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Chapter 4: Variability
Chapter Outline
4.1 Overview (The Purpose for Measuring Variability)
4.2 The Range
4.3 Standard Deviation and Variance for a Population
Formulas for Population Variance and Standard Deviation
Final Formulas and Notation
4.4 Standard Deviation and Variance for Samples
Sample Variability and Degrees of Freedom
4.5 More about Variance and Standard Deviation
Presenting the Mean and Standard Deviation in a Frequency Distribution Graph
Sample Variance as an Unbiased Statistic
Standard Deviation and Descriptive Statistics
Transformations of Scale (Adding a Constant or Multiplying by a Constant)
In the Literature: Reporting the Standard Deviation
Variance and Inferential Statistics
Learning Objectives and Chapter Summary
1. Students should understand the general purpose for measuring variability and they should be
able to recognize the difference between scores with high variability versus scores with low
variability.
Variability recognizes that not all of the scores are the same, there are differences (unless
there is zero variability), and then measures how big the differences are. Note that
variability is a measure of distance.
2. Students should be able to define and calculate the range, but they should also realize that this
is a relatively crude measure of variability.
The range is defined as the distance from the largest score to the smallest score and thus
is completely determined by only two scores. Technically, the range is measured as the
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distance from the lower real limit of the lowest score to the upper real limit of the highest
score, but is often calculated as simply the highest score minus the lowest score. Note
that the range does not measure variability for the entire set of scores, but rather is
determined entirely by the two extreme scores.
3. Students should understand the concept of standard deviation as a measure of the standard
distance from the mean.
Some scores are close to the mean and other scores are far away. The standard deviation
provides a measure of the standard or “average” distance from the mean. In a frequency
distribution graph showing either a sample or a population, students should be able to
estimate the location of the mean and estimate the size of the standard deviation. If
students are given values for the mean and standard deviation, they should be able to
sketch a frequency distribution corresponding to the given values.
Although the concept of variance is not as concrete as the concept of standard deviation,
students should be able to provide a verbal definition for variance: standard deviation is a
measure of distance from the mean, so variance is a measure of squared distance.
4. Students should be able to calculate SS (sum of squared deviations), variance, and standard
deviation for a sample and for a population. In addition, they should understand the concept of
an unbiased statistic and the correction for bias that is used in the formula for sample variance.
The three values (SS, variance, and standard deviation) are always calculated in order:
First you find the sum of the squared deviations, then find the average of the squared
deviations, and finally you take the square root to find the standard deviation. If students
can provide verbal definitions for these three values, then most of the formulas are selfevident. (The notable exception is the computational formula for SS.)
Note that the correction for bias (dividing by n – 1) is made only when calculating sample
variance. SS is computed exactly the same for samples and populations, and standard
deviation is always the square root of variance for both samples and for populations.
Other Lecture Suggestions
1. The same demonstration that was used in Chapter 3 to show how adding a constant (or
multiplying by a constant) affects the mean, can be used again to show how these actions affect
the standard deviation.
Sketch a simple histogram and label the values along the X axis using 1, 2, 3, 4, and 5.
Locate the mean using a vertical line at M = 3 and identify the standard deviation using arrows
that point from the mean up to X = 4 and from the mean down to X = 2 (showing a standard
deviation of one point). Now ask the students what will happen to the distribution if you add 10
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points to every score. Answer: The whole distribution moves 10 points to the right. (You can
keep the same sketch, simply re-label the values on the X axis to 11, 12, 13, 14, and 15.) Note
that the mean (middle) has shifted 10 points to the right but the standard deviation arrows are
still exactly one point. Adding a constant does not change the standard deviation.
Go back to the original distribution and ask what would happen if every score were
multiplied by 10. This time the 1s become 10s, the 2s become 20s, and so on. Again, you can
keep the same sketch, simply re-label the values on the X axis to 10, 20, 30, 40, and 50. After
multiplying, the mean (that was M = 3) is now located at M = 30 (10 times bigger), and the
arrows that used to extend one point from the mean now extend 10 points. Multiplying by a
constant multiplies both the mean and the standard deviation.
