Section 8.1 assumed the population standard deviation was known, which in practice it almost never is. Replacing sigma with the sample standard deviation adds a second source of uncertainty, and for small samples the normal distribution no longer describes the resulting statistic. William Gosset, working at the Guinness brewery where experiments yielded very few samples, found the distribution that does, and published it under the pen name Student. The t distribution is symmetric about zero like the normal but has more probability in its tails and less in its centre, and its exact shape depends on the degrees of freedom, which equal the sample size minus one because the deviations used to compute the sample standard deviation must sum to zero. As the degrees of freedom grow the t curve approaches the normal, so the price of not knowing sigma is large for a small sample and negligible for a large one.
Subject: Statistics · 65 slides · symbolic lesson
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Title
Statistics · Chapter 8 — Confidence Intervals
A Single Population Mean using the Student t Distribution
Objectives
Five outcomes, and the first is a recognition problem rather than a calculation.
OpenStax Introductory Statistics 2e, §8.2 A Single Population Mean using the Student t Distribution §8.2, pp. 416-420 — the section these objectives are drawn from
Warm-up
Section 8.1 built intervals as the sample mean plus and minus a z-score times sigma over the square root of n.
Discussion prompt
A sample of 15 acupuncture subjects gives a mean sensory rate of 8.2267. The population standard deviation is not known, but the sample's is 1.6722. Can section 8.1's method simply be reused with 1.6722 in place of sigma?
Hint: Ask what is now uncertain that was not before.
Answer:
Not quite. In section 8.1 the only uncertain quantity was the sample mean; sigma was a fixed known number. Now s is itself an estimate from the same 15 observations, so there are two things being estimated rather than one.
That extra uncertainty has to be paid for somewhere, and it is paid for by using a wider multiplier. For 15 observations at 95 percent, the multiplier is 2.145 rather than 1.96 — about ten percent wider.
The book records the history: statisticians once used the normal for large samples and reserved this correction for samples of at most 30, but with calculators the practice now is to use the Student t distribution whenever s is used as an estimate for sigma.
Concept
If you draw a simple random sample of size n from a population that is approximately normal with unknown standard deviation, and calculate the t-score as the sample mean minus mu over s divided by the square root of n, then those t-scores follow a Student t distribution with n minus one degrees of freedom. The t-score has the same interpretation as the z-score: it measures how far the sample mean is from mu.
Student t distribution — The distribution of the standardised sample mean when the population standard deviation is estimated from the sample. Written T follows t sub df, with df equal to n minus one, and there is a different curve for every sample size.
\[ t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \;\sim\; t_{n-1} \]
Gosset's problem is worth knowing because it explains the shape. His experiments with hops and barley produced very few samples, and simply replacing sigma with s did not give accurate confidence intervals — he found that the actual distribution depends on the sample size. The pen name Student was required because Guinness did not allow its employees to publish, and the distribution has carried it ever since.
Figure (svg): A dashed standard normal curve with three t curves of increasing degrees of freedom, the lowest sitting lower in the centre and higher in the tails
OpenStax Introductory Statistics 2e, §8.2 A Single Population Mean using the Student t Distribution §8.2, pp. 416-417
Section
Section 1
Concept
In practice we rarely know the population standard deviation. Using the sample standard deviation s in its place worked well enough for large samples but caused inaccuracies when the sample was small, because s is itself an estimate and varies from sample to sample.
the t-score — The sample mean minus mu, divided by s over the square root of n. It differs from a z-score only in that the denominator is estimated rather than known, and that difference is what makes the distribution wider.
\[ z = \frac{\bar{x}-\mu}{\sigma/\sqrt{n}} \quad\text{against}\quad t = \frac{\bar{x}-\mu}{s/\sqrt{n}} \]
The extra width has an intuitive reading. A z-score's denominator is a fixed number, so all its variability comes from the numerator. A t-score's denominator wobbles too, and a sample that happens to give a small s produces an unusually large t — which is why the t distribution has more probability far from zero than the normal does.
Figure (svg): Two columns contrasting when to use the normal distribution with when to use the t distribution
OpenStax Introductory Statistics 2e, §8.2 A Single Population Mean using the Student t Distribution §8.2, pp. 416-417 — Gosset, and the modern practice
Picture it
The decision is about sigma, not about the sample size.
Figure (svg): Two columns contrasting when to use the normal distribution with when to use the t distribution
The last row of each column is the practical summary. Section 8.1's situation — a genuinely known population standard deviation — is rare outside textbook problems and quality control, so the t is the ordinary case and the z is the exception.
Worked example
Two problems that look alike and are not.
\[ \text{(a) } \sigma = 3 \text{ given}; \quad \text{(b) } s = 1.6722 \text{ computed from the data} \]
Read what the spread is
Why: Given, or computed?
Case (a)
Why: sigma is stated.
