scipy.stats — distributions, hypothesis tests: Worked examples

The scenario matrix

Every question this topic throws at you falls into one of these cells. The worked examples below each carry a [cell] tag so you can see the whole grid gets covered.

#	Case class	What makes it distinct	Covered by
A	Left tail — `.cdf`	want $P (X \leq x)$ , area to the left	Ex 1
B	Right tail — `.sf`	want $P (X > x)$ , upper tail	Ex 2
C	Middle band — cdf minus cdf	want $P (a \leq X \leq b)$	Ex 3
D	Inverse / quantile — `.ppf`	given probability, find the cutoff value	Ex 4
E	Degenerate: zero effect	$\overset{x}{ˉ} = μ_{0}$ , so $t \approx 0$ , $p \to 1$	Ex 5
F	Two-sample difference	clear gap → tiny $p$	Ex 6
G	Categorical counts — chi-square	discrete faces, not a continuous $x$	Ex 7
H	Boundary probabilities	$p = 0$ and $p = 1$ give $\pm \infty$	Ex 8
I	Word problem + one-sided	translate English → $H_{0}$ , pick tail	Ex 9
J	Exam twist: non-normal `scale`	`expon` where `scale = 1/λ`, not λ	Ex 10

We use the standard normal picture repeatedly, so pin it down first.

Figure — scipy.stats — distributions, hypothesis tests

Prerequisites worth a glance: Normal Distribution, p-values and Significance, Chi-square Distribution, Maximum Likelihood Estimation.

[cell A — left tail] IQ scores, $μ = 100, σ = 15$ . What fraction score at most 115? Forecast: 115 is $μ + 1 σ$ . The 68–95 rule says ~68% sit within $\pm 1 σ$ , so ~34% between 100 and 115, plus 50% below 100 → guess ≈ 84%.

Set up the object: stats.norm(loc=100, scale=15). Why this step? loc is the centre $μ$ , scale is the spread $σ$ — for norm they literally are $μ, σ$ .
We want $P (X \leq 115)$ — the mint left zone of figure s01 → use .cdf (this is $F (115)$ ). Why this step? "at most" = "up to and including" = left tail = the C in the CDF mnemonic.
Compute the z-score to place 115 on the standard bell: $z = \frac{115 - 100}{15} = 1$ — exactly one spread right of centre. Why this step? Standardizing (defined above) lets us read the answer off the one standard bell instead of a custom one.
So stats.norm(100,15).cdf(115) = area left of $z = 1$ = 0.8413. Why this step? The mint zone up to one spread right of centre is always 0.8413.

Verify: $z = 1$ exactly; my 84% forecast matched. Sanity: answer is between 0 and 1 ✓, and above 0.5 because 115 is above the mean ✓.

Worked example (4)

[cell D — inverse / quantile] What score marks the top 5% (the 95th percentile)? Forecast: the famous "1.645 for one-sided 5%" on the standard scale → back on IQ scale, $100 + 1.645 \times 15 \approx 124.7$ .

We show it two equivalent ways; pick one lane and stay in it.

Lane A — work on the scaled object directly. Given probability 0.95 below, ask for the value → use .ppf, the lavender reverse-arrow of s01: stats.norm(100,15).ppf(0.95) → 124.67. Why this step? .ppf(0.95) on the IQ-scaled object answers "which IQ score has 95% below it?" directly — no hand-standardizing.
Lane B — standardize first, then un-standardize. On the standard bell, stats.norm.ppf(0.95) = $z^{*} = 1.6449$ . Then invert the z-formula $z = \frac{x - μ}{σ}$ to get $x = μ + z^{*} σ = 100 + 1.6449 \times 15 = 124.67$ . Why this step? This exposes why the answer is 124.67 — it is exactly 1.6449 spreads right of centre.

Both lanes give the same 124.67; never mix a standard-bell ppf with a scaled-object one in the same line.

Verify: feed Lane A back — .cdf(124.67) should return ≈ 0.95 ✓ (round-trip check in VERIFY). Units: IQ points, plausibly above the mean ✓.

Worked example (5)

[cell E — degenerate, zero effect] One-sample t-test, data [5.1,4.9,5.3,5.0,4.8], claim mean $= 5.0$ . Forecast: the values scatter roughly symmetrically around 5.0, so $\overset{x}{ˉ}$ is basically 5.0 → $t \approx 0$ → $p \approx 1$ (no surprise at all).

