Statistics · S.IC.B.4

Conditions for one mean

An interval or a test for a mean will run on any list of numbers. What must be true of the data before its answer means what it says?

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Conditions for one mean

The t distribution and degrees of freedom One sample interval for a mean and sample size One sample test for a mean Interval for a difference of means ▾ Show all 7 (2 more)

Why this exists

Jiwoo runs the weighing shed at a ginseng farm in Korea (made-up data throughout this lesson). One field holds about 5{,}000 four-year roots, and she wants \mu, the mean fresh weight of a root there. She digs up a simple random sample of 12 roots and weighs each, in grams:

38,\ 42,\ 45,\ 47,\ 49,\ 51,\ 52,\ 55,\ 58,\ 61,\ 66,\ 74

Their mean is \bar{x} \approx 53.17 grams and their standard deviation s_x \approx 10.31 grams.

The next lessons turn numbers like these into a confidence interval and a significance test for \mu. Both rest on the The t distribution and degrees of freedom: the standardized mean

t = \frac{\bar{x} - \mu}{s_x/\sqrt{n}}

follows a t distribution with n - 1 degrees of freedom when the values are independent draws from a normal population. The arithmetic runs just as smoothly when that is false. The conditions for one mean say when it is close enough to true for the stated confidence level to be trusted.

Definition

Inference about a population mean \mu with t procedures needs three conditions.

Random. The data come from a random sample of the population, or from a randomized experiment, one that uses Random assignment and confounding.
Ten percent. When the sample is drawn without replacement, n \le 0.10\,N.
Normal or large sample. At least one holds: the population is known to be close to normal; or n \ge 30, the usual guide; or, for a smaller sample, a plot of the data shows no strong skew and no outliers.

The third is the normal condition of the Sampling distribution of a sample mean, where the guide by sample size was set out. What is new here is checking all three together, from the data in hand, before any interval or test is computed.

Intuition

Each condition guards one part of the t calculation.

Random guards the center. A random method favors no root, so \bar{x} is centered on \mu. Roots picked by eye can sit off center by an amount nobody knows, and a margin of error measures only chance variation, never bias.

Ten percent guards the spread. The standard error \displaystyle s_x/\sqrt{n} assumes independent draws. Without replacement from a small population the true spread is smaller by the factor \displaystyle \sqrt{\frac{N-n}{N-1}} (the The ten percent condition), which is close to 1 when n \le 0.10\,N.

Normal or large guards the shape. The t curve is exact for a normal population. For any other, the The central limit theorem pulls the sampling distribution of \bar{x} toward normal as n grows. A small sample needs the population itself to be close to normal, and the sample’s own plot is the only evidence of that.

Figure 1Jiwoo’s random sample of 12 roots (mild skew, no outliers) and Chaewon’s random sample of 45 roots (moderate skew), made-up data.

In Figure 1, Jiwoo’s 12 roots have a short right tail and no dot standing apart. Chaewon’s 45 lean more strongly right, but there are 45 of them.

Mild skew is tolerated; outliers are not. Procedures whose real confidence level stays close to the stated one when a condition is only roughly met are called robust. The t procedures are fairly robust to mild skew: no value is far from the rest, and averaging already smooths the shape of \bar{x}. They are not robust to outliers. One far value pulls \bar{x} toward itself and inflates s_x too, so the center and the width of an interval both rest on a single observation.

Figure 2One heavy root in Doyun’s sample of 10 moves the mean from 50.0 to 59.0 grams (made-up data).

In Figure 2, one root of 140 grams moves Doyun’s mean from 50.0 to 59.0 grams, as heavy as the heaviest of the other nine.

Results

The three conditions and what each protects.

