is a field of study concerned with collection, organization, summarization and analysis of, Summaries of Statistics

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Measures of Central Tendency and i Dispersion Objectives At the end of this lecture students will be able to explain:  Central tendency, mean, median and mode.  Explain the measures of dispersion  Find the variance, standard deviation, quartiles, and coefficient of variation of a data set. 2 Arithmetic Mean  Best descriptive measure for quantitative data that are normally distributed  Sensitive to any change in the value of any observation  Very sensitive to extreme value /outlier 5 Arithmetic Mean cont.  Calculation method  Sum up all of the values in data set  Divide the sum by the number of observations (n) 6 ace | Formula & example — ” x= Liza Xi . n 8 29 Mean= 605/20 =30.25 i 5 Calculation method  Arrange the observations in order  Find middle position as (n + 1) / 2  Identify the value at the middle 10 4 Example n=19 median=n+1/2 median=19+1/2=10 median age=30 years O;/OIN IDA) AWIN| = Even number of observation n=20 median=n+1/2 median=20+1/2=10.5 Median age = Average value between10th and 11th observation 30+30/2=30 years 12 Mode  Method for identification  Arrange data into frequency distribution showing the values and the frequency with which each value occurs  Identify the value that occurs most often 15 Example n Age 1 27 2 27 3 28 4 28 5 28 6 29 7 29 8 29 9 29 10 30 11 30 12 30 13 30 14 30 15 31 16 31 17 32 18 34 19 36 20 37 Age Frequency 27 2 28 3 29 4 30 5 31 2 32 1 33 0 34 1 35 0 36 1 37 1 Total 20 Mode 16 Histogram Mode = 30 | Frequency yn wo k A OD N = 27 28 29 30 31 Age 32 33 34 35 36 37 (years) 17 Is Mode Sensitive to 4 Outliers? Number of patients Oo = N oOo fb Oo OD 0 10 20 30 40 50 60 70 80 90 100110 120 130 140 150 Nights of stay 20 Mean Uses All Data, So Sensitive to Outliers ORFNWOA THD OANHW HYD ly sa Number of patients - Mean = 12.0 25 30 35 40 45 50 Mean = 15.3 = oOo = a Nh oO Oo 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 Nights of stay 21 Exercise  We wish to obtain the mean age of sample of 36 subjects presenting below  25, 26, 24, 21, 18, 36, 32, 27, 28, 22, 23, 27, 28, 21, 38, 40, 41, 18, 19, 22, 25, 28, 34, 21, 32, 22, 27, 29, 22, 18, 39, 42, 32, 33, 34, 40 22 4 Answers Var A_VarB_ Var C Sum: 55 55 55 Mean: 5 5 5 Median: 5 5 5 Mode: 1,9 45,6 none 25 Measures of Dispersion  The dispersion of a set of observations refers to the variety that they exhibit.  Measures of dispersion are descriptive statistics that describe how similar a set of scores (values) are to each other 26 Measures of Dispersion  If all values are the same, there is no dispersion  The amount of dispersion may be small when the values are close to each other  Synonymous terms are variation, spread, scatter 27 Measures of dispersion  The variance  Standard deviation  The coefficient of variation  The range  Percentile and Quartiles 30 Variance Variance is the average squared deviation from the mean of a set of data. It is used to find the standard deviation. 31 To obtain variance 1. Find the mean of the data. 2. Subtract the mean from each value – the result is called the deviation from the mean. 3. Square each deviation of the mean. 4. Find the sum of the squares. 5. Divide the total by the number of items. 32 4 To obtain standard deviation 1. Find the variance. 2. Take the square root of the variance. viex (Xi ~~ x)? n-1 _ vinx(%i — 1)? o= a 35 Find the variance and standard deviation The math test scores of five students are: 92,88,80,68 and 52. 1) Find the mean: (92+88+80+68+52)/5 = 76 2) Find the deviation from the mean: 92-76=16 88-76=12 80-76=4 68-76=-8 52-76=-24 36 3) Square the deviation from the mean: 2 ( 8) 64  2 (16) 256 2 (12) 144 2 (4) 16 2 ( 24) 576  Find the variance and standard deviation The math test scores of five students are: 92,88,80,68 and 52. 37 Standard Deviation A different math class took the same test with these five test scores: 92,92,92,52,52. Find the standard deviation for this class. 40 Solve: Answer Now A different math class took the same test with these five test scores: 92,92,92,52,52. Find the standard deviation for this class. 41 The math test scores of five students are: 92,92,92,52 & 52 1) Find the mean:(92+92+92+52+52)/5 = 76 2) Find the deviation from the mean: 92-76=16 92-76=16 92-76=16 52-76= -24 52-76= -24 4) Find the sum of the squares: 256+256+256+576+576= 1920 2 2 2(16) 256 (16) 256 (16) 256           3) Square the deviation from the mean: 42 Coefficient of variation  The ratio measure of dispersion/ inequality is called coefficient of variation, which is the simply standard deviation divided by the mean.  It answers the question: how big is the SD of the distribution relative to the mean of the distribution. 45 Exercise Sample 1 Sample 2 Age 25 years 11 years Mean Weight 145 pounds 80 pounds Standard deviation 10 pounds 10 pounds 46 7 Exercise cont. « For 25 year we have the CV as 10 _ ° - €.V.= =, (100) = 6.9% = For 11year we have the CV as . C.V.= = (100) = 12.5% 47 Interquartile range (IQR)  The interquartile range is defined as the difference of the first and third quartiles in a set of data.  The first quartile is the 25th percentile  The third quartile is the 75th percentile  IQR = Q3 - Q1 50 Example  What is the IQR for the data to the right?  25 % of the scores are below 5  5 is the first quartile  25 % of the scores are above 25  25 is the third quartile  IQR = Q3 - Q1 = 25 - 5= 20 2 4  5 = 25th %tile 6 8 10 12 14 20  25 = 75th %tile 30 60 51 Interquartile range (IQR)  A large IQR indicates a large amount of variability among the relevant observations.  Such statement is vague, it is more informative to compare the interquartile range with the range for the entire data set.  A comparisons may be made by forming the ration of the IQR to the range (R) and multiplying by 100  100(IQR/R) 52 Range in a Histogram 14 124 104 Number of States te | tnd ea od RANGE = 18% -4% = 14% PERCENT OF POPULATION AGED 65 AND OVER . ¥ 7 ™ ¥. ¥ —— a 9 10 41 $712 135 14°15 16 17 Fe 19 55 Consider both sets of scores. Both classes have the same mean, 76. However, each class does not have the same scores. Thus we use the standard deviation to show the variation in the scores. With a standard variation of 14.53 for the first class and 19.6 for the second class, what does this tell us? Question Answer Now 56 Question Class A: 92,88,80,68,52 Class B: 92,92,92,52,52 With a standard variation of 14.53 for the first class and 19.6 for the second class, the scores from the second class would be more spread out than the scores in the first class. 57