Algebra 2 · S.ID.A.4

Checking whether data look normal

A normal model makes everything easy to calculate, but only if the data really are close to normal. How can you tell, from the data alone, whether they are?

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The normal distributionAlgebra 2 PercentilesAlgebra 1

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Sampling distribution of a sample meanStatistics

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The normal distribution

Checking whether data look normal

Density curves: area, mean and median Sampling distributions7 z-scores and the empirical rule ▾ Show all 8 (3 more)

When you reach for it

The The normal distribution turns two numbers, a mean and a standard deviation, into a whole curve, and from the curve you can read the share of values in any range. That is a great deal of power from very little, and it is only honest when the data are close to normal. Many of the methods that come later, such as the conditions for inference about a mean, ask you to check exactly this before you lean on a normal model.

The lesson on the normal distribution already rules some variables out by thinking about them: a mean far from the median, or a hard floor at 0 only a standard deviation or so below the mean, means the model cannot fit. This lesson goes further and puts the data themselves next to the model. You reach for it whenever you hold a list of values and want to know whether a normal curve describes them, and you want the answer from the data, not from a hope.

There are three checks, from rough to sharp:

the shape of a histogram (or dot plot),
the share of values within one and two standard deviations of the mean, compared with the 68\% and 95\% a normal curve gives, and
a normal probability plot, which puts every value next to the value a normal model would place at the same percentile.

The procedure

Looking at the shape

A normal curve is one symmetric hump that falls away smoothly on both sides. So the first thing to look for in a histogram is one peak, roughly the same tail on each side, and no value standing apart from the rest. A histogram that is clearly skewed, has two peaks, or has an isolated bar far from the others already says no.

The weakness of this check is the histogram itself. With a small data set the bars are short and ragged, and moving the bin edges can change the picture (as Making a histogram and choosing the bin width shows), so a histogram can rule normality out but rarely gives a confident yes.

Counting within one and two standard deviations

On any normal curve about 68\% of the area lies within one standard deviation of the mean and about 95\% within two. Where those figures come from is the empirical rule in z-scores and the empirical rule; here they are a yardstick. Find the mean \bar{x} and the standard deviation s of the data (as usual, dividing by n - 1), count the values between \bar{x} - s and \bar{x} + s and between \bar{x} - 2s and \bar{x} + 2s, and compare the shares with 68\% and 95\%.

This check is quick and uses only summaries, but it sees only two numbers. Very different shapes can give similar counts, and with ten values a share can only be 60\%, 70\%, 80\% and so on, so one value either side of a boundary moves it by ten points. Only a large gap means much.

Building a normal probability plot

The sharpest check compares every value at once. The idea comes straight from percentiles: if the data were normal, the value at each percentile would be where a normal curve puts that percentile.

Sort the data and give each value a percentile. Write the n values in order. The i-th smallest value is given the percentile

p_i = \frac{i - 0.5}{n}.

The reason for the half: split the share 0 to 1 into n equal slices, one per value, and put each value in the middle of its own slice. With n = 10 the slices are 0 to 0.1, 0.1 to 0.2 and so on, so the values sit at 0.05, 0.15, \dots, 0.95. Using \displaystyle \frac{i}{n} instead would put the largest value at the 100th percentile, which no normal curve has: its tails never end.

Turn each percentile into a normal score. The normal score z_i of the i-th value is the number on the standard normal curve with share p_i of the area to its left. Finding such a number is a normal calculation (a table, a calculator’s inverse normal key, or software). For the two sizes used in this lesson, here they are, to two decimal places:

i (of 10)	p_i	z_i
1	0.05	-1.64
2	0.15	-1.04
3	0.25	-0.67
4	0.35	-0.39
5	0.45	-0.13
6	0.55	0.13
7	0.65	0.39
8	0.75	0.67
9	0.85	1.04
10	0.95	1.64

i (of 5)	p_i	z_i
1	0.1	-1.28
2	0.3	-0.52
3	0.5	0
4	0.7	0.52
5	0.9	1.28

The scores are symmetric about 0 because the normal curve is symmetric about its mean.

Plot each value against its normal score. Put the normal score z_i across and the data value up, one point per value. The result is a normal probability plot (also called a normal quantile plot).