Sample mean: $\overset{x}{ˉ} = (5.1 + 4.9 + 5.3 + 5.0 + 4.8) /5 = 25.1/5 = 5.02$ . Why this step? The t-statistic needs $\overset{x}{ˉ}$ to measure the gap from $μ_{0} = 5.0$ .
Sample standard deviation $s$ , defined as $s = \frac{1}{n - 1} \sum_{i} (x_{i} - \overset{x}{ˉ})^{2}$ . Deviations from $5.02$ are ${0.08, - 0.12, 0.28, - 0.02, - 0.22}$ ; their squares sum to $0.0064 + 0.0144 + 0.0784 + 0.0004 + 0.0484 = 0.148$ . Divide by $n - 1 = 4$ : $0.037$ ; square-root: $s = 0.037 \approx 0.1924$ . Why this step? $s$ estimates the spread $σ$ we don't know. Why divide by $n - 1$ , not $n$ ? We already "spent" one fact computing $\overset{x}{ˉ}$ from the data, so only $n - 1$ deviations are free to vary — this is the degrees of freedom $df = n - 1 = 4$ , and dividing by it removes bias.
Statistic $t = \frac{x ˉ - μ _{0}}{s / n} = \frac{0.02}{0.1924/ 5} = \frac{0.02}{0.0860} \approx 0.2325$ . Why this step? This is the z-trick with $s$ replacing $σ$ ; the tiny numerator forces $t$ near 0. (scipy reports $t \approx 0.2325$ ; the $\pm$ sign depends only on which side of 5.0 the mean falls.)
stats.ttest_1samp(data, 5.0) → $t \approx 0.2325$ , $p \approx 0.8278$ . Why this step? Two-sided $p = 2 P (T_{4} > ∣ t ∣)$ with $df = 4$ ; a small $∣ t ∣$ leaves almost all the mass beyond it.

Verify: $p \approx 0.83 ≫ 0.05$ → fail to reject — exactly what "the data agrees with the claim" should give. (The parent's -0.0 / 1.0 was a stylized illustration; here $\overset{x}{ˉ} = 5.02 \neq = 5.0$ so $t$ is small but nonzero.) VERIFY confirms $t, p$ .

Worked example (6)

[cell F — two-sample difference] Group a=[20,22,19,24,25], b=[28,27,30,26,29]. Do they differ? Forecast: the group means are $\overset{a}{ˉ} = 22$ and $\overset{ˉ}{b} = 28$ , a clean gap of $6$ , and every $b$ value beats every $a$ value → huge separation relative to within-group wobble → expect big $∣ t ∣$ , tiny $p$ .

stats.ttest_ind(a,b) compares two independent means. Why this step? Two separate samples, not one sample vs a fixed number → the independent two-sample test.
Pooled standard deviation. With equal-size groups, pool the two variances: $s_{a}^{2} = 6.5$ , $s_{b}^{2} = 2.5$ , so the pooled variance is $s_{p}^{2} = \frac{( n _{a} - 1 ) s _{a}^{2} + ( n _{b} - 1 ) s _{b}^{2}}{n _{a} + n _{b} - 2} = \frac{4 \cdot 6.5 + 4 \cdot 2.5}{8} = 4.5$ , giving $s_{p} = 4.5 \approx 2.121$ . Why this step? The pooled (default equal_var=True) test assumes both groups share one true spread, so it averages their sample variances to estimate it.
Statistic $t = \frac{a ˉ - b ˉ}{s _{p} 1/ n _{a} + 1/ n _{b}} = \frac{22 - 28}{2.121 1/5 + 1/5} = \frac{- 6}{2.121 \times 0.6325} = \frac{- 6}{1.3416} \approx - 4.47$ , with $df = n_{a} + n_{b} - 2 = 8$ . Why this step? Standardize the mean-gap by the pooled standard error; the sign is negative because $\overset{a}{ˉ} < \overset{ˉ}{b}$ .
Edge case — the variance assumption. If the spreads were doubtful you would pass equal_var=False for Welch's correction, which adjusts the degrees of freedom instead of pooling. Why this step? Here both groups have modest, similar spread so the default pooled test is fine; on wildly unequal spreads the pooled p-value can mislead.
Result (pooled default): $t \approx - 4.47$ , two-sided $p \approx 0.0021$ . Why this step? Two-sided $p = 2 P (T_{8} > ∣ t ∣)$ ; a $∣ t ∣$ of 4.47 on $df = 8$ leaves very little tail mass.

Verify: $p \approx 0.0021 < 0.05$ → reject "equal means" ✓. Direction: $t < 0$ says group b is higher, matching the eyeball ✓. Gap $\overset{ˉ}{b} - \overset{a}{ˉ} = 6$ ✓. VERIFY checks $t \approx - 4.47$ , $p \approx 0.0021$ , and the means.