Case (b)
Why: s comes from the sample.
Note what does NOT decide it
Why: The sample size.
Figure (svg): The solution to Worked example deciding between z and t shown as a ladder of expressions, one row per legal move
\[ \sigma \text{ known} \to z; \qquad s \text{ estimated} \to t \]
Verify: confirm the modern rule against the older one
Why: Older practice used a z whenever n exceeded 30, which is why some tables and textbooks still say so. The book is explicit that the practice now is to use the t whenever s is used as an estimate for sigma — and since the t approaches the normal for large df, following the modern rule on a large sample costs almost nothing anyway. Following the old rule on a small sample, by contrast, understates the interval.
OpenStax Introductory Statistics 2e, §8.2 A Single Population Mean using the Student t Distribution §8.2, p. 417
Sorting
Ask whether the standard deviation is given or computed.
Sort into buckets
Sort each situation.
Item (e) is the one that tests the modern rule: 500 observations is plenty large, but s is still an estimate, so a t is correct. The practical difference there is tiny — t with 499 df is 1.965 against 1.96 — but the rule is simpler for having no exception.
Worked example
The 95 percent multiplier at four sample sizes.
\[ \text{CL} = 0.95; \; n = 5, 15, 30, 1000 \]
At n = 5
Why: df = 4.
\[ t = 2.776 \]
At n = 15
Why: df = 14.
\[ t = 2.145 \]
At n = 30
Why: df = 29.
\[ t = 2.045 \]
At n = 1000
Why: df = 999.
\[ t = 1.962 \]
Figure (svg): The solution to Worked example how much the correction matters shown as a ladder of expressions, one row per legal move
\[ 2.776 \to 2.145 \to 2.045 \to 1.962 \]
Verify: confirm the direction of the convergence and what it means practically
Why: Every t exceeds 1.96 and the excess shrinks with n, so a t interval is always wider than the corresponding z interval and the gap closes. At n equal to 5 the interval is 42 percent wider; at n equal to 1000 it is a tenth of a percent wider. That is why the old thirty-observation rule was tolerable — the error it made was small — and why the modern rule costs nothing to follow.
OpenStax Introductory Statistics 2e, §8.2 A Single Population Mean using the Student t Distribution §8.2, pp. 417-418
Trap
\[ n = 60 \text{ and } s \text{ estimated} \;\Rightarrow\; \text{use } z = 1.96 \]
Apply the old rule that a large sample licenses the normal
Why: Many older tables and texts say so.
\[ t_{59} = 2.001 \text{, not } 1.96 \]
The interval comes out about 2 percent too narrow, which is small but avoidable and always in the same direction.
\[ \text{use } t_{59} = 2.001 \text{ whenever } s \text{ replaces } \sigma \]
Let the source of the spread decide, not the sample size
Why: The book's modern rule.
The error here is genuinely minor for a large sample, which is exactly what makes the habit worth forming on easy cases: the same reasoning applied to a sample of five would understate the interval by nearly a third. Asking where the spread came from is a one-second check that never gives the wrong answer.
Two truths and a lie
All three concern why the t exists.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false as a rule, though it describes older practice. The deciding question is whether sigma is known, not how large the sample is — a sample of 500 with s estimated still calls for a t, and a sample of 10 with sigma genuinely known calls for a z.
Prediction
Commit before reasoning.
Predict first
Why does the t distribution have more probability in its tails than the normal?
Correct: Because the denominator varies too.
Why: A z-score's denominator is fixed, so all its variability comes from the numerator; a t-score's denominator is itself estimated and wobbles, and a sample that happens to give a small s produces a large t. That extra source of variation is what fattens the tails. The t IS symmetric, so the last option is simply false.
Fill the middle
When the t distribution should be used.
Fill in the blanks
\textsigma s \text___ ___
Why: Sigma. The rule refers to the source of the standard deviation rather than to the sample size, which makes it simpler than the older thirty-observation convention and never wrong.
Section
Section 2
Concept
The degrees of freedom, n minus one, come from the calculation of the sample standard deviation. That calculation needs n deviations from the mean, and because those deviations sum to zero, the last one is determined once the other n minus one are known.
degrees of freedom — The number of deviations that can vary freely, which is n minus one. Each sample size has its own t distribution, indexed by this number.
\[ \sum_{i=1}^{n}(x_i - \bar{x}) = 0 \;\Longrightarrow\; \text{df} = n - 1 \]
The book derives the number rather than stating it, and the derivation generalises: every parameter estimated from the data costs one degree of freedom. That rule reappears in chapter 11, where a goodness-of-fit test has one fewer degree of freedom than it has categories, and in chapter 12, where a regression costs two because both a slope and an intercept are estimated.