2. An alternative demonstration of the effects of adding or multiplying by a constant is simply to
note that standard deviation is a measure of distance. To determine how the standard deviation
is affected, you can simply pick any two scores and determine how the distance between them is
affected. For example, you have a quiz score of 7 and your friend has a score of 5. What
happens to the distance between these scores if (a) five points are added to every score in the
distribution? (b) every score is multiplied by 3?
3. To test students’ understanding of standard deviation, give them parameters for a distribution
of scores (for example, µ = 70 and σ = 15) and ask if a specific score such as X = 80 is an
extreme score, out in the tail of the distribution. To answer the question, sketch a distribution (a
pile of scores) with the appropriate parameters and locate the approximate position of the
specific score. Note: If you use scores that are within one standard deviation of the mean (not
extreme) or more than two or three standard deviations from the mean (extreme) you probably
will not need to define “extreme.”
4. Example 4.7 and Table 4.1 can be repeated in class to demonstrate several concepts and
calculations concerning variability. First, you can calculate variance for the original population
showing the population formulas. Second, you can compute SS and variance for one or two of
the samples to demonstrate the sample formulas. Finally, you can use the set of nine sample
variances to demonstrate the concept of an unbiased statistic. Note that none of the samples has
a variance exactly equal to the population variance, however the average of the sample variances
is exactly equal to the population variance.
5. The concept of biased samples and representative samples can be used as an analogy to help
explain the concept of biased and unbiased statistics. If you were interested in studying
elementary school children, for example, you probably would not recruit your sample from a
science and computers camp for child geniuses. These kids would be a biased sample that is not
representative of the general population. Similarly, calculating sample variance without making
some correction, produces a biased value that is not representative of the general population. In
each case, the goal is to obtain a sample that is representative of the population. (Note: Be sure
to differentiate the concept of a biased sample and a biased statistic. Even if you have a good
representative sample, the sample variance will be a biased statistic unless you make a correction
in the formula.)
6. The following values produce whole-number answers for classroom examples.
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Population Variance and Standard Deviation
Definitional formula
Computational formula
X
7
1
7
9
SS = 36
σ2 = 9
σ =3
X
1
6
1
1
1
SS = 20
σ2 = 4
σ=2
Sample Variance and Standard Deviation
Definitional formula
Computational formula
X
5
1
5
5
SS = 12
s2 = 4
s=2
X
1
7
1
1
SS = 27
s2 = 9
s=3
Exam Items for Chapter 4
Multiple-Choice Questions
1. (www) In a population of N = 10 scores, the smallest score is X = 8 and the largest score is
X = 20. Using the concept of real limits, what is the range for this population?
a. 11
b. 12
c. 13
d. cannot be determined without more information
2. A sample consists of n = 16 scores. How many of the scores are used to calculate the range?
a. 2
b. 4
c. 8
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d. all 16
3. For a sample of n = 16 scores, how many scores are used to calculate the sample variance?
a. 2
b. 8
c. 15
d. all 16
4. What is the value of SS (sum of squared deviations) for the following sample?
Sample: 2, 3, 4, 7
a. 14/3 = 2.67
b. 14
c. 72
d. 78
5. (www) What is the value of SS (sum of squared deviations) for the following population?
Population: 1, 1, 1, 5
a. 3
b. 7
c. 12
d. 28
6. What is the value of SS (sum of squared deviations) for the following sample?
Sample: 1, 1, 1, 3
a. 0
b. 1
c. 3
d. 12
7. What is the value of SS (sum of squared deviations) for the following population?
Population: 2, 3, 0, 5
a. 13
b. 38
c. 13/4 = 3.25
d. 38/4 = 9.50
8. What is the value of SS for the following set of scores? Scores: 8, 3, 1.
a. 26
b. 29
c. 74
d. 144
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9. A population of N = 5 scores has ΣX = 20 and ΣX2 = 100. For this population, what is the
value of SS?
a. 20
b. 80
c. 100
d. 380
10. (www) A population has SS= 100 and 2 = 4. How many scores are in the population?
a. 25
b. 26
c. 200
d. 400
11. A population has SS= 100 and 2 = 4. What is the value of (X – ) for the population?
a. 0
b. 25
c. 100
d. 400
12. What is the value of SS for the following set of scores?
a. 18
b. 10
c. 9
d. 6
Scores: 1, 1, 4, 0
13. A sample of n = 4 scores has ΣX = 8 and ΣX2 = 40. What is the value of SS for this sample?
a. 6
b. 8
c. 24
d. 40
14. A sample of n = 5 scores has ΣX = 20 and ΣX2 = 120. For this sample, what is the value of
SS?