Worked example (7)

[cell G — categorical counts] Is a die fair? 60 rolls give obs=[8,9,19,5,8,11], fair expects [10]*6. Forecast: face 3 shows up 19 times (expected 10) — that's a big overshoot. Suspect the die is loaded → smallish $p$ .

Statistic $\sum (O - E)^{2} / E$ over all six faces. Why this step? Here $x$ isn't a continuous value — it's counts in categories, so we compare observed vs expected counts, not tail areas of a bell. That's the chi-square test's job.
Term-by-term: $\frac{( 8 - 10 ) ^{2}}{10} + \frac{( 9 - 10 ) ^{2}}{10} + \frac{( 19 - 10 ) ^{2}}{10} + \frac{( 5 - 10 ) ^{2}}{10} + \frac{( 8 - 10 ) ^{2}}{10} + \frac{( 11 - 10 ) ^{2}}{10}$ = $0.4 + 0.1 + 8.1 + 2.5 + 0.4 + 0.1 = 11.6$ . Why this step? The $(19 - 10)^{2} /10 = 8.1$ term dominates — face 3 is the culprit.
stats.chisquare([8,9,19,5,8,11],[10]*6) → statistic $= 11.6$ , $p \approx 0.0407$ , df $= 5$ . Why this step? df = categories − 1 = 5; p is the upper-tail area of the Chi-square Distribution beyond 11.6.

Verify: my hand-sum 11.6 equals scipy's statistic ✓. $p \approx 0.041 < 0.05$ → die looks loaded ✓. VERIFY checks statistic and $p$ .

Worked example (9)

[cell I — word problem, one-sided] A factory claims bolts are at least 50 mm. A sample of 8 has $\overset{x}{ˉ} = 49.6$ , $s = 0.5$ . At $α = 0.05$ , are bolts too short? Forecast: $\overset{x}{ˉ}$ is below the claim by 0.4 mm with SE $= 0.5/ 8 \approx 0.177$ , so $t \approx - 2.26$ . One-sided → guess $p$ around 0.03, borderline.

Frame it: $H_{0} : μ \geq 50$ vs $H_{1} : μ < 50$ — a one-sided (left) test because we only care if bolts are too short. Why this step? The English "at least" and the worry "too short" pick a direction; two-sided would waste half our sensitivity.
$t = \frac{49.6 - 50}{0.5/ 8} = \frac{- 0.4}{0.1768} = - 2.2627$ , with $df = n - 1 = 7$ . Why this step? Standardize the gap by the estimated SE $s / n$ ; the given $s = 0.5$ plays the role of the sample sd from Ex 5.
One-sided p $= P (T_{7} < t) =$ stats.t.cdf(-2.2627, df=7) → 0.0292. Why this step? Left tail only, so use .cdf (not doubled) — we test just the "too short" direction.

Verify: $p \approx 0.029 < 0.05$ → reject $H_{0}$ : bolts are significantly short ✓. Two-sided would give $2 \times 0.0292 = 0.0584 > 0.05$ (fail!) — showing why choosing the correct side matters. VERIFY checks the one-sided $p$ .

Worked example (10)

[cell J — exam twist, non-normal scale] Bus waits are exponential with rate $λ = 0.5$ buses/min. What fraction of waits exceed 3 minutes? Forecast: exponential mean $= 1/ λ = 2$ min; 3 min is 1.5 means, moderately long → guess ≈ 0.22.

Build the object: stats.expon(scale=2) because for expon the scale argument is $1/ λ = 1/0.5 = 2$ , not $λ$ . Why this step? scipy parametrizes expon by the mean-like scale; only when scale $= 1/ λ$ does its survival match $e^{- λ x}$ . Setting scale=0.5 would silently model the wrong distribution.
We want $P (X > 3)$ — "exceed" is an upper tail, the same coral-zone idea as Ex 2 but on an exponential curve → use .sf, giving stats.expon(scale=2).sf(3). Why this step? "exceed 3" = right tail = survival function; .cdf(3) would answer the opposite (waits shorter than 3).
Evaluate: survival of an exponential is the clean formula $P (X > x) = e^{- x / scale} = e^{- 3/2} = e^{- 1.5}$ , so stats.expon(scale=2).sf(3) → 0.2231. Why this step? The exponential's memoryless shape gives this exact closed form; plugging $x = 3, scale = 2$ lands on $e^{- 1.5}$ .

Verify: hand formula $e^{- 1.5} = 0.2231$ matches scipy ✓, and it sits between 0 and 1 as a probability must ✓. If you'd wrongly set scale=0.5 you'd get $e^{- 6} = 0.0025$ — off by ~90×, the classic exam mistake. VERIFY checks 0.2231.

Recall Quick self-quiz over the matrix

Which method for $P