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DATA ANALYSIS

DATA ANALYSIS. Module Code :CA660 (Application Areas: Bio-, Business, Social, Environment etc.). STRUCTURE of Investigation/DA . 1,2, many samples E.D., Reg n. , C.T. Replication, Assays, Counts. Estimation/H.T. H.T. Study techniques. Lab. techniques. Non-Parametric.

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DATA ANALYSIS

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  1. DATA ANALYSIS Module Code :CA660 (Application Areas: Bio-, Business, Social, Environment etc.)

  2. STRUCTURE of Investigation/DA 1,2, many samples E.D., Regn., C.T. Replication, Assays, Counts Estimation/H.T. H.T. Study techniques Lab. techniques Non-Parametric Parametric Distributional Assumptions, Probability , Estimation properties Basis: Size/Type of Data Set/Tools Level of Measurement 2

  3. Probability & Statistics Primer -overviewNote:Short overview. Other statistical distributions in lectures

  4. Summary Statistics- DescriptiveIn analysis of practical sets of data, useful to define a small number of values that summarise main features present. We derive (i) representative values, (ii) measures of spread and (iii) measures of skewness and other characteristics.Representative ValuesSometimes called measures of location or measures of central tendency.1. Random ValueGiven a set of data S = { x1, x2, … , xn }, we select a random number, say k, in the range 1 to n and return the value xk. This method of generating a representative value is straightforward, but it suffers from the fact that extreme values can occur and successive values could vary considerably from one another.2. Arithmetic Mean For the set S above, the arithmetic mean (or just mean) is x = {x1 + x2 + … + xn }/ n.If x1 occurs f1 times, x2 occurs f2 times and so on, we get the formula x = { f1 x1 + f2 x2 + … + fnxn } / { f1 + f2 + … + fn } ,written

  5. Example 1. Data are student marks in an examination. Find the average mark for the class. Note 1: Marks are given as ranges, so care MarkMid-Point Numberneeded in range interpretation of Range of Students All intervals must be of equal rank and therexififi ximust be no gaps in the classification0 - 19 10 2 20 We interpret the range 0 - 19 to contain marks 21 - 39 30 6 180greater than 0 and less than or equal to 20. 40 - 59 50 12 600Thus, mid-point is 10. The other intervals are 60 - 79 70 25 1750 are interpreted accordingly. 80 - 99 90 5 450 Sum - 50 3000 The arithmetic mean is x = 3000 / 50 = 60 marks. Note 2: Pivot. If weights of size fi are suspended x1x2 x xnfrom a metre stick at the points xi, then the average is the centre of gravity of thedistribution. Consequently, it is very sensitive f1 f2 fnto outlying values. Note 3: Population should be homogenous for average to be meaningful. For example, if assume that typical height of girls in a class is less than that of boys, then average height of all students is neither indicative of the girls nor of the boys.

  6. Frequency 3. The ModeThis is the value that occurs mostfrequently. By common agreement,it is calculated from the histogram using linear interpolation on the modal class.The various similar triangles in the diagram generate the common ratios. In our case, the mode is 60 + 13 / 33 (20) = 67.8 marks.4. The MedianThe middle point of the distribution. If { x1, x2, … , xn } are marks of studentsin a class, arranged in nondecreasingorder, then the median is the mark of the (n + 1)/2 student.Often use the ogiveorcumulative frequency Diagram to calculate. In our case,the median is 60 + 5.5 / 25 (20) = 64.4 marks. 50 13 20 25 13 20 12 6 5 2 20 40 60 80 100 Cumulative Frequency 50 25.5 20 40 60 80 100

  7. Measures of Dispersion or ScatteringExample 2. The distribution shown has the same Marks Frequencyarithmetic mean as Example 1, but values are morexjfjfjxjdispersed. Illustrates that an average value alone may not adequately describe statistical 10 6 60distributions.30 8 240 50 6 300To devise a formula that captures degree to which a 70 15 1050distribution is concentrated about the average, we 90 15 1350consider the deviations of the values from the average. Sums 50 3000If distribution is concentrated around the mean, then deviations will be small, while if it is very scattered,then deviations will be large. The average of the squaresof the deviations is called the variance and this is used as a measure of dispersion. The square root of the variance is the standard deviation, has same units of measurement as the original values and is the preferred measure of dispersion in many applications. x6 x5 x4 x3 x2 x1 x

  8. Variance & Standard Deviation s2 = VAR[X] = Average of the Squared Deviations = S f { Squared Deviations } / S f= Sfi{ xi - x } 2 / S fi = S f xi 2 / S f - x 2 , called the product moment formula. s = Standard Deviation = Ö VarianceExample 1 Example 2 f x f x f x2 f x f x f x22 10 20 200 6 10 60 6006 30 180 5400 8 30 240 720012 50 600 30000 6 50 300 1500025 70 1750 122500 15 70 1050 735005 90 450 40500 15 90 1350 12150050 3000 198600 50 3000 217800VAR [X] = 198600 / 50 - (60) 2 VAR [X] = 217800 / 50 - (60)2 = 372 marks2 = 756 marks2

