In the previous chapter we talked about measures of central tendency. But there is another crucial characteristic of distributions that we need to know in addition to the center: how widely scattered or closely clustered scores are around it. That’s what we mean by variability. We’ll start with the range, then build up the standard deviation from deviations, sums of squares, and variance. Along the way we’ll meet another important concept: degrees of freedom.
To start with, let’s consider a few small sets of scores. Imagine you check the reviews of three movies you’re thinking about watching, scored from 0 to 10.
Fine, Actually: 5, 5, 5, 5, 5, 5
The Good, the Bad and the Okay: 0, 4, 5, 5, 6, 10
The Polarizing Express: 0, 0, 1, 9, 10, 10
Comparing movie ratings
How might you describe the differences among these sets of ratings? Don’t worry about calculating any new statistics yet. Start with the ideas from the previous chapter and think about what they capture or leave out.
There’s no single required description. If you noticed that the sets can resemble one another in some ways and look quite different in others, you’ve found the issue to keep hold of. We’ll put it into more precise terms after the next question.
Suppose you’re about to watch each movie and add your own review. What rating would you predict you’ll give each one? Explain what in the existing ratings would make you more or less confident in those predictions.
A reasonable prediction is 5 for each movie because all three have a mean and median of 5. But I would be most confident about Fine, Actually, where every rating so far is 5, and least confident about The Polarizing Express, where every rating is far from 5. The Good, the Bad and the Okay sits somewhere in between.
Using the tools we explored in the previous chapter, we find that the median and mean are 5 for all three movies. The mode is also 5 for the first two. The Polarizing Express has two modes, 0 and 10, which already gives us a clue that something different is going on. But none of these measures quantifies just how closely the ratings cluster around the center. If we plot the ratings as block histograms, we can see what they’re leaving out.
The ratings for Fine, Actually do not vary at all: every reviewer so far gave it a 5. Most of the ratings for The Good, the Bad and the Okay cluster in the middle, aside from the two extremes. The ratings for The Polarizing Express sit much farther towards the extremes, with none in the middle of the scale at all. So Fine, Actually has no variability, The Polarizing Express has relatively high variability, and The Good, the Bad and the Okay sits somewhere in between. The mean gives us the same prediction for all three, but their variability tells us how far individual ratings tend to depart from it.
How do we quantify that difference? Like measures of central tendency, measures of variability use a single number to summarize a more complex dataset. But this time we’re describing spread rather than center. Greater variability means scores are more widely dispersed, so the center by itself tells us less about what any individual score is likely to look like. And, as with central tendency, there is more than one way to quantify it.
Remember Hipparchus? I mentioned him in Chapter 3 as an early astronomical pioneer, gazing at the heavens, taking measurements, and realizing that his data were not always perfectly consistent. His solution (according to later sources; his own original work is mostly lost) was to take the midrange: the halfway point between the lowest and highest observations (Stahl, 2006).
The utility of this to the early astronomers was to smooth over the discrepant measurements; once the midrange was calculated it could be taken as providing a reliable answer, and the extremes could be largely ignored. But those two values also offer a simple and intuitive way of quantifying variability: the range, calculated by subtracting the minimum score from the maximum.
Here are those movie ratings from before:
Fine, Actually: 5, 5, 5, 5, 5, 5
Range: \(5 - 5 = 0\)The Good, the Bad and the Okay: 0, 4, 5, 5, 6, 10
Range: \(10 - 0 = 10\)The Polarizing Express: 0, 0, 1, 9, 10, 10
Range: \(10 - 0 = 10\)
The range captures one important difference: it is zero for Fine, Actually because its ratings don’t vary. But it gives us the same answer for The Good, the Bad and the Okay and The Polarizing Express. There’s nothing wrong about that. The range is telling us precisely what it is designed to, but it misses the obvious difference between how the ratings are distributed between the extremes.
This is the main limitation of the range as a measure of variability: it does not consider all the observations. Only the two extremes contribute to its value. As a result, it is not affected by how observations are distributed between them. This characteristic also makes it highly susceptible to outliers. One unusual observation can drastically change the range. It also tends to grow with sample size, simply because larger samples have more opportunities to include extreme values.