a. 20
b. 40
c. 100
d. 120
15. A population of N = 6 scores has ΣX = 12 and ΣX2 = 54. What is the value of SS for this
population?
a. 5
b. 9
c. 30
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d. 54
16. Which of the following symbols identifies the sample variance?
a. s
b. s2
c. σ
d. σ 2
17. Which of the following symbols identifies the population standard deviation?
a. s
b. s2
c. σ
d. σ 2
18. A population of N = 100 scores has µ = 30 and σ = 4. What is the population variance?
a. 2
b. 4
c. 8
d. 16
19. A sample of n = 25 scores has M = 20 and s2 = 9. What is the sample standard deviation?
a. 3
b. 4.5
c. 9
d. 81
20. A set of 10 scores has SS = 90. If the scores are a sample, the sample variance is ____ and if
the scores are a population, the population variance is ____.
a. s2 = 9, σ2 = 9
b. s2 = 9, σ2 = 10
c. s2 = 10, σ2 = 9
d. s2 = 10, σ2 = 10
21. (www) The sum of the squared deviation scores is SS = 20 for a population of N = 5 scores.
What is the variance for this population?
a. 4
b. 5
c. 80
d. 100
22. (www) The sum of the squared deviation scores is SS = 20 for a sample of n = 5 scores.
What is the variance for this sample?
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a.
b.
c.
d.
4
5
80
100
23. (www) A population has µ = 50 and σ = 5. If 10 points are added to every score in the
population, then what are the new values for the mean and standard deviation?
a. µ = 50 and σ = 5
b. µ = 50 and σ = 15
c. µ = 60 and σ = 5
d. µ = 60 and σ = 15
24. A population of scores has µ = 50 and σ = 5. If every score in the population is multiplied
by 3, then what are the new values for the mean and standard deviation?
a. µ = 50 and σ = 5
b. µ = 50 and σ = 15
c. µ = 150 and σ = 5
d. µ = 150 and σ = 15
25. A sample of n = 8 scores has SS = 50. If these same scores were a population, then the SS
value for the population would be _____.
a. 50
b. greater than 50
c. less than 50
d. impossible to determine without additional information
26. A sample of n = 9 scores has a variance of s2 = 18. If the scores were a population, what
value would be obtained for the population variance.
a. 2 = 14
b. 2 = 16
c. 2 = 24
d. 2 = 144
27. (www) What are the values for SS and variance for the following sample of n = 3 scores?
Sample: 1, 4, 7
a. SS = 18 and variance = 6
b. SS = 18 and variance = 9
c. SS = 66 and variance = 22
d. SS = 66 and variance = 33
28. (www) What are the values for SS and variance for the following sample of n = 4 scores?
Sample: 1, 1, 0, 4
a. SS = 9 and variance = 3
b. SS = 9 and variance = 2.25
c. SS = 18 and variance = 6
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d. SS = 18 and variance = 9
29. What is the value of SS for the following set of scores? Scores: 0, 1, 4, 5
a. 17
b. 18
c. 42
d. Cannot answer without knowing whether it is a sample or a population.
30. What is the variance for the following population of scores? Scores: 5, 2, 5, 4
a. 6
b. 2
c. 1.5
d. 1.22
31. Which of the following is true for most distributions?
a. Around 30% of the scores will be located within one standard deviation of the
mean.
b. Around 50% of the scores will be located within one standard deviation of the
mean.
c. Around 70% of the scores will be located within one standard deviation of the
mean.
d. Around 90% of the scores will be located within one standard deviation of the
mean.
32. Which set of scores has the smallest standard deviation?
a. 11, 17, 31, 53
b. 5, 11, 42, 22
c. 145, 143, 145, 147
d. 27, 105, 10, 80
33. For a particular sample, the largest distance (deviation) between a score and the mean is 11
points. The smallest distance between a score and the mean is 4 points. Therefore, the standard
deviation _____.
a. will be less than 4
b. will be between 4 and 11
c. will be greater than 11
d. It is impossible to say anything about the standard deviation.
