scipy.stats — distributions, hypothesis tests: Visual walkthrough

Definition What the whole derivation assumes (spell it out first)

The exact Student-t result below is not free — it rides on two assumptions about the raw numbers $x_{i}$ :

Independence ::: each measurement is drawn without influencing any other — one box's weight tells you nothing extra about the next. (Think: fresh, separate draws from numpy.random, not a running total.)
Normality ::: the population the $x_{i}$ come from is a Normal $(\mu,\sigma^2)$ — a symmetric bell with true centre $μ$ and true spread $σ$ .

WHY it matters: only under independent draws does the variance of the mean shrink cleanly to $σ^{2} / n$ (Step 4), and only under normal draws does the standardized gap follow exactly a Student-t (Step 5–6). If $n$ is large, the Central Limit Theorem rescues the normality part; independence you must earn by how you collect the data.

Step 1 — Start with a bag of numbers and one claim

WHAT. We have five measured values — say the weight of cereal boxes labelled "5.0 kg": $x_{1}, x_{2}, x_{3}, x_{4}, x_{5} .$ Each $x_{i}$ is one number the machine spat out, an independent draw from a Normal population (the two assumptions above). Someone claims the true average weight is exactly $5.0$ . We call that claim the null hypothesis and write it $H_{0} : μ = μ_{0}$ , with $μ_{0} = 5.0$ .

WHY. Before we can measure "how surprising" our data is, we need something to be surprised relative to. $μ_{0}$ is that reference line. Without a claim, there is nothing to test.

PICTURE. The dots are our five numbers scattered on a number line. The plum line is the claim $μ_{0}$ . The question the whole page answers: are these dots too far from the plum line to be luck?

Step 2 — Collapse the bag to one summary: the sample mean

WHAT. Add all the numbers and divide by how many: $\overset{x}{ˉ} = \frac{1}{n} (x_{1} + x_{2} + \dots + x_{n}) .$

Reading the symbols where they sit: $\overset{x}{ˉ}$ (say "x-bar") is our best single guess for the centre; the $\frac{1}{n}$ out front shares the total equally among the $n$ values; the sum inside piles them up.

WHY. We cannot compare five dots to one claim directly. So we squeeze the whole bag into a single balance point — the mean. Now it's one number ( $\overset{x}{ˉ}$ ) versus one number ( $μ_{0}$ ).

PICTURE. The orange triangle is the balance point where the dots would sit level on a seesaw. That balance point is $\overset{x}{ˉ}$ . The gap between it and the plum $μ_{0}$ line is the raw "effect" we're worried about.

Step 3 — Measure the spread, because a gap only matters relative to noise

WHAT. Compute the sample standard deviation $s$ : $s = \frac{1}{n - 1} \sum_{i = 1}^{n} (x_{i} - \overset{x}{ˉ})^{2} .$

Term by term: $(x_{i} - \overset{x}{ˉ})$ is how far one dot sits from the balance point; squaring $(\cdot)^{2}$ makes every distance positive and punishes big misses harder; $\sum$ adds those squared misses up; $\frac{1}{n - 1}$ averages them (we divide by $n - 1$ , not $n$ , because one degree of "wiggle" was already spent computing $\overset{x}{ˉ}$ ); the outer brings the units back from "squared kg" to "kg".

WHY. A gap of $0.2$ is huge if the dots are tightly bunched, but trivial if they're scattered wildly. We literally cannot judge the gap until we know the typical spread. $s$ is that yardstick. It estimates the true, unknown $σ$ from Step 1.

PICTURE. The teal band shows $\overset{x}{ˉ} \pm s$ : the typical wobble of a single measurement. Notice how the same gap looks alarming against a narrow band and harmless against a wide one.

Step 4 — Shrink the noise of the mean itself: the standard error

WHAT. The mean $\overset{x}{ˉ}$ is far steadier than any single dot. Its own spread is the standard error: $SE = \frac{s}{n} .$

Here $s$ is the spread of one measurement (Step 3), and $n$ in the denominator shrinks it because we averaged $n$ of them.