That’s not to say looking at the range isn’t useful for some purposes. Researchers often begin by checking the minimum, maximum, and range to get a quick sense of the values in their data. Sometimes you just don’t know what to expect, so it’s helpful to see the extremes and how far apart they are. It can also reveal a problem. Suppose we write a new extraversion item that allows responses numbered one to five. A range of three might make us wonder why part of the response scale went unused. Maybe asking people “Are you the world’s most extraverted person?” was a bit too much and restricted the range of responses. A range of ten, on the other hand, tells us to check for misrecorded values.
Useful as the range is, we also want a measure of variability that considers every observation, not just the two extremes. For that, we need a more mathematically sophisticated measure.
In Chapter 3 we introduced deviations: the differences between each score and the mean. The mean is the point at which those deviations balance, but the deviations themselves also give us a way to quantify how far the scores spread around it.
Astronomers realized this too. Deviations offered a way to quantify the imprecision of their instruments or observations. For statisticians applying the same ideas to human measurements, they captured a different kind of variation. In Quetelet’s measurements of soldiers’ chests, a deviation did not necessarily indicate a mistake of the measuring tape; it described how one person’s body differed from another’s. In either case, if we think of the mean as a prediction, a deviation is the prediction error: how far, and in which direction, the actual score fell from it.
The next step was to find a way to summarize those deviations in a single number. Should be easy, no? If we want a “standard deviation,” meaning a typical deviation from the mean, an obvious solution would be to calculate the average of all the deviations. Before we go any further, think through what answer that calculation would give us.
Why can’t we simply average the raw deviations from the mean?
The mean is the balance point of the distribution, so the negative deviations below it and the positive deviations above it balance each other out. The signed deviations therefore always add up to zero. Their average would be zero too, no matter how much the scores varied.
So we need a different strategy. One option would be to take the absolute value of every deviation, dropping the minus sign from the negative ones, and then find the mean absolute deviation. But the strategy we’re going to focus on is squaring each deviation. Squaring solves the problem by turning the negative deviations into positive values, so they no longer cancel out the deviations above the mean when we sum them.1
Having squared the deviations, we can continue with one connected chain of calculations. Add the squared deviations to get \(SS\), the sum of squared deviations (or “sum of squares” for short). Divide \(SS\) by the number of observations to get the population variance, the mean squared deviation. Then take its square root to get the standard deviation. The square root returns us to the original units, so we can interpret the standard deviation as a typical deviation from the mean.2
To complicate life slightly, populations and samples use slightly different calculations. We’ll start with a population and return to samples shortly.
The table puts the steps into mathematical notation. Population variance is represented by lowercase sigma squared (\(\sigma^2\)), and population standard deviation by lowercase sigma (\(\sigma\)).
| Step | Calculation |
|---|---|
| 1. Find the deviation for each score | \(X - \mu\) |
| 2. Square the deviations | \((X - \mu)^2\) |
| 3. Sum the squared deviations | \(SS = \Sigma(X - \mu)^2\) |
| 4. Calculate variance (mean squared deviation) | \(\sigma^2 = \frac{SS}{N}\) |
| 5. Calculate standard deviation (root-mean-square deviation) | \(\sigma = \sqrt{\sigma^2}\) |
Equations in hand, let’s return to the two movies whose ratings had the same range. For now, we’ll treat each complete set of ratings as its own population. We’ll work through The Good, the Bad and the Okay together, following that sequence of five steps; then you can try the same procedure for The Polarizing Express.
When I’m doing these calculations by hand, I find it easiest to put the scores in the first column of a table, labeled \(X\). We’ll add to this table as we go along. So here are the original ratings for The Good, the Bad and the Okay.
First find the population mean. The ratings add to 30, and there are six of them, so \(\mu = 30/6 = 5\). This will be the reference point for every deviation in the next column.
Now add a column for the deviations, \(X-\mu\). Subtracting the mean from each score, the deviations are \(-5, -1, 0, 0, 1,\) and \(5\). The signs tell us which side of the mean the scores fall on, while the sizes tell us how far away they are. If you simply add up these six values, you get an answer of zero.
Next add a column for the squared deviations. Taking the values in the deviation column and multiplying each value by itself gives us \(25, 1, 0, 0, 1,\) and \(25\). Squaring makes the negative values turn positive, so when we add the squared deviation up in the next step we won’t get zero.
Now we add up all the squared deviations to get the sum of squares: \(SS = 25+1+0+0+1+25 = 52\).
