  Office Microsoft Excel 2010 : Calculating the Standard Deviation and Variance (part 2) - Population Parameters and Sample Statistics

1/25/2015 7:29:24 PM

Squaring the Deviations

Why square each deviation and then take the square root of their total? One primary reason is that if you simply take the average deviation, the result is always zero. Suppose you have three values: 8, 5, and 2. Their average value is 5. The deviations are 3, 0, and −3. The deviations total to zero, and therefore the mean of the deviations must equal zero. The same is true of any set of real numbers you might choose.

Because the total deviation is always zero, regardless of the values involved, it’s useless as an indicator of the amount of variability in a set of values. Therefore, each deviation is squared before totaling them. Because the square of any number is positive, you avoid the problem of always getting zero for the total of the deviations.

It is possible, of course, to use the absolute value of the deviations: that is, treat each deviation as a positive number. Then the sum of the deviations must be a positive number, just as is the sum of the squared deviations. And in fact there are some who argue that this figure, called the mean deviation, is a better way to calculate the variability in a set of values than the standard deviation.

Population Parameters and Sample Statistics

You normally use the word parameter for a number that describes a population and statistic for a number that describes a sample. So the mean of a population is a parameter, and the mean of a sample is a statistic.

It’s traditional to use Greek letters for parameters that describe a population and to use Roman letters for statistics that describe a sample. So, you use the letter s to refer to the standard deviation of a sample and σ to refer to the standard deviation of a population.

With those conventions in mind—that is, Greek letters to represent population parameters and Roman letters to represent sample statistics—the equation that defines the variance for a sample that was given above should read differently for the variance of a population. The variance as a parameter is defined in this way: The equation shown here is functionally identical to the equation for the sample variance given earlier. This equation uses the Greek σ, pronounced sigma. The lowercase σ is the symbol used in statistics to represent the standard deviation of a population, and σ2 to represent the population variance.

The equation also uses the symbol μ. The Greek letter, pronounced mew, represents the population mean, whereas the symbol , pronounced X bar, represents the sample mean. (It’s usually, but not always, related Greek and Roman letters that represent the population parameter and the associated sample statistic.)

The symbol for the number of values, N, is not replaced. It is considered neither a statistic nor a parameter.

Another issue is involved with the formula that calculates the variance (and therefore the standard deviation). It stays involved when you want to estimate the variance of a population by means of the variance of a sample from that population. I

Suppose now that you have a sample of 100 piston rings taken from a population of, say, 10,000 rings that your company has manufactured. You have a measure of the diameter of each ring in your sample, and you calculate the variance of the rings using the definitional formula: You’ll get an accurate value for the variance in the sample, but that value is likely to underestimate the variance in the population of 10,000 rings. In turn, if you take the square root of the variance to obtain the standard deviation as an estimate of the population’s standard deviation, the underestimate comes along for the ride.

Samples involve error: in practice, their statistics are virtually never precisely equal to the parameters they’re meant to estimate. If you calculate the mean age of ten people in a statistics class that has 30 students, it is almost certain that the mean age of the ten student sample will be different from the mean age of the 30 student class.

Similarly, it is very likely that the mean piston ring diameter in your sample is different, even if only slightly, from the mean diameter of your population of 10,000 piston rings. Your sample mean is calculated on the basis of the 100 rings in your sample. Therefore, the result of the calculation which uses the sample mean , is different from, and smaller than, the result of this calculation: which uses the population mean μ.

Bear in mind that when you calculate deviations using the mean of the sample’s observations, you minimize the sum of the squared deviations from the sample mean. If you use any other number, such as the population mean, the result will be different from, and larger than, when you use the sample mean.

Therefore, any time you estimate the variance (or the standard deviation) of a population using the variance (or standard deviation) of a sample, your statistic is virtually certain to underestimate the size of the population parameter.

There would be no problem if your sample mean happened to be the same as the population mean, but in any meaningful situation that’s wildly unlikely to happen.

Is there some correction factor that can be used to compensate for the underestimate? Yes, there is. You would use this formula to accurately calculate the variance in a sample: But if you want to estimate the value of the variance of the population from which you took your sample, you divide by N − 1: The quantity (N − 1) in this formula is called the degrees of freedom.

Similarly, this formula is the definitional formula to estimate a population’s standard deviation on the basis of the observations in a sample (it’s just the square root of the sample estimate of the population variance): If you look into the documentation for Excel’s variance functions, you’ll see that VAR() or, in Excel 2010, VAR.S() is recommended if you want to estimate a population variance from a sample. Those functions use the degrees of freedom in their denominators.

The functions VARP() and, in Excel 2010, VAR.P() are recommended if you are calculating the variance of a population by supplying the entire population’s values as the argument to the function. Equivalently, if you do have a sample from a population but do not intend to infer the population variance—that is, you just want to know the sample’s variance—you would use VARP() or VAR.P(). These functions use N, not the N − 1 degrees of freedom, in their denominators.

The same is true of STDEVP() and STDEV.P(). Use them to get the standard deviation of a population or of a sample when you don’t intend to infer the population’s standard deviation. Use STDEV() or STDEV.S() to infer a population standard deviation. Others  Top 10    Popular tags Video Tutorail Microsoft Access Microsoft Excel Microsoft OneNote Microsoft PowerPoint Microsoft Project Microsoft Visio Microsoft Word Active Directory Biztalk Exchange Server Microsoft LynC Server Microsoft Dynamic Sharepoint Sql Server Windows Server 2008 Windows Server 2012 Windows 7 Windows 8 Adobe Indesign Adobe Flash Professional Dreamweaver Adobe Illustrator Adobe After Effects Adobe Photoshop Adobe Fireworks Adobe Flash Catalyst Corel Painter X CorelDRAW X5 CorelDraw 10 QuarkXPress 8 windows Phone 7 windows Phone 8 BlackBerry Android Ipad Iphone iOS
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