean of standard scores is 0 Z=0 26 Demonstration Zi Z 0 n as Zi 0 0 Z 0 n 27 3 The sum of the squares of standard scores is equal to n: Zi = n 2 28 Demonstration 2 ( X X ) 1 i 2 2 Zi ( X X ) 2 2 i S S 1 n 2 2 ( X X ) ( X X ) n i i 2 2 ( X i X) ( X i X) n 29 4 The standard deviation and variance of standard scores is equal to 1. Zi n 2 = -Z = =1 n n 2 SZ 2 30 Demonstration 2 Z S Z n but, as Z 0 y Z2 n n 2 SZ 1 y S Z 1 1 n 2 Z 2 i 31 Standard scores properties summary 1. The sum of typical scores is 0: Z=0 2. The mean of standard scores is 0: Z=0 3. The sum of the squares of standard scores is equal to n: Zi =n 2 4. The standard deviation and variance of standard 2 scores is equal to 1: 2 Z n 2 i SZ = -Z = =1 n n Example The following scores: 4,85 1,68 1,12 0,56 0 -0,56 -1,12 -1,68 -4,85 Answer justification: A) Can they be the difference scores of a data set? B) Can they be the standard scores of a data set? 33 Typical scores and normal curve 34 6 5 4 The limit of the bar chart: to increase indefinitely the number of bars, reduces its size and thin bars are more and more. In the limit, the broken line is identified with the curve (Normal or Gaussian). 20 18 16 14 12 3 2 1 0 10 8 6 4 2 0 35 Among all probability distributions from absolutely continuous random variables, this is the most important. These distributions are knowing as Laplace-Gauss, but there were had already been used by De Moivre. 36 Normal curve 1) The sample size grows. 2) The bar size decreases. For each pair of values of and we’ll have one normal curve. 37 Normal curve characteristics 38 1ª 0 - + This distribution depends on 2 parameters : and 39 2ª Mo = Mdn = The normal curve has a single maximum. 40 3ª 1 1 The normal curve has 2 inflection points 41 4ª The normal curve is asymptotic to the abscissa. 42 5ª g1 = 0 It is a symmetrical distribution(g1 = 0) 43 6ª g2 = 0 It is a mesokurtic distribution (g2 = 0). 44 X N ( ; ) The parameter gives the distribution center and the parameter variability , verifying the following equalities : , the 45 -3 -2 - 68 % 2 3 95,5 % 99,7 % 46 Case 1 A B C A B C 47 A B C A A B A C B 48 Applications to the typified normal curve 49 The characteristics of the normal distribution have many distributions, each with their means and standard deviations. Implies the existence of infinite normal distributions. Solution: convert raw scores into typical. Implies to transform the normal distribution in a standardized normal distribution. It has always mean 0 and standard deviation 1. Generally, standardized distances in themselves are not usually of our interest, but as an intermediate product between raw scores and proportions. Translation of raw scores into proportions. Translation of proportion to raw scores. Translation: use the Tables ¿Qué distancia estandarizada representa el 33.4% de los datos, inmediatamente por encima de la media? Attending to raw scores and table use, we’ll going to use one of the following 2 choices: 1. Translations since raw scores to proportions: Translation since raw scores to standard scores. Look for the area in the table. We work with the areas if it’s necessary to obtain the result. Ex. In a normal distribution with mean 100 and standard deviation 15, how much data proportion do the values between 70 and 130 have? 2. Translations since proportions to raw scores: We work with the area. Translation the area (to the mean) in typical scores. Translation this score in a raw score. Ex. In a normal distribution with mean 100 and standard deviation 15, what score defines the top 10% of the data?