Week 14 · Chapter

Chapter 14 — Testing Claims About Means

Introduction to Statistics Generic evergreen edition

Course: Introduction to Statistics (18-week generic edition)
The module's primary reading · ~30–40 minutes · Objective 7 (means portion; reinforces Objective 6) · Read this before (or alongside) the week's lecture — everything else in the module builds on it.

Why you already care about this

Companies make promises about averages all day long, and they make them to you. "Average delivery: 50 minutes." "Up to 40 mpg highway — 40 on average in our testing." "Members typically spend 45 minutes per session." An average is a wonderfully slippery thing to promise, because no single experience can contradict it. Your order took an hour? One order proves nothing — that's just the spread. And the companies know it.

But a sample can contradict an average. If you time 25 deliveries and they average 52 minutes against a promised 50, you hold actual evidence — and this week you learn the machinery that weighs it. Last week you learned the logic of hypothesis testing: the null hypothesis gets presumed innocent, the p-value measures surprise, and the only verdicts are reject and fail to reject. Every p-value in Week 13, though, was handed to you. This week you compute the evidence yourself, for the most common claim in the wild: a claimed mean.

By the end of this chapter you will be able to: run a complete one-sample t-test — hypotheses, conditions, test statistic, verdict — and state the conclusion in honest English; choose between a one-sided and two-sided alternative and explain why the choice must come before the data; recognize paired data (before/after, same individual twice) and test it with one subtraction and the machinery you already own; and use the test ↔ interval duality to read a confidence interval as a list of every claim that would survive testing.

One line to carry through the week: a t-statistic counts standard errors between the data and the claim.

Where we are

Everything this week is assembled from parts you already have. Week 10 built the ruler: a sample mean varies from sample to sample with standard error σ∕√n. Week 11 swapped in s for the unknown σ, which is exactly why the t-distribution and its table — the same table reprinted below — entered the course. Week 13 supplied the courtroom: hypotheses, α, p-values, Type I and II errors, and the discipline of "fail to reject ≠ accept." This week the ruler goes to court. (Next week, the same trial runs on percentages and on two-group comparisons.)

1. The one-sample t-test, start to finish

Start with the plain idea. Someone claims the population mean is some particular number — call it μ₀, the claimed value. Your sample mean x̄ will never equal μ₀ exactly, even if the claim is perfectly true; Week 10 taught us that honest samples wobble. So the real question is never "is x̄ different from μ₀?" — it always is. The question is: "is x̄ too far from μ₀ to blame on sampling wobble?"

"Too far" is measured the only way this course ever measures distance: in standard errors. That measurement is the test statistic:

t = (x̄ − μ₀) ⁄ SE, where SE = s ⁄ √n — sample mean minus claimed mean, divided by the standard error.

A t of 0.3 says the sample landed a third of a standard error from the claim — utterly ordinary if the claim is true. A t of 4 says the sample landed four standard errors out — the kind of thing that essentially never happens by luck. Somewhere in between sits a cutoff, and the cutoff comes from the t-distribution with df = n − 1, the same one Week 11 used, for the same reason: we're using s as a stand-in for σ, and small samples must pay for that substitution with wider cutoffs.

The full test is five moves, and the recipe never changes:

State the hypotheses. H₀: μ = μ₀ (the claim, presumed innocent) vs. Hₐ: μ ≠ μ₀ — or μ < μ₀ or μ > μ₀, chosen before the data (Section 2).
Check the conditions (Section 5).
Compute SE, then t. Two lines, in that order. The √n in the denominator is where your sample size earns its keep.
Get the cutoff t* from the course t-table at df = n − 1 (or a p-value from technology).
Verdict, in context. |t| beyond t* → reject H₀; otherwise fail to reject — then say what that means about the claim, in words.

Memory hook: "Claim → ruler → distance → cutoff → verdict."

Here is the table this course runs on — the same one embedded in Week 11, reprinted so you never have to page back.

