CHLD-3301.783 Introductory Statistics Option A. Cite one example in which the mean, median, and mode can be used effectively to reduce groups of data and illustrate measures of central tendency and va

= µ= ∑ = = ∑ 2. When to use the mode, median and mean. To a large extent, the scale of measurement determines the appropriate measure of central tendency. With a nominal scale of measurement, the mode is the only useful measure of central tendency, since the categories cannot be ordered in a meaningful manner. Page 8 With an ordinal scale of measurement, although the mode may be useful at times, the median is the preferred measure of central tendency since it conveys more information than the mode. The mean is not very useful with ordin al data because the mean relies on calculating the sum. A sum is meaningful only when the intervals between successive categories are equal. With interval/ratio data, the mode and median are both quick and meaningful but do not convey information about t he balance point. The mean does. The mean is very useful with interval/ratio data since the intervals between categories are equal. The mean also has the advantage that it is much easier to manipulate and simplify in mathematical derivations and formulas. General recommendations for the most appropriate measure of central tendency to use with different scales of measurement are: Preferred measure Scale of measurement of central tendency nominal mode ordinal median interval / ratio mean These preferences need to be tempered somewhat, however. For example, if you have interval/ratio data and want a quick and easy measure to see how the data are stacking up, you might choose the mode. At other times, with interval /ratio data the mean may give a misleading idea of where a distribution of scores stacks up. This will happen with skewed distributions. For instance, consider the distributions in Figures 3 and 4. Notice that the mean is closer to Figure 3. The loca tion of the mean, median, and mode in a negatively skewed distribution. Page 9 Figure 4. Hypothetical distribution of the annual incomes of heads of household in the United States following the 1980 census . This distribution shows the location of the mean, median, and mode in a positively skewed distribution. the tail of the distribution than the median for both negative (Figure 3) and positive (Figure 4) skews. The mean is shifted in the direction of the skew because it takes into account the magnitudes o f the extreme but infrequent scores in the tail of the distribution. Since the median is located more with the bulk of the scores, it is a more representative measure of central tendency when a distribution is extremely skewed 2. Notice that the distributio n of the annual incomes of heads of household in Figure 4 are positively skewed. The choice of whether to use the mean or the median as a measure of central tendency might be guided by the impression of economic prosperity you want to convey. A country de siring Western immigrants and industrial ization may use the mean to paint a glowing picture, whereas a small crowded country may choose the median. 3. Introduction to Measures of Variability. Measures of variability are as important as measures of central tendency in describing a distribution. When a measure of central tendency and a measure of variability are both found, you can make more elaborate comparisons than with only a measure of central tendency. For example, assume your IQ is 130 and you have just been accepted into an exclusive 2 When a distribution is symmetrical, th e mean and median are equal . Thus you might conclude that the most representative descriptive measure of central tendency is always the median. In a sense this is true, because the median is the measure of central tendency closest to all the scores in the group. However, when it is appropriate to calculate the mean, the mean is preferable because of its relative ease of manipulation in mathematical formulas. Page 10 honor society that has a mean IQ of 120. How do you compare with the rest of the society's members? You are 10 IQ points above the mean, but that is only half the story. Imagine that the society's distribution of IQ scores looks like the one on the left in Figure 5. Compare it with the one on the right. Both hypothetical Figure 5. Comparing an IQ score of 130 in two hypothetical distributions with means of 120. distributions have means of 120. With the distri bution on the left, your IQ is higher than more than half of the members, but with distribution on the right, your IQ is the