  9. Other Summary StatisticsSkewnessAn important attribute of a statistical distribution is its degree of symmetry. The “skew” means a tail, so distributions with a large tail of outlying values on the right-hand-side are positively skewed or skewed to the right. The notion of negative skewnessis defined similarly. A simple formula for skewness isSkewness = ( Mean - Mode ) / Standard Deviationwhich for Example 1 is:Skewness = (60 - 67.8) / 19.287 = - 0.4044.Coefficient of VariationThis formula was devised to ‘standardise’ the arithmetic mean so comparisons can be drawn between different distributions. Not universally used. Coefficient of Variation = Mean / Standard Deviation. Semi-Interquartile RangeThe Median is the mid or 0.5 point in a distribution. The quartiles Q1, Q2, Q3 correspond to the 0.25, 0.50 and 0.75 points. An alternative measure of dispersion is thus Semi-Interquartile Range = ( Q3 - Q1 ) / 2.Geometric MeanFor data that grow geometrically, e.g. economic data with high inflation effect, another mean is sometimes used. The G.M. is defined for a product of frequencies, where N = S f G. M. = NÖ x1f1 x2f2 … xkfk

  10. Regression[Example 3.] As a motivating example, suppose we model sales data over time.SALES 3 5 4 5 6 7 TIME 1990 1991 1992 1993 1994 1995 Want the straight line “Y = m X + c” that best approximates the data. “Best” in this caseis the line which minimizes the sum of squaresof vertical deviations of points fromthe line:SSQ = SS = ( Yi - [ mXi + c ] )2 Setting partial derivatives of SS w.r.t. m and c to zero leads to the “Normal Equations”Y = m X + n c X Y= m X2+ c X , where n = # points Let 1990 correspond to Year 0.X.X X X.Y Y Y.Y 0 0 0 3 9 1 1 5 5 25 4 2 8 4 16 9 3 15 5 25 16 4 24 6 36 25 5 35 7 49 55 15 87 30 160 Yi = m Xi + c Y Yi m Xi + c 0 Xi X Sales 10 5 0 5 Time

  11. Example 3 - WorkingThe normal equations are: 30 = 15 m + 6 c => 150 = 75 m + 30 c 87 = 55 m + 15 c 174 = 110 m + 30 c=> 24 = 35 m => 30 = 15 (24 / 35) + 6 c => c = 23/7Thus the regression line of Y on X is Y = (24/35) X + (23/7)and to plot the line just need two points, so X = 0 => Y = 23/7 and X = 5 => Y = (24/35) 5 + 23/7 = 47/7. Easy to see that ( X, Y ) satisfies the normal equations, so that the regression line of Y on X passes through “Centre of Gravity” of the data. By expanding terms, getS( Yi - Y ) 2 = S( Yi - [ m Xi + c ] ) 2+ S( [ m Xi + c ] - Y ) 2Total Sum Error Sum Regression Sumof Squares of Squares of SquaresSST = SSE + SSRDistinguish the independent and dependent variables (X and Y respectively) Y Yi mXi +C Y Y X X

  12. CorrelationThe coefficient of determination r2 ( which takes values in the range 0 to 1) is a measure of the proportion of the total variation that is associated with the regression process: r2 = SSR/ SST = 1 - SSE / SST.The coefficient of correlation ‘r’ (values in the range -1 to +1 ) is a more common measure of the degree to which a mathematical relationship exists between X and Y. It can be calculated as: r =  ( X - X ) ( Y - Y ) ( X - X )2 ( Y - Y ) 2 = n  X Y -  X  Y[{ n X 2 - ( X ) 2 } { n  Y 2 - ( Y ) 2 }]Example. In our case,r = {6(87) - (15)(30)}/ { 6(55) - (15)2 } { 6(160) - (30)2 } = 0.907. r = - 1 r = 0 r = + 1

  13. CollinearityFor correlation coefficient value > 0.9 or < - 0.9, we would take this to mean that there is a mathematical relationship between the variables. Does not imply that a cause-and-effect relationship exists.E.g. consider a country with a slowly changing population size, where a certain political party retains a relatively stable percentage of the poll in elections. Let X = Number of people that vote for the party in an election Y = Number of people that die of a given disease in a year Z = Population size.Then, correlation coefficient between X and Y is ~1, indicating a mathematical relationship between them (i.e.) X is a function of Z and Y is a function of Z also. It would clearly be silly to suggest that the incidence of disease is caused by the number of people that vote for the given political party. This is known as the problem of collinearity. Spotting hidden dependencies is non-trivial. Statistical experimentation can only be used to disprove hypotheses, or to lend evidence to support the view that reputed relationships between variables may be valid. Thus, the fact of a high correlation coefficient between deaths due to heart failure in a given year with the number of cigarettes consumed twenty years earlier does not establish a cause-and-effect relationship, though may be useful to guide research.