34. The smallest score in a population is X = 5 and the largest score is X = 10. Based on this
information, you can conclude that ______.
a. the population mean is somewhere between 5 and 10.
b. the population standard deviation is smaller than 6.
c. the population mean is between 5 and 10, and the standard deviation is less
than 6.
d. None of the other choices is correct.
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35. If sample variance is computed by dividing SS by n, then the average value of the sample
variances from all the possible random samples will be _______ the population variance.
a. smaller than
b. larger than
c. exactly equal to
d. unrelated to
36. If sample variance is computed by dividing SS by df = n – 1, then the average value of the
sample variances from all the possible random samples will be _______ the population variance.
a. smaller than
b. larger than
c. exactly equal to
d. unrelated to
37. For a population with  = 60, which of the following values for the population standard
deviation would cause X = 68 to have the most extreme position in the distribution?
a. σ = 1
b. σ = 2
c. σ = 3
d. σ = 4
38. There is a 6-point difference between two sample means. If the two samples have the same
variance, then which of the following values for the variance would make the mean difference
easiest to see in a graph showing the two distributions.
a. s2 = 2
b. s2 = 8
c. s2 = 16
d. s2 = 64
39. (www) On an exam with a mean of μ = 70, you have a score of X = 75. Which of the
following values for the standard deviation would give you the highest position within the class?
a. σ = 1
b. σ = 5
c. σ = 10
d. cannot determine from the information given
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40. (www) On an exam with a mean of μ = 70, you have a score of X = 65. Which of the
following values for the standard deviation would give you the highest position within the class?
a. σ = 1
b. σ = 5
c. σ = 10
d. cannot determine from the information given
True/False Questions
41. The range and the standard deviation, are both measures of distance.
42. Using the concept of real limits, the range is 8 points for a set of scores that range from a
high of X = 16 to a low of X = 8.
43. The range is usually considered to be a relatively crude measure of variability.
44. For a population of scores, the sum of the deviation scores is equal to N.
45. A population of N = 5 scores has SS = 20 and σ2 = 4. If the 5 scores were a sample, the
value of SS would still be 20 but the variance would be s2 = 5.
46. The value for SS is always greater than or equal to zero.
47. A positive deviation always indicates a score that is less than the mean.
48. After a researcher adds 5 points to every score in a sample, the standard deviation is found to
be s = 10. The original sample had a standard deviation of s = 5.
49. After a researcher multiplies every score in a sample by 2, the standard deviation is found to
be s = 10. The original sample had a standard deviation of s = 5.
50. Multiplying every score in a sample by 3 will not change the value of the standard deviation.
51. If the population variance is 5, then the population standard deviation is σ = 25.
52. A sample with a variance of 25 has a standard deviation equal to 5 points.
53. If the scores in a population range from a low of X = 5 to a high of X = 14, then the
population standard deviation must be less than 15 points.
54. A sample of n = 25 scores is selected from a population with a variance of σ2 = 16. The
sample variance probably will be smaller than 16.
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55. For a population, a deviation score is computed as X – μ.
56. If the population variance is 4, then the standard deviation will be σ = 16.
57. A sample of n = 7 scores has SS = 42. The variance for this sample is s2 = 6.
58. For a sample of n = 6 scores with X = 30 and X2 = 200, SS = 20.
59. For a population of N = 4 scores with X = 10 and X2 = 30, SS = 5.
60. A population with SS = 90 and a variance of 9 has N = 10 scores.
61. A sample with SS = 40 and a variance of 8 has n = 5 scores.
62. To calculate the variance for a sample, SS is divided by df = n – 1.
63. To calculate the variance for a population, SS is divided by N.
64. In a population with a mean of μ = 40 and a standard deviation of σ = 8, a score of
X = 46 would be an extreme value, far out in the tail of the distribution.
65. If you have a score of X = 66 on an exam with  = 70 you should expect a better grade if σ =
10 than if σ = 5.
66. If you have a score of X = 76 on an exam with  = 70 you should expect a better grade if σ =
10 than if σ = 5.
67. For a sample with M = 20 and s = 1, a score of X = 17 would be considered an extremely
low score.
68. For a sample with M = 40 and s = 4, about 95% of the individuals will have scores between
X = 32 and X = 48.
69. For a population with µ = 70 and σ = 5, about 95% of the individuals will have scores
between X = 65 and X = 75.