Look for a straight line. Here is why straightness is the test. If the data came from N(\mu, \sigma), the value at the percentile p would be about \mu + \sigma z, where z is the standard normal value at the same percentile, because every normal curve is the standard one slid by \mu and stretched by \sigma. So the points would lie close to the line

\text{value} = \mu + \sigma z,

a straight line whose height at z = 0 is the mean and whose slope is the standard deviation. That holds whatever \mu and \sigma are, so you do not need to know them to use the plot: any normal data give a straight line, and data far from normal give a line that bends. A useful reference line to draw is \bar{x} + s z, the normal model with the data’s own mean and standard deviation.

Reading the bends

Each way of departing from normal bends the plot in its own way, because the plot shows how fast the values climb as the percentile rises.

Right skew (a long tail of large values): the small values are bunched, so the points start nearly flat; the large values are spread out, so the points climb steeply at the right. The plot curves upward, like the inside of a bowl.
Left skew (a long tail of small values): the reverse. The points climb steeply on the left and level off on the right, curving downward like the top of a hill.
An outlier: one point well above the line at the right end, or well below it at the left end, with the rest in line.
Tails shorter than normal (a flat, uniform shape, say): the ends are flatter than the middle, an S lying on its side. Tails longer than normal: the ends are steeper than the middle.

Real data always wobble a little about the line. The question is whether the points follow a curve or show a gap, not whether they hit the line exactly.

Worked examples

Example 1

The shape and the share for a symmetric set

Seraphine makes soap by hand and weighs ten bars from one batch, in grams:

\begin{gathered} 134,\ 139,\ 143,\ 145,\ 149, \\ 151,\ 154,\ 158,\ 161,\ 166. \end{gathered}

Draw a histogram with bins of width 10 starting at 130, find the mean and standard deviation, and compare the shares within one and two standard deviations of the mean with 68\% and 95\%.

Solution

Using the bins 130 to 140, 140 to 150, 150 to 160 and 160 to 170, with a value on an edge going to the bar on its right, the counts are 2, 3, 3, 2: one hump, symmetric, no gaps. It is the first panel of the figure.

Figure 1The three data sets of the worked examples. The soap bars make one symmetric hump, the drives pile up at the left with a tail to the right, and the eggs make one hump with a single value far out on its own.

The values add to 1500, so \bar{x} = 150. The deviations from 150 are -16, -11, -7, -5, -1, 1, 4, 8, 11, 16, whose squares add to 910. Dividing by n - 1 = 9:

s = \sqrt{\frac{910}{9}} \approx 10.06 \text{ g}.

One standard deviation either side of the mean runs from 150 - 10.06 = 139.94 to 160.06. The values inside are 143, 145, 149, 151, 154, 158: six of ten, or 60\%. (139 and 161 are just outside.) Two standard deviations either side runs from 129.88 to 170.12, which holds all ten: 100\%.

Against 68\% and 95\%, these are about as close as ten values allow: 68\% of ten is 6.8 and 95\% is 9.5, a count must be a whole number, and each count here is within one value of those targets. The shape and the shares both fit a normal model, though with only ten values neither check can say much more.

Example 2

The shape and the share for a skewed set

A food bank records how far ten of its volunteers drive to reach it, in kilometers:

2,\ 2,\ 3,\ 3,\ 4,\ 5,\ 6,\ 8,\ 11,\ 16.

Describe the histogram (bins of width 4 from 0), find the mean, median and standard deviation, compare the shares with 68\% and 95\%, and find what share a normal model with this mean and standard deviation would put below 0 kilometers.

Solution

The bins 0 to 4, 4 to 8, 8 to 12, 12 to 16 and 16 to 20 hold 4, 3, 2, 0, 1 drives. The tallest bar is at the left and the bars fall away to a long tail on the right: skewed right, as in the second panel of the figure above.

The values add to 60, so \bar{x} = 6. With ten values the median is halfway between the fifth and sixth, \displaystyle \frac{4 + 5}{2} = 4.5. The mean is pulled above the median by the long right tail, a sign of skew on its own. The deviations -4, -4, -3, -3, -2, -1, 0, 2, 5, 10 have squares adding to 184, so

s = \sqrt{\frac{184}{9}} \approx 4.52 \text{ km}.

Within one standard deviation, 1.48 to 10.52, lie eight values: 80\%. Within two, -3.04 to 15.04, lie nine: 90\%. Too many are packed near the middle and one is far out, which is what a pile of short drives with a few long ones looks like.

