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Variance and Standard Deviation: Measuring Data Dispersion, Study notes of Statistics

The concept of variance and standard deviation as measures of data dispersion. It provides examples of how to calculate variance and standard deviation from a set of data, and discusses their significance for teachers and researchers. The document also includes practice problems for the reader.

Typology: Study notes

2021/2022

Uploaded on 09/12/2022

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VARIANCE AND STANDARD DEVIATION
Recall that the range is the difference between the upper and lower limits of the data. While this is important, it does
have one major disadvantage. It does not describe the variation among the variables. For instance, both of these sets of
data have the same range, yet their values are definitely different.
90, 90, 90, 98, 90 Range = 8
1, 6, 8, 1, 9, 5 Range = 8
To better describe the variation, we will introduce two other measures of variationโ€”variance and standard deviation
(the variance is the square of the standard deviation). These measures tell us how much the actual values differ from the
mean. The larger the standard deviation, the more spread out the values. The smaller the standard deviation, the less
spread out the values. This measure is particularly helpful to teachers as they try to find whether their studentsโ€™ scores
on a certain test are closely related to the class average.
To find the standard deviation of a set of values:
a. Find the mean of the data
b. Find the difference (deviation) between each of the scores and the mean
c. Square each deviation
d. Sum the squares
e. Dividing by one less than the number of values, find the โ€œmeanโ€ of this sum (the variance*)
f. Find the square root of the variance (the standard deviation)
*Note: In some books, the variance is found by dividing by n. In statistics it is more useful to divide by n -1.
EXAMPLE
Find the variance and standard deviation of the following scores on an exam:
92, 95, 85, 80, 75, 50
SOLUTION
First we find the mean of the data:
Mean = 92+95+85+80+75+50
6 = 477
6 = 79.5
Then we find the difference between each score and the mean (deviation).
Score
Score - Mean
Difference from mean
92
92 โ€“ 79.5
+12.5
95
95 โ€“ 79.5
+15.5
85
85 โ€“ 79.5
+5.5
80
80 โ€“ 79.5
+0.5
75
75 โ€“ 79.5
-4.5
50
50 โ€“ 79.5
-29.5
Next we square each of these differences and then sum them.
Difference
Difference Squared
+12.5
156.25
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VARIANCE AND STANDARD DEVIATION

Recall that the range is the difference between the upper and lower limits of the data. While this is important, it does have one major disadvantage. It does not describe the variation among the variables. For instance, both of these sets of data have the same range, yet their values are definitely different. 90, 90, 90, 98, 90 Range = 8 1, 6, 8, 1, 9, 5 Range = 8 To better describe the variation, we will introduce two other measures of variationโ€” variance and standard deviation (the variance is the square of the standard deviation). These measures tell us how much the actual values differ from the mean. The larger the standard deviation, the more spread out the values. The smaller the standard deviation, the less spread out the values. This measure is particularly helpful to teachers as they try to find whether their studentsโ€™ scores on a certain test are closely related to the class average. To find the standard deviation of a set of values: a. Find the mean of the data b. Find the difference (deviation) between each of the scores and the mean c. Square each deviation d. Sum the squares e. Dividing by one less than the number of values, find the โ€œmeanโ€ of this sum (the variance *) f. Find the square root of the variance (the standard deviation ) *Note: In some books, the variance is found by dividing by n. In statistics it is more useful to divide by n - 1. EXAMPLE Find the variance and standard deviation of the following scores on an exam: 92, 95, 85, 80, 75, 50 SOLUTION First we find the mean of the data: Mean = 92 + 95 + 85 + 80 + 75 + 50 6

477 6

Then we find the difference between each score and the mean (deviation). Score Score - Mean Difference from mean 92 92 โ€“ 79.5 +12. 95 95 โ€“ 79.5 +15. 85 85 โ€“ 79.5 +5. 80 80 โ€“ 79.5 +0. 75 75 โ€“ 79.5 - 4. 50 50 โ€“ 79.5 - 29. Next we square each of these differences and then sum them. Difference Difference Squared +12.5 156.

Sum of the squares โ†’ 1317. The sum of the squares is 1317.50. Next, we find the โ€œmeanโ€ of this sum (the variance).

  1. 50 5

Finally, we find the square root of this variance. โˆš 263. 5 โ‰ˆ 16. So, the standard deviation of the scores is 16.2; the variance is 263.5. EXAMPLE Find the standard deviation of the average temperatures recorded over a five-day period last winter: 18, 22, 19, 25, 12 SOLUTION This time we will use a table for our calculations. Temp Temp โ€“ mean = deviation Deviation squared 18 18 โ€“ 19.2 = - 1.2 1. 22 22 โ€“ 19.2 = 2.8 7. 19 19 โ€“ 19.2 = - 0.2 0. 25 mean 25 โ€“ 19.2 = 5.8 33. 12 โ†“ 12 - 19.2 = - 7.2 51. 96 รท 5 = 19.2 94.80 โ† sum of squares To find the variance, we divide 5 โ€“ 1 = 4.

  1. 8 4 = 23. Finally, we find the square root of this variance. (^) โˆš 23. 7 โ‰ˆ 4. So the standard deviation for the temperatures recorded is 4.9; the variance is 23.7. Note that the values in the second example were much closer to the mean than those in the first example. This resulted in a smaller standard deviation. We can write the formula for the standard deviation as s = โˆš โ…€(๐‘ฅ๐‘– โˆ’ ๐‘ฅฬ… )^2 ๐‘›โˆ’ 1 where โ…€ means โ€œthe sum ofโ€ ๐‘ฅ๐‘– represents each value x in the date ๐‘ฅฬ… is the mean of the ๐‘ฅ๐‘– values n is the total of ๐‘ฅ๐‘– values