Chapter 3 Section 2

1. Chapter 3Section 2 Measures of Dispersion

2. Chapter 3 – Section 2 • Comparing two sets of data • Comparing two sets of data • The measures of central tendency (mean, median, mode) measure the differences between the “average” or “typical” values between two sets of data • Comparing two sets of data • The measures of central tendency (mean, median, mode) measure the differences between the “average” or “typical” values between two sets of data • The measures of dispersion in this section measure the differences between how far “spread out” the data values are

3. Chapter 3 – Section 2 • The range of a variable is the largest data value minus the smallest data value • The range of a variable is the largest data value minus the smallest data value • Compute the range of 6, 1, 2, 6, 11, 7, 3, 3 • The range of a variable is the largest data value minus the smallest data value • Compute the range of 6, 1, 2, 6, 11, 7, 3, 3 • The largest value is 11 • The smallest value is 1 • The range of a variable is the largest data value minus the smallest data value • Compute the range of 6, 1, 2, 6, 11, 7, 3, 3 • The largest value is 11 • The smallest value is 1 • Subtracting the two … 11 – 1 = 10 … the range is 10

4. Chapter 3 – Section 2 • The range only uses two values in the data set – the largest value and the smallest value • The range is not resistant • The range only uses two values in the data set – the largest value and the smallest value • The range is not resistant • If we made a mistake and 6, 1, 2 was recorded as 6000, 1, 2 • The range only uses two values in the data set – the largest value and the smallest value • The range is not resistant • If we made a mistake and 6, 1, 2 was recorded as 6000, 1, 2 • The range is now ( 6000 – 1 ) = 5999

5. Chapter 3 – Section 2 • The variance is based on the deviation from the mean • ( xi – μ ) for populations • ( xi – ) for samples • The variance is based on the deviation from the mean • ( xi – μ ) for populations • ( xi – ) for samples • To treat positive differences and negative differences, we square the deviations • ( xi – μ )2 for populations • ( xi – )2 for samples

6. Chapter 3 – Section 2 • The populationvariance of a variable is the sum of these squared deviations divided by the number in the population • The populationvariance of a variable is the sum of these squared deviations divided by the number in the population • The populationvariance of a variable is the sum of these squared deviations divided by the number in the population • The population variance is represented by σ2 • Note: For accuracy, use as many decimal places as allowed by your calculator

7. Chapter 3 – Section 2 • Compute the population variance of 6, 1, 2, 11 • Compute the population variance of 6, 1, 2, 11 • Compute the population mean first μ = (6 + 1 + 2 + 11) / 4 = 5 • Compute the population variance of 6, 1, 2, 11 • Compute the population mean first μ = (6 + 1 + 2 + 11) / 4 = 5 • Now compute the squared deviations (1–5)2 = 16, (2–5)2 = 9, (6–5)2 = 1, (11–5)2 = 36 • Compute the population variance of 6, 1, 2, 11 • Compute the population mean first μ = (6 + 1 + 2 + 11) / 4 = 5 • Now compute the squared deviations (1–5)2 = 16, (2–5)2 = 9, (6–5)2 = 1, (11–5)2 = 36 • Average the squared deviations (16 + 9 + 1 + 36) / 4 = 15.5 • The population variance σ2 is 15.5

8. Chapter 3 – Section 2 • The samplevariance of a variable is the sum of these squared deviations divided by one less than the number in the sample • The samplevariance of a variable is the sum of these squared deviations divided by one less than the number in the sample • The sample variance is represented by s2 • We say that this statistic has n – 1 degrees of freedom

9. Chapter 3 – Section 2 • Compute the sample variance of 6, 1, 2, 11 • Compute the sample variance of 6, 1, 2, 11 • Compute the sample mean first = (6 + 1 + 2 + 11) / 4 = 5 • Compute the sample variance of 6, 1, 2, 11 • Compute the sample mean first = (6 + 1 + 2 + 11) / 4 = 5 • Now compute the squared deviations (1–5)2 = 16, (2–5)2 = 9, (6–5)2 = 1, (11–5)2 = 36 • Compute the sample variance of 6, 1, 2, 11 • Compute the sample mean first = (6 + 1 + 2 + 11) / 4 = 5 • Now compute the squared deviations (1–5)2 = 16, (2–5)2 = 9, (6–5)2 = 1, (11–5)2 = 36 • Average the squared deviations (16 + 9 + 1 + 36) / 3 = 20.7 • The sample variance s2 is 20.7

10. Chapter 3 – Section 2 • Why are the population variance (15.5) and the sample variance (20.7) different for the same set of numbers? • Why are the population variance (15.5) and the sample variance (20.7) different for the same set of numbers? • In the first case, { 6, 1, 2, 11 } was the entire population (divide by N) • Why are the population variance (15.5) and the sample variance (20.7) different for the same set of numbers? • In the first case, { 6, 1, 2, 11 } was the entire population (divide by N) • In the second case, { 6, 1, 2, 11 } was just a sample from the population (divide by n – 1) • Why are the population variance (15.5) and the sample variance (20.7) different for the same set of numbers? • In the first case, { 6, 1, 2, 11 } was the entire population (divide by N) • In the second case, { 6, 1, 2, 11 } was just a sample from the population (divide by n – 1) • These are two different situations