Figure (svg): A card explaining that the degrees of freedom equal n minus one because the deviations must sum to zero
OpenStax Introductory Statistics 2e, §8.2 A Single Population Mean using the Student t Distribution §8.2, p. 417 — the derivation of n minus one
Picture it
The constraint that costs the degree of freedom.
Figure (svg): A card explaining that the degrees of freedom equal n minus one because the deviations must sum to zero
It is worth checking the claim on a tiny example. For the values 2, 4 and 9 the mean is 5, and the deviations are minus 3, minus 1 and 4 — which total zero. Knowing any two of them fixes the third, so only two can vary freely, and the degrees of freedom are 2.
Worked example
Checking the derivation on three values.
\[ \text{the values } 2, 4, 9 \]
Find the mean
Why: Fifteen over three.
\[ 5 \]
List the deviations
Why: Each value minus 5.
\[ -3, -1, 4 \]
Add them
Why: Their total.
\[ 0 \]
Ask how many are free
Why: The third is forced.
\[ 2 = n - 1 \]
Figure (svg): The solution to Worked example the constraint on a small sample shown as a ladder of expressions, one row per legal move
\[ -3 + (-1) + 4 = 0, \qquad \text{df} = 3 - 1 = 2 \]
Verify: confirm the constraint is a consequence of the mean, not a coincidence
Why: The deviations sum to the total of the values minus n times their mean, and n times the mean IS the total — so the sum is always exactly zero for any data set whatever. The constraint is therefore built into the definition of the mean rather than being a property of these three numbers, which is why the degree of freedom is always lost.
OpenStax Introductory Statistics 2e, §8.2 A Single Population Mean using the Student t Distribution §8.2, p. 417
Faded example
A sample of 20 items.
Fill in the blanks
\text1 = 20 - 19 = ___
Why: Nineteen, because the twenty deviations from the mean must sum to zero, so only nineteen of them can vary freely. The book writes this distribution as T following t sub 19.
Worked example
Examples 8.8 and 8.9.
\[ n = 15 \text{ and } n = 20 \]
For the 15 subjects
Why: Fifteen minus one.
\[ d f = 14 \]
Its 95 percent t
Why: invT at 0.975 with 14 df.
\[ 2.145 \]
For the 20 infants
Why: Twenty minus one.
\[ d f = 19 \]
Its 90 percent t
Why: invT at 0.95 with 19 df.
\[ 1.729 \]
Figure (svg): The solution to Worked example df for the book's two samples shown as a ladder of expressions, one row per legal move
\[ t_{0.025, 14} = 2.145, \qquad t_{0.05, 19} = 1.729 \]
Verify: confirm each t exceeds its z counterpart
Why: The 95 percent z is 1.96 against a t of 2.145, and the 90 percent z is 1.645 against a t of 1.729 — both larger, as they must be. A t-score smaller than the corresponding z is impossible for any finite df, so it is a reliable check that the right table row and column were read.
OpenStax Introductory Statistics 2e, §8.2 A Single Population Mean using the Student t Distribution §8.2, pp. 418-419
Error analysis
n is 15 at a 95 percent level; the correct value is 2.145.
Annotate
On: \( \begin{aligned} &(1)\; \text{invT}(0.975, 15) = 2.131 \\ &(2)\; \text{invT}(0.95, 14) = 1.761 \\ &(3)\; \text{invNorm}(0.975) = 1.96 \\ &(4)\; \text{invT}(0.975, 14) = 2.145 \end{aligned} \)
Errors (1) and (3) both understate the interval, and error (2) understates it further still — all three failures push the same way, which is worth knowing. A t interval that comes out narrower than the z interval for the same data has one of these three faults.
Two truths and a lie
All three concern degrees of freedom.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false: they are n minus one. Using n instead gives a t-score slightly too small, and the error grows as the sample shrinks — at n equal to 5 it is the difference between 2.776 and 2.571.
Estimation
Compare the t and z multipliers at 95 percent.
Predict first
At which sample size is the t interval most different from the z interval?
Correct: n = 5.
Why: At n equal to 5 the t is 2.776 against a z of 1.96 — 42 percent wider. By n equal to 30 the gap is about 4 percent, and by n equal to 1000 it is a tenth of a percent. The correction exists for small samples, which is exactly Gosset's problem with his few brewery experiments.
Prediction
Commit before reasoning.
Predict first
Why does estimating a parameter from data cost a degree of freedom?
Correct: Because it imposes a constraint.
Why: Estimating the mean forces the deviations to sum to zero, which removes one degree of freedom. The same logic recurs throughout the book — a goodness-of-fit test in chapter 11 loses one for the total, and a regression in chapter 12 loses two for the slope and intercept — so it is worth understanding rather than memorising per case.
