1. Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 3 Section 2 – Slide 1 of 27 Chapter 3 Section 2 Measures of Dispersion

2. Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 3 Section 2 – Slide 2 of 27 Chapter 3 – Section 2 ●Learning objectives  The range of a variable  The variance of a variable  The standard deviation of a variable  Use the Empirical Rule  Use Chebyshev’s inequality 1 2 3 5 4

3. Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 3 Section 2 – Slide 3 of 27 Chapter 3 – Section 2 ●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

4. Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 3 Section 2 – Slide 4 of 27 Chapter 3 – Section 2 ●Learning objectives  The range of a variable  The variance of a variable  The standard deviation of a variable  Use the Empirical Rule  Use Chebyshev’s inequality 1 2 3 5 4

5. Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 3 Section 2 – Slide 5 of 27 Chapter 3 – Section 2 ●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

6. Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 3 Section 2 – Slide 6 of 27 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

7. Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 3 Section 2 – Slide 7 of 27 Chapter 3 – Section 2 ●Learning objectives  The range of a variable  The variance of a variable  The standard deviation of a variable  Use the Empirical Rule  Use Chebyshev’s inequality 1 2 3 5 4

8. Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 3 Section 2 – Slide 8 of 27 Chapter 3 – Section 2 ●The variance is based on the deviation from the mean  ( x i – μ ) for populations  ( x i – ) for samples ●The variance is based on the deviation from the mean  ( x i – μ ) for populations  ( x i – ) for samples ●To treat positive differences and negative differences, we square the deviations  ( x i – μ ) 2 for populations  ( x i – ) 2 for samples

9. Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 3 Section 2 – Slide 9 of 27 Chapter 3 – Section 2 ●The population variance of a variable is the sum of these squared deviations divided by the number in the population ●The population variance is represented by σ 2 ●Note: For accuracy, use as many decimal places as allowed by your calculator

10. Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 3 Section 2 – Slide 10 of 27 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 σ 2 is 15.5

11. Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 3 Section 2 – Slide 11 of 27 Chapter 3 – Section 2 ●The sample variance 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 s 2 ●We say that this statistic has n – 1 degrees of freedom

12. Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 3 Section 2 – Slide 12 of 27 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 s 2 is 20.7

13. Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 3 Section 2 – Slide 13 of 27 Chapter 3 – Section 2 ●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

14. Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 3 Section 2 – Slide 14 of 27 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

15. Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 3 Section 2 – Slide 15 of 27 Chapter 3 – Section 2 ●Learning objectives  The range of a variable  The variance of a variable  The standard deviation of a variable  Use the Empirical Rule  Use Chebyshev’s inequality 1 2 3 5 4

16. Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 3 Section 2 – Slide 16 of 27 Chapter 3 – Section 2 ●The standard deviation is the square root of the variance ●The population standard deviation  Is the square root of the population variance (σ 2 )  Is represented by σ ●The standard deviation is the square root of the variance ●The population standard deviation  Is the square root of the population variance (σ 2 )  Is represented by σ ●The sample standard deviation  Is the square root of the sample variance (s 2 )  Is represented by s

17. Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 3 Section 2 – Slide 17 of 27 Chapter 3 – Section 2 ●If the population is { 6, 1, 2, 11 }  The population variance σ 2 = 15.5  The population standard deviation σ = ●If the population is { 6, 1, 2, 11 }  The population variance σ 2 = 15.5  The population standard deviation σ = ●If the sample is { 6, 1, 2, 11 }  The sample variance s 2 = 20.7  The sample standard deviation s = ●If the population is { 6, 1, 2, 11 }  The population variance σ 2 = 15.5  The population standard deviation σ = ●If the sample is { 6, 1, 2, 11 }  The sample variance s 2 = 20.7  The sample standard deviation s = ●The population standard deviation and the sample standard deviation apply in different situations

18. Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 3 Section 2 – Slide 18 of 27 Chapter 3 – Section 2 ●Learning objectives  The range of a variable  The variance of a variable  The standard deviation of a variable  Use the Empirical Rule  Use Chebyshev’s inequality 1 2 3 5 4

19. Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 3 Section 2 – Slide 19 of 27 Chapter 3 – Section 2 ●The standard deviation is very useful for estimating probabilities

20. Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 3 Section 2 – Slide 20 of 27 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

21. Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 3 Section 2 – Slide 21 of 27 Chapter 3 – Section 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 ●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 (less than 6.8) and a value of 33.2 (greater than 27.2) would both be very unusual

22. Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 3 Section 2 – Slide 22 of 27 Chapter 3 – Section 2 ●Learning objectives  The range of a variable  The variance of a variable  The standard deviation of a variable  Use the Empirical Rule  Use Chebyshev’s inequality 1 2 3 5 4

23. Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 3 Section 2 – Slide 23 of 27 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) ●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

24. Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 3 Section 2 – Slide 24 of 27 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

25. Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 3 Section 2 – Slide 25 of 27 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

26. Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 3 Section 2 – Slide 26 of 27 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

27. Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 3 Section 2 – Slide 27 of 27 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