highest. Knowing how spread out the scores are in these two distributions makes a ig difference in how you rate with the rest of t he members. Whereas a measure of central tendency is the point at which a distribution of scores stacks up, a measure of variability indicates how the stack is spread out. Three different measures of variability are the range, the variance, and the standar d deviation. The range is the easiest measure of variability to calculate. However, there is are two ways to calculate the range, and hence a bit of confusion may arise if you have had a statistics class before or otherwise encountered this term on the internet or workplace . Most commonly fou nd in recent texts , the internet , and statistical computer programs is where the range is defined and calculated simply as the difference between the largest score minus the smallest score . For examples, if the X scores in a distri bution were 5, 5, 5, 6, 7, 7, 8, 8, 9, 10, and 10, the range would be 10 – 5 = 5 . This is actually the “exclusive ” range, but it is rarely labeled as such in reference materials. In the present book, however, we want you to calculate the “inclusive ” range , defined as the upper real limit of the largest score minus the lower real limit of the smallest score. Conceptually, the inclusive range renders the “number ” of values in whole units that are spread between the largest and smallest scores. For example, with the X scores of 5, 5, 5, 6, 7, 7, 8, 8, 9, 10, and 10, the using Formula 5 below, the range is 10 - 5 + 1 = 6 . This is Page 11 because X values of 5, 6, 7, 8, 9, and 10, six values, can fit in this range of scores. Or, concep tually speaking , when one thinks of whole values or integers, the value of 10 really represents numbers ½ unit above and below 10 , from 9.5 to 10.5, where 10.5 is ca lled the upper real limit of a valu e of 10 , and 9.5 is the lower limit . The value of 5 represents numbers ½ unit above 5, from 4.5 to 5.5, where 4.5 is the lower limit of 5 , and 5.5 is the upper limit . Now, if one subtracts the lower real limi t of 4.5 from the upper real limit of 10.5, then you get the rang e of 6 : 10.5 – 4.5 = 6 . To simplify this process using Formula 5 below, one ad ds “1” after subtracting the smallest from the largest whole score to take into account the ½ unit below the 5 ( its lower real limit) and the ½ above the ten (it s upper real limit). It’s unfortunate that there are two methods , but please calculate the range using Formula 5 below for this course. Henceforth, we will refer to the range calculated using Formula 5 as simply the “range ” and cease using the term “inclusive ”. To use another example, consider the freshman anxiety scores . You can calculate the range by subtracting the smallest score, 10, from the biggest score, 86, and then adding 1. The result is 77, which is the range. Again, the reason you add 1 is to reflect the fact that your scores are rounded off to the nearest whole number. Any scores between 85 .5 and 86 .5 would be round e d to 86, a nd a ny sc ore from 9 .5 to 10 .5 w ould be rounde d to 10. Thus the range would be from the lowe st possible score of 9 .5 to the highest possible score of 86 .5, which is actually 77 points. You can say "the spread of the scores is 77," or, "the range is 77." In simple formula form, the range is: Formula 5: = − + The range is useful because it indicates the span between the lowest and highest scores. However, like the mode, the range is very unstable because all the scores except the two extreme sco res are ignored. For example, changing one of the 86s in the freshman data to 200 drastically changes the range to 19 1. This value of 19 1 is not representative of the spread of the bulk of the scores, and is misleading since most readers would assume that there are more high scores than actually occur. With interval/ratio data it would be nice to have a measure of variability that takes into account the magnitude of each score in the distribution, not just the two extremes. A reasonable thing to do might be to calculate the distance , or devia tion, eac h score is from the mea n a nd the n find the average (mean) of these deviations. The more spread out the distribution, the larger should be the mean deviation from the mean. For example, with the freshman anxiet y data, Page 12 subtract the mean of 49 from each score to find the deviation from the mean . This operation is − µ: X − µ 86 37 44 -5 43 -6 40 -9 30 -19 86 37 69 20 45 -4 37 -12 10 -39 1 0 —39 Now, total the X — µ, column to find the sum of all the deviations from the