Condition	Check	Protects
Random	random sample, or random assignment	\bar{x} centered on \mu
Ten percent	n \le 0.10\,N, when sampling without replacement	the standard error \displaystyle s_x/\sqrt{n}
Normal or large	normal population, or n \ge 30, or a plot with no strong skew and no outliers	a shape close to the t curve

When one fails, say so; never compute as if it held. A failed random condition has no repair: the data describe the roots measured and no others. A failed ten percent condition makes \displaystyle s_x/\sqrt{n} too large, so an interval is wider than needed; report that, or correct with \displaystyle \sqrt{\frac{N-n}{N-1}}. A failed normal condition calls for a closer look: correct an outlier that is a recording error; report a real one, with results shown with and without it, clearly labeled; remove strong skew with a transformation such as the logarithm of each value; or take a larger sample.

Why it is true

Why shape matters at small sizes. For a normal population, \displaystyle \frac{\bar{x} - \mu}{s_x/\sqrt{n}} follows the t distribution with n-1 degrees of freedom exactly, for every n. Otherwise this is only approximate, improving as n grows. How large is large enough depends on how far the population is from normal, the very thing not known; so 30 is a guide, not a theorem.

How far one outlier reaches. Take an extreme case: n - 1 values equal to m and one equal to m + d. The mean is

\bar{x} = \frac{(n-1)m + (m + d)}{n} = m + \frac{d}{n}.

The equal values each sit \displaystyle \frac{d}{n} below \bar{x}, and the odd one sits \displaystyle \frac{(n-1)d}{n} above it. So

\begin{aligned} s_x^2 &= \frac{1}{n-1}\left[(n-1)\frac{d^2}{n^2} + \frac{(n-1)^2 d^2}{n^2}\right] \\ &= \frac{d^2}{n^2}\,(1 + n - 1) = \frac{d^2}{n}, \end{aligned}

and \displaystyle s_x = \frac{d}{\sqrt{n}}. The standard deviation, 0 without the odd value, is set entirely by it and grows without bound as d does. One far value controls both \bar{x} and s_x; under mild skew, none pulls much.

Worked examples

Example 1

Jiwoo’s twelve roots

Check the conditions for a t interval for \mu from Jiwoo’s simple random sample of 12 roots from the field of about 5{,}000 (Figure 1, top).

Solution

Random. A simple random sample, so \bar{x} has no bias from how roots were chosen.

Ten percent. She digs without replacement, so it applies: 12 \le 0.10(5000) = 500. Met.

Normal or large. With n = 12 the sample is small, and nothing says root weights are normal (a weight cannot be negative, and big roots may trail off to the right), so the plot decides. The median \displaystyle \frac{51 + 52}{2} = 51.5 sits a little below the mean 53.17, matching the short right tail: mild skew. For outliers, the lower six values give \displaystyle Q_1 = \frac{45 + 47}{2} = 46 and the upper six \displaystyle Q_3 = \frac{58 + 61}{2} = 59.5, so \text{IQR} = 13.5 and the 1.5 \times \text{IQR} fences are 46 - 20.25 = 25.75 and 59.5 + 20.25 = 79.75. Every root lies between them. Met.

All three hold, so a t interval from \bar{x} \approx 53.17 and s_x \approx 10.31 can be trusted to have about its stated confidence level.

Example 2

One heavy root

Doyun takes a simple random sample of 10 roots from a hillside bed of about 900:

41,\ 44,\ 46,\ 48,\ 50,\ 52,\ 54,\ 56,\ 59,\ 140

Check the conditions (Figure 2), and say how much the heaviest root matters.

Solution

Random: met. Ten percent: 10 \le 90, met.

Normal or large: the sample is small, so the plot decides. The lower five values give Q_1 = 46, the upper five Q_3 = 56, so \text{IQR} = 10 and the upper fence is 56 + 15 = 71. The 140 gram root lies far past it. The condition fails.

How much it matters:

	\bar{x}	s_x	\displaystyle s_x/\sqrt{n}
all 10 roots	59.0	28.99	9.17
without the 140	50.0	5.85	1.95

One root moves the mean 9.0 grams and makes the standard error about 4.70 times as large. Any interval would be centered and sized by it.

Doyun first checks the record: a slip (say 14.0 written as 140) is corrected, and the conditions checked again. If the root is real, the bed sometimes grows a giant, a long right tail that a sample of 10 cannot describe. He reports the root, may show both rows above, labeled, and needs a larger sample for a trustworthy interval. Dropping it silently would change the population to “roots that are not giants”.