Most telling of all, 0 is only \displaystyle \frac{6}{4.52} \approx 1.33 standard deviations below the mean. The normal model N(6, 4.52) puts about 9\% of its area below 0: almost one volunteer in ten driving less than no distance. A distance cannot be negative, so the model is wrong in a way no amount of data could fix. A normal model does not suit these drives.

Example 3

A normal probability plot for the soap

Make a normal probability plot of Seraphine’s ten soap bars from the first example, and compare each value with the value the model N(150, 10.06) would put at the same percentile.

Solution

The data are already in order. Each value gets the percentile \displaystyle \frac{i - 0.5}{10} and the normal score from the table. The model’s value at that percentile is 150 + 10.06z, since every normal curve is the standard one stretched by s and slid to \bar{x}. In the table, “model” is 150 + 10.06z_i and “gap” is the value minus the model.

i	value	z_i	model	gap
1	134	-1.64	133.5	0.5
2	139	-1.04	139.5	-0.5
3	143	-0.67	143.3	-0.3
4	145	-0.39	146.1	-1.1
5	149	-0.13	148.7	0.3
6	151	0.13	151.3	-0.3
7	154	0.39	153.9	0.1
8	158	0.67	156.7	1.3
9	161	1.04	160.5	0.5
10	166	1.64	166.5	-0.5

Plotting each value against its z_i gives the figure below, with the line 150 + 10.06z dashed.

Figure 2The soap bars plotted against their normal scores. The dashed line is the normal model with the data’s own mean and standard deviation, 150 + 10.06z. Every dot is within about 1.3 grams of it.

Every point is within about 1.3 g of the line, and the differences go up and down with no pattern: no run of positives at both ends, no single large gap. The points follow a straight line, so a normal model fits these bars well. The table is the plot in numbers: a difference column with no pattern and no large entries is a straight plot.

Example 4

A normal probability plot for the drives

Make a normal probability plot of the ten volunteer drives from the second example and describe its shape. What does the shape say about the data?

Solution

The distances are in order, so the i-th one goes with the i-th normal score:

\begin{gathered} (-1.64, 2),\ (-1.04, 2),\ (-0.67, 3),\ (-0.39, 3), \\ (-0.13, 4),\ (0.13, 5),\ (0.39, 6), \\ (0.67, 8),\ (1.04, 11),\ (1.64, 16). \end{gathered}

Figure 3The volunteers’ drives. The dots bend upward away from the dashed line 6 + 4.52z, above it at both ends, below it in the middle. That bowl shape is the mark of a right skew.

Read the climb from point to point. Across the first five points the normal score rises by about 1.5 while the distances rise by only 2 km. Across the last two the score rises by 0.6 while the distances jump 5 km. The plot starts flat and gets steeper and steeper: it bends upward like a bowl.

Against the reference line 6 + 4.52z the same thing shows as a pattern in the gaps. At z = -1.64 the line gives 6 - 7.41 = -1.41, below the actual 2; at z = 0.13 it gives 6.59, above the actual 5; at z = 1.64 it gives 13.41, below the actual 16. The points are above the line at both ends and below it in the middle.

A bowl means the small values are crowded together and the large ones spread out: the data are skewed right, which agrees with the histogram and with the mean of 6 sitting above the median of 4.5. A normal model is not reasonable.

Example 5

An outlier

The masses of ten eggs collected from one henhouse on one morning, in grams, are

61,\ 64,\ 66,\ 67,\ 68,\ 69,\ 70,\ 71,\ 74,\ 90.

Their mean is 70 and their standard deviation is about 7.92. Compare the shares within one and two standard deviations with 68\% and 95\%, then make a normal probability plot and say what it shows that the shares do not.

Solution

From 70 - 7.92 = 62.08 to 77.92 lie eight values, 80\%; from 54.16 to 85.84 lie nine, 90\%. These are some way off 68\% and 95\%, but the counts alone do not say why.