mean and then divide this sum by N. Since the "sum of the deviations from the me an" is symbolized ∑( − µ), the mean deviation from the mean is: ∑( − µ) = 0 10 = 0 The sum of the deviations is zero, so the mean deviation from the mean is zero. 3 In fact, the sum of the deviations from the mean will always be zero. This zero -sum problem reflects the status of the mean as the balance point. The scores above the mean have positive deviations and are counterbalanced with scores below the mean that have negative deviations. Consequently, to generate a measure of variability that do es, indeed, increase as the spread of a distribution increases, you must eliminate the zero -sum problem The best solution to the zero -sum problem is to square each deviation from the mean. This creates positive values. Squaring also has the advantage that it is easy to manipulate mathematically. Now, if you find how much each score deviates from the mean and square these deviations before finding their mean, what will you call your answer? It is called the mean squared deviation from the mean , or varian ce . The symbols ( X - µ)2 are abbreviated “squared deviations from the mean” and by dividing by N one can calculate the mean squared deviation from the mean as follows: Page 13 X X - µ (X - µ)2 86 37 1369 44 -5 25 43 -6 36 40 -9 81 30 -19 361 86 37 1369 69 20 400 45 -4 16 37 -12 144 10 -39 1521 ∑( − µ)2= 5322 ℎ = ∑( − µ)2 = 5322 10 = 532 .2 However, researchers do not have to go around saying "mean squared deviation from the mean" all the time (try repeating this a few dozen times); they simply say variance. The variance of a population is abbreviated σ2, and is read "sigma squared." The definitional (or, “conce ptual”) formula for the variance of a population is : variance of a population = 2 = ∑(−µ)2 [Note, the definitional/conceptual formula for the population variance above is shaded in blue and is not the best formula to use with a hand calculator. The best formulas to use with a hand calculator are what are called the computational formulas. The computational formulas will always be highlighted in yellow and will be numbered formulas for reference. The reader is respon sible for knowing how the concept of variance was derived with the definitional/conceptual formulas, but not the calculations.] The variance of a sample can be calculated the same way. We will not give a different abbreviation for the variance of a samp le right now because inferential statistics treats sample variances in a special way to be discussed later . If you wish to desc ribe the va riance of a sample, treat it as a population and compute the variance using the same formula as you would for the pop ulation. Page 14 The variance is a useful measure of variability, but it has two slight drawbacks. First, the large deviations of scores farthest from the mean are emphasized more than the small deviations of scores nearest the mean. Squaring a large deviation g ives a disproportionately bigger value than squaring a small deviation. Second, the units of measurement in the variance are squared units. For example, if you measure the heights of players on a basketball team (in inches) and calculate the variance, it w ill be in square -inch units. This is a measure of area, not height! The drawbacks of the variance can be somewhat alleviated by taking its square root. This gives a value in terms of the original unit of measurement, and it somewhat deemphasizes large de viations from the mean. The value obtained by taking the square root of the variance is called the standard deviation. For example, the standard deviation for the freshman data is 23.07: ℎ = √532 .2 = 23 .07 Symbolically, the square root of the variance is √2 , which simplifies to . Thus the symbol for the standard deviation of a population is and is defined by the following formula: standard deviation of a population = = √∑(−µ)2 The standard deviation is the "square root of the mean squared deviation from the mean." To get a better understanding of how useful the standard deviation is, consider its role in the normal distribution. A normal distribution is a theoretical distr ibution that is symmetrical, unimodal, and bell -shaped as shown in Figure 6. Page 15 Figure 6. Properties of the standard deviation in a normal distribution. The standard deviation units are shown in relation to raw -score equivalents (IQ scores). The deri vation of the raw -score equivalents is shown above the curve. The approximate percentage of raw scores lying above and below the mean are shown below the abscissa . Suppose