  14. Overview of Probability TheoryIn statistical theory, an experiment is any operation that can be replicated infinitely often and gives rise to a set of elementary outcomes, which are deemed to be equally likely. The sample space S of the experiment is the set of all possible outcomes of the experiment. Any subset E of the sample space is called an event.An event E occurs whenever any of its elements is an outcome of the experiment. The probability of occurrence of E is P {E} = Number of elementary outcomes in E Number of elementary outcomes in SThe complement E of an event E is the set of all elements that belong to S but not to E. The union of two events E1 E2 is the set of all outcomes that belong to E1or to E2or to both. The intersection of two events E1 E2 is the set of all events that belong to both E1and E2.Two events are mutually exclusiveif occurrence of either precludes occurrence of the other (i.e) their intersection is the empty set . Two events are independent if occurrence of either is unaffected by occurrence or non-occurrence of the other event.Theorem of Total Probability. P {E1E2} = P{E1} + P{E2} - P{E1 E2}Proof. P{E1 E2} = (n1, 0 + n1, 2 + n0, 2) / n = (n1, 0 + n1, 2) / n + (n1, 2 + n0, 2) / n - n1, 2 / n = P{E1} + P{E2} - P{E1 E2}Corollary.If E1 and E2 are mutually exclusive, P{E1 E2} = P{E1} + P{E2} - see Axioms and Addition Rule S E S E1 E2 n1, 0 n1, 2 n0, 2 n0, 0 n = n0, 0 + n1, 0 + n0, 2 + n1, 2

  15. The probability P{E1 | E2} thatE1 occurs, given that E2 has occurred (or must occur) is called the conditional probability of E1. Note : only possible outcomes of the experiment are confined to E2 and not to S.Theorem of Compound ProbabilityMultiplication Rule.P{E1 E2} = P{E1 | E2} P{E2}.Proof. P{E1 E2} = n1, 2 / n = {n1, 2 / (n1, 2 + n0, 2) } { n1, 2 + n0, 2) / n} CorollaryIf E1 and E2 are independent, P{E1 E2} = P{E1} P{E2}. Special caseof Multiplication RuleNote: If Eitself compound, expands further = Chain Rule: P{E7E8 E9} =P{E7 (E8 E9)}Countingpossible outcomes of an event is crucial to calculating probabilities.Apermutation of size r of n different items, is an arrangement of r of the items, where orderof arrangement is important. If orderis not important, the arrangement is called a combination.Example. There are 54 permutations and 54 / (21) combinations of size 2 of A, B, C, D, EPermutations: AB, BA, AC, CA, AD, DA, AE, EACD, DC, CE, EC BC, CB, BD, DB, BE, EBDE, EDCombinations: AB, AC, AD, AE, BC, BD, BE, CD, CE, DEStandard reference books on probability theory give a comprehensive treatment of how these ideas are used to calculate the probability of occurrence of the outcomes of games of chance. S E2 E1 n1, 0 n1, 2 n0, 2 n0, 0

  16. Bayes’ Rule (Theorem): For a series of mutually exclusive and exhaustive events Br, where union of the Br = B1 B2 B3 …….Br = all possibilities for B, Then: Where the denominator is the Total probability of A occurring. Ex. Paternity indices: based on actual genotypes of mother, child, and alleged father. Before collection of any evidence, have a prior probability of paternity P{C}. So, what is the situation after the genetic evidence ‘E’ is in? From Bayes’: P {man is father | E} P[E | man is father} P{man is father} P{man not father | E} P{E | man not father} P{man not father} Written in terms of ratio of posterior probs. (= LHS), paternity index (L say) and ratio of prior probs. (RHS). Rearrange and substitute in above to give prob. of an alleged man with particular genotype ‘C’ being the true father NB: L is a way of ‘weighting’ the genetic evidence; the issue is setting a prior. = 

  17. Statistical Distributions- CharacterisationIf a statistical experiment only gives rise to real numbers, the outcome of the experiment is called a random variable. If a random variable X takes values X1, X2, … , Xn with probabilities p1, p2, … , pnthen the expected(average) value of X is defined to be E[X] = pjXjand its variance is VAR[X] = E[X2] - E[X]2 = pjXj2 - E[X]2. Example. Let X be a random variable measuring Prob. Distancethe distance in Kilometres travelled by children pjXjpjXjpj Xj2to a school and suppose that the following data applies. Then the mean and variance are 0.15 2.0 0.30 0.60 E[X] = 5.30 Kilometres 0.40 4.0 1.60 6.40 VAR[X] = 33.80 - 5.302 =5.71 km2 0.20 6.0 1.20 7.20 0.15 8.0 1.20 9.60 Similar concepts apply to continuous distributions. 0.10 10.0 1.00 1.00 The distribution function is defined by 1.00 - 5.30 33.80 F(t) = P{ X t} and its derivative is the frequency function f(t) = d F(t) / dtso that F(t) = f(x) dx.

