Describing, Exploring, and Comparing Data: Measures of Center and Variation

Measures of Center

  • Definition of Measure of Center: A measure of center is a value at the center or middle of a data set. It provides a representative value to summarize the entire data set.

The Mean

  • Definition: The mean (or arithmetic mean) of a set of data is the measure of center found by adding all of the data values and dividing the total by the number of data values. It is the most important numerical measurement used to describe data and corresponds to what is commonly called an average.
  • Important Properties of the Mean:
    • Sample means drawn from the same population tend to vary less than other measures of center.
    • The mean of a data set uses every single data value.
    • Disadvantage (Sensitivity to Outliers): A single extreme value (outlier) can substantially change the value of the mean. Therefore, the mean is not resistant.
  • Definition of Resistance: A statistic is resistant if the presence of extreme values (outliers) does not cause it to change very much.
  • Notation and Formulas for the Mean:
    • Σ\Sigma denotes the sum of a set of data values.
    • xx is the variable used to represent individual data values.
    • nn represents the total number of data values in a sample.
    • NN represents the total number of data values in a population.
    • English letters represent sample statistics; Greek letters represent population parameters.
    • Sample Mean Formula (Formula 3-1):xˉ=Σxn=sum of all data valuesnumber of data values\bar{x} = \frac{\Sigma x}{n} = \frac{\text{sum of all data values}}{\text{number of data values}}         (Pronounced "x-bar").
    • Population Mean Formula:μ=ΣxN\mu = \frac{\Sigma x}{N}         (Denoted by lower-case Greek letter mu, μ\mu).
Worked Example 1: Calculating the Mean
  • Problem: Find the mean of the first five male pulse rates from a body data set: 8484, 7474, 5050, 6060, 5252 (all in beats per minute, or BPM\text{BPM}).
  • Calculation:xˉ=Σxn=84+74+50+60+525=3205=64.0 BPM\bar{x} = \frac{\Sigma x}{n} = \frac{84 + 74 + 50 + 60 + 52}{5} = \frac{320}{5} = 64.0\,\text{BPM}
  • Result: The mean pulse rate is 64.0 BPM64.0\,\text{BPM}.

The Median

  • Definition: The median of a data set is the measure of center that is the middle value when the original data values are arranged in order of increasing (or decreasing) magnitude. Approximately half of the data values are less than the median and half are greater than the median.
  • Important Properties of the Median:
    • The median does not change significantly when extreme values are included; therefore, the median is a resistant measure of center.
    • The median does not directly use every data value (e.g., changing the largest value to a much higher value leaves the median unchanged).
  • Notation:
    • Sample median is denoted by x~\tilde{x} (pronounced "x-tilde"), MM, or MedMed.
    • There is no universally accepted notation or special symbol for a population median.
  • Procedure for Calculating the Median:
    1. Sort the data values in ascending order.
    2. Odd number of values (nn is odd): The median is the exact middle number in the sorted list.
    3. Even number of values (nn is even): The median is the mean of the two middle numbers in the sorted list.
Worked Example 2: Median with an Odd Number of Data Values
  • Problem: Find the median of five male pulse rates: 8484, 7474, 5050, 6060, 5252\text{ BPM}.
  • Step 1: Arrange in ascending order: 50,52,60,74,8450, 52, 60, 74, 84.
  • Step 2: Count n=5n = 5 (odd). The middle (3rd) value is 60.0 BPM60.0\,\text{BPM}.
  • Result: Median = 60.0 BPM60.0\,\text{BPM} (Note: this differs from the mean of 64.0 BPM64.0\,\text{BPM}).
Worked Example 3: Median with an Even Number of Data Values
  • Problem: Find the median after adding a sixth pulse rate (62 BPM62\,\text{BPM}) to the dataset: 8484, 7474, 5050, 6060, 5252, 6262\text{ BPM}.
  • Step 1: Arrange in ascending order: 50,52,60,62,74,8450, 52, 60, 62, 74, 84.
  • Step 2: Count n=6n = 6 (even). Locate the two middle numbers: 6060 and 6262
  • Step 3: Calculate the mean of the two middle numbers:     Median=60+622=1222=61.0 BPM\text{Median} = \frac{60 + 62}{2} = \frac{122}{2} = 61.0\,\text{BPM}
  • Result: Median = 61.0 BPM61.0\,\text{BPM}.