Pair the values with the normal scores for ten values:

\begin{gathered} (-1.64, 61),\ (-1.04, 64),\ (-0.67, 66), \\ (-0.39, 67),\ (-0.13, 68),\ (0.13, 69), \\ (0.39, 70),\ (0.67, 71),\ (1.04, 74), \\ (1.64, 90). \end{gathered}

A reference line built from \bar{x} and s would be pulled about by the very value in question, so draw the line through the third and eighth points instead, (-0.67, 66) and (0.67, 71), which mark off the middle half of the data. Its slope is \displaystyle \frac{71 - 66}{1.34} \approx 3.73 and its height at z = 0 is \displaystyle \frac{66 + 71}{2} = 68.5. At z = 1.64 it gives 68.5 + 3.73 \times 1.64 \approx 74.6.

Figure 4The eggs. The dashed line passes through the third and eighth values, the middle half of the data, which one extreme value cannot move. Nine dots follow it; the largest stands far above it.

Nine points lie close to the line; the tenth, 90, is about 15 g above where the line puts the largest of ten values. That is one outlier, not a skew: the rest of the data look normal. The Outliers and the 1.5 IQR rule agrees. The lower half 61, 64, 66, 67, 68 gives Q_1 = 66 and the upper half 69, 70, 71, 74, 90 gives Q_3 = 71, so the upper fence is 71 + 1.5 \times 5 = 78.5 and 90 is beyond it, as a Box plots that show outliers would show.

The outlier also explains the shares. It inflates the standard deviation to 7.92, more than twice the slope of 3.73 the middle of the data has, so the band \bar{x} \pm s is wide enough to take in eight of the other nine. The shares saw that something was wrong; the plot shows what. Whether the 90 is a misreading, a double-yolked egg, or simply a very large one is a question about the data, which the plot raises but cannot answer.

Example 6

Reading the mean and standard deviation off a straight plot

Ianto times his cycle ride to work on five days, in minutes: 18, 21, 23, 25, 28. Make a normal probability plot. Use it to estimate the mean and standard deviation of a normal model, and compare with \bar{x} and s.

Solution

With five values the normal scores are -1.28, -0.52, 0, 0.52 and 1.28, giving the points

\begin{gathered} (-1.28, 18),\ (-0.52, 21),\ (0, 23), \\ (0.52, 25),\ (1.28, 28). \end{gathered}

On a straight plot the points lie near \text{value} = \mu + \sigma z, so the height at z = 0 estimates the mean and the slope estimates the standard deviation.

The middle point is at height 23, so the mean is about 23 minutes. The slope from the first point to the last is

\frac{28 - 18}{1.28 - (-1.28)} = \frac{10}{2.56} \approx 3.91,

and from the second to the fourth it is \displaystyle \frac{25 - 21}{1.04} \approx 3.85. The two slopes nearly agree, which is what “straight” means, and both say the standard deviation is about 3.9 minutes.

From the data directly, \displaystyle \bar{x} = \frac{115}{5} = 23, and the deviations -5, -2, 0, 2, 5 have squares adding to 58, so \displaystyle s = \sqrt{\frac{58}{4}} \approx 3.81. The plot’s estimates are close.

With only five points, though, the plot is weak evidence. Five values drawn from a skewed population can easily happen to line up; a straight plot from five points says only that nothing is obviously wrong.

Boundaries

A plot can reject, never prove. A straight plot says the data are consistent with a normal model; it cannot show that the population they came from is normal. A clear bend or a gap is real evidence against normality; a straight line is only the absence of that evidence.

Small data sets say little. With five or ten values, random wobble can hide a moderate skew or fake a mild bend, and the share check moves in steps of 10 or 20 points. Judge only clear patterns, and use what you know about the variable as well: lengths of time, incomes and other quantities with a floor at 0 and no ceiling are often skewed right.

The plotting position is a convention. This lesson gives the i-th of n values the percentile \displaystyle \frac{i - 0.5}{n}. Other books and software use slightly different rules, such as \displaystyle \frac{i}{n + 1}. They move the points a little, especially at the ends, but a straight plot stays straight and a bend stays a bend.

Axes may be swapped. Some books and software put the data across and the normal scores up. Then a straight line still means normal, but every bend turns over: right skew bends the other way. Read the axes before reading the shape.

Repeated values. Equal values get consecutive percentiles and so sit at the same height side by side, making short flat steps (as in the plot of the drives). Data recorded to a coarse unit give a staircase; that comes from the rounding, not from the shape of the distribution.

Normal enough is relative to the use. A slight bend may not matter for one purpose and matter a great deal for another, especially one that relies on the far tails. The plot shows how the data differ from normal; deciding whether that difference matters depends on what the model is for.