  18. Sums and Differences of Random VariablesDefine the covariance of two random variables to be COVAR [ X, Y] = E [(X - E[X]) (Y - E[Y]) ] = E[X Y] - E[X] E[Y].If X and Y are independent, COVAR [X, Y] = 0.Lemma E[ XY] = E[X] + E[Y] VAR [ X Y] = VAR [X] VAR [Y] 2 COVAR [X, Y]E[ k. X] = k .E[X] VAR[ k. X] = k2 .E[X] for a constant k.Example. A company records the journey time X X= 123 4Totalsof a lorry from a depot to customers and Y =1 75 4 4 20 the unloading times Y, as shown. 2 2 6 8 3 19E[X] = {1(10)+2(13)+3(17)+4(10)}/50 = 2.543 1 2 5 311E[X2] = {12(10+22(13)+32(17)+42(10)}/50 = 7.5 VAR[X] = 7.5 - (2.54)2 = 1.0484 Totals 10 13 17 10 50E[Y] = {1(20)+2(19)+3(11)}/50 = 1.82 E[Y2] = {12(20)+22(19)+32(11)}/50 = 3.9VAR[Y] = 3.9 - (1.82)2 = 0.5876 E[X+Y] = { 2(7)+3(5)+4(4)+5(4)+3(2)+4(6)+5(8)+6(3)+4(1)+5(2)+6(5)+7(3)}/50 = 4.36 E[(X + Y)2] = {22(7)+32(5)+42(4)+52(4)+32(2)+42(6)+52(8)+62(3)+42(1)+52(2)+62(5)+72(3)}/50 = 21.04VAR[(X+Y)] = 21.04 - (4.36)2 = 2.0304 E[X Y] = {1(7)+2(5)+3(4)+4(4)+2(2)+4(6)+6(8)+8(3)+3(1)+6(2)+9(5)+12(3)}/50 = 4.82COVAR (X, Y) = 4.82 - (2.54)(1.82) = 0.1972VAR[X] + VAR[Y] + 2 COVAR[ X, Y] = 1.0484 + 0.5876 + 2 ( 0.1972) = 2.0304

  19. Standard Statistical DistributionsMost elementary statistical books provide a survey of commonly used statistical distributions. Importantly, we can characterise them by their expectation and variance (as for random variables) and by the parameters on which these are based; (see lecture notes for those we refer to). So, e.g. for a Binomial distribution, the parameters are p the probability of ‘success in an individual trial’ and n the No. of trials. The probability of success remains constant – otherwise, another distribution applies.Use of the correct distribution is core to statistical inference– I.e. estimating what is happening in the population on the basis of a (correctly drawn, probabilistic)sample.The sample is then representativeof the population.Fundamental to statistical inference is the Normal (or Gaussian), with parameters,  the mean (or formallyexpectationof the distribution) and  (SD) or variance (2). For small samples, or when 2 not known but must be estimated from a sample,a slightly more conservative distribution -the Student’s T or just ‘t’ distribution, applies. Introduces the degrees of freedomconcept.

  20. Student’s t Distribution A random variable X has a t distribution with n degrees of freedom ( tn ) .The t distribution is symmetrical about the origin, with E[X] = 0 VAR [X] = n / (n -2). For small values of n, the tn distribution is very flat. As n is increased the density assumes a bell shape. For values of n ³ 25, the tn distribution is practically indistinguishable from the Standard Normal curve. O If X and Y are independent random variables If X has a standard normal distribution and Y has a cn2 distribution then X has a tn distribution(Y / n)O If x1, x2, … , xn is a random sample from a normal distribution, with mean m and variance s2 and if we define s2 = 1 / ( n - 1) å ( xi - x ) 2 then ( x - m ) / ( s /  n) has a tn- 1 distributionEstimated Sample variance - see calculators ,tables etc. + Many other standard distributions

  21. Sampling TheoryTo draw a random sample from a distribution, assign numbers 1, 2, … to the elements of the distribution, use random number tabes or generated set to decide which elements are included in the sample. If the same element can not be selected more than once, we say that the sample is drawn without replacement; otherwise, the sample is said to be drawn with replacement.Usual convention in sampling is that lower case letters designate the sample characteristics, with capital letters used for the (finite) parent population and greek letters for the infinite. Thus if sample size =n, its elements are designated, x1, x2, …, xn, its mean is x and its modified variance is s2= å (xi - x )2 / (n - 1). Corresponding parent population characteristics = N, X and S2 or (,  and 2)Suppose we repeatedly draw random samples of size n (with replacement) from a distribution with mean  and variance 2. Let x1, x2, … be the collection of sample means and let xi’ = xi - m(i = 1, 2, … )s / Ö nThe collection x1’, x2’, … is called the sampling distribution of means, (usual U or Z)Central Limit Theorem.In the limit, as sample size n tends to infinity,the sampling distribution of meanshas a Standard Normal distribution. Basis for Statistical Inference.