The Mode

  • Definition: The mode of a data set is the value(s) that occurs with the greatest frequency.
  • Important Properties of the Mode:
    • The mode can be computed for both quantitative data and qualitative (categorical) data consisting of names, labels, or categories.
    • It is the only measure of center that can be used with qualitative data.
    • A data set can have no mode, one mode, or multiple modes.
  • Classifications of Mode:
    • Single Mode: One value occurs most frequently.
    • Bimodal: Two values occur with the same maximum frequency; each value is a mode.
    • Multimodal: More than two values occur with the same maximum frequency; each value is a mode.
    • No Mode: No data value is repeated.
Worked Examples: Mode
  • Example 4a (Single Mode): Dataset 58,58,58,58,60,60,62,6458, 58, 58, 58, 60, 60, 62, 64
    • Mode = 58 BPM58\,\text{BPM} (occurs most frequently, 4 times).
  • Example 4b (Two Modes / Bimodal): Dataset 58,58,58,60,60,60,62,6458, 58, 58, 60, 60, 60, 62, 64
    • Modes = 58 BPM58\,\text{BPM} and 60 BPM60\,\text{BPM} (both occur 3 times).
  • Example 4c (No Mode): Dataset 58,60,64,68,7258, 60, 64, 68, 72
    • Mode = No mode (no value is repeated).

Calculating the Mean from a Frequency Distribution

  • When data are summarized in a frequency table, the original raw values are unknown. Therefore, the calculated mean is an approximation.
  • Formula 3-2 (Mean from a Frequency Distribution):xˉ=Σ(f⋅x)Σf\bar{x} = \frac{\Sigma (f \cdot x)}{\Sigma f}     Where:
    • ff = frequency of each class.
    • xx = class midpoint of each class.
    • Σf=n\Sigma f = n = sum of frequencies (total sample size).
    • Σ(f⋅x)\Sigma (f \cdot x) = sum of the products of each class frequency and midpoint.
Worked Example: Mean from Frequency Distribution
Pulse Rate (BPM)Frequency (ff)Class Midpoint (xx)f⋅xf \cdot x
40–5440\text{--}541515474715×47=70515 \times 47 = 705
55–6955\text{--}696363626263×62=390663 \times 62 = 3906
70–8470\text{--}846262777762×77=477462 \times 77 = 4774
85–9985\text{--}991111929211×92=101211 \times 92 = 1012
100–114100\text{--}114221071072×107=2142 \times 107 = 214
TotalsΣf=153\Sigma f = 153Σ(f⋅x)=10,611\Sigma (f \cdot x) = 10,611
  • Calculation:xˉ=Σ(f⋅x)Σf=10,611153=69.4 BPM\bar{x} = \frac{\Sigma (f \cdot x)}{\Sigma f} = \frac{10,611}{153} = 69.4\,\text{BPM}
  • Comparison: The result 69.4 BPM69.4\,\text{BPM} is an approximation. The actual mean calculated using all raw individual male pulse rates is 69.6 BPM69.6\,\text{BPM}.

Calculating a Weighted Mean

  • When data values (xx) are assigned different weights (ww), a weighted mean is calculated.
  • Formula 3-3 (Weighted Mean):xˉ=Σ(w⋅x)Σw\bar{x} = \frac{\Sigma (w \cdot x)}{\Sigma w}
  • Procedure: Multiply each weight ww by its corresponding value xx, sum these products, and divide by the total sum of the weights Σw\Sigma w
Worked Example: Grade-Point Average (GPA)
  • Problem: Compute the first-semester Grade-Point Average (GPA) for a student taking five courses:
    • Course 1: Grade A (33 credits)
    • Course 2: Grade A (44 credits)
    • Course 3: Grade B (33 credits)
    • Course 4: Grade C (33 credits)
    • Course 5: Grade F (11 credit)
  • Quality Points Assignment: A=4A = 4, B=3B = 3, C=2C = 2, D=1D = 1, F=0F = 0
  • Weights (ww): Credits = 3,4,3,3,13, 4, 3, 3, 1; Total Σw=3+4+3+3+1=14\Sigma w = 3 + 4 + 3 + 3 + 1 = 14
  • Values (xx): Points = 4,4,3,2,04, 4, 3, 2, 0
  • Calculation:xˉ=(3×4)+(4×4)+(3×3)+(3×2)+(1×0)3+4+3+3+1\bar{x} = \frac{(3 \times 4) + (4 \times 4) + (3 \times 3) + (3 \times 2) + (1 \times 0)}{3 + 4 + 3 + 3 + 1}xˉ=12+16+9+6+014=4314=3.07\bar{x} = \frac{12 + 16 + 9 + 6 + 0}{14} = \frac{43}{14} = 3.07
  • Result: GPA = 3.073.07 (typically rounded to two decimal places).