Next divide \(SS\) by the population size, \(N\), to get the variance: \(\sigma^2 = 52/6 = 8.67\). Because we’ve squared the deviations, variance is measured in squared rating points. That makes it useful mathematically, but not especially intuitive on the scale of the original scores.
Finally, take the square root of the variance: \(\sigma = \sqrt{8.67} = 2.94\). This returns us to the original units, so we can say that a typical deviation from the mean rating for The Good, the Bad and the Okay is around three points.
Now calculate variability for The Polarizing Express
Follow the whole procedure for the ratings of The Polarizing Express: 0, 0, 1, 9, 10, and 10. Treating these six ratings as a population, what is the standard deviation?
Start by finding the mean. Then make columns for \(X - \mu\) and \((X - \mu)^2\).
Add the squared deviations, divide by \(N = 6\), and take the square root.
4.69
The mean rating is \(\mu = 5\). Here is the whole calculation:
Taking the square root of 22 gives a standard deviation of 4.69. This is larger than the 2.94 for The Good, the Bad and the Okay, capturing the greater spread that the range missed.
Allow me for a moment to impress upon you the importance of squares. \(SS\) is not just an intermediate answer that we can forget about as soon as we’ve found the standard deviation. Later in this book we’ll find ourselves splitting variability into different sources, each with its own sum of squares. By the time we finish ANOVA and regression near the end of the semester, I like to joke with my students that the whole course has been nothing but squares all the way down.
The tables above followed the definitional formula for \(SS\): find each deviation, square it, and add the squared deviations.
\[ SS = \Sigma(X - \mu)^2 \]
There is an algebraically equivalent shortcut called the computational formula:
\[ SS = \Sigma X^2 - \frac{(\Sigma X)^2}{N} \]
Order of operations is important here. For the first term, \(\Sigma X^2\), square every score and sum the results. Then subtract the correction term: the sum of the scores squared, divided by the number of scores. For The Good, the Bad and the Okay, \(\Sigma X=30\) and \(\Sigma X^2=202\), so:
\[ SS = 202 - \frac{30^2}{6} = 202 - 150 = 52 \]
That is the same answer we found in Activity 4.1 following the definitional approach. The shortcut likewise gives \(SS=132\) for The Polarizing Express, so the rest of both calculations remains unchanged.
We’ll stick with the definitional approach for now, but tuck this computational arrangement in the back of your mind; it will become our preferred method when we get to the ANOVA section of the book.
So far we’ve treated the movie ratings as populations. I mentioned that we have to do things a little differently for samples. Don’t worry, the difference in the math is tiny, but the reason for it is important.
Like I mentioned in Chapter 1, there’s overlap between descriptive and inferential statistics. Here, we want to know our sample’s variance, but not because we especially care about the sample itself. Rather, we want to use it as an estimate of the population variance. The trouble is that if we used precisely the calculation above, dividing \(SS\) by \(n\), the estimate would be biased.
Bias does not mean that any one sample estimate is wrong: a sample’s variance can over- or underestimate its population variance. Bias is a long-run tendency. An unbiased estimator is sometimes too high, sometimes too low, but those errors balance out in the long run. A biased estimator systematically leans in one direction. If we repeatedly took samples and calculated variance by dividing \(SS\) by \(n\), the average of all the estimates would be less than the actual population variance; it would underestimate the true value.
The bias comes from using the sample mean, \(M\), in place of the unknown \(\mu\). Each sample’s \(M\) is fitted to the scores in that particular sample, designed to minimize their sum of squared deviations. As a result, the sum of squared deviations around \(M\) can never be larger than the sum of squared deviations of those same scores around \(\mu\), and will be smaller unless \(M = \mu\).
Here we have a happy little population, a symmetrical mound of colorful dots. It has a fixed mean, \(\mu\), and standard deviation, \(\sigma\). The length of the black line underneath shows \(\mu \pm \sigma\). We’ll use that fixed span as our reference length, so in this picture \(\sigma =\) ....
Now we draw a sample of ... dots from the population. We want to compute the variability of the sample, but how?
Because this is a simulation, we know \(\mu\). So we can calculate the sample’s squared deviations from \(\mu\), divide by \(n\) to get variance, and take the square root to get SD. The dashed blue line shows the answer relative to the population SD: .... This is the SD we would get using the fixed population mean.