Section
Section 3
Concept
When the population standard deviation is not known, the error bound is the t-score with area alpha over two to its right, at n minus one degrees of freedom, multiplied by the sample standard deviation over the square root of n. The interval is the sample mean plus and minus that.
the t interval — Sample mean plus and minus t times s over root n. Only two things change from section 8.1: s replaces sigma, and the multiplier is indexed by degrees of freedom.
\[ \bar{x} \pm t_{\alpha/2}\left(\frac{s}{\sqrt{n}}\right) \]
The book states the assumptions carefully at this point: the underlying population of individual observations is assumed to be normally distributed with unknown mean and unknown standard deviation, and random sampling is assumed but is a completely separate assumption from normality. Both matter. Non-normality is a problem mainly for small samples, since chapter 7's theorem rescues larger ones; a non-random sample is fatal at any size.
Figure (svg): The procedure for a confidence interval when sigma is unknown
OpenStax Introductory Statistics 2e, §8.2 A Single Population Mean using the Student t Distribution §8.2, pp. 417-419 — the error bound formula and Example 8.8
Picture it
The section 8.1 procedure with s and t substituted in.
Figure (svg): The procedure for a confidence interval when sigma is unknown
It is worth noticing how little has changed. The shape of the interval, the meaning of the confidence level, the interpretation template and the effects of raising the level or the sample size all carry over unaltered from section 8.1 — only the multiplier and the spread estimate differ.
Worked example
Example 8.8, worked from the 15 raw sensory rates.
\[ n = 15, \; \text{CL} = 0.95 \]
Compute the sample statistics
Why: Mean and sample sd.
\[ 8.2267\text{ and } 1.6722 \]
Set the degrees of freedom
Why: Fifteen minus one.
\[ 14 \]
Find the t-score
Why: invT at 0.975 with 14 df.
\[ 2.145 \]
Compute the error bound
Why: 2.145 times 1.6722 over root 15.
\[ 0.926 \]
Figure (svg): The solution to Worked example the acupuncture study shown as a ladder of expressions, one row per legal move
\[ 8.2267 \pm (2.145)\left(\frac{1.6722}{\sqrt{15}}\right) = (7.30, 9.15) \]
Verify: reconcile the book's two printed answers
Why: The book's step-by-step working gives an error bound of 0.9240 and endpoints of 7.30 and 9.15, while its calculator output gives (7.3006, 9.1527). Carrying full precision gives an error bound of 0.9261 and reproduces the calculator interval exactly, so the printed 0.9240 is an intermediate rounding artefact rather than a different method. The difference is about 0.002 and changes nothing, but the calculator interval is the more accurate of the two.
OpenStax Introductory Statistics 2e, §8.2 A Single Population Mean using the Student t Distribution §8.2, pp. 418-419
Faded example
Twelve subjects give a mean of 8.9833 hours and s = 1.3159. Use 95 percent.
Fill in the blanks
\text11 = 0.836, \quad t = 2.201, \quad \text___ = 2.201\left(\frac______}\right) \approx ___
Why: Eleven degrees of freedom, and an error bound of about 0.836 hours — so the interval runs from about 8.15 to 9.82 hours of sleep.
Worked example
Example 8.9, at a 90 percent level.
\[ n = 20, \; \text{CL} = 0.90 \]
Compute the sample statistics
Why: From the 20 counts.
\[ 127.45\text{ and } 25.965 \]
Set the degrees of freedom
Why: Twenty minus one.
\[ 19 \]
Find the t-score
Why: invT at 0.95 with 19 df.
\[ 1.729 \]
Compute the error bound
Why: 1.729 times 25.965 over root 20.
\[ 10.038 \]
Figure (svg): The solution to Worked example industrial chemicals in cord blood shown as a ladder of expressions, one row per legal move
\[ 127.45 \pm 10.038 = (117.412, 137.488) \]
Verify: confirm the interval against the data's own spread
Why: Individual counts in this sample run from 79 to 160, a range of 81, while the interval for the MEAN spans only about 20 — a quarter of that. That contrast is the point of an interval for a mean: it locates the centre far more precisely than any individual observation could, and an interval as wide as the data itself would signal that s had been used in place of the standard error.
OpenStax Introductory Statistics 2e, §8.2 A Single Population Mean using the Student t Distribution §8.2, pp. 419-420
Trap
\[ \text{EBM} = (2.145)(1.6722) = 3.587 \]
Multiply the t-score by the sample standard deviation
Why: s is the spread the problem supplies.
\[ (4.64, 11.81) \text{ instead of } (7.30, 9.15) \]
That interval describes where individual sensory rates fall, not where their mean lies.
\[ \text{EBM} = (2.145)\left(\frac{1.6722}{\sqrt{15}}\right) = 0.926 \]
Divide s by the square root of n to get the standard error
Why: The parameter estimated is a mean.
This is section 8.1's trap with s in place of sigma, and it is worth stating twice because the error is identical and equally common. The check is the same: an interval for a mean should be much narrower than the data's own range, and here it is about a quarter as wide.