The table below gives critical values t* by degrees of freedom and confidence level.

df	t* for 90%	t* for 95%	t* for 99%
9	1.833	2.262	3.250
15	1.753	2.131	2.947
24	1.711	2.064	2.797

Reading it for tests: a two-sided test at α = 0.05 uses the 95% column (the middle 95% of the t-curve is "ordinary"; the outer 5%, split between two tails, is "too far"). A two-sided test at α = 0.01 uses the 99% column. A one-sided test at α = 0.05 uses the 90% column, because the 90% value leaves exactly 5% in a single tail. Any df not on the table, technology supplies.

Worked example. A courier service advertises: "Average delivery: 50 minutes." A city consumer-affairs office times a random sample of 25 deliveries and finds x̄ = 52 minutes with s = 5 minutes. Test the claim at α = 0.05.

Hypotheses: H₀: μ = 50 vs. Hₐ: μ ≠ 50. The office is auditing the promise in both directions — two-sided.
SE = 5 ⁄ √25 = 5 ⁄ 5 = 1 minute. (Read that aloud: honest 25-delivery averages typically wobble about a minute.)
t = (52 − 50) ⁄ 1 = 2.0. The sample sits two standard errors above the promise.
Cutoff: df = 24, two-sided at α = 0.05 → t* = 2.064.
Verdict: |2.0| < 2.064, so fail to reject H₀. Technology sharpens the picture: p ≈ 0.057, a whisker above 0.05. In words: the sample does not provide convincing evidence that the true average delivery time differs from 50 minutes.

Notice how carefully that last sentence is worded — and notice what it does not say. It does not say the average is 50. Hold that thought for Section 4, where the confidence interval will show exactly how much doubt survives the verdict.

Misconception check. "t = (52 − 50)/5 = 0.4 — why divide by SE instead of s?" Because the thing on trial is a sample mean, and sample means wobble by s∕√n, not by s. Dividing by s alone shrinks every t by a factor of √n, and nothing would ever get rejected. Compute SE first, on its own line, every single time.

Misconception check. "Can't I just use 1.96 like the z-table?" Not with a small sample. Using s in place of σ costs certainty, and the t-table charges for it: 2.064 at df 24, 2.262 at df 9. Slipping 1.96 in at n = 10 is borrowing precision you don't have — and it can flip a verdict (a t of 2.1 "rejects" against 1.96 but not against 2.262).

2. One tail or two? Choosing the alternative

The null hypothesis is always the same kind of statement: μ equals the claimed value. The alternative is where the question lives, and there are two honest shapes it can take.

Two-sided — Hₐ: μ ≠ μ₀. "Is the claim wrong, in either direction?" This is the default for auditing a promise: a courier that averages 47 minutes is just as much not-50 as one that averages 53.
One-sided — Hₐ: μ < μ₀ (or μ > μ₀). "We suspect a specific direction, and that direction was our question from the start." A watchdog that suspects inflated mpg claims cares about less than advertised; a training program is supposed to raise scores.

Mechanically the difference is one move: a one-sided test pours all of α into one tail, so its cutoff is closer in. At α = 0.05 that means the 90% column of the table (1.833, 1.753, 1.711) instead of the 95% column — and, when the data lean the suspected way, a one-sided p-value is half the two-sided one.

Worked example. A compact car model advertises 40 mpg highway. A consumer magazine — which stated up front that it suspects the real figure is lower — runs 16 standardized test drives: x̄ = 38.4 mpg, s = 3.2.

SE = 3.2 ⁄ √16 = 3.2 ⁄ 4 = 0.8. t = (38.4 − 40) ⁄ 0.8 = −2.0. df = 15.
One-sided test (Hₐ: μ < 40): cutoff −1.753. Since −2.0 falls beyond it, reject H₀ — technology gives p ≈ 0.032. Convincing evidence the true average is below 40 mpg.
Two-sided test (Hₐ: μ ≠ 40): cutoff ±2.131. Since |−2.0| < 2.131, fail to reject — p ≈ 0.064.

Same data, different question, different verdict. That is not a flaw; it's the reason the question must be locked in before the data arrive. The magazine's one-sided test was legitimate because the direction was the question, in writing, from the start.