Example 3

Forty-five skewed roots

Chaewon takes a simple random sample of 45 roots from the north field, about 3{,}000 roots (Figure 1, bottom). Their mean is \bar{x} \approx 48.33 grams, median 45, and s_x \approx 11.03 grams. Check the conditions.

Solution

Random: met. Ten percent: 45 \le 300, met.

Normal or large: the plot is moderately skewed right, a pile at 40 to 50 grams with a tail to 75, and the mean is above the median. With Q_1 = 40 and Q_3 = 55, the upper fence is 55 + 22.5 = 77.5, and the heaviest root, 75, is inside. Moderate skew, no outliers, and n = 45 past the guide of 30: the central limit theorem makes \bar{x} close to normal. Met.

The condition is about \bar{x}. A single root’s weight is still skewed, and a statement about single roots could not use a normal or t model.

Example 4

The display box

Doyun weighs the 20 roots in the farm shop’s display box, chosen each morning as the best-looking roots of the harvest. Their plot is roughly symmetric with no outliers. Can they give an interval for the mean weight of the farm’s roots?

Solution

No. Random fails: display roots are chosen for size and shape, so their mean is likely above the farm’s \mu, by an amount nothing in the data reveals. The other two conditions concern chance variation and cannot help with bias; a neat plot from a biased sample is still biased. Doyun can report the mean weight of these 20 display roots as a description, and needs a random sample of the farm’s roots to estimate \mu.

Example 5

A small trial bed

Chaewon takes a simple random sample of 40 of the 250 roots in a trial bed under a new shade cloth. The plot shows mild skew and no outliers. Check the conditions, and say what follows.

Solution

Random: met, for this bed only. Normal or large: met, n = 40 with a mild plot. Ten percent: \displaystyle \frac{40}{250} = 0.16 > 0.10. Fails.

The true standard deviation of \bar{x} is the usual one times \displaystyle \sqrt{\frac{250 - 40}{250 - 1}} = \sqrt{\frac{210}{249}} \approx 0.9184. So \displaystyle s_x/\sqrt{n} is about \displaystyle \frac{1}{0.9184} \approx 1.09 times too large, and an interval is about 9\% wider than needed. This errs on the safe side: such intervals capture \mu more often than stated. Chaewon can report the usual interval with that note, or use the corrected standard error.

Example 6

Strong skew and a transformation

Jiwoo takes a simple random sample of 8 one-year roots from a nursery bed of about 2{,}000:

12,\ 15,\ 18,\ 22,\ 30,\ 41,\ 60,\ 95

Check the normal condition in grams, then for \log_{10} of each weight (Figure 3).

Solution

Grams. The values crowd between 12 and 30 and spread far apart above, and the mean \displaystyle \frac{293}{8} \approx 36.6 sits well above the median \displaystyle \frac{22 + 30}{2} = 26. Strong right skew with n = 8: the condition fails.

Logs. To two places, the logs are

1.08,\ 1.18,\ 1.26,\ 1.34,\ 1.48,\ 1.61,\ 1.78,\ 1.98

spread fairly evenly, with mean 1.4623 and median 1.41, close together, and no value apart. The condition is reasonable for the log weights.

Figure 3Jiwoo’s 8 one-year roots in grams (strong right skew) and as \log_{10} of the weight (close to symmetric), made-up data.

The cost is a change of question. A t interval from the logs estimates the mean of \log_{10} of weight. Back-transformed, 10^{1.4623} \approx 29.0 grams is the geometric mean, not the mean weight. Because the logs are roughly symmetric, it sits near the median weight; in a right-skewed population the mean weight lies above it.

Boundaries

Meeting the conditions does not make an estimate right: a 95\% interval built correctly still misses \mu for about 5\% of samples.
A random sample licenses inference only to the population actually sampled.
Ten percent applies only to sampling without replacement, not to random assignment.
The guide n \ge 30 is a convention. Strong skew can need more, and size alone never makes a far outlier harmless.
A small sample’s plot cannot prove the population normal; it can only fail to show a strong departure. Knowledge of the context counts as evidence too.
The conditions support statements about \mu, not about single values.