Discrimination
For a t interval with n = 15.
Sort into buckets
Sort each quantity by whether it belongs in the error bound.
Two truths and a lie
All three concern the construction.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false, and the book says so directly: random sampling is assumed, but that is a completely separate assumption from normality. A perfectly normal population sampled non-randomly gives an interval that is wrong in a way no amount of data will fix.
Estimation
Twenty observations range from 79 to 160, with s about 26.
Predict first
Roughly how wide should a 90 percent interval for the MEAN be?
Correct: About 20.
Why: The standard error is 26 over the square root of 20, about 5.8, and the error bound is about 1.73 times that, roughly 10 — giving a width near 20. The option 81 is the data's own range, which is what using s undivided would produce, and it is about four times too wide.
Section
Section 4
Concept
For every finite number of degrees of freedom the t-score exceeds the corresponding z-score, so a t interval is always wider than the z interval that would be built from the same data. As the degrees of freedom increase, the t distribution becomes more like the standard normal and the gap closes.
the width penalty — The extra width a t interval carries over a z interval, which is the price of not knowing sigma. It is about 42 percent at n equal to 5, about 4 percent at n equal to 30, and negligible beyond a few hundred.
\[ t_{\alpha/2, \,df} > z_{\alpha/2} \;\text{ for all finite df}, \quad t \to z \text{ as } df \to \infty \]
The convergence explains the historical practice the book describes. Using a normal approximation for samples over 30 was tolerable because the error was around four percent — noticeable but rarely decisive. It also explains why the correction is not optional for small samples: at n equal to 5 an interval built with a z is nearly a third too narrow, which will overstate what the data supports.
Figure (svg): A four-column table showing the t multiplier for a ninety-five percent interval falling from two point seven eight at a sample of five toward one point nine six at a sample of a thousand
OpenStax Introductory Statistics 2e, §8.2 A Single Population Mean using the Student t Distribution §8.2, pp. 417-418 — the properties of the t distribution
Picture it
The 95 percent t against the normal's 1.96.
Figure (svg): A four-column table showing the t multiplier for a ninety-five percent interval falling from two point seven eight at a sample of five toward one point nine six at a sample of a thousand
The last column falls fast at first and then very slowly, which is the same square-root-like behaviour seen throughout these chapters. Most of the benefit of a larger sample for this particular purpose is realised by about thirty observations, even though the standard error keeps shrinking well beyond that.
Worked example
Example 8.8, built correctly and then with a z.
\[ \bar{x} = 8.2267, \; s = 1.6722, \; n = 15 \]
The correct t interval
Why: t of 2.145.
\[ (7.30, 9.15) \]
An incorrect z interval
Why: z of 1.96.
\[ (7.38, 9.07) \]
Compare the widths
Why: 1.852 against 1.692.
\[ \text{about } 9 \% \]
Say which is right
Why: sigma is unknown.
Figure (svg): The solution to Worked example the same data both ways shown as a ladder of expressions, one row per legal move
\[ \text{width}_t = 1.852, \qquad \text{width}_z = 1.692 \]
Verify: confirm the direction of the error is always the same
Why: The t always exceeds the z, so using a z always produces an interval that is too narrow — never too wide. That one-sidedness matters: the mistake systematically overstates precision and makes results look more conclusive than the data warrants, which is a worse kind of error than random noise would be.
OpenStax Introductory Statistics 2e, §8.2 A Single Population Mean using the Student t Distribution §8.2, pp. 417-419
Faded example
At 95 percent with 14 degrees of freedom.
Fill in the blanks
t = 2.145 \text1.96 z = wider, \text___ ___
Why: The t of 2.145 exceeds the z of 1.96, so the interval built from it is wider — about 9 percent wider here, which is the cost of estimating sigma from 15 observations.
Worked example
The same comparison at a large sample.
\[ n = 1000, \; \text{CL} = 0.95 \]
Find the t
Why: df = 999.
\[ 1.962 \]
Recall the z
Why: For 95 percent.
\[ 1.960 \]
Compare
Why: The difference.
\[ 0.002 \]
As a percentage
Why: Of the width.
\[ \text{about } 0.1 \% \]
Figure (svg): The solution to Worked example when the choice stops mattering shown as a ladder of expressions, one row per legal move
\[ t_{999} = 1.962 \quad\text{against}\quad z = 1.960 \]
Verify: confirm this justifies the rule rather than undermining it
Why: Because the difference is negligible at large n, following the modern rule costs nothing there — so a rule with no exception is strictly better than one with a threshold to remember. The old thirty-observation convention existed because t tables were finite and awkward, not because the t was wrong above thirty, and calculators removed that reason.