Misconception check. "My x̄ came out below μ₀, so I'll run the one-sided 'less than' test — it's more likely to reject." That move — choosing the tail after peeking — secretly doubles your Type I error rate: you'd take the half-sized p whichever way the data happened to lean. The cure is a habit: pick your tail before you peek, and when in doubt, go two-sided. Reviewers, auditors, and instructors all read a one-sided test as a claim that you pre-registered the direction.

3. Paired data: subtract first

Now a new shape of data. An office of 10 staff members takes a two-week typing course, and everyone's speed is measured before and after. Two columns of numbers — but not two separate groups. The same ten people appear in both columns, and each before is linked to its own after. That's paired data: two measurements on the same individual (before/after, with/without, left/right), or on deliberately matched pairs.

Paired data gets a beautifully simple treatment. For each pair, compute the difference:

d = after − before — then run an ordinary one-sample t-test on the differences, testing H₀: μd = 0 ("no change on average").

Memory hook: "Pairs? Subtract first — then it's one sample." Everything from Section 1 applies verbatim, with d̄ playing x̄, sd playing s, and 0 playing μ₀ (df is still n − 1, where n is the number of pairs).

Why subtract? Because each person is their own control. Fast typists stay fast and slow typists stay slow; that person-to-person spread is huge, and it has nothing to do with whether the course worked. Subtracting cancels it, leaving only the change. Ignore the pairing and that spread floods back in, burying the signal.

Worked example. The office's words-per-minute, before and after:

The table below lists each staffer's speeds and the difference d = after − before.

Staffer	1	2	3	4	5	6	7	8	9	10
Before	38	45	36	50	41	44	39	47	42	48
After	47	44	44	50	47	46	43	51	46	52
d	9	−1	8	0	6	2	4	4	4	4

d̄ = (9 − 1 + 8 + 0 + 6 + 2 + 4 + 4 + 4 + 4) ⁄ 10 = 40 ⁄ 10 = 4 wpm.
sd: the squared deviations from 4 sum to 90, so sd² = 90 ⁄ 9 = 10, and sd = √10 ≈ 3.162.
SE = 3.162 ⁄ √10 = 1 exactly (√10 divided by √10). t = (4 − 0) ⁄ 1 = 4.0, at df = 9.
Verdict: two-sided at α = 0.05, t* = 2.262 → 4.0 is far beyond → reject H₀ (technology: p ≈ 0.003). In words: convincing evidence the course changed average typing speed — and the change is an improvement of about 4 words per minute.

That last clause matters. A verdict without an effect size is half a report: "significant" tells you the change is real; "about 4 wpm" tells you whether anyone should care.

Misconception check. "Two columns means a two-sample comparison." Not when the columns are linked. One question decides everything: is each number in column one attached to one specific number in column two? Attached (same person, same van, same matched pair) → paired → subtract first. Genuinely separate groups → the two-sample procedures, which arrive next week. Running the wrong one isn't a style error — it throws away (or invents) information.

4. Tests and intervals: two tools, one answer

Here is the connection that makes the whole inference half of the course feel like one idea instead of four. Week 11's confidence interval and this week's two-sided test are the same machine, run in opposite directions:

The 95% confidence interval lists every value of μ the data find plausible.
The two-sided test at α = 0.05 asks whether one particular value, μ₀, is on that list.

The duality: μ₀ inside the 95% CI ⟺ fail to reject H₀: μ = μ₀ at α = 0.05 (two-sided). μ₀ outside ⟺ reject. Matching confidence level and α, same s, same √n, same t* — the two can never disagree.

Worked example. Back to the courier: x̄ = 52, SE = 1, df = 24, t* = 2.064. The 95% confidence interval is 52 ± 2.064(1) = 52 ± 2.064 → (49.936, 54.064), or about 49.94 to 54.06 minutes.

Is the promised 50 on the list? Yes — so the two-sided test had to fail to reject, and it did. But now read what the interval adds that the bare verdict hides: 54.06 is on the list too. The data that "let the promise survive" are equally consistent with deliveries averaging four minutes slower than promised. That's the width of the reasonable doubt — and it is exactly why a fail-to-reject verdict is not a confirmation. The interval doesn't just repeat the test's answer; it shows how much the test couldn't see.

