Lecture 3Central TendencyDistributions: Try to Find the “Center”The MeanConceptualizing the MeanSlide 6Example with a frequency distribution tableThe Weighted MeanAn alternative procedureCharacteristics of the MeanSlide 11Let’s Try One:Slide 13The MedianSlide 15The MedianSlide 17Characterizing the MedianLet’s Try a Couple…Reconceptualizing the Middle: The Mean vs. The MedianThe ModeLet’s try a fewCharacterizing the ModeWhich to Use?Symmetrical DistributionsPositively Skewed DistributionNegatively Skewed DistributionsGroup ActivityVariabilityWhy is Variability Important?Slide 31Measures of VariabilityRangeSlide 34PercentilesQuartilesA Special Case: Interquartiles and Semi-InterquartilesComputing QuartilesInterquartile and Semi-Interquartile rangeExampleStandard DeviationHomework - Chapter 3Lecture 3Central Tendency And VariabilityCentral TendencyCentral tendency = statistical measure that identifies a single score as representative of an entire distribution. –Median –Mean–ModeMORE ABOUT DISTRIBUTIONS - Measures of central tendency help us to summarize and understand the distribution better ** The important thing to remember about central tendency is that these measures describe and summarize a group of individuals rather than any single person. IN FACT, the average person may not exist!Distributions: Try to Find the “Center”f1 2 3 4 5 6 7 98f1 2 3 4 5 6 7 98f1 2 3 4 5 6 7 98The MeanArithmetic Average: computed by adding all scores in the distribution and dividing by the number of scores.Population mean = , Greek letter muSample mean = M or X, read “x-bar”Computed: = X / N M = X / nConceptualizing the MeanThe amount each individual would get if the total (X) were divided equally amongst the individuals (n)–10 friends plan to pitch in for lotto tickets. They want to buy $200 dollars worth of tickets. What is the average amount each person would have to put in?X = 200, n = 10, so M = ?–The friends are lucky and won the jackpot. They each got a share of 500 K. Given this information what is the total amount of money won?•M = 500 K, n = 10, X = ?M = X / nConceptualizing the MeanA balance point for a distribution…Consider a population consisting of N scores:2, 4, 8, 10Balance by distance Always must occur between hi and low score = X / N1 2 3 4 5 6 7 8 9 10Score Distance from  X=2 4 pts below M X=4 2 pts below M X=9 3 pts above M X=9 3 pts above MExample with a frequency distribution tableX f fx10987654262512040184814305155* If only 1 of each score x, then M = (10 + 9 + 8 + 7 + 6 + 5) / 6 = 7.5 * From ungrouped frequency distribution M = x/n x = fx M = (40 + 18 + 48 + 14 + 30 + 5) / 20 M = 7.75The Weighted MeanHow do you combined 2 sets of scores and find the overall mean? A mean of a group of means?e.g. 4 statistics classes with mean exam scores 75, 78, 72, 80?–If each class has the same number of people: (75 + 78 + 72 + 80) / 4 = 76.3–If each class has different # of people you must account for it:Class X f fX75787280304025501452250312018004000111170(1) Find total n in groups = 145(2) Find x in each group and then sum them x = fx = 11170(3) M = x/n M = 11170/ 145 = 77An alternative procedureClass X f fX75787280304025501452250312018004000111170M = (30/145)(75) + (40/145)(78) + (25/145)(72) + (50/145)(80)M = .21(75) + .28(78) + .17(72) + .34(80)M = 15.75 + 21.84 + 12.24 + 27.2M = 77.03* For yet another way to compute weighted means see Box 3.1 in your book (pg. 77).Characteristics of the MeanMost commonly used measure of central tendency–Because it uses every score in the distribution it tends to be a good representativeMay or may not be an occurring scoreVery sensitive to outliers–e.g. 1, 3, 4, 5, 5, 6, 10, 100 (M = 16.75)Affected by skewed distributions - pulled toward the extremesClosely related to the variance and standard deviation…makes it good for use with inferential statistics–Summed deviations = 0 (more on this later) Adding, Subtracting or Changing a score will change the mean. e.g. 5, 6, 7, 10 (M = 7) vs. 5, 6, 7, 8 (M = 6.5) What if we removed the data point X = 8 from the second set (M = 6) UNLESS the altered score is the same value as the mean: e.g. 5, 6, 7, 7, 10 (M = 7) Adding, Subtracting, Multiplying or Dividing each score in a distribution by a constant will change the mean in the same way by the same amount. e.g. 5, 6, 7, 10 (M = 7) vs. 7, 8, 9, 12 (M = 9) e.g. 5, 6, 7, 10 (M = 7) vs. 10, 12, 14, 20 (M = 14)Characteristics of the MeanLet’s Try One:X f fx(1) Finish the table by calculated fx.(2) Find f = n and fx = x.(3) First calculate the mean and pretend that we have no f column: x/n(4) Next, calculate a weighted mean taking the different frequencies for each score into account. 90.5 4 36289.6 3 26986.2 6 51784 6 50482.5 8 66080.2 10 80278 8 62474.5 15 111872.2 16 115571 20 142066 10 66065.4 8 52360.3 5 30258.2 2 11656 2 11252.2 5 2619404.8/128 = 73.48Admin wants to know the mean grade for all stats classes for the year 2003-2004 at the UA. Below are the mean grades and the number of people in the class. Grade Class size fx 75 34 2550 80 42 3360 82 200 16400 68 156 10608 88 75 6600 77 222 17094 72 144 10368 83 172 1427681256/1045 =77.76The Median Median - Divides the distribution into exactly 2 parts (by area). The median is equal to PR 50.No sanctioned abbreviation for the median; some use Mdn.The definition and computations for the median are identical for populations and samples.Computation:–Method 1: When N is an odd number•Rank order the scores from lowest to highest•Mdn is the middle score 3 5 7 8 10 12 13 14 18 –Method 2: When N is an Even number•Rank order the scores from lowest to highest3 5 7 8 10 12 13 14 18 18 •Select the middle 2 scores, add them, and divided by 2.•10 + 12 / 2 = 11The Median 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18The MedianComputation continued:–Method 3: When you are using a grouped frequency distributionReal limits f cumf68.5 - 71.565.5 - 68.562.5 - 65.559.5 - 62.556.5 - 59.553.5 - 56.513152028196101887353256Find the row closest to the middle of the distribution.Median =