Introduction to Statistics inPsychologyPSY 201Professor Greg FrancisLecture 05central tendencyWanna bet?DISTRIBUTION USESsummarize dataindicate most frequent data valuesindicate amount of variation acrossdata valuesallows us to interpret a single score inthe context of other scoreswe are exploring quantitative methodsto describe distributions2LIMITATIONSLast time we discussed percentiles andpercentile ranksvery useful for comparing a score to adistribution of scoresnot so good for talking about adistribution overallwant to quantify ideas of centraltendency (most of scores, averagescore,...)• mode• median• meanand variation (how variable scores arein a distribution)3MODEthe most frequent data value (score)easy to find from a table of frequencyscoresExactLimits 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 0.564MODEtop of a hill on a frequencydistribution grapheasy to find25 30 35 40 45 50 55 60 65 70Score010203040Frequency5MODEwe actually lo ok for a modalinterval and consider the midpoint ofthe interval to be the modeunimodal distribution: whenthere is a single mode. (single hill)multimodal distribution: whenthere are several modes. (many hills)bimodal distribution: when thereare two modes. (two hills)NOTE: the use of the terms are notquite consistent!6BIMODALthis distribution might be calledbimodal, even though there is reallyonly one mode!25 30 35 40 45 50 55 60 65 70Score010203040Frequencynot very useful for mathematics!7MEDIANthe point below which 50% of scoresfallthe 50th percentileMdn = P50= ll +n(0.5) − cffi(w)25 30 35 40 45 50 55 60 65 70Score020406080100PERCENTAGE8CALCULATIONSfor our data set• ll = 44.5; exact lower limit of theinterval containing the n(0.5) score• n = 180; total number of scores• 0.5 = 50/100, proportion correspond-ing to 50th percentile (decimal form)• cf = 50; cumulative frequency of scoresbelow the interval containing then(0.5) score• fi= 42; frequency of scores in the in-terval containing the percentile point• w =5; width of class intervalMdn = P50= 44.5+180(0.5) − 5042(5) = 49.269CALCULATIONSwhen the raw scores are used (insteadof class intervals)1. Arrange the scores in ascending or-der (from lowest to highest).2. If there is an odd number of scores,the median is the middle score.3. If there is an even number of scoresthe median is halfway between thetwo middle scores.Will this always give the same value asfor the frequency distributionapproach?10CALCULATIONSName Sex ScoreAime´e Female 94Greg Male 95Ian Male 89Jim Male 92scores: 89, 92, 94, 95 (even number of scores)the median is: halfway between 92 and 94 = 93Name Sex ScoreAime´e Female 94Greg Male 95Ian Male 89Jim Male 92Bob Male 83scores: 83, 89, 92, 94, 95 (odd number ofscores)the median is: the middle score = 9211MEANarithmetic average of scores in adistributionmean of a population is designatedas µmean of a sample is designated as XCalculated as:X =1nn'i=1XiXi= the ith scoren = total number of scoressometimes just written asX =1n'Xi12CALCULATIONSName Sex ScoreAime´e Female 94Greg Male 95Ian Male 89Jim Male 92X =1n'XiX =14[X1+ X2+ X3+ X4]X =14[95 + 89 + 94 + 92] =3704= 92.513COMPARISON• mean can only be used on interval or ratiodata.• mode and median can be used on any data• mean can be manipulated mathematically• mean can be sensitive to extreme scoressuppose a manufacturing company is getting atax break from the local governmentthe tax break is up for renewal and thecompany argues it should continue because itprovides an average salary of $47,684 for 38peoplea local group opposed to the renewal pointsout that the modal salary is $22,000.a councilman notes that the median salary isalso $22,000they have all done the correct calculations14COMPARISONHere’s the data for the companyPosition Number SalaryPresident 1 $500,000Ex. vice pres. 1 150,000Vice pres. 2 120,000Controller 1 60,000Senior sales 3 45,000Junior sales 4 36,000Foreman 1 33,000Machinists 12 22,000If you just averaged the salaries, you would get$120,750, but this is not correct.15MEAN OF MEANSName Sex ScoreGreg Male 95Ian Male 89Aime´e Female 94Jim Male 92X =(XinX =X1+ X2+ X3+ X44X =95 + 89 + 94 + 924=3704= 92.516COMBINED GROUPSthe mean of means is not the samething as the mean of all the scores inall the groups!!!Consider our small data setXF=941= 94.0XM=95 + 89 + 923= 92.0the mean of the means is:XF+ XM2= 93.0but we already found that the mean ofall the scores wasX = 92.5too much weight on the “female” group17COMBINED GROUPScorrect calculation goes likeX =nFXF+ nMXMnF+ nMwhereXFis the mean for the femalesXMis the mean for the malesnFis the number of femalesnMis the number of malesX =(1)(94.0) + (3)(92.0)1+3=94 + 2764= 92.5same as direct calculation of X!18COMBINED GROUPSin general, givenXi= individual group meansni= number of observations inindividual groupsN =(ni= total number ofobservations in all groupsX =(niXiN19WANNA BET?the properties of averages make forsome odd conclusions that you cantake advantage ofbatting averagesX =number of hitsnumber of at batsconsider two players who arecompeting for the batting title (highestaverage)over the first half of the season theyget:Jim JohnAt bats 45 12Hits 12 3Average 0.267 0.25020WANNA BET?over the second half of the season theyget:Jim JohnAt bats 15 60Hits 5 19Average 0.333 0.317but over the entire season they getJim JohnAt bats 60 72Hits 17 22Average 0.283 0.3065you have to look at how many “atbats” take place at different times21PROPERTIES OF MEAN1. The sum of deviations of all scoresfrom the mean is zero.2. The sum of squares of the deviationsfrom the mean is smaller than thesum of squares of deviations from anyother value.deviations: data value minus meanxi= Xi− Xpluses and minuses cancel each otherout!'xi='(Xi− X)= (95−92.5)+(89−92.5)+(94−92.5)+(92−92.5)= (2.5) + (−3.5) + (1.5) + (−0.5) = 022SUM OF SQUARESwe ignore the direction of deviation,and consider the squared magnitude ofdeviation'x2i='(Xi− X)2= (95−92.5)2+(89−92.5)2+(94−92.5)2+(92−92.5)2= (2.5)2+(−3.5)2+ (1.5)2+(−0.5)2=6.25 + 12.25 + 2.25 + 0.25 = 21.0sum of squared deviations from anyother value is larger'(Xi−