70. It is easier to see the mean difference between two samples if the sample variances are small.
Other Exam Items
71. For each of the following samples:
Sample #1: 1, 0, 3, 6
a. Compute the mean
Sample #2: 1, 3, 5, 4, 7
b. Determine whether it would be better to use the computational or the
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definitional formula for SS.
c. Compute SS.
72. Using the definitional formula, compute SS, variance and the standard deviation for the
following sample of scores. Scores: 3, 6, 1, 6, 5, 3
73. (www) For the following sample, use the computational formula to calculate SS. Then,
compute the sample variance and standard deviation. Scores: 1, 3, 1, 1
74. Calculate the variance and the standard deviation for the following sample data.
Scores: 10, 7, 9, 1, 2, 0, 6
75. Without some correction, sample variability is said to be "biased." Define the term biased,
and explain how this bias is corrected in the formula for sample variance.
Answers for Multiple-Choice Questions (with section and page numbers from the text)
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
c, 4.2, p. 106
a, 4.2, p. 106
d, 4.4, p. 115
b, 4.4, p. 115
c, 4.3, p. 111
c, 4.4, p. 115
a, 4.3, p. 111
a, 4.3, p. 111
a, 4.3, p. 112
a, 4.3, p. 113
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
a, 4.3, p. 108
c, 4.3, p. 111
c, 4.4, p. 115
b, 4.4, p. 115
c, 4.3, p. 112
b, 4.4, p. 115
c, 4.3, p. 113
d, 4.3, p. 108
a, 4.4, p. 115
c, 4.4, p. 115
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
a, 4.3, p. 113
b, 4.4, p. 115
c, 4.5, p. 122
d, 4.5, p. 123
a, 4.4, p. 115
b, 4.4, p. 115
b, 4.4, p. 115
a, 4.4, p. 115
a, 4.4, p. 115
c, 4.3, p. 113
31.
32.
33.
34.
35.
36.
37.
38.
39.
40.
c, 4.5, p. 121
c, 4.5, p. 121
b, 4.4, p. 116
c, 4.5, p. 121
a, 4.5, p. 120
c, 4.5, p. 120
a, 4.5, p. 122
a, 4.5, p. 125
a, 4.5, p. 122
c, 4.5, p. 122
Answers for True/False Questions (with section and page numbers from the text)
41.
42.
43.
44.
45.
46.
47.
48.
49.
50.
T, 4.1, p. 105
F, 4.2, p. 106
T, 4.2, p. 106
F, 4.3, p. 108
T, 4.4, p. 115
T, 4.3, p. 111
F, 4.3, p. 107
F, 4.5, p. 122
T, 4.5, p. 123
F, 4.5, p. 123
51.
52.
53.
54.
55.
56.
57.
58.
59.
60.
F, 4.3, p. 108
T, 4.4, p. 115
T, 4.3, p. 110
F, 4.5, p. 120
T, 4.3, p. 107
F, 4.3, p. 108
F, 4.4, p. 115
F, 4.4, p. 115
T, 4.3, p. 112
T, 4.3, p. 113
61.
62.
63.
64.
65.
66.
67.
68.
69.
70.
F, 4.4, p. 115
T, 4.4, p. 117
T, 4.3, p. 113
F, 4.5, p. 122
T, 4.5, p. 122
F, 4.5, p. 122
T, 4.5, p. 122
T, 4.5, p. 121
F, 4.5, p. 121
T, 4.5, p. 125
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Answers for Other Exam Items
71. a. For sample #1, M = 2.5. For sample #2, M = 4.
b. The computational formula is better for sample #1 and the definitional formula for
sample #2.
c. For sample #1, SS = 21. For sample #2, SS = 20.
72. SS = 20; s 2 = 4; s = 2.
73. ΣX = 6 and ΣX2 = 12. SS = 3, σ2 = 1and σ = 1.
74. SS = 96; s2 = 16; s = 4
75. Without some correction, sample variability tends to be smaller than the population
variability. Whenever a statistic consistently underestimates (or overestimates) the
corresponding population parameter, the statistic is said to be biased. The bias in sample
variability is corrected by dividing the sum of squared deviations (SS) by n – 1 (instead of n) in
the formula for sample variance.
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