Where this goes next

With the conditions checked, One sample interval for a mean and sample size builds \displaystyle \bar{x} \pm t^* \frac{s_x}{\sqrt{n}}, and One sample test for a mean tests a claimed \mu under the same three conditions. The parallel list for proportions is Conditions for one proportion. Other ways to judge shape are in Checking whether data look normal, and Outliers, leverage and influential points follows the effect of one far value into regression.

Exercises

Checking the three conditions

Jiwoo takes a new simple random sample of 15 roots from the field of about 5{,}000. The dot plot is roughly symmetric with no outliers. Check the three conditions.
A trial bed holds 360 roots. What is the largest simple random sample from it that meets the ten percent condition?
A simple random sample of 7 roots from the field of 5{,}000 weighs 44, 47, 50, 51, 53, 56 and 60 grams. Check the normal condition, using the median and the 1.5 \times \text{IQR} fences.
A simple random sample of 8 roots from the hillside bed weighs 36, 39, 41, 42, 44, 45, 47 and 88 grams. Check the normal condition.
Chaewon takes a simple random sample of 60 roots from a field of about 4{,}000. The histogram is moderately skewed right, with no value far from the others. Check the conditions.
A random sample of 35 roots has all weights between 30 and 90 grams except one recorded as 400 grams. Does n \ge 30 settle the normal condition?

Outliers and robustness

For the 8 roots of Exercise 4, find \bar{x} and s_x with and without the 88 gram root.
For the same roots, find \displaystyle s_x/\sqrt{n} with and without the 88 gram root, and how many times as large the first is.
Find the median of the 8 roots of Exercise 4 with and without the 88 gram root. Why does the median’s steadiness not rescue a t interval?
In a random sample of 12 roots, the weights run from 40 to 70 grams except one entered as 520. The scale’s printed log shows 52.0 for that root. What should be done?
Two random samples of 12 roots: plot A has a mild right tail and no outliers; plot B is symmetric except for one value far above the rest. Which is the bigger problem for a t interval, and why?

When a condition fails

Doyun weighs the 25 roots on top of a harvest crate, where workers set the largest roots so they are not crushed. Which condition fails, and what can he report?
Chaewon takes a simple random sample of 30 of the 200 roots in a second trial bed. Check ten percent, and find by what factor the usual standard error overstates the true one.
A simple random sample of 9 one-year roots weighs 10, 13, 16, 19, 25, 33, 44, 58 and 80 grams. Judge the normal condition in grams and for \log_{10} of each weight, and back-transform the mean log.
Doyun finds the 140 gram root of Example 2 is real. He plans a new sample. What should he aim for, and why might size alone not settle the condition?
Chaewon randomly assigns 12 of 24 volunteer plants to a new fertilizer and 12 to the old one. Which conditions apply, and how is each checked?

Reasoning

Jiwoo says her 12 roots look normal, so the field’s roots are normal. Explain what the plot can and cannot show.
Explain why a population known to be normal meets the third condition for every sample size.
Explain why t procedures tolerate mild skew in a moderate sample but not a single far outlier.
Doyun proposes dropping the 140 gram root without mentioning it, so the conditions pass. Explain what that does to his conclusion.
In the extreme case of “Why it is true” (n - 1 values equal to m, one equal to m + d), show that \displaystyle \frac{\bar{x} - m}{s_x/\sqrt{n}} = 1 for every d > 0, and say what this suggests about an outlier as evidence.