Memory hook: "The interval lists every claim that would survive; the test tries one." Professional practice reports both: the verdict (with its p-value) and the interval (with its units).

Misconception check. "We failed to reject, so we've shown the mean is 50." Look at the interval: 49.94 to 54.06. Failing to reject 50 also means the test would fail to reject 51, 52.5, and 54 — you can't have "proven" all of them. Not guilty ≠ innocent: the evidence didn't clear the bar, and that is the entire content of the verdict. When a company turns "fail to reject" into "independent study confirms our claim," it is buying an alibi from a test that never issues certificates of innocence. (This week's discussion hands you a real version of this move to argue about.)

5. The fine print: conditions

The t-test inherits Week 11's conditions word for word — because it's the same machinery:

Random & representative. The sample came from a chance-based method aimed at the population the claim is about. A t-test on a convenience sample is precise arithmetic on a biased number — Week 1 never leaves.
Independent observations. One measurement doesn't influence another.
Roughly normal population, or a large enough n. The t procedures tolerate mild non-normality, and the Central Limit Theorem (Week 10) covers larger samples. But at n = 10 or 16, a strongly skewed histogram or a single wild outlier can manufacture — or destroy — a rejection all by itself. For paired data, these conditions apply to the differences.

The working habit: before any small-sample t-test, make a quick dot plot or histogram of the data (or of the differences) and look. If one value dominates, say so in your report and think hard about where it came from; don't run the test silently and let the p-value launder it.

In practice (your tools). Build the courier test in a spreadsheet one cell at a time: =AVERAGE(A2:A26) → 52, =STDEV(A2:A26) → 5, =STDEV(A2:A26)/SQRT(25) → 1 for the SE, then the t by subtraction and division. The p-value is one function: =T.DIST.2T(2, 24) → 0.0569 (feed it |t| and df); one-sided uses =T.DIST(-2, 15, TRUE) → 0.0320 for a left tail. The course t-table itself is =T.INV.2T(0.05, 24) → 2.064, three rows at a time. Paired data collapses to one call: =T.TEST(B2:B11, C2:C11, 2, 1) → 0.003 — tails = 2, type = 1 (paired) — which should match your hand-built t = 4.0. A Desmos-class tool's tdist(24) will draw the curve and shade beyond ±2 so you can see the 0.057. And a warning that is graded in this week's lab: chatbots asked to run t-tests routinely divide by s instead of s∕√n, or quietly use 1.96 instead of the t cutoff. The tool drafts; you judge.

Key terms

One-sample t-test — the procedure for testing a claim H₀: μ = μ₀ about a population mean, using t = (x̄ − μ₀)/SE with df = n − 1.
Claimed value (μ₀) — the number the null hypothesis asserts for the population mean.
Test statistic (t) — the distance between x̄ and μ₀, measured in standard errors.
Standard error (SE = s∕√n) — the typical sample-to-sample wobble of x̄, estimated from the data.
Degrees of freedom (df = n − 1) — the row of the t-table the test uses; for paired data, pairs minus one.
Critical value (t*) — the cutoff |t| must beat for a rejection at the chosen α.
Two-sided alternative (μ ≠ μ₀) — "wrong in either direction"; the honest default.
One-sided alternative (μ < μ₀ or μ > μ₀) — a direction-specific question, legitimate only when chosen before the data.
Paired data — two linked measurements per individual (or matched pair), analyzed as one sample of differences.
Mean difference (μd) — the population average of the paired differences; H₀ usually says μd = 0.
Test ↔ interval duality — μ₀ inside the 95% CI ⟺ fail to reject at α = 0.05 (two-sided); outside ⟺ reject.
Conditions for inference — random & representative, independent, roughly normal or large n (applied to differences for paired data).