Measures of Variation

Measure of Dispersion

  • Variation is a foundational concept in statistics. Measures of center (mean, median, mode) fail to capture differences in data spread.
  • Example Visualizing Spread: Consider pulse rates of two groups (Treatment vs. Placebo):

Dotplots of Pulse Rates

  • Both groups have identical measures of center:
    • Mean = 70.2 BPM70.2\,\text{BPM}
    • Median = 70.0 BPM70.0\,\text{BPM}
    • Mode = 70 BPM70\,\text{BPM}
  • However, the Treatment group displays much greater variation (spread) than the Placebo group.

The Range

  • Definition: The range of a set of data values is the difference between the maximum data value and the minimum data value.
  • Formula:Range=(maximum data value)−(minimum data value)\text{Range} = (\text{maximum data value}) - (\text{minimum data value})
  • Important Properties of the Range:
    • Uses only the maximum and minimum data values.
    • Extremely sensitive to extreme values; therefore, it is not resistant.
    • Does not take every data value into account and fails to reflect the overall variation among all data values.
Worked Example: Range
  • Problem: Find the range of male pulse rates: 8484, 7474, 5050, 6060, 5252\text{ BPM}.
  • Calculation:Range=84−50=34.0 BPM\text{Range} = 84 - 50 = 34.0\,\text{BPM}

Standard Deviation of a Sample

  • Definition: The standard deviation of a set of sample values, denoted by ss, is a measure of how much data values deviate away from the mean.
  • Notation:
    • ss = sample standard deviation.
    • σ\sigma = population standard deviation (lowercase Greek sigma).
  • Formulas:
    • Formula 3-4 (Definitional Formula):s=Σ(x−xˉ)2n−1s = \sqrt{\frac{\Sigma (x - \bar{x})^2}{n - 1}}
    • Formula 3-5 (Shortcut / Computational Formula):s=n(Σx2)−(Σx)2n(n−1)s = \sqrt{\frac{n (\Sigma x^2) - (\Sigma x)^2}{n (n - 1)}}(Formulas 3-4 and 3-5 are algebraically equivalent).
    • Population Standard Deviation Formula:σ=Σ(xi−μ)2N\sigma = \sqrt{\frac{\Sigma (x_i - \mu)^2}{N}}
  • Important Properties of Standard Deviation:
    • Measures the average departure of data values from the mean.
    • The value of ss is never negative. It equals zero only when all data values are identical.
    • Larger values of ss indicate greater variation.
    • The value of ss can increase dramatically with outliers (it is not resistant).
    • The units of standard deviation (e.g.,  BPM\,\text{BPM},  feet\,\text{feet},  seconds\,\text{seconds}) are identical to the original units.
Step-by-Step Calculation using Formula 3-4
  • Sample Data Set: 84,74,50,60,5284, 74, 50, 60, 52
Step NumberProcedureCalculation for Sample Data
Step 1Compute the sample mean xˉ\bar{x}.xˉ=84+74+50+60+525=3205=64.0\bar{x} = \frac{84 + 74 + 50 + 60 + 52}{5} = \frac{320}{5} = 64.0
Step 2Subtract the mean from each sample value to obtain deviations (x−xˉ)(x - \bar{x}).84−64=2084 - 64 = 20\74−64=1074 - 64 = 10\50−64=−1450 - 64 = -14\60−64=−460 - 64 = -4\52−64=−1252 - 64 = -12
Step 3Square each deviation (x−xˉ)2(x - \bar{x})^2.202=40020^2 = 400\102=10010^2 = 100\(−14)2=196(-14)^2 = 196\(−4)2=16(-4)^2 = 16\(−12)2=144(-12)^2 = 144
Step 4Sum all squared deviations to get Σ(x−xˉ)2\Sigma (x - \bar{x})^2.400+100+196+16+144=856400 + 100 + 196 + 16 + 144 = 856
Step 5Divide total by n−1n - 1 (11 less than sample size).8565−1=8564=214\frac{856}{5 - 1} = \frac{856}{4} = 214
Step 6Take the square root of the result.s=214≈14.6287→14.6 BPMs = \sqrt{214} \approx 14.6287 \rightarrow 14.6\,\text{BPM}
Step-by-Step Calculation using Formula 3-5
  • Components needed for Formula 3-5:
    • n=5n = 5
    • Σx=84+74+50+60+52=320\Sigma x = 84 + 74 + 50 + 60 + 52 = 320
    • Σx2=842+742+502+602+522=7056+5476+2500+3600+2704=21,336\Sigma x^2 = 84^2 + 74^2 + 50^2 + 60^2 + 52^2 = 7056 + 5476 + 2500 + 3600 + 2704 = 21,336
  • Substitution into Formula 3-5:s=5(21,336)−(320)25(5−1)=106,680−102,4005(4)=428020=214=14.6 BPMs = \sqrt{\frac{5(21,336) - (320)^2}{5(5 - 1)}} = \sqrt{\frac{106,680 - 102,400}{5(4)}} = \sqrt{\frac{4280}{20}} = \sqrt{214} = 14.6\,\text{BPM}