you administer an IQ test to a population of 1000 people and find that the empiri cal distribution of scores is approximately normally distributed as in Figure 6, with a mean of 100 (µ = 100) and standard deviation of 15 (σ = 15). Think of the standard deviation as one chunk or block of successive IQ scores lying on the abscissa, with a spread of 15 IQ points, such as the chunk from 100 to 115. Conversely, you can think of a range of 15 IQ points as one standard deviation unit (1 σ = 15, or equivalently, σ = 15). Then a range of 30 IQ points is two standard deviation units (2 times 15 equals 30, or 2 σ = 30), and 45 IQ points is three standard deviation units (3 times 15 equals 45, or 3 σ = 45). Now you can think of being some number of standard deviation units away from the mean, suc h as one, two, or three standard deviation units above the mean (represented + 1 σ, +2 σ, +3 σ), or one, two, or three standard deviation units below the mean (represented -1σ, -2σ, -3σ). The IQ scores lying within one standard deviation above and below the mean ( -1σ to + 1σ) are the scores from 85 to 115. The score 85 is the raw -score equivalent of -1σ, obtained by subtracting one standard deviation unit from the mean (µ - 1σ, or 100 - 15 = 85). The score 115 is the raw -score equivalent of +lσ, obtained by adding one standard deviation unit to the mean (µ + 1 σ, or 100 + 15 = 115) . The derivation of more raw -score equivalents is shown in Figure 3 - 6. The role of the standard deviation can be best be seen in light of the properties of the normal distribution. For example , one property of the normal distributio n is that 68.26% of the scores lie within one standard deviation above and below Page 16 the mean. This indicates that in the approximately normal distribution of IQ scores in Figure 6, about 68.26% of the IQ scores must be within one standard deviation above and below the mean. Thus, if there are 1000 people represented in Figure 6, you will find about 683 of them (68.26%) with IQs between 85 and 115. Another way to see this is to remember that the approximately normal distribution shown in Figure 6 is a frequency polygon, in which the height of the curve above a particular IQ score corresponds to the number of people who have that IQ. You can think of the people with a given IQ as being "stacked up" over that IQ score in the graph, as in Figure 7. Thus you could directly count Figure 7. This approximately normal distribution of IQ scores is a frequency polygon in which the height of the curve above a particular IQ score corresponds to the number of people who have that IQ. For example, if there are 1000 sc ores, you will find about 21 people stacked up with IQs of 111, and about 683 people with IQs between 85 and 115. the number of people with IQs between 85 and 115. As seen in Figure 6, f urther properties of the normal distribution are that 95.44% of all scores lie within two standard deviations above and below the mean, and 99.74% of all scores lie within three standard deviations above and below the mean. Translated into raw scores, 68.26% of the people have IQs between 85 and 115, 95.44% of the people have IQs between 70 and 130, and 99.74% have IQs between 55 and 145. Even when a distribution is not very well approximated by a normal distribution, there is often close to 68% of the scores lying within ±1 σ, close to 95% lying within ± 2σ , and close to 99% lying within ±3 σ. In general, regardless of the shape of the distribution, the standard deviation serves as a useful measure of variability that takes into account the magnitude of each score. Consider the following sets of raw scores along with their means and calculated standard deviations: Page 17 Set Raw Scores ___________ µ σ__ A 2, 3, 4, 4, 5, 8 4.33 1.89 B 2, 4, 6, 6, 8, 14 6.67 3.77 C 52, 53, 54, 54, 55, 58 54.33 1.89 D 52, 54, 56, 56, 58, 64 56.67 3.77 E 3, 4, 4, 4, 5, 6, 7, 7 5 1.41 F 1, 3, 3, 3, 5, 7, 9, 9 5 2.83 G 53, 54, 54, 54, 55, 56, 57, 57 55 1.41 H 51, 53, 53, 53, 55, 57, 59, 59 55 2.83 Notice that the broader the distribution the larger the standard deviation. The formulas for the variance and standard deviati on given above in boxes are technically referred to as definitional formulas, since they serve to precisely define the mathematical operations corresponding to the reasoning that went into developing them. In most applications, however, the computational formulas below provide an easier method of calculation. By applying a little mathe matical finesse to the definitional formula for the variance, you can derive the following computational