Introduction to Statistics inPsychologyPSY 201Professor Greg FrancisLecture 04describing distributionsWhat all that stuff on the SATsmeant.DISTRIBUTIONSAs we saw last time, a well-drawngraph conveys a lot of usefulinformation...but a poorly drawn graph can misleadand confuse.We would like a quantitativemethod of describing distributionsmay not entirely avoid msinformation,but at least the limitations will beidentifiable2FREQUENCYDISTRIBUTIONSfrequency versus score class interval(from data set in book)25 30 35 40 45 50 55 60 65 70Score010203040Frequency3CUMULATIVE25 30 35 40 45 50 55 60 65 70Score050100150Frequency25 30 35 40 45 50 55 60 65 70Score020406080100PERCENTAGE4TABLE FORMATExactLimits Midpoint f cf % c%64.5–69.5 67 6 180 3.33 10059.5–64.5 62 15 174 8.33 96.6754.5–59.5 57 37 159 20.56 88.3449.5–54.5 52 30 122 16.67 67.7844.5–49.5 47 42 92 23.33 51.1139.5–44.5 42 22 50 12.22 27.7834.5–39.5 37 18 28 10.00 15.5629.5–34.5 32 7 10 3.89 5.5624.5–29.5 27 2 3 1.11 1.6719.5–24.5 22 1 1 0.56 0.565DISTRIBUTION USESsummarize dataindicate most frequent data valuesindicate amount of variation acrossdata valuesallows us to interpret a single score inthe context of other scoreswe will explore quantitative methodsto describe distributions6PERCENTILEpoint in a distribution at (or below)which a given percentage of scores isfoundwritten asPpercentage28th percentile is written as P2899th percentile is written as P99...7PERCENTILEwhat are the data values for the lowest60% of the population?several steps1. Find out how many data values makeup 60% of the population.2. Find the lowest class interval in thecumulative frequency distribution thatincludes at least that many data val-ues.3. Estimate how far into the class inter-val you must go to reach exactly thepercentile.works for any percentage!8CALCULATIONSfind P60using the above data set ofscores(1) number of scores making up 60% ofstudent scores is(180)(0.60) = 108In general, calculate(n)(p)where n is the size of the population(number of scores)and p is the percentage in decimal form9CALCULATIONS(2) lowest class interval in the cfincluding 108 scores is with midpoint52ExactLimits Midpoint f cf % c%64.5–69.5 67 6 180 3.33 10059.5–64.5 62 15 174 8.33 96.6754.5–59.5 57 37 159 20.56 88.3449.5–54.5 52 30 122 16.67 67.7844.5–49.5 47 42 92 23.33 51.1139.5–44.5 42 22 50 12.22 27.7834.5–39.5 37 18 28 10.00 15.5629.5–34.5 32 7 10 3.89 5.5624.5–29.5 27 2 3 1.11 1.6719.5–24.5 22 1 1 0.56 0.5610CALCULATIONSso we know that the percentile issomewhere between 49.5 and 54.5. Wewant a more precise estimatewe need to know• width of class interval (5)• frequency of scores in the class in-terval containing the percentile point(30)• exact lower limit of class interval con-taining the percentile point (49.5)• cf of scores below the class intervalcontaining the percentile point (92)• remaining number of scores in classinterval containing the percentile point(108 - 92 = 16)11CALCULATIONSestimate of percentile pointgo into the interval the remaining(unaccounted for) percentage25 30 35 40 45 50 55 60 65 70Score020406080100PERCENTAGE12CALCULATIONSPX= ll +np − cffi(w)ll = exact lower limit of the intervalcontaining the percentile pointn = total number of scoresp = X/100, proportion correspondingto percentile (decimal form)cf = cumulative frequency of scoresbelow the interval containing thepercentile pointfi= frequency of scores in the intervalcontaining the percentile pointw = width of class interval13PERCENTILE RANKgiven a particular data value, whatpercentage of data values are smaller?e.g. given a score on a test, whatpercentage of scores were lower?sort of the reverse of percentilefor a data value of 39, we write thepercentile rank asPR39(Used on achievement tests!)14OGIVEplot cumulative frequency percentageagainst score class interval(gives percentile rank)25 30 35 40 45 50 55 60 65 70Score020406080100PERCENTAGE15CALCULATIONSPRX=cf +(fi)(X − ll)/wn(100)X = score for which percentile rank isto be determinedcf = cumulative frequency of scoresbelow the interval containing thescore Xll = exact lower limit of the intervalcontaining Xw = width of class interval containingXfi= frequency of scores in the intervalcontaining Xn = total number of scores16CALCULATIONSPRX=cf +(fi)(X − ll)/wn(100)PR39=10 + (18)(39 − 34.5)/5180(100)PR39= 14.556ExactLimits Midpoint f cf % c%64.5–69.5 67 6 180 3.33 10059.5–64.5 62 15 174 8.33 96.6754.5–59.5 57 37 159 20.56 88.3449.5–54.5 52 30 122 16.67 67.7844.5–49.5 47 42 92 23.33 51.1139.5–44.5 42 22 50 12.22 27.7834.5–39.5 37 18 28 10.00 15.5629.5–34.5 32 7 10 3.89 5.5624.5–29.5 27 2 3 1.11 1.6719.5–24.5 22 1 1 0.56 0.5617LIMITATIONSpercentiles help describe a data value relativeto its frequency distributionbut they have some drawbackspercentiles use an ordinal scaleequal differences in percentiles do not indicateequal differences in raw scores!class intervals with higher frequency cover abroader range of percentiles(steeper part of ogive)25 30 35 40 45 50 55 60 65 70Score020406080100PERCENTAGE18LIMITATIONSpercentiles exaggerate differences inscores when lots of people have similarscoresunderestimate actual differences whenlots of people have very different scoresdifferences in percentiles should notbe compared across differentdistributions!!!(only provide information on relativeranking of scores: ordinal scale!)cannot be meaningfully averaged,summed, multiplied,...fixing these problems requires additional termsfor describing distributions (central tendency)19CONCLUSIONSpercentilespercentile ranks20NEXT TIMEcentral tendancy (mode, median,mean)Wanna