Variance

  • Definition: The variance of a set of values is a measure of variation equal to the square of the standard deviation.
  • Symbols and Formulas:
    • Sample Variance (s2s^2): Square of the sample standard deviation sss2=Σ(x−xˉ)2n−1s^2 = \frac{\Sigma (x - \bar{x})^2}{n - 1}
    • Population Variance (σ2\sigma^2): Square of the population standard deviation σ\sigmaσ2=Σ(x−μ)2N\sigma^2 = \frac{\Sigma (x - \mu)^2}{N}
  • Important Properties of Variance:
    • Units: Units of variance are the squares of the original units (e.g., if original data are in feet, variance units are ft2\text{ft}^2; if seconds, sec2\text{sec}^2).
    • Increases dramatically with inclusion of outliers (not resistant).
    • Value is never negative; equals zero only when all data values are identical.
    • Estimation Bias: The sample variance s2s^2 is an unbiased estimator of the population variance σ2\sigma^2 (it targets the true population value). In contrast, sample standard deviation ss is a biased estimator of σ\sigma

Coefficient of Variation (CV)

  • Definition: The coefficient of variation (or CVCV) for a set of non-negative sample or population data, expressed as a percentage, describes the standard deviation relative to the mean.
  • Formulas:
    • Sample Coefficient of Variation:CV=sxˉ⋅100%CV = \frac{s}{\bar{x}} \cdot 100\%
    • Population Coefficient of Variation:CV=σμ⋅100%CV = \frac{\sigma}{\mu} \cdot 100\%
  • When to Use CV:
    • To compare variation from two samples/populations with different units or scales (e.g., comparing pulse rates in BPM\text{BPM} to heights in cm\text{cm}).
    • To compare variation between groups with vastly different means.
    • Note: Direct comparison of standard deviations is appropriate only when sample means are approximately equal and measurement scales/units are identical.
Worked Example: Comparing Variation (Pulse Rates vs. Heights)
  • Problem: Compare variation between 153153 male pulse rates (xˉ=69.6 BPM\bar{x} = 69.6\,\text{BPM}, s=11.3 BPMs = 11.3\,\text{BPM}) and their heights (xˉ=174.12 cm\bar{x} = 174.12\,\text{cm}, s=7.10 cms = 7.10\,\text{cm}).
  • Male Pulse Rates CV:CV=11.3 BPM69.6 BPM⋅100%=16.2%CV = \frac{11.3\,\text{BPM}}{69.6\,\text{BPM}} \cdot 100\% = 16.2\%
  • Male Heights CV:CV=7.10 cm174.12 cm⋅100%=4.1%CV = \frac{7.10\,\text{cm}}{174.12\,\text{cm}} \cdot 100\% = 4.1\%
  • Conclusion: Male pulse rates (CV=16.2%CV = 16.2\%) exhibit significantly greater variation than male heights (CV=4.1%CV = 4.1\%).