11. Chapter 3 – Section 2 • Why do we use different formulas? • The reason is that using the sample mean is not quite as accurate as using the population mean • If we used “n” in the denominator for the sample variance calculation, we would get a “biased” result • Bias here means that we would tend to underestimate the true variance

12. Chapter 3 – Section 2 • The standarddeviation is the square root of the variance • The standarddeviation is the square root of the variance • The populationstandarddeviation • Is the square root of the population variance (σ2) • Is represented by σ • The standarddeviation is the square root of the variance • The populationstandarddeviation • Is the square root of the population variance (σ2) • Is represented by σ • The samplestandarddeviation • Is the square root of the sample variance (s2) • Is represented by s

13. Chapter 3 – Section 2 • If the population is { 6, 1, 2, 11 } • The population variance σ2 = 15.5 • The population standard deviation σ = • If the population is { 6, 1, 2, 11 } • The population variance σ2 = 15.5 • The population standard deviation σ = • If the sample is { 6, 1, 2, 11 } • The sample variance s2 = 20.7 • The sample standard deviation s = • If the population is { 6, 1, 2, 11 } • The population variance σ2 = 15.5 • The population standard deviation σ = • If the sample is { 6, 1, 2, 11 } • The sample variance s2 = 20.7 • The sample standard deviation s = • The population standard deviation and the sample standard deviation apply in different situations

14. Chapter 3 – Section 2 • The standard deviation is very useful for estimating probabilities

15. Chapter 3 – Section 2 • The empirical rule • If the distribution is roughly bell shaped, then • The empirical rule • If the distribution is roughly bell shaped, then • Approximately 68% of the data will lie within 1 standard deviation of the mean • The empirical rule • If the distribution is roughly bell shaped, then • Approximately 68% of the data will lie within 1 standard deviation of the mean • Approximately 95% of the data will lie within 2 standard deviations of the mean • The empirical rule • If the distribution is roughly bell shaped, then • Approximately 68% of the data will lie within 1 standard deviation of the mean • Approximately 95% of the data will lie within 2 standard deviations of the mean • Approximately 99.7% of the data (i.e. almost all) will lie within 3 standard deviations of the mean

16. Chapter 3 – Section 2 • For a variable with mean 17 and standard deviation 3.4 • For a variable with mean 17 and standard deviation 3.4 • Approximately 68% of the values will lie between(17 – 3.4) and (17 + 3.4), i.e. 13.6 and 20.4 • For a variable with mean 17 and standard deviation 3.4 • Approximately 68% of the values will lie between(17 – 3.4) and (17 + 3.4), i.e. 13.6 and 20.4 • Approximately 95% of the values will lie between(17 – 2  3.4) and (17 + 2  3.4), i.e. 10.2 and 23.8 • For a variable with mean 17 and standard deviation 3.4 • Approximately 68% of the values will lie between(17 – 3.4) and (17 + 3.4), i.e. 13.6 and 20.4 • Approximately 95% of the values will lie between(17 – 2  3.4) and (17 + 2  3.4), i.e. 10.2 and 23.8 • Approximately 99.7% of the values will lie between(17 – 3  3.4) and (17 + 3  3.4), i.e. 6.8 and 27.2 • For a variable with mean 17 and standard deviation 3.4 • Approximately 68% of the values will lie between(17 – 3.4) and (17 + 3.4), i.e. 13.6 and 20.4 • Approximately 95% of the values will lie between(17 – 2  3.4) and (17 + 2  3.4), i.e. 10.2 and 23.8 • Approximately 99.7% of the values will lie between(17 – 3  3.4) and (17 + 3  3.4), i.e. 6.8 and 27.2 • A value of 2.1 and a value of 33.2 would both be very unusual

17. Chapter 3 – Section 2 • Chebyshev’s inequality gives a lower bound on the percentage of observations that lie within k standard deviations of the mean (where k > 1) • Chebyshev’s inequality gives a lower bound on the percentage of observations that lie within k standard deviations of the mean (where k > 1) • This lower bound is • An estimated percentage • The actual percentage for any variable cannot be lower than this number • Chebyshev’s inequality gives a lower bound on the percentage of observations that lie within k standard deviations of the mean (where k > 1) • This lower bound is • An estimated percentage • The actual percentage for any variable cannot be lower than this number • Therefore the actual percentage must be this value or higher

18. Chapter 3 – Section 2 • Chebyshev’s inequality • For any data set, at least of the observations will lie within k standard deviations of the mean, where k is any number greater than 1

19. Chapter 3 – Section 2 • How much of the data lies within 1.5 standard deviations of the mean? • From Chebyshev’s inequality so that at least 55.6% of the data will lie within 1.5 standard deviations of the mean

20. Chapter 3 – Section 2 • If the mean is equal to 20 and the standard deviation is equal to 4, how much of the data lies between 14 and 26? • 14 to 26 are 1.5 standard deviations from 20 so that at least 55.6% of the data will lie between 14 and 26

21. Summary: Chapter 3 – Section 2 • Range • The maximum minus the minimum • Not a resistant measurement • Variance and standard deviation • Measures deviations from the mean • Not a resistant measurement • Empirical rule • About 68% of the data is within 1 standard deviation • About 95% of the data is within 2 standard deviations