Chapter 02
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  • 1. Chapter TwoThe mean, the variance,and the standard deviation1. The Mean and other Measures of Central TendencyThe arithmetic mean is the average of a set of numbers. It can besymbolised as M and its formula is: N ∑X ∑X Arithmetic Mean = M = i=1 = (2 :1) N NIt is the most important of three measures of central tendency.The other two are the median and the mode. The mode is definedsimply as the value which occurs most frequently. The median isthe value below which exactly fifty percent of cases fall, and abovewhich are exactly fifty percent of cases. It is the point which splitsthe set of scores into two equal parts.ProblemsFind the mean, mode and median of the following sets of scores. A. 1, 2, 2, 3, 4, 5, 6. B. 10, 11, 12, 14, 15, 15. C. 7, 7, 8, 8, 10, 10.
  • 2. Philip Ley. Quantitative Aspects of Psychological Assessment 11AnswersA. Mean 3.3; mode 2; median 3.B. Mean 12.8; mode 15; median 13 (by convention half way between the two mid-most scores when there is an even number of scores).C. Mean 8.3; mode 8; median 8.From this point onwards subscripts will be used only whennecessary.Returning now to the formula for the mean it can be shown that: ∑ X = NM (2:2)Proof ⎛∑ X ⎞(1) NM = N ⎜ ⎟. ⎝ N ⎠ N∑ X(2) = . N(3) The N’s cancel one another out leaving ∑X.This general principle that the sum of a set of values equals Ntimes the mean of that set will be useful at several later points.It is also true that: ∑ (X − M ) = 0 (2:3)© 1972, 2007 Philip Ley Text re-typed for computer by Irene Page
  • 3. Philip Ley. Quantitative Aspects of Psychological Assessment 12Proof(1) ∑ (X − M ) = ∑ X − ∑ M by Summation Rule 1.(2) Further by Summation Rule 3, as the mean is a constant: ∑ M = NM , so (1) becomes ∑ X − NM .(3) But we have just shown in equation (2:2) that NM = ∑ X so we obtain ∑ X − ∑ X = 0.The value X i − M x , the deviation of a score from the mean, iscalled a deviation score and is sometimes symbolised as x i .Similarly Yi − M y is symbolised as yi . As demonstrated in (2:3) ∑ x = 0; ∑y=02. The VarianceThe variance is a measure of dispersion. It tells us somethingabout the scatter of scores around the mean. It is defined as themean squared deviation from the mean, and symbolised by asmall sigma squared - σ . Its formula is: 2 ∑ (X − M ) 2 Variance = σ x = (2:4) 2 Nor using x for X – M ∑x 2 σx = 2 (2:5) N It follows from this formula that:© 1972, 2007 Philip Ley Text re-typed for computer by Irene Page
  • 4. Philip Ley. Quantitative Aspects of Psychological Assessment 13 ∑ x = ∑ (X − M ) = Nσ 2 (2:6) 2 2 x(2:6) is obtained from (2:5) by multiplying both sides of theequation by N.Another variant of the formula for the variance is: ∑x 2 σx = 2 −M2 (2:7) NProof σ x2 = ∑ (X − M ) / N . 2(1)(2) = ∑ (X 2 + M 2 − 2XM )/ N.(3) Using Summation Rules 1 and 3 this becomes: (∑ X 2 + NM 2 − 2∑ XM )/ N(4) but we have shown in (2:2) that ∑ X = NM so we can write: σ x2 = (∑ X 2 + NM 2 − 2NMM )/ N(5) but 2NMM = 2NM so (4) becomes: 2 σ x2 = (∑ X 2 − NM 2 )/ N(6) Dividing by N this gives: ∑X 2 σx = 2 −M2 N© 1972, 2007 Philip Ley Text re-typed for computer by Irene Page
  • 5. Philip Ley. Quantitative Aspects of Psychological Assessment 14The numerator (top part) of equation (2:4) for the variance is thesum of squared deviations from the mean. This sum is usuallycalled the sum of squares. Sum of squares = SS = ∑ (X − M ) (2:8) 2An alternative formula for this value is: SS = ∑ X 2 − (∑ X ) 2 (2:9) NProof(1) In the proof of (2:7) at (5) it has been shown that ∑ (X − M ) = ∑ X − NM 2 2 2 ⎛ ∑ X ⎞⎛ ∑ X ⎞(2) But NM = N ⎜ ⎟⎜ ⎟ 2 ⎝ N ⎠⎝ N ⎠ N∑ X∑ X(3) Multiplying this becomes: N2(4) Dividing numerator and denominator by N gives: (∑ X ) 2 N(5) Substituting this in (1) we obtain: ∑ (X − M ) = ∑ X 2 2 − (∑ X ) 2 N© 1972, 2007 Philip Ley Text re-typed for computer by Irene Page
  • 6. Philip Ley. Quantitative Aspects of Psychological Assessment 151. The Standard DeviationThe standard deviation is the square root of the variance and issymbolised by a small Greek sigma - σ . Its formula is the squareroot of any of the formulae for the variance, e.g. ∑x 2 σx = (2:10) NThe mean, the variance and the standard deviation are importantin psychometrics because of their relationships to the normalcurve. These relationships will be discussed in the next chapter.ProblemsGiven the following set of scores: 1, 2, 3, 4, 5, 6, 7.Find:A. The mean.B. The variance.C. The standard deviation.D. ∑ (X − M ).AnswersA. 4; B. 4; C. 2; D. 0.© 1972, 2007 Philip Ley Text re-typed for computer by Irene Page