Answers

All three are met: a simple random sample; 15 \le 500; and although n = 15 is small, the plot shows no strong skew and no outliers.
36 roots, since 0.10 \times 360 = 36.
Met. The median 51 and mean \displaystyle \frac{361}{7} \approx 51.57 are close. Leaving out the median, Q_1 = 47 and Q_3 = 56, so \text{IQR} = 9 and the fences are 33.5 and 69.5; every value is inside.
Fails. \displaystyle Q_1 = \frac{39 + 41}{2} = 40 and \displaystyle Q_3 = \frac{45 + 47}{2} = 46, so the upper fence is 46 + 9 = 55, and the 88 gram root is an outlier in a sample of only 8.
All three are met: a simple random sample; 60 \le 400; and n = 60 \ge 30 with moderate skew and no far values.
No. A 400 gram root among weights of 30 to 90 is a far outlier and would dominate \bar{x} and s_x even with 35 values. Check the record; if real, report it and show results with and without it, or take a larger sample.
With it: \displaystyle \bar{x} = \frac{382}{8} = 47.75 and s_x \approx 16.63 grams. Without it: \displaystyle \bar{x} = \frac{294}{7} = 42.00 and s_x \approx 3.74 grams.
With it: \displaystyle \frac{16.63}{\sqrt{8}} \approx 5.88. Without it: \displaystyle \frac{3.74}{\sqrt{7}} \approx 1.41. The first is about 4.16 times as large.
With it, \displaystyle \frac{42 + 44}{2} = 43; without it, 42. The median barely moves, but a t interval is built from \bar{x} and s_x, and both move a great deal.
Correct the entry to 52.0, a known typing slip, and check the conditions again. This is fixing an error, not dropping an outlier.
Plot B. Mild skew pulls each value only a little, and t procedures are fairly robust to it. One far value pulls \bar{x} and inflates s_x, so the interval’s center and width both depend on it.
Random fails: the largest roots are put on top on purpose, so the mean is biased upward by an unknown amount. He can report the mean of those 25 roots, not an estimate of the farm’s mean.
It fails: \displaystyle \frac{30}{200} = 0.15 > 0.10. The factor is \displaystyle \sqrt{\frac{170}{199}} \approx 0.9243, so the usual standard error is about \displaystyle \frac{1}{0.9243} \approx 1.08 times the true one, and an interval about 8\% wider than needed.
In grams it fails: the mean \displaystyle \frac{298}{9} \approx 33.1 is well above the median 25, strong right skew in a small sample. The logs are 1.00, 1.11, 1.20, 1.28, 1.40, 1.52, 1.64, 1.76, 1.90, evenly spread, with mean 1.42 and median 1.40: reasonable. Back-transformed, 10^{1.4248} \approx 26.6 grams, the geometric mean: near the median weight since the logs are roughly symmetric, and below the mean weight.
At least 30 roots by the guide, so the central limit theorem applies. But a bed that grows occasional giants has a long right tail, which may need far more than 30, and the new plot must still be checked.
Random: met by random assignment, so a difference can be credited to the fertilizer, for plants like these. Ten percent: not needed, since nothing is sampled without replacement. Normal or large: each group has only 12 plants, so each group’s plot is checked for strong skew and outliers. The comparison belongs to Test for a difference of means.
A plot of 12 values can show a strong departure, such as heavy skew or an outlier, but cannot prove normality; many shapes give samples like hers. Passing makes a normal population a reasonable working model, not a known fact.
Then \bar{x} is exactly normal for every n, and the standardized mean follows the t distribution exactly, with no help needed from the central limit theorem.
Under mild skew no value is far out, so each pulls \bar{x} and s_x only a little, and averaging brings the shape of \bar{x} close to normal. An outlier pulls \bar{x} and inflates s_x at once, so one value decides the interval.
It changes the population the conclusion describes, from the bed’s roots to the bed’s roots other than giants, without telling the reader. The honest course is to report the root and show results with and without it.
From “Why it is true”, \displaystyle \bar{x} - m = \frac{d}{n} and \displaystyle s_x = \frac{d}{\sqrt{n}}, so \displaystyle \frac{s_x}{\sqrt{n}} = \frac{d}{n} and the ratio is \displaystyle \frac{d/n}{d/n} = 1. However far out, the outlier inflates s_x as fast as it shifts \bar{x}, so one far value cannot by itself make a strong case that the mean differs from m.