But usually we can’t do that: we don’t know \(\mu\)! So instead we find the sample mean, \(M\), and calculate the squared deviations around it. The solid red line shows that answer, ..., centered on \(M\) because that’s how we calculated it.
Now that we’ve calculated it, let’s slide the red span over so all three SD lines are centered on \(\mu\) and we can compare them. The red span fits inside the dashed blue one; computing the sample’s SD around \(M\) rather than \(\mu\) gave a smaller answer.
A new sample gives us a different \(M\), and different standard deviations computed around \(\mu\) or \(M\) (the actual population \(\sigma\) is fixed and doesn’t change). Here the red sample SD is larger than the actual population \(\sigma\). An individual sample can still overestimate variability. But the red \(M\)-based span is still no wider than the dashed \(\mu\)-based span.
For every new sample, the red line will be narrower than the blue one (or they might be the same if the sample mean \(M\) happens to be equal to \(\mu\)).
Keep clicking to take as many new samples as you like. Both computed standard deviations will change, and either can land above or below the population SD. But the \(M\)-based answer can never be larger than the \(\mu\)-based answer for the same observations, because it is always fitted to the new sample mean—a moving target. That built-in shrinkage is what creates the long-run bias.
So now we know what the problem is, here’s the solution: divide \(SS\) by \(n - 1\) rather than \(n\). If only all of life’s problems were so easily solved.
The adjusted denominator makes sample variance, \(s^2\), an unbiased estimator of population variance. Taking its square root gives the usual sample standard deviation, \(s\), often labeled \(SD\) when results are reported in psychology.3
Here is the table of steps and equations again, with a new column applying them to samples. There are differences in notation—the Greek symbols \(\mu\) and \(\sigma\) and uppercase \(N\) for populations; \(M\), \(s\), and \(n\) for samples. But aside from the different denominator in step 4, the process is the same.
| Step | Population | Sample |
|---|---|---|
| 1. Find the deviation for each score | \(X - \mu\) | \(X - M\) |
| 2. Square the deviations | \((X - \mu)^2\) | \((X - M)^2\) |
| 3. Sum the squared deviations | \(SS = \Sigma(X - \mu)^2\) | \(SS = \Sigma(X - M)^2\) |
| 4. Calculate variance | \(\sigma^2 = \frac{SS}{N}\) | \(s^2 = \frac{SS}{df} = \frac{SS}{n - 1}\) |
| 5. Calculate standard deviation | \(\sigma = \sqrt{\sigma^2}\) | \(s = \sqrt{s^2}\) |
From population to sample
Now suppose the six ratings for The Good, the Bad and the Okay are a sample from the larger population of ratings that viewers might give it. Earlier, you found that \(SS = 52\). What is the sample standard deviation?
You do not need to calculate the deviations or \(SS\) again. Pick up at the step where the sample calculation differs from the population calculation.
For \(n = 6\), the degrees of freedom are \(n - 1 = 5\). First calculate \(s^2 = 52 / 5\), then take the square root.
3.22
\(s^2 = SS / (n - 1) = 52 / 5 = 10.4\). Its square root, the sample standard deviation, is about 3.22.
Well, it’s not quite as simple as I made it sound there. To fully understand the issue and why this solution works, we need to know about \(df\), which I snuck into Table 4.2. That refers to degrees of freedom. For ordinary sample variance, \(df = n - 1\). More generally, a variance is a sum of squares divided by the degrees of freedom associated with it, and \(df\) is not always \(n - 1\) as it is in this case. Later, in ANOVA, we’ll split variability into several sums of squares, each with its own \(df\).
So why does the correction here use exactly \(n - 1\)? It is a result of the constraint imposed by calculating deviations from the sample mean. Doing so costs us one degree of freedom.
To illustrate, suppose I put a sample of three playing cards face down in front of you and tell you that I already calculated the mean of their values, \(M=5\). I point to the first card and ask you what number you think it is. There is no way for you to know, of course, so you shrug and I show you that it is a three. I point to the second card. You still can’t know what it is; I turn it over and reveal a five. Now I point to the last card. Can you tell me its number? You jump up and exclaim, “eureka!” Given the mean and the other two values, there’s only one thing this last one can be: it has to be seven.