  22. Attribute and Proportionate SamplingIf sample elements are a measurement of some characteristic, then have attribute sampling. However, if all sample elements are 1 or 0 (success/failure,agree/ do-not agree), we have proportionate sampling. For proportionate sampling, the sample average x and the sample proportion p are synonymous, (just as for mean m and proportion P for the parent population). From our results on the Binomial distribution, the sample variance is p (1 - p) and the variance of the parent distribution is P (1 - P) in the proportionate case.The ‘sampling distribution’ of means concept generalizes to get the sampling distribution of any statistic. We say that a sample characteristic is an unbiased estimator of the parent population characteristic, i.e.the expectation of the corresponding sampling distribution is equal to the parent characteristic. Lemma. The sample average (proportion ) is an unbiased estimator of the parent average (proportion): E [ x] = m; so E [p] = P.The quantity Ö ( N - n) / ( N - 1) is called the finite population correction (fpc). If the parent population is infinite or we have sampling with replacement the fpc = 1.Lemma. E [s] = S fpcfor estimated sample S.D. with fpc

  23. Confidence IntervalsFrom the statistical tables for a Standard Normal (Gaussian)distribution, we note that Area Under From To Density Function 0.90 -1.64 1.64 0.95 -1.96 1.96 0.99 -2.58 2.58 From the central limit theorem, if x and s2 are the mean and variance of a random sample of size n (with n greater than 25) drawn from a large parent population, size N , then the following statement ,about the unknown parent mean m, appliesProb { -1.64 £ x - m£ 1.64) » 0.90s / Ö ni.e. Prob { x - 1.64 s / Ö n £ m£ x + 1.64 s / Ö n } » 0.90The range x 1.64 s / Ön is called a 90% confidence interval for the parent mean .Example [ Attribute Sampling]A random sample of size 25 has x = 15 and s = 2. Then a 95% confidence interval for m is 15 1.96 (2 / 5) (i.e.) 14.22 to 15.78Example [ Proportionate Sampling]A random sample of size n = 1000 has p = 0.40 Þ 1.96 Ö p (1 - p) / (n - 1) = 0.03.A 95% confidence interval for P is 0.40 0.03 (i.e.) 0.37 to 0.43. N(0,1) 0.95 -1.96 0 +1.96

  24. Small Sampling TheoryFor reference purposes, it is useful to regard the expression x 1.96 s / nas “default formula” for a confidence interval and to modify it for particular circumstances. O If we are dealing with proportionate sampling, the sample proportion is the sample mean and the standard error (s.e.) term s / n simplifies as follows: x -> p and s / n -> p(1 - p) / (n -1). (Also n-1 -> n) O A 90% confidence interval will bring about the swap 1.96 -> 1.64. O For sample size n less than 25, the Normal distribution must be replaced by Student’s t n - 1 distribution. O For sampling without replacement from a finite population, a fpc term must be used.The width of the confidence interval band increases with the confidence level.Example. A random sample of size n = 10, drawn from a large parent population, has mean x = 12 and a standard deviation s = 2. Then a 99% confidence interval for the parent mean is x 3.25 s / Ö n (i.e.) 12 3.25 (2)/3 (i.e.) 9.83 to 14.17and a 95% confidence interval for the parent mean is x 2.262 s / Ö n (i.e.) 12 2.262 (2)/3 (i.e.) 10.492 to 13.508.Note: For n = 1000, 1.96 Ö p (1 - p) / n » 0.03 for values of pbetween 0.3 and 0.7. This gives rise to the statement that public opinion polls have an “inherent error of 3%”. Simplifies calculations in the case of public opinion polls for large political parties.

  25. Tests of Hypothesis[Motivational Example]. It is claimed that average grade of all 12 year olds in a country in a particular aptitude test is 60%. A random sample of n= 49students gives a mean x = 55% with standard deviation s = 2%. Is the sample finding consistent with the claim? The original claim regarded as anull hypothesis (H0) which is tentatively accepted as TRUE: H0 : m = 60.If the null hypothesis is true, the test statistic TS= x - ms / Ö nis a Random Variable with a Normal(0, 1) i.e. Standardised Normal Z(0,1) (or U(0,1)) distribution. Thus 55 - 60 = - 35 / 2 = - 17.5 2/ Ö 49 rejection regionsis a random value from Z(0, 1). But this lies outside the 95% confidence interval (falls in the rejection region), so either (i) The null hypothesis is incorrector (ii) An event with a probability of at most 0.05 has occurred.Consequently, rejectthe null hypothesis, knowinga probability of 0.05 exists that we arein error. Technically: reject the null hypothesis at the 0.05 level of significance.The alternative to rejecting H0, is to declare the test to be inconclusive.This means that there is some tentative evidence to support the view that H0 is approximately correct. Z(0,1) 0.95 -1.96 1.96

  26. ModificationsBased on the properties of the Normal , Student ‘t’ and other distributions, we can generalise these ideas. If the sample size n < 25, a t n-1 distribution should be used; the level of significance of the test may also be varied or the test applied to a proportionate sampling environment.Example. 40% of a random sample of 1000 people in a country indicate satisfaction with government policy. Test at the 0.01 level of significance if this consistent with the claim that 45% of the people support government policy?Here, H0: P = 0.45 p = 0.40n = 1000 so Ö p (1-p) / n = 0.015 test statistic = (0.40 - 0.45) / 0.015 = - 3.3399% critical value = 2.58 so H0 is rejected at the 0.01 level of significance.One-Tailed TestsIf the null hypothesis is of the form H0 : P > 0.45 then arbitrary large values of p are acceptable, so that the rejection region for the test statistic lies in the left hand tail only.Example. 40% of a random sample of 1000 people in a country indicate satisfaction with government policy. Test at the 0.05 level of significance if this consistent with the claim that at least 45% of the people support government policy?Here the critical value is -1.64, so the null hypothesis H0: P ³ 0.45 is rejected at the 0.05 level of significance N(0,1) 0.95 -1.64 Rejection region