formula for the population variance : 4. Computational Formulas for the Variance and Standard Deviation . The formulas highlighted in blue above for the variance and standard deviation are technically referred to as definitional or conceptual formulas, since they serve to precisely define the mathematical operations cor responding to the reasoning that went into developing them. In most applications, however, the computational formulas below highlighted in yellow and numbered provide an easier method of calculation. By applying a little mathe matical finesse to the def initional /conceptual formula for the variance, you can derive the following computational formula for the population variance : : ℎ 2 = ∑ 2− (∑ )2 Page 18 The computational formula for the population standard deviation is: Formula 7: The computational formula for the standard deviation = √∑ 2− (∑ )2 The difference between the definitional /conceptual and computational formulas lies in the numerator. The computational formulas require you to square each score to get X2, sum the squared scores to get ∑ 2, and then subtract the square of the sum of scores, (∑ )2, divided by . The best way to keep track of each operation when using the computa tional formulas is to list each score and its square i n two columns, labeled X and X 2, respectively. Next, sum each column and square the sum of the X column. Be sure to label all the numerical values with their symbolic terms, as with the freshman anxiety data below: X X2 86 7396 44 1936 43 1849 40 1600 30 900 N = 10 86 7396 69 4761 45 2025 37 1369 10 100 ∑ = 490 ∑ 2= 29332 (∑ ) 2 = (490 )2= 240 ,100 Then, substitute into the computational Formula 6 and complete the calcula tions. Thus the population variance is 532.2: 2= ∑ 2− (∑ )2 = 29 ,332 − 240 ,100 10 10 Page 19 = 29 ,332 − 24 ,010 10 = 5322 10 = 532 .2 To find the standard deviation, simply take the square root of the variance: = √∑ 2− (∑ )2 = √532 .2= 23 .07 The computational formulas are superior to the definitional /conceptual formulas wh en N is fairly large because you do not have to perform many "error -inducing" subtractions when finding X — µ. The computational formulas are far superior when using hand -held calculators. 5. When to use the range, variance and standard deviation. The scale of measurement guides the selection of measures of variability. With nominal data there is no useful measure of variability since the categories of scores cannot be meaningfully ordered. With ordinal data, the range is the most usefu l measure of variability. Calculating the variance and standard deviation requires you to subtract the mean from each score and then add the deviations. This operation is useful only when you can assume that equivalent entities are being added. This is tru e only with interval/ratio data. With interval/ratio data, although the range may be used, the variance and standard deviation are preferred because they convey more information.

To summarize these preferences: Scale of measurement Preferred m easure of variability nominal none possible ordinal range only interval/ratio variance and standard deviation Page 20 SUMMARY of KEY CONCEPTS and FORMULAS for WEEK 2 1. Measures of central tendency are single scores that describe where the scores i n a distribution stack up. Three measures of central tendency are the mode, median, and mean. 2. The mode is the most frequently occurring score . There may be more than one modal score. It is used more often with nominal data, although it may be used as a quick measure of central tendency with ordinal and interval/ratio data. The mode is unstable. 3. The median is the middle score in a distribution. The median is most appropriate with ordinal data and with interval/ratio data that are extremely skewed.  When N is odd an odd number: median = the middle score in the distribution  When N is an even number : median = the value halfway between the two Middle scores 4. Measures of variability are single numbers that describe the spread of a distributi on. Three measures of variability are the range, variance, and standard deviation. 