Section 5.2: Measures of Variation

Section 5.2: Measures of Variation

Another component of describing a data set is how much “Spread” there is in the data set. In other words, how much the data in the distribution vary from one another. It may seem like once we know the center of a data set, we know everything there is to know. The first example will demonstrate why we need measures of variation (or spread).

There are several ways to measure this “Spread” of the data. The three most common measures are the range, standard deviation, and quartiles. In this section we will learn about the range and standard deviation. We will discuss quartiles in the following section.

We will focus first on the simplest measure of spread, called the range.

Range

The range is the difference between the maximum value and the minimum value of the data set.

Example 8

Consider these three sets of quiz scores. Find the range of each.

Section A: 5 5 5 5 5 5 5 5 5 5
Section B: 0 0 0 0 0 10 10 10 10 10
Section C: 4 4 4 5 5 5 5 6 6 6

Solution: All three of these sets of data have a mean of 5 and median of 5. If we only calculated a measure of center for each set of scores, we would say the three sets are all identical, yet the sets of scores are clearly quite different. Calculating a measure of variability (or spread) will help identify how they are different.

For section A, the range is 0 since both maximum and minimum are 5 and 5 – 5 = 0
For section B, the range is 10 since 10 – 0 = 10
For section C, the range is 2 since 6 – 4 = 2

Standard Deviation

The standard deviation is a measure of variation based on measuring how far, on average, each data value deviates, or is different, from the mean. A few important characteristics:

1) Standard deviation is always positive. Standard deviation will be zero if all the data valuesare equal, and will get larger as the data spreads out.
2) Standard deviation has the same units as the original data.
3) Standard deviation, like the mean, can be highly influenced by outliers.

Using the data from Section D: 0 5 5 5 5 5 5 5 5 10, we could compute for each data value the difference between the data value and the mean. This will give us an idea of “how far” each value in the data set lies away from the mean.

Data Value Deviation: Data Value – Mean
0 0 – 5 = – 5
5 5 – 5 = 0
5 5 – 5 = 0
5 5 – 5 = 0
5 5 – 5 = 0
5 5 – 5 = 0
5 5 – 5 = 0
5 5 – 5 = 0
5 5 – 5 = 0
10 10 – 5 = 5

We would like to get an idea of the “average” deviation from the mean, but if we find the average of the values in the second column the negative and positive values cancel each other out (this always happens), so instead we square every value in the second column:

Data Value Deviation: Data Value – Mean Deviation Squared
0 0 – 5 = – 5 (-5)2
5 5 – 5 = 0 02
5 5 – 5 = 0 02
5 5 – 5 = 0 02
5 5 – 5 = 0 02
5 5 – 5 = 0 02
5 5 – 5 = 0 02
5 5 – 5 = 0 02
5 5 – 5 = 0 02
10 10 – 5 = 5 52

We then add the squared deviations up to get 25 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 25 = 50. Ordinarily we would then divide by the number of scores, n, (in this case, 10) to find the mean of the
deviations. But we only do this if the data set represents a population; if the data set represents a sample (as it almost always does), we instead divide by n – 1 (in this case, 10 ─ 1 = 9).

So in our example, we would have [latex]\frac{50}{10}[/latex] = 5 if section D represents a population and [latex]\frac{50}{9}[/latex] = about 5.56 if section D represents a sample. These values (5 and 5.56) are called, respectively, the population variance and the sample variance for section D.

Variance can be a useful statistical concept, but note that the units of variance in this instance would be points – squared since we squared all of the deviations. What are points – squared? Good question. We would rather deal with the units we started with (points in this case), so to convert back we take the square root and get:

Population standard deviation: [latex]\sqrt\frac{50}{10}[/latex] ≈ 2.2

Sample standard deviation: [latex]\sqrt\frac{50}{9}[/latex] ≈ 2.4

If we are unsure whether the data set is a sample or a population, we will usually assume it is a sample, and we will round answers to one more decimal place than the original data, as we have done above.

To Compute the Standard Deviation

1) Find the deviation of each data from the mean. In other words, subtract the mean from the data value.
2) Square each deviation.
3) Add the squared deviations.
4) Divide by n, the number of data values, if the data represents a whole population; divide by n – 1 if the data is from a sample.
5) Compute the square root of the result.

Example 9

Computing the standard deviation for Section B above.

Solution: We first calculate that the mean is 5. Using a table can help keep track of your computations for the standard deviation:

Data Value Deviation: Data Value – Mean Deviation Squared
0 0 – 5 = -5 (-5)2 = 25
0 0 – 5 = -5 (-5)2 = 25
0 0 – 5 = -5 (-5)2 = 25
0 0 – 5 = -5 (-5)2 = 25
0 0 – 5 = -5 (-5)2 = 25
10 10 – 5 = 5 (5)2 = 25
10 10 – 5 = 5 (5)2 = 25
10 10 – 5 = 5 (5)2 = 25
10 10 – 5 = 5 (5)2 = 25
10 10 – 5 = 5 (5)2 = 25

Assuming this data represents a population, we will add the squared deviations, divide by 10, the number of data values, and compute the square root:

[latex]\sqrt\frac{25~+~25~+~25~+~25~+~25~+~25~+~25~+~25~+~25~+~25}{10}[/latex] = [latex]\sqrt\frac{250}{10}[/latex] = 5

Notice that the standard deviation of this data set is much larger than that of section D since the data in this set is more spread out.

For comparison, the standard deviations of all four sections are:

Section A: 5 5 5 5 5 5 5 5 5 5 Standard deviation: 0
Section B: 0 0 0 0 0 10 10 10 10 10 Standard deviation: 5
Section C: 4 4 4 5 5 5 5 6 6 6 Standard deviation: 0.8
Section D: 0 5 5 5 5 5 5 5 5 10 Standard deviation: 2.2

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MAT112 Mathematical Concepts & Applications Copyright © by Leah Polanski. All Rights Reserved.

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