OpenStax Introductory Statistics 2e, §8.2 A Single Population Mean using the Student t Distribution §8.2, p. 417
Trap
\[ \text{a t interval should be tighter, since it uses the sample's own } s \]
Reason that data-derived numbers are more specific
Why: s came from the actual observations.
\[ t > z \text{ always, so the interval is WIDER} \]
Using the sample's own s adds uncertainty rather than removing it, because s is itself an estimate.
\[ \text{a t interval is always wider than the corresponding z interval} \]
Read the extra width as the price of not knowing sigma
Why: Two estimated quantities cost more than one.
The intuition to correct is that using real data must be better than using an assumption. Here the comparison is not between data and assumption but between knowing sigma and estimating it — and knowing is genuinely better. The t interval is the honest one; the z interval built on an unknown sigma is simply wrong.
Two truths and a lie
All three concern the comparison.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false and has the direction backwards. Since the z is the smaller multiplier, using it produces an interval that is too NARROW — overstating precision. The error always runs in that one direction, which is what makes it worth guarding against.
Estimation
A 95 percent interval from a sample of 5.
Predict first
Roughly how much wider is the t interval than a z interval would be?
Correct: About 42 percent.
Why: The t with 4 degrees of freedom is 2.776 against a z of 1.96, a ratio of about 1.42. That is a substantial difference and it is why Gosset needed the distribution at all — his brewery experiments produced samples of exactly this size.
Explain it
A classmate expects the t interval to be narrower because it uses the sample's real standard deviation rather than an assumed one.
Discussion prompt
In two sentences or fewer, correct them.
Hint: Ask how sure they are of the value of s.
Answer:
Point out that s is itself estimated from the same fifteen observations, so it could easily have come out somewhat larger or smaller with a different sample.
That second layer of uncertainty has to be paid for with a wider interval, which is exactly what the larger t multiplier does — the z interval would be pretending sigma was known.
Section
Section 5
Concept
The interpretation follows section 8.1's template unchanged: we estimate with a stated confidence that the true population mean, described in the words of the problem, lies between two values. What differs is the set of assumptions standing behind it.
the method's assumptions — The population of individual observations is assumed approximately normal, and the sample is assumed random. The book stresses these are separate assumptions, and neither follows from the other.
\[ \bar{x} \pm t_{\alpha/2}\,\frac{s}{\sqrt{n}}, \quad \text{population approximately normal} \]
The normality assumption deserves more attention here than in section 8.1, because it bites hardest exactly where the t is most needed. For a large sample chapter 7's theorem makes the sample mean normal regardless of the population, so a moderate departure from normality is harmless; for a sample of fifteen there is no such rescue, and a strongly skewed population would make the interval unreliable.
Figure (svg): A number line showing the acupuncture sample mean with its error bound either side
OpenStax Introductory Statistics 2e, §8.2 A Single Population Mean using the Student t Distribution §8.2, pp. 417-419 — the assumptions and the interpretations
Picture it
Example 8.8's result on a number line.
Figure (svg): A number line showing the acupuncture sample mean with its error bound either side
Stated in full: we estimate with 95 percent confidence that the true population mean sensory rate is between 7.30 and 9.15. The sentence names the level, the parameter, both endpoints and the context, which is what section 8.1's template asked for and what makes the result usable by someone who did not do the calculation.
Worked example
Example 8.9, in the book's own words.
\[ (117.412, 137.488) \text{ at } 90 \text{ percent} \]
Name the level
Why: Ninety percent.
\[ \text{we estimate with } 90 \%\text{ confidence} \]
Name the parameter
Why: A population mean.
Give the context
Why: Cord blood in the US.
Give both endpoints
Why: With units implied.
\[ \text{between } 117.412\text{ and } 137.488 \]
Figure (svg): The solution to Worked example writing the interpretation shown as a ladder of expressions, one row per legal move
\[ (117.412, 137.488) \]
Verify: confirm the sentence describes a mean rather than an individual
Why: It says the MEAN number found in cord blood, not the number found in any one infant — and the distinction is large, since individual counts in this sample ran from 79 to 160 while the interval spans only 117 to 137. A reader who took the interval as a range for individuals would badly understate the variation between infants.
OpenStax Introductory Statistics 2e, §8.2 A Single Population Mean using the Student t Distribution §8.2, p. 420
Two truths and a lie
All three concern interpretation.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false and is one of the most consequential errors in applied statistics. A larger biased sample estimates the wrong quantity more precisely — the interval narrows around a value that is not the population mean, which makes the error harder to spot rather than easier.
Worked example
What has to be true for Example 8.8's interval to mean anything.
\[ n = 15 \text{ sensory rates} \]
Normality
Why: Population approximately normal.
\[ \text{matters at } n = 15 \]
Random sampling
Why: A simple random sample.
Note they are separate
Why: Neither implies the other.
Say which is rescuable
Why: Large n helps normality only.