Where this goes next

Checking normality is the first condition in much of what follows. The conditions for inference about a mean ask for data with no strong skew or outliers when the sample is small, which is exactly what these plots look for; the conditions for regression inference ask the same of the residuals, and a normal probability plot of the residuals is the standard way to check it.

The The central limit theorem explains why a skewed population is less of a problem for averages: the sampling distribution of a sample mean becomes close to normal as the sample grows, whatever the population looks like. A normal probability plot of many simulated sample means is a good way to see that happen.

Exercises

Counting within one and two standard deviations

The lengths of ten courgettes, in centimeters, are 11, 14, 16, 18, 19, 21, 22, 24, 26, 29. Their mean is 20 cm and their standard deviation is about 5.5 cm. What share lie within one standard deviation of the mean, and what share within two? Do the shares fit a normal model?
The prices of ten second-hand guitars, in dollars, are 260, 270, 280, 300, 310, 330, 360, 420, 550, 920, with mean 400 dollars and standard deviation about 202 dollars. Find the shares within one and two standard deviations, and the value of \bar{x} - 2s. Does a normal model suit these prices?
A data set of 200 values has 160 within one standard deviation of the mean and 190 within two. Compare with a normal model.
A data set of 400 values has 270 within one standard deviation of the mean and 382 within two. Compare with a normal model.

Normal scores

A data set has 4 values. Find the percentile \displaystyle \frac{i - 0.5}{n} given to each.
A data set has 20 values. What percentile is given to the smallest, and to the 15th smallest?
A data set of ten values has \bar{x} = 30 and s = 4. Using the table of normal scores, what value does the model N(30, 4) place at the percentile of the 3rd smallest value, and at that of the largest?
Write the five points of the normal probability plot for 8, 11, 12, 14, 15, and estimate the standard deviation from the slope between the first and last points. Compare with s.

Reading a normal probability plot

The four plots below each show ten values against their normal scores, with the data value up.

Figure 5Four normal probability plots for Exercises 9 to 12.

Which plot is consistent with a normal distribution?
Which plot shows data skewed to the right? Explain how you can tell.
Which plot shows data skewed to the left? Explain how you can tell.
Which plot shows an outlier, and is it a very large or a very small value?

Making and judging plots

Ten trays of seeds had these percentages germinate: 62, 78, 84, 88, 90, 91, 93, 94, 95, 97. Find the mean and median, describe the shape of their normal probability plot, and say whether a normal model is reasonable.
Suppose an eleventh soap bar of 210 g were added to Seraphine’s ten. Describe how the normal probability plot would change.
Make a normal probability plot of 40, 46, 50, 54, 60. Estimate the mean from the middle point and the standard deviation from the slope between the first and last points, and compare with \bar{x} and s.
A normal probability plot is close to a straight line through (0, 85) and (1, 93). Estimate the mean and standard deviation, and the value a normal model puts at the 90th percentile (whose standard normal value is 1.28).

Explaining

Explain why data from any normal distribution, whatever its mean and standard deviation, give a normal probability plot close to a straight line.
Explain why right-skewed data give a normal probability plot that bends upward, when the data are plotted up and the normal scores across.
Explain why comparing the share within one standard deviation with 68\% is a weak check for a data set of ten values.
In the egg example, eight of ten values were within one standard deviation of the mean, more than 68\%. Explain how a single large value can make that share larger rather than smaller.
A normal probability plot of eight values is close to a straight line. Explain why this does not show that the population they came from is normal.