We could shuffle the same three cards and repeat the process; the order the cards are turned over in doesn’t matter. And we could have more than three cards. Whatever the last card is, we’ll always be able to determine its value without turning it over. Knowing the mean imposes a constraint. Only two of the three values are free to vary. The same applies to deviations around the sample mean: once \(M\) has been calculated, the deviations have to sum to zero. Any \(n - 1\) deviations can vary, but their values determine the last one. So sample variance has \(df = n - 1\) degrees of freedom. (Later we’ll meet calculations that fit more means and therefore have different degrees of freedom.)
Note that this problem is specific to the sample mean, \(M\). If the population mean, \(\mu\), were known and we measured the sample’s deviations around that fixed population center, we would not lose a degree of freedom. That known \(\mu\) is a constant, not something we estimate from the sample data, and so it does not impose the constraint of deviations summing to zero.
One last thing. Now that you know about statistical bias, you might be worried about other statistics being biased. What about the mean? We spent most of Chapter 3 talking about it, but never mentioned bias. It shouldn’t get off the hook, right? We ought to check if it is biased or unbiased. And turning back to variance, Activity 4.2 illustrated the cause of its bias, but it only showed one sample at a time. Like we said, bias operates in the aggregate.
It would help to be more precise about the term error. An estimate can differ from the true parameter for different reasons. Sometimes it is thrown off by a genuine mistake or a bad design choice. But even under ideal circumstances, a statistic calculated from one sample will usually not be exactly equal to the corresponding population parameter. This kind of naturally-occuring discrepancy between the estimate and the true parameter, is called sampling error. It’s not an error in the colloquial sense of something we did wrong and could have avoided. It is inherent to the process of sampling from a population.
Bias, by contrast, is a long-run property. Across repeated samples, do a statistic’s errors balance around zero, or do they systematically lean in one direction? Being unbiased doesn’t mean every estimate gets the answer exactly right; sampling error means each individual estimate can still be above or below the true parameter. But it makes the overestimates and underestimates balance out in the long run. To see the difference between sampling error and bias, we need to look at many estimates. Let’s check for bias in the mean, as well as in the two variance calculations.
Here’s our happy little population again. We’ll use it to track long-run bias in a couple of statistics. First we’ll check whether the sample mean, \(M\), is biased or unbiased. Then we’ll compare the two variance calculations we learned, dividing \(SS\) either by \(n\) or by \(n - 1\) to compute sample variance.
Here’s our first sample. Its mean is a little below the true population mean, because of sampling error. But again, bias is not about any one sample over- or underestimating its parameter. We need to track the errors across a bunch of samples.
The population will stay the same while we draw sample after sample, so let’s shift it out of the way and collect the individual errors on the timeline below. Each square’s vertical position is that sample’s estimation error, \(M-\mu\): below zero is an underestimate and above zero is an overestimate. The first sample’s mean is a little below the population mean, so it goes just below zero.
Now we draw nine more samples, one at a time. The ten squares are the ten sample means we’ve seen so far. We’ve got some below zero, some above it, and some very close to zero, where the sample mean was very close to \(\mu\).
This second tracker summarizes the first. After each sample, it averages all the errors collected so far. So the end of the line is the average of all ten means’ errors so far. Bias is the long-run position of this average. But ten is not much of a long run. Let’s make room for a lot more.
Now we have space for 1,000 samples. (Including the ten we already took, squished over on the left.) When you click next, watch as a whole lot more samples are taken, the errors in their means added to the first tracker, and the cumulative average error updates in the second tracker.
Once all 1,000 sample means have been collected, the average error is very close to zero. The individual errors on the top tracker scatter evenly above and below the 0 line. The cumulative average error line—the running average of all those means—has meandered up and down, but it has no long-run tendency to stay above or below zero. The sample mean is an unbiased estimator of the population mean.
Now let’s estimate population variance by dividing \(SS\) by \(n\), without the correction. The black span is the true population variance; the red span is this sample’s estimate. I’ve lined them up at the same center so we can compare their lengths directly (though the sample variance uses deviations from the sample mean). One estimate can miss high or low, so it still cannot establish bias. This one, even though it uses the wrong denominator, happens to be almost exactly right, so sits at 0 on the error tracker.
Now let’s repeat that calculation across 1,000 samples. The individual red estimates scatter above and below zero, but their cumulative average settles below it. The errors do not balance out: dividing by \(n\) systematically underestimates population variance.