Figure (svg): The solution to Worked example checking the assumptions shown as a ladder of expressions, one row per legal move
\[ \text{normal population} \;+\; \text{random sample} \]
Verify: confirm why sample size does not fix a non-random sample
Why: Chapter 7's theorem describes what happens to the mean of a RANDOM sample as n grows; it says nothing about a sample selected in a biased way. Averaging a thousand observations from a badly chosen group estimates that group's mean precisely, not the population's — the interval will be narrow and centred in the wrong place, which is worse than being wide.
OpenStax Introductory Statistics 2e, §8.2 A Single Population Mean using the Student t Distribution §8.2, p. 417
Trap
\[ (117.412, 137.488) \;\Rightarrow\; \text{most infants have } 117 \text{ to } 137 \text{ chemicals} \]
Read the interval as describing the data
Why: It is in the same units as the observations.
\[ \text{but the sample itself ran from } 79 \text{ to } 160 \]
The interval estimates the population MEAN, and means vary far less than individual values do.
\[ \text{the MEAN number lies between } 117.412 \text{ and } 137.488 \]
Say mean explicitly in the interpretation
Why: The parameter is a mean, not an individual value.
The width comparison makes this checkable without any calculation: an interval for a mean is narrower than the data's own range by roughly the square root of n, and one that matched the data's range would be describing something else entirely. Here 20 against 81 is about the factor of 4.5 that a sample of 20 predicts.
Faded example
Twenty counts range from 79 to 160; the interval for the mean spans 117.4 to 137.5.
Fill in the blanks
\text81 = 4, \quad \text___ \approx 20, \quad \text___ \approx ___
Why: The data spans 81 and the interval about 20, a ratio near 4 — close to the square root of 20, which is about 4.5. An interval for a mean is always much narrower than the data it came from.
Sorting
Each situation threatens one of the two assumptions.
Sort into buckets
Sort each by which assumption it threatens.
The right-hand column is the more dangerous one, because a larger sample fixes the left and not the right. Item (c) is the subtlest: dropping people who did not complete treatment is a selection made after the fact, and it is how many otherwise careful studies go wrong.
Prediction
Commit before reasoning.
Predict first
Which problem is relieved by increasing the sample size?
Correct: A non-normal population.
Why: Chapter 7's theorem makes the sample mean approximately normal for a large enough random sample, whatever the population's shape — so non-normality becomes harmless. Bias is not relieved at all: a larger biased sample simply pins down the wrong number more tightly, which is why the two assumptions have to be checked separately.
Comparison
Fill the blanks. Two rows differ and the rest carry over unchanged.
Comparison matrix
| Section 8.1 (z) | Section 8.2 (t) | |
|---|---|---|
| Spread used | sigma, known | s, estimated from the sample |
| Multiplier | z, one value for every n | t, indexed by n minus one df |
| Interval shape | point estimate plus and minus a bound | the same |
| Width, same data | narrower | always wider |
The third row is worth noticing as much as the first two. Everything about the meaning of a confidence interval — the level, the interpretation, the effect of changing n — carries over unaltered, so this section is a substitution rather than a new idea.
Pattern
Six steps, and the second is the one that distinguishes this section.
Two checks: the t must exceed the matching z, and the interval must be much narrower than the data's own range — by roughly the square root of n.
OpenStax Introductory Business Statistics 2e, §8.2 A Confidence Interval When the Population Standard Deviation Is Unknown and Small Sample Case §8.2 A Confidence Interval When the Population Standard Deviation Is Unknown and Small Sample Case
Check
Choosing the distribution.
Check your understanding
A sample of 200 gives s = 4.2; the population standard deviation is unknown. Which multiplier is correct?
Answer: A
Why: The modern rule is to use a t whenever s estimates sigma, and the degrees of freedom are n minus one.
Check
Degrees of freedom.
Check your understanding
Why are the degrees of freedom n minus one?
Answer: A
Why: Computing s requires n deviations, and because they must total zero, only n minus one can vary freely.
Check
Comparing widths.
Check your understanding
For the same data, how does a t interval compare with a z interval?
Answer: A
Why: The t-score exceeds the corresponding z-score for every finite degrees of freedom, so the t interval is always wider.
Real world
A small manufacturer tests the breaking strength of a new cable. Eight specimens give a mean of 1,240 newtons with a sample standard deviation of 95 newtons. An engineer builds a 95 percent interval using a z of 1.96, gets 1,174 to 1,306, and certifies the cable for loads up to 1,150 newtons on the grounds that the whole interval clears it.
Discussion prompt
Rebuild the interval correctly, assess the certification, and say what else is wrong with the reasoning.
Hint: Two separate errors, and the second is the more serious.