WHY. We're comparing $\overset{x}{ˉ}$ to $μ_{0}$ , so the ruler must describe how much $\overset{x}{ˉ}$ jitters, not how much a single $x_{i}$ jitters. Averaging cancels ups against downs: because the draws are independent, the variance of a mean of $n$ values is $σ^{2} / n$ , so its standard deviation is $σ / n$ . Estimating $σ$ by $s$ gives $s / n$ . (If the draws were not independent, the cross-terms would not vanish and this clean $n$ shrinkage would break.) This is exactly the $n$ shrinkage the Central Limit Theorem promises.

PICTURE. Left: the wide teal band, the spread of one measurement. Right: the much narrower orange band, the spread of the mean — same data, shrunk by $n$ . More data ⇒ tighter guess.

Step 5 — Standardize: turn the gap into "how many rulers away"

WHAT. Divide the gap by the standard error: $t = \frac{x ˉ - μ _{0}}{s / n} = \frac{x ˉ - μ _{0}}{SE} .$

Top ( $\overset{x}{ˉ} - μ_{0}$ ): the raw gap between our balance point and the claim. Bottom ( $SE$ ): the size of one "ruler" of mean-jitter. The whole fraction $t$ answers: how many standard errors does the gap span?

WHY. A gap in kilograms can't be judged on its own — is $0.2$ kg a lot? By dividing by SE we get a unit-free number. This is the identical z-trick from the parent note (center, then scale), applied to the mean. The result $t$ means the same thing for cereal boxes, exam scores, or reaction times.

PICTURE. The gap is now drawn as a count of SE-rulers laid end to end. Here the gap is a small fraction of one ruler — a tiny $t$ .

Step 6 — Put $t$ on its distribution and read the tail

WHAT. Under the claim $H_{0}$ , the value $t$ we just built is one draw from $T_{n - 1}$ . The two-sided p-value is the total area in both tails beyond $∣ t ∣$ : $p = 2 \cdot P (T_{n - 1} > ∣ t ∣) = 2 * stats.t.sf(abs(t), df=n-1) .$

$∣ t ∣$ (absolute value) folds a negative $t$ to positive, because "the mean is too low" is just as surprising as "too high." The $sf$ is the upper-tail area (survival = $1 - CDF$ ); the factor $2$ counts the mirror-image lower tail.

WHY. The p-value asks: if the claim were true, how often would pure luck hand us a $t$ this far from zero or farther? That "at least this extreme" is exactly the shaded tail area. Small area ⇒ our data lives out in the boonies ⇒ the claim looks wrong. This is p-values and Significance made literal.

PICTURE. The Student-t bell centred at $0$ . Our $t$ sits just off centre; its mirror $- t$ on the left. Both burnt-orange shaded slivers together are $p$ . Two-sided means both slivers count.

Definition One-sided vs two-sided — pick your tail

before looking stats.ttest_1samp always returns the two-sided p-value (both tails). Sometimes your question is directional — "is the mean greater than $μ_{0}$ ?" — and you want a one-sided test:

Right-tailed ( $H_{1} : μ > μ_{0}$ ) ::: $p_{right} = P (T_{n - 1} > t) = stats.t.sf(t, df=n-1)$ — shade only the right tail beyond the signed $t$ .
Left-tailed ( $H_{1} : μ < μ_{0}$ ) ::: $p_{left} = P (T_{n - 1} < t) = stats.t.cdf(t, df=n-1)$ — shade only the left tail.
Link ::: when $t$ falls on the side your alternative predicted, the one-sided p is exactly half the two-sided p; it is more powerful but you must commit to the direction before seeing the data, or you cheat.

res = stats.ttest_1samp(data, popmean=5.0)   # two-sided
p_right = res.pvalue/2 if res.statistic > 0 else 1 - res.pvalue/2
# (SciPy ≥1.6 also accepts alternative="greater"/"less" directly)

Worked example Run it on real numbers

from scipy import stats
data = [5.1, 4.9, 5.3, 5.0, 4.8]
stats.ttest_1samp(data, popmean=5.0)
# statistic = 0.2182..., pvalue = 0.8380...

$\overset{x}{ˉ} = 5.02$ , $s \approx 0.2280$ , $SE \approx 0.1020$ , so $t = (5.02 - 5.0) /0.1020 \approx 0.218$ . The big p-value ( $0.84$ ) means: a gap this small is completely ordinary luck. Fail to reject — no evidence the boxes miss $5.0$ kg.

Step 7 — Edge & degenerate cases (never leave the reader stranded)

WHAT & WHY, case by case. The formula has a denominator and an area — both can misbehave.