We can go back through those same 1,000 samples and recompute the variance using the corrected \(n - 1\) denominator. We’ll start again with the first sample. Above the dots, the green corrected span replaces the red uncorrected one. In the trackers, we’ll keep all the red results where they are and begin adding the green results. The samples are identical. With the new, correct denominator, this variance happens to overestimate the true value. But again, one sample can’t establish bias.
Now we replay the same 1,000 samples. The green corrected squares are added on top of the existing red estimates, and the green cumulative line gets drawn as we average those new values together. Dividing by \(n - 1\) inflates each estimate just enough, on average, for the green line to settle around zero while the red line remained below it. The individual estimates still scatter around the true value due to sampling error, but the correction eliminates the long-run bias.
So phew, by taking a whole bunch of samples and aggregating their errors, we can see that the sample mean is an unbiased estimator of the population mean.4 It doesn’t always get the answer exactly right. Individual samples can overestimate or underestimate the true population mean, because of ever-present sampling error. But in the long run those random errors balance out and the mean is unbiased.
Sample variance can be an unbiased estimator of population variance too, as long as we divide \(SS\) by \(df\) (\(n - 1\) in this situation). Dividing by \(n\) does not stop the estimates from varying randomly; it shifts their long-run center downward. Individual estimates can still be too high or too low, but their errors average below zero, as we saw in the cumulative average of the 1,000 variance estimates calculated with the uncorrected denominator. The green line, where we divided by \(n - 1\), fluctuated along the way but made its way close to zero in the long run.
Learning Checks
Two datasets both have a range of 10. Which conclusion can you draw from that information alone?
Only the maximum-minus-minimum difference is the same.
The range uses only the distance between the minimum and maximum. It does not tell us where those extremes fall or how any of the scores between them are distributed.
Is it possible for a sample to have a standard deviation of zero?
True
A standard deviation of zero occurs when every score has the same value.
A set of response times is measured in seconds. What units would its variance and standard deviation use?
Variance in seconds squared; standard deviation in seconds
Variance is the mean squared deviation, so its units are squared too. Taking the square root gives standard deviation in the original units.
A sample of four scores has \(SS = 24\). What is the variance?
Sample variance is \(SS\) divided by its degrees of freedom.
For a sample, the number of degrees of freedom is \(n - 1\).
8
For a sample, \(s^2 = SS / (n - 1) = 24 / 3 = 8\). Dividing by \(n\) instead of \(n - 1\) would give the biased answer of 6.
A random opinion poll estimates 48% support for a proposal, while the true population value is 50%. The same polling procedure produces estimates centered on 50% across many random samples. How should we describe the two-percentage-point discrepancy in the first poll?
It is sampling error, but the polling estimator is unbiased.
One random sample can miss its population parameter even when nothing has gone wrong. That discrepancy is sampling error. Bias is a long-run tendency, and the estimates balancing around 50% tell us that this estimator is unbiased.
Why do we divide \(SS\) by \(n - 1\) rather than \(n\) when estimating population variance from a sample?
The sample mean, \(M\), is fitted from the same scores and minimizes their sum of squared deviations. Dividing that reduced \(SS\) by \(n\) therefore underestimates population variance on average. Dividing by \(n - 1\) corrects the bias.
Missing rare or extreme scores cannot by itself explain systematic bias: some samples overestimate variability, and others underestimate it. The downward bias comes from centering every sample on its own fitted mean. The \(n - 1\) denominator is the degrees-of-freedom correction for sample variance.
The mean absolute deviation from the mean is a perfectly legitimate measure of variability, and it is less affected by extreme scores because it does not square their distances. But squaring has some other mathematical properties that will become useful when we get around to partitioning variance later in the book. As usual, statistics don’t invent and apply themselves: we still have to think about which measure makes the most sense for what we’re trying to do.↩︎
More precisely, the population standard deviation is the root-mean-square deviation: we square the deviations, find their mean, and take the square root. “Typical deviation” is a useful interpretation, but it isn’t the literal arithmetic mean of the raw deviations or their absolute values. I don’t hear social scientists use the term “root-mean-square deviation” very often, so I’m just going to follow the convention of calling it the standard deviation.↩︎
Strictly speaking, taking the square root introduces a small downward bias, especially with very small samples. But \(s\) is much less biased than an uncorrected standard deviation based on \(SS/n\).↩︎
More precisely, the sample mean is an unbiased estimator of the population mean under the random-sampling setup used here. Bias can be introduced in other ways, but that’s an issue for another time.↩︎