Answer:
The multiplier is wrong. With sigma unknown and n equal to 8, the t has 7 degrees of freedom and a 95 percent value of 2.365 rather than 1.96 — about 21 percent larger. The error bound becomes 2.365 times 95 over the square root of 8, about 79.4 newtons, and the interval runs from about 1,161 to 1,319 rather than 1,174 to 1,306.
\[ \text{EBM} = 2.365\left(\frac{95}{\sqrt{8}}\right) \approx 79.4, \qquad (1160.6,\; 1319.4) \]
The certification still clears 1,150 on this interval, but only just, and that is not what matters. The deeper error is that the interval estimates the MEAN breaking strength, while a safe working load depends on the WEAKEST cables, not the average one. Individual specimens vary with a standard deviation of 95 newtons, so a cable two standard deviations below the mean would break near 1,050 — well under the certified load.
The right quantity is a lower bound on individual strength, not on the mean. With s of 95 newtons, roughly 2.5 percent of cables would be expected below about 1,054 newtons even if the mean is exactly 1,240. Engineering practice handles this with a tolerance interval or a lower percentile bound, and typically with a safety factor on top, precisely because the mean is not the quantity that fails.
Two further points belong in a careful answer. Eight specimens is a very thin basis for estimating variability at all — s itself is uncertain, which is what the wide t multiplier is acknowledging — so more testing would narrow both the mean and the spread estimate. And the normality assumption matters here: breaking strengths often have a longer left tail than a normal, which would make the weak end worse than the calculation suggests rather than better.
Commit first
Answer, then rate your confidence honestly.
Predict first
When should the Student t distribution be used instead of the normal?
Correct: Whenever s is used as an estimate for sigma.
\[ t = \frac{\bar{x}-\mu}{s/\sqrt{n}} \sim t_{n-1}, \qquad \text{df} = n-1 \]
Why: This is the book's stated modern practice, and it refers to where the standard deviation came from rather than to how many observations there are. The thirty-observation rule is older convention, tolerable because the t approaches the normal but unnecessary now that calculators supply any t. Non-normality is a separate assumption that the t does not address, and the confidence level does not affect the choice at all.
Explain it
They built a 95 percent interval from 15 observations using invT(0.975, 15), getting 2.131.
Discussion prompt
In two sentences or fewer, locate the error.
Hint: Ask how many deviations can vary freely.
Answer:
They used the sample size as the degrees of freedom, but computing s forces the fifteen deviations to sum to zero, so only fourteen vary freely.
The right call is invT(0.975, 14), which gives 2.145 — a slightly larger multiplier, and the gap grows for smaller samples.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the first, ask whether the spread was given or computed. For the second, the deviations sum to zero, so one is determined. For the third, divide s by root n before multiplying by t. For the fourth, normality is relieved by a large sample and randomness never is. Do five problems of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Twenty minutes.
Draw it
At the top, draw a standard normal curve as a dashed line and three t curves over it, for 2, 5 and 30 degrees of freedom, and label the two features that distinguish them: thicker tails and a shorter centre. Beneath, write the t-score formula beside the z-score formula and circle the one difference between them. Then write out the degrees-of-freedom argument in full: take the three values 2, 4 and 9, compute the mean, list the three deviations, add them to get zero, and say how many can vary freely. In the middle of the page, work Example 8.8 completely from the fifteen sensory rates: sample mean, sample standard deviation, degrees of freedom, t-score, error bound and interval, ending with the interpreting sentence. Beside it, rebuild the same interval using a z of 1.96, write both widths, and say which is correct and why the wrong one errs in the direction it does. Below that, make a table of the 95 percent multiplier at n equal to 5, 15, 30, 100 and 1000, with a final column giving the excess over 1.96. At the bottom, write the two assumptions the method makes, and beside each write one situation that threatens it and whether a larger sample would help.
Check the middle section by confirming your t interval is wider than your z interval — if it is not, the degrees of freedom or the halving of alpha has gone wrong. Check the table by confirming the last column falls steadily toward zero and never becomes negative, since the t exceeds the z at every finite df.
Recap
Five things, and the first is a decision rather than a computation.
| If you see | Then |
|---|---|
| sigma given in the problem | Use a z, as in section 8.1 |
| s computed from the data | Use a t, whatever the sample size |
| A sample of n | df is n minus one |
| A t interval narrower than the z one | An error: df, alpha or the multiplier |
| A small sample | The t correction is large and matters |
| A sample of several hundred | The correction is negligible but still correct |
| A biased sample | No sample size repairs it |
Section 8.3 changes the parameter rather than the multiplier. When the data are categorical — how many people own a smartphone, how many students are registered to vote — the quantity of interest is a proportion, its point estimate is the sample proportion, and the standard error takes a different form built from the binomial of section 4.3.
OpenStax Introductory Statistics 2e, §8.2 A Single Population Mean using the Student t Distribution §8.2, pp. 416-420 — everything on these slides traces back here
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