Definition The corners of the formula

Perfect fit, $\overset{x}{ˉ} = μ_{0}$ ::: numerator $= 0$ so $t = 0$ ; the tail area is the whole half on each side ⇒ $p = 1$ . Maximum "no surprise." This is the parent's statistic=-0.0, pvalue=1.0.
Zero spread, $s = 0$ (all values identical) ::: SE $= 0$ , so $t = \pm \infty$ if the shared value $\neq = μ_{0}$ (infinite surprise, $p \to 0$ ), and is undefined ( $0/0$ ) if it equals $μ_{0}$ . scipy returns nan for the truly indeterminate case.
$n = 1$ ::: then $n - 1 = 0$ degrees of freedom and $s$ can't even be computed ( $0/0$ ). One data point can never test a spread — the test is undefined. You need at least two.
Sign of $t$ ::: $t > 0$ means $\overset{x}{ˉ}$ sits above the claim, $t < 0$ below. Two-sided $p$ uses $∣ t ∣$ , so both signs give the same p-value — the mirror symmetry of the bell. A one-sided test, by contrast, depends on the sign.
Assumptions broken ::: if the draws are dependent (Step 4's $n$ fails) or strongly non-normal with small $n$ (Step 5's exact t-shape fails), the printed p-value is untrustworthy — reach for a non-parametric or resampling test instead.
Huge $n$ ::: SE $\to 0$ , so even a microscopic gap yields a giant $t$ and tiny $p$ . Statistically significant need not mean practically important — always look at the raw gap $\overset{x}{ˉ} - μ_{0}$ too.

PICTURE. Four mini-panels: (a) $t = 0$ centred, all area, $p = 1$ ; (b) $s = 0$ arrow flying to the edge, $p \to 0$ ; (c) equal-and-opposite $t$ shading the same total area; (d) a needle-thin bell from huge $n$ where a tiny gap already reaches the tail.

The one-picture summary

Everything on this page is one pipeline: bag of independent Normal numbers → one gap → one ruler → one standardized $t$ → one tail area → the p-value.

Recall Feynman retelling — say it to a 12-year-old

A cereal box says "5 kilos." You weigh five boxes — each weighed on its own, so one reading never leans on another (that's the independence assumption), and the factory's weights wobble in a nice symmetric bell (the normal assumption). First you find their average (the balance point of the dots) — that's your one honest guess for the truth. It's a little off from 5. But "a little off" means nothing until you know how jumpy the scale is, so you measure the spread of the five weights. Then here's the clever bit: the average of five weights wobbles much less than any single weight, so you shrink the spread by dividing by the square root of five — that's your real ruler, the standard error. Now you ask: how many rulers is my average away from 5? That count is $t$ . Finally you look at a bell curve of "how often plain luck gives a $t$ this big," and you shade the far tails. That shaded sliver is the p-value. Big sliver: "eh, luck, boxes are fine." Tiny sliver: "no way that's luck — the factory is cheating." That is the entire test, and stats.ttest_1samp just does these five squeezes for you.

Flashcards

Why do we divide the gap by SE instead of by

s

We're judging the wobble of the mean

\overset{x}{ˉ}

, not of a single value; the mean's spread is

s / n

, so SE is the correct ruler.

What is

t

when

\overset{x}{ˉ} = μ_{0}

, and the resulting p-value?

t = 0

; the entire half-bell lies beyond it on each side, so the two-sided

p = 1

(maximum no-surprise).

What does

T_{ν}

denote?

A random variable following a Student-t distribution with

ν

degrees of freedom — a heavier-tailed bell that approaches the normal as

ν \to \infty

How do you get a one-sided p-value from a two-sided one?

When

t

falls on the side your alternative predicted, the one-sided p is exactly half the two-sided p; otherwise it is

1

minus that half.

Why does a two-sided p-value use

∣ t ∣

A mean too high and a mean too low are equally surprising; folding to

∣ t ∣

and doubling the upper tail counts both mirror-image extremes.

Why can't a one-sample t-test run on

n = 1

With one value

s

is undefined and there are

n - 1 = 0

degrees of freedom; a single point carries no information about spread.

What two assumptions make the exact Student-t valid?

The

x_{i}

are independent draws from a Normal $(μ, σ^{2})$ population; independence gives the clean

σ^{2} / n

variance, normality gives the exact t-shape.