Summary

A claimed average gets a full trial: H₀: μ = μ₀, then t = (x̄ − μ₀)/(s∕√n) counts standard errors between data and claim, then the t-table at df = n − 1 supplies the cutoff, then the verdict is stated in context. Claim → ruler → distance → cutoff → verdict.
Two-sided tests at α = 0.05 read the 95% column (2.262 / 2.131 / 2.064); one-sided tests at 5% read the 90% column (1.833 / 1.753 / 1.711). The alternative is chosen before the data — pick your tail before you peek.
Paired data (same individual, twice) → subtract first: a one-sample t-test on the differences, df = pairs − 1. Pairing cancels person-to-person spread; ignoring it buries the signal.
Duality: the 95% CI is the list of every μ₀ a two-sided α = 0.05 test would let survive. Inside ⟺ fail to reject. The interval also shows the doubt's width — which is why fail to reject ≠ accept.
Conditions before conclusions: random, independent, roughly normal or n large enough — checked on the differences for paired data. Software verifies none of it.

Check your understanding

Work these before looking at the answers — they're ungraded, and the answers (with reasoning) are at the bottom of the page.

A phone-repair shop advertises that its average screen replacement takes 30 minutes. A skeptical reviewer times a random sample of 16 replacements: x̄ = 33 minutes, s = 8. At α = 0.05, test the shop's claim (two-sided). Show SE, t, df, the cutoff, and the verdict in context.
In Question 1, the shop's owner reads your verdict and announces: "An independent review confirmed our 30-minute average." In two or three sentences, explain what's wrong with that announcement — and name the number from your test that makes the point sharpest.
A warehouse measures 16 order-pickers' items-per-hour before and after a new scanner layout. The 16 differences (after − before) have d̄ = 2.5 and sd = 4. (a) Why is this a paired analysis? (b) Run the two-sided test at α = 0.05 and state the conclusion with the effect size.
A streaming-device maker claims its box completes setup in 24 seconds on average. From a random sample of setups, a 95% confidence interval for the true mean setup time is (18.9, 23.5) seconds. Without computing anything new: (a) what verdict does the two-sided α = 0.05 test of "the average is 24" give, and why? (b) What about a claimed average of 20 seconds?

Answers (worked reasoning — read after attempting)

Hypotheses: H₀: μ = 30 vs. Hₐ: μ ≠ 30. SE = 8 ⁄ √16 = 8 ⁄ 4 = 2. t = (33 − 30) ⁄ 2 = 1.5, with df = 15. Two-sided cutoff at α = 0.05: t* = 2.131. Since |1.5| < 2.131, fail to reject H₀ (technology gives p ≈ 0.154). In context: the sample does not provide convincing evidence that the true average replacement time differs from 30 minutes.
"Fail to reject" is a not-guilty verdict, not a certificate of innocence: the evidence didn't clear the bar, which is different from proving the claim. The sharpest number is the interval the same data produce: 33 ± 2.131(2) = (28.7, 37.3) minutes — a true average of 35 or even 37 minutes is also perfectly consistent with this sample, so the data "confirm" 30 only in the sense that they can't rule it out (along with many slower averages).
(a) Each picker appears twice — their after is linked to their own before — so the analysis runs on each picker's difference; the person-to-person speed spread cancels out. (b) SE = 4 ⁄ √16 = 1. t = (2.5 − 0) ⁄ 1 = 2.5, df = 15. Two-sided cutoff 2.131 → 2.5 > 2.131 → reject H₀ (technology: p ≈ 0.025). Convincing evidence the layout changed average picking speed — an improvement of about 2.5 items per hour.
(a) Reject. 24 lies outside the 95% CI (24 > 23.5), and a value outside the 95% interval is exactly a value a two-sided α = 0.05 test rejects — the duality, no arithmetic needed. The data are not consistent with a 24-second average (every plausible mean is faster). (b) Fail to reject. 20 lies inside (18.9 ≤ 20 ≤ 23.5), so 20 is on the list of plausible means — the test lets it survive. Note what you did not conclude: not that the mean is 20 — the list also contains 19.5 and 23.

Next up in this module: the slide deck if you want the lecture's visual walk-through, then the Readings & Resources page for optional deeper dives — and when you're ready to practice, Lecture Tutorial 14 turns your chatbot into your personal t-test coach.