Answers

60\% and 100\%. One standard deviation either side is 14.5 to 25.5, holding 16, 18, 19, 21, 22, 24 (six values); two either side is 9 to 31, holding all ten. Each count is within one value of 6.8 and 9.5, as close as ten values can be expected to come, so the shares fit a normal model.
90\% and 90\%. One standard deviation either side is about 198 to 602, holding every price but 920; two either side is about -4 to 804, which still leaves out 920. So \bar{x} - 2s \approx -4, a negative price. Too many prices crowd the middle, one is far out, and the model would put area on prices below 0: the prices are skewed right and a normal model does not suit them.
80\% and 95\%: \displaystyle \frac{160}{200} = 0.8 and \displaystyle \frac{190}{200} = 0.95. The share within two fits, but 80\% within one is well above 68\%: the data are more packed in the middle than a normal curve, so the shares count against a normal model.
67.5\% and 95.5\%: \displaystyle \frac{270}{400} = 0.675 and \displaystyle \frac{382}{400} = 0.955. Both are very close to 68\% and 95\%, consistent with a normal model (though not proof of one).
0.125, 0.375, 0.625, 0.875: these are \displaystyle \frac{0.5}{4}, \displaystyle \frac{1.5}{4}, \displaystyle \frac{2.5}{4} and \displaystyle \frac{3.5}{4}.
0.025 and 0.725: \displaystyle \frac{1 - 0.5}{20} = 0.025 and \displaystyle \frac{15 - 0.5}{20} = 0.725.
About 27.3 and 36.6. The 3rd of ten has z = -0.67, giving 30 + 4 \times (-0.67) = 27.32; the largest has z = 1.64, giving 30 + 4 \times 1.64 = 36.56.
The points are (-1.28, 8), (-0.52, 11), (0, 12), (0.52, 14), (1.28, 15). The slope is \displaystyle \frac{15 - 8}{2.56} \approx 2.73, so the standard deviation is about 2.7. From the data, \bar{x} = 12 and the squared deviations add to 30, so \displaystyle s = \sqrt{7.5} \approx 2.74.
B. Its points lie along the line with no bend and no gap.
A. Its points start nearly flat and climb more and more steeply, a bowl shape: the small values are bunched and the large values spread out, which is a long right tail.
D. Its points climb steeply at the left and level off at the right, a hill shape: the small values are spread out and the large ones bunched, which is a long left tail.
C. The point at the far left is far below the line while the other nine follow it, so the outlier is a very small value.
Mean 87.2, median 90.5. The plot climbs steeply at the left (from 62 to 78 to 84) and levels off at the right (93, 94, 95, 97), bending downward like the top of a hill: the data are skewed left, as the mean below the median also says. A percentage cannot pass 100, which crowds the large values together. A normal model is not reasonable.
With eleven values every point gets a new percentile and normal score, but the ten old values still lie close to a line. The new value is the largest, so it is plotted at the right end, and at 210 g it sits far above the line through the others: the plot would show a single outlier at the top. The mean and standard deviation would also be pulled up by it.
Points (-1.28, 40), (-0.52, 46), (0, 50), (0.52, 54), (1.28, 60). Mean about 50; slope \displaystyle \frac{60 - 40}{2.56} \approx 7.81, so the standard deviation is about 7.8. From the data, \bar{x} = 50 and \displaystyle s = \sqrt{\frac{232}{4}} = \sqrt{58} \approx 7.62.
Mean about 85, standard deviation about 8, and 90th percentile about 95.2. The height at z = 0 is the mean; the rise from z = 0 to z = 1 is the slope, 93 - 85 = 8, which is the standard deviation; and 85 + 8 \times 1.28 = 95.24.
Every normal curve N(\mu, \sigma) is the standard normal curve stretched by \sigma and slid by \mu, so the value at any percentile is \mu + \sigma z, where z is the standard normal value at that percentile. Normal data have their i-th value near the percentile \displaystyle \frac{i - 0.5}{n}, so the points sit near \text{value} = \mu + \sigma z_i, a straight line. Changing \mu or \sigma only moves the line or changes its slope.
In right-skewed data the small values are close together and the large values are far apart. Moving from one normal score to the next, the data value rises a little at the left end and a lot at the right end. A rise that keeps getting steeper is a curve bending upward.
With ten values the share within one standard deviation can only be a whole number of tenths, so it can never equal 68\%, and one value on either side of a boundary changes it by ten points. Data from a normal population can easily give 50\% or 80\%, and quite different shapes can give 70\%. A small difference from 68\% therefore tells you almost nothing.
The standard deviation measures distances from the mean, and a value far out adds a very large squared deviation. That makes s much bigger than the typical spread of the other values, so the band from \bar{x} - s to \bar{x} + s becomes wide and takes in nearly all of them. The outlier lands outside the band, but it has stretched the band around everything else.
A plot shows only these eight values. Eight values from a skewed population can line up by chance, and a population can depart from normal in its far tails, where eight values rarely reach. A straight plot shows the data give no evidence against normality; it cannot prove the population is normal.