5. The range is the distance between the highest score and the lowest score and is obtained by subtracting the lowest score from the highest score and adding one. The range is most appropriate for ordinal data, although it may be used as a quick measure of variability for interval/ratio data. The range is unstable. = − + 6. The population variance = o2 and is the mean squared deviation from the mean. It is most easily calculated with the computational formula below: 2= ∑2− (∑)2 Page 21 7. The population standard deviation = and is the square root of the variance: = √ 2 Taking the square root of the variance converts back to the original units of measurement. The population standard deviation is most easily calculated with the computational formula below: = √∑ 2− (∑ )2 8. An interpretation of the standard deviation is seen in the normal distribution: 68.26% of the scores will lie within one standard deviation above and belo w the mean; 95.44% will lie within two standard deviations above and below the mean; and 99.74% will lie within three standard deviations above and below the mean. 9. Raw scores may be interpreted as a number of standard deviation units above or belo w the mean, and vice versa. Practice Problems for Week Two 1. Calculate the mode, median, and mean for each of the following groups of scores: a. 7, 4, 3, 2, 5, 8, 3, 1 b. 118, 4, 3, 2, 4, 8, 3, 1 c. 11, 6, 4, 0, 0, 12, 1, 2, 0 d. 11, 22, 22, 10, 18, 15, 13, 15, 15, 12, 10, 7, 12 e. 14, 21, 16, 3, 14, 22, 14, 9, 9, 18 f. 1.6, 2.5, 2.5, 3.8, 3.6, 5.0, 4.7, 3.2, 4.2, 4.2, 2.7 2. Calculate the mode, median, and mean of the data in the frequency distribution below . Remember, to calculate the median and t he mean, one must “take apart” the frequency distribution and list each score, and list a score multiple times if it was scored two, three, four, or more times. Page 22 X f 60 1 61 1 62 0 63 1 64 2 65 3 66 2 67 1 68 2 69 4 70 5 71 1 72 1 3. For each of the following hypothetical research situations, identify the scale of measurement, and calculate the preferred measure of central tendency. a. You measure the maximum speed (in miles per hour) of 15 automobiles traveling through a school zone where t he maximum speed is posted at 20 miles per hour. The speeds are as follows: 21 55 19 17 20 15 15 22 20 16 21 18 18 17 47 b. There were 16 airline personnel who loaded cargo for one year without wearing protective devices over their ears. You classify them as having high . medium, low, or no hearing impairment. Your data are as follows: high high medium low medium medium low high medium no high medium high low medium medium c You examine the school psychologist's diagnoses of ten children classified during the first month of school as requiring special education.

Each child's diagnosis is given below: learning disorder impulse control disorder impulse control disorder impulse control disorder Page 23 attention deficit disorder anxiety disorder lea rning disorder impulse control disorder impulse control disorder learning disorder d. In a relaxation study, prior to initiating any specific relaxation training you record the systolic blood pressure of eight persons with hypertension.

The systolic blood pressures were: 162.5 178.7 159.0 152.1 187.8 164.1 190.7 174.3 e. You observe 20 children in several classroom leadership roles. After observing all the children in all the roles, you rate each child on a scale from 1 to 5, based on your judgmen t of their classroom leadership. You use ratings of 1, 2, 3, 4, or 5. Children you judge very high in classroom leadership are given a rating of 5, those judged very low are given a rating of 1, and those judged as having intermediate levels of leadership are assigned intermediate ratings. The ratings are as follows: 2 1 3 1 4 2 1 1 2 2 1 5 1 2 3 1 1 3 1 1 4. Problem #4 has been deleted. 5. Calculate the range for each of the following groups of scores: a. 7, 4, 3, 2, 4, 6, 3, 1 b. 12, 10, 2, 4, 5, 3 , 0, 12, 1, 7 c. 7 , 4 , 3 , 2 , 4 , 6 , 3 , 4 2 d. 42, 26, 25, 25, 38, 32, 38, 44 e. 1 4 , 0 , -5, -2 , 8 , 7 , 6 , 6 , 1 3 , 1 6 , 1 4 6. Calculate the mean and the variance and standard deviation using the computational formulas for each of the following groups of data: a. 7, 4, 3, 2, 4, 6, 3, 1 b. 12, 10, 2, 4, 5, 3, 0, 12, 1, 7 c. 37, 34, 33, 32, 34, 36, 33, 31 d. 92, 90, 82, 84, 85, 83, 80, 92, 81, 87 e. 10, 5, 5, 8, 6, 8, 7, 9, 5, 8, 6 f. 14, 0, -5, -2, 8, 7, 6, 6, 13, 16, 14 g. 70, 65, 65, 68, 66, 68, 67, 69, 65, 68, 66 Page 24 h. 54, 40, 35, 38, 48, 47, 46, 46, 53, 56, 54 ANSWERS TO PROBLEMS FOR WEEK ONE : 2. mode = 70, median = 68, mean = 67.125 3. a. interval/ratio; the median is preferred since these data are very skewed; median = 19 b. ordinal; median = medium c. nominal; mode = impulse control disorder d. interval/ratio; mean = 171.15 e. ordinal; median = 1.5 4. mode = 1000; median = 850; mean = 854.167; mode was reported; either median or mean would be more accurate. 5.. a. range = 7 b. range = 13 c. range = 41 d. range = 20 e. range = 22 Page 25 © Pfeiffer & Olson 201 4