  27. Testing Differences between Means Suppose that x1 x2 … xm is a random sample, mean x and standard deviation s1drawn from a distribution with mean m1 and y1 y2 … yn is a random sample, mean y and standard deviation s2drawn from a distribution with mean m2. Suppose that we wish to test thenull hypothesis that both samples are drawn from the same parent population (i.e.) H0: m1 = m2.The pooled estimate of the parent variance is s*2 = sp2 = { (m - 1) s12 + (n - 1) s22 } / ( m + n - 2)and the variance of ( x – y), is the variance of the difference of two independent random variables, i.e.sdiff2 = sp2 / m + sp2/ n.This allows us to construct the test statistic, which under H0has a tm+n-2 distribution.Example. A random sample of size m = 25 has mean x = 2.5 and standard deviation s1 = 2, while a second sample of size n = 41 has mean y = 2.8 and standard deviation s2 = 1. Test at the 0.05 level of significance if the means of the parent populations are identical.Here H0 : m1 = m2 x - y = - 0.3 and sp2 = {24(4) + 40(1)} / 64 = 2.125so the test statistic is - 0.3 / Ö (2.125 / 25 + 2.125 / 41) = - 0.811The 0.05 critical value for Z(0, 1) is 1.96, so the test is inconclusive.

  28. Paired TestsIf the sample values ( xi , yi ) are paired, such as the marks of students in two examinations, then let di= xi - yi be their differences and treat these values as the elements of a sample to generate a test statistic for the hypothesis H0: m1 = m2.The test statistic d / sd /Ö n has a tn-1 distribution if H0 is true.Example. In a random sample of 100 students in a national examination their examination mark in English is subtracted from their continuous assessment mark, giving a mean of 5 and a standard deviation of 2. Test at the 0.01 level of significance if the true mean mark for both components is the same.Here n = 100, d = 5, sd /Ö n = 2/10 = 0.2so the test statistic is then 5 / 0.2 = 10.the 0.01 critical value for a Z(0, 1) distribution is 2.58, so H0 is rejected at the 0.01 level of significance.Tests for the Variance.For normally distributed random variables, given H0: s2 = k, a constant, then (n-1) s2 / k has a c 2n - 1 distribution.Example. A random sample of size 30 drawn from a normal distribution has variance s2 = 5.Test at the 0.05 level of significance if this is consistent with H0 : s2 = 2 .Test statistic = (29) 5 /2 = 72.5, while the 0.05 critical value for c 229 is 45.72, so H0 is rejected at the 0.05 level of significance.

  29. Chi-Square Test of Goodness of FitCan be used to test the hypothesis H0 that a set of observations is consistent with a given probability distribution. Given a set of categories with observed (Oj ) and expected (Ej ) number of observations (frequency) in each category. Under H0Test StatisticS (Oj -Ej)2 / Ej has a c 2n - 1 distribution, with n the number of categories.Example. A pseudo random number generator is used to used to generate 40 random numbers in the range 1 - 100. Test, at the 0.05 level of significance, if the results are consistent with the hypothesis that the outcomes are randomly distributed.Range 1-25 26 - 50 51 - 75 76 - 100 TotalObserved Number 6 12 14 8 40Expected Number 10 10 10 10 40Test statistic = (6-10)2/10 + (12-10)2/10 + (14-10)2/10 + (8-10)2/10 = 4.The 0.05 critical value of c 23 = 7.81, so the test is inconclusive.Chi-Square Contingency TestTo test that two random variables are statistically independent, a set of observations can be tabled, with m rows corresponding to categories for one random variable and n columns for the other. Under H0, the expected number of observations for the cell in row i and column j = appropriate (row total  column total)  (Grand total). Under H0 Test StatisticSS(Oij -Eij)2 / Eij has a c 2(m -1)(n-1) distribution.

  30. Chi-Square Contingency Test - ExampleIn the following table, the Results Maths History Geography Totalsfigures in brackets are the Honours 100 (50) 70 (67) 30 (83) 200expected values. Pass 130 (225) 320 (300) 450 (375) 900 Fail 70 (25) 10 (33) 20 (42) 100The test statistic is Totals 300 400 500 1200S [ (Oij -Eij)2 / Eij] = (100-50)2/ 50 + (70 - 67)2/ 67 + (30-83)2/ 83 + (130-225)2/ 225+ (320-300)2/ 300 + (450-375)2/375 + (70-25)2/ 25 + (10-33)2/ 33 + (20-42)2/ 42 = 248.976The 0.05 critical value for c 22 * 2 is 9.49 so H0 is rejected at the 0.05 level of significance.Note: In general, chi-squared tests tend to be very conservative vis-a-vis other tests of hypothesis, (i.e.) they tend to give inconclusive results.The meaning of the term “degrees of freedom” . In simplified terms, as the chi-square distribution is the sum of, say k, squares of independent random variables, it is defined in a k-dimensional space. When we impose a constraintof the type that the sum of observed and expected observations in a column are equal or estimate a parameter of the parent distribution, we reduce the dimensionality of the space by 1. In the case of the chi-square contingency table, with m rows and n columns, the expected values in the final row and column are predetermined, so the number of degrees of freedom of the test statistic is (m-1)(n-1).

  31. Analysis of Variance/Experimental Design-Many samples, Means and Variances • Analysis of Variance (AOV or ANOVA) was • originally devised for agricultural statistics • on crop yields etc. Typically, row and column format, = small plots of a fixed size. The yield • yi, j within each plot was recorded. One Way classificationModel: yi, j = + i + i, j , i ,j ~ N (0, s2) in the limitwhere = overall mean as sample size largei= effect of the ith factori, j = error term.Hypothesis: H0: 1 = 2 = … = m y1, 1 y1, 2 y1, 3 y1, 4 y1, 5 1 y2, 1 y2, 2 y2, 3 2 y3, 1 y3, 2 y3, 3 3

  32. Totals MeansFactor 1 y1, 1 y1, 2 y1, 3 y1, n1 T1 = y1, j y1. = T1 / n1 2 y2, 1 y2,, 2 y2, 3 y2, n2 T2 = y2, j y2 . = T2 / n2 m ym, 1ym, 2ym, 3 ym, nm Tm = ym, jym. = Tm / nmOverall mean y = yi, j / n, where n = niDecomposition (Partition) of Sums of Squares: (yi, j - y )2 = ni (yi.- y )2 + (yi, j - yi. )2 Total Variation (Q) = Between Factors (Q1) + Residual Variation (QE )Under H0 : Q / (n-1) -> 2n - 1, Q1 / (m - 1) -> 2m - 1, QE / (n - m) -> 2n - m Q1 / ( m - 1 ) -> Fm - 1, n - m QE / ( n - m ) AOV Table:Variation D.F. Sums of Squares Mean Squares F Between m -1 Q1= ni( yi. - y )2 MS1 = Q1/(m - 1) MS1/ MSE Residual n - m QE= (yi, j - yi .)2 MSE = QE/(n - m) Total n -1 Q= (yi, j. - y )2 Q/( n - 1)

  33. Two-Way Classification Factor I MeansFactor II y1, 1 y1, 2 y1, 3 y1, n y1. : : : :ym, 1ym, 2ym, 3 ym, n ym.Means y. 1 y. 2 y. 3 y . n y. .So we write as y Partition SSQ: (yi, j - y )2 = n (yi. - y )2 + m (y . j - y )2 + (yi, j - yi. - y. j + y )2 Total Between Between Residual Variation Rows Columns VariationModel:yi, j = + i+ j + i, j with i, j~ N ( 0, s2)H0: All iare equal. H0: all j are equalAOV Variation D.F. Sums of Squares Mean Squares F Betweenm -1 Q1= n (yi. - y )2 MS1 = Q1/(m - 1) MS1/ MSE Rows Between n -1 Q2= m (y.j - y )2 MS2 = Q2/(n - 1) MS2/ MSE Columns Residual (m-1)(n-1) QE= (yi, j - yi. - y. j + y)2 MSE = QE/(m-1)(n-1) Total mn -1 Q= (yi, j. - y )2 Q/( mn - 1)

  34. Two-Way Example ANOVA outline Factor I1 2 3 4 5 Totals Means Variation d.f. SSQ MSQ F Fact II 1 20 18 21 23 20 102 20.4 Rows 3 76.95 25.65 18.86**2 19 18 17 18 18 90 18.0 Columns 4 8.50 2.13 1.573 23 21 22 23 20 109 21.8 Residual 12 16.304 17 16 18 16 17 84 16.8 Totals 79 73 78 80 75 385 Total 19 101.75 Means 19.75 18.25 19.50 20.00 18.75 19.25FYI software such as R,SAS,SPSS, MATLAB is designed for analysing these data, e.g. SPSS as spreadsheet recorded with variables in columns and individual observations in the rows. Thus the ANOVA data above would be written as a set of columns or rows, e.g. Var. value 20 18 21 23 20 19 18 17 18 18 23 21 22 23 20 17 16 18 16 17Factor 1 1 1 1 1 1 2 2 2 2 2 3 3 3 3 3 4 4 4 4 4Factor 2 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4

  35. DA APPLICATIONS CONTEXT e.g. BIO • GENETICS : 5 branches; aim = ‘Laws’ of Chemistry, Physics, Maths. for Biology • GENOMICS : Study of Genomes (complete set of DNA carried by Gamete) by integration of 5 branches of Genetics with ‘Informatics and Automated systems’ • PURPOSE of GENOME RESEARCH : Info. on Structure, Function, Evolution of all Genomes – past and present • Techniques of Genomics from molecular, quantitative, population genetics: Concepts and Terminology from Mendelian genetics and cytogenetics 35

  36. CONTEXT: GENETICS - BRANCHES • Classical Mendelian – Gene and Locus, Allele, Segregation, Gamete, Dominance, Mutation • Cytogenetics – Cell, Chromasome, Meiosis and Mitosis, Crossover and Linkage • Molecular – DNA sequencing, Gene Regulation and Transcription, Translation and Genetic Code Mutations • Population – Allelic/Genotypic Frequencies, Equilibrium, Selection, Drift, Migration, Mutation • Quantitative – Heritability/Additive, Non-additive Genetic Effects, Genetic by Environment Interaction, Plant and Animal Breeding

  37. CONTEXT+ : GENOMICS -LINKAGES Mendelian Cytogenetics Molecular GENOMICS Genetic markers DNA Sequences Linkage/Physical Maps Gene Location QTL Mapping Population Quantitative

  38. GENOMICS – some KEY QUESTIONS • HOW do Genes determine total phenotype? • HOW MANY functional genes necessary and sufficient in a given system? • WHAT are necessary Physical/Chemical aspects of gene structure? • IS gene location in Genome specific? • WHAT DNA sequences/structures are needed for gene-specific functions? • HOW MANY different functional genes in whole biosphere? • WHAT MEASURES of essential DNA sameness in different species?

  39. ‘DATA’ : STATISTICALGENOMICS • Some UNUSUAL/SPECIAL FEATURES • Size – databases very large e.g. molecular marker and DNA / protein sequence data; unreconciled, Legacy • Mixtures of variables - discrete/continuous e.g. combination of genotypes of genetic markers (D) and values quantitative traits (C) • Empirical Distributions needed for some Test Statisticse.g. QTL analysis, H.T. of locus order • Intensive Computation e.g. Linkage Analysis, QTL and computationally greedy algorithms in locus ordering, derivation of empirical distributions, sequence match etc. • Likelihood Analysis - Linear Models typically insufficient alone 39

  40. DA APPLICATIONS CONTEXT e.g. BUSINESS/ FINANCE • http://big.computing.dcu.ie/; http://sci-sym.dcu.ie • Data-rich environments – under-utilisation of resources • RAW DATA into useful information and knowledge • Similar underpinning: (‘Laws’)– based on analysis • Purpose – Informed decision-making • Techniques – quantitative. Concepts & Nature – Pervasive, Dynamic, ‘Health’ subject to Internal/External environments. Key elements - Systems and people • Forecasting/Prediction/Trigger 40

  41. CONTEXT+ : FACTORS Supply Chain Capital Knowledge & Systems HEALTH of ENTERPRISE (governmental, corporate, educational, non-profit) Adaptability Labour Globalisation, technology 41

  42. FRAMEWORK • Status: Huge array of information systems &product software. • Challenges: include development, delivery, adoption, and implementation of IT solutions into usable and effective systems that mimic/support organisational processes. ‘KS alignment with work practice.’ (Toffler & Drucker – 80’s : organisations of 20thCentury -> knowledge- based. Greater autonomy, revised management structures). • Opportunities: KM popularity grew through 90’s, spawned ideas of 'KM models', ‘KM strategy', concepts of 'organisational learning', 'knowledge /practice networks', 'knowledge discovery', ‘intellectual capital‘). • Objectives: To Plan, develop, implement, operate, optimise, cost information /communication systems and interpret use. • Starting point :understanding ICT opportunities requires both technological and organisational perspective + understanding of benefits associated with data capture and analysis.

  43. Data Mining & KM • The Knowledge Discovery Process • Classification e.g. clusters, trees • Exploratory Data Analysis • Models (including Bayesian Networks), Graphical or other. • Frequent Pattern Mining and special groups/subgroups Key Features: • ‘Learning models’ from data can be an important part of building an intelligent decision support system. • Sophistication of analyses – computationally expensive data mining methods, complexity of algorithms, interpretation and application of models.

  44. Hot Topics in BI • Business Process Management and Modelling • Supply Chain Management and Logistics • Innovation and ICT • Analytical Information Systems, Databases and Data Warehousing • Knowledge Management and Discovery • Social Networks and Knowledge Communities • Performance Indicators &Measurement systems/Information Quality • Data Analytics, Integration and Interpretation • Cost-benefit and Impact Analysis • Reference Models and Modelling • Process Simulation and Optimization • Security and Privacy • IT and IS Architectures/Management • Info. Sys. development, Tools and Software Engineering

  45. ExampleQuestions • What are the characteristics of internet purchases for a given age-group? How can this be used to develop further E-business? • What are key risk factors for profit/loss on a product on the basis of historical data and demographic variables? • Can we segment into/identify groups of similar on the basis of their characteristics and purchase behaviour? • Which products are typically bought together in one transaction by customers? • What are financial projections, given market volatility and knock-on for recent shock? • What data should an in-house information system collect? What design principles are involved for a large database? • What is involved in modelling and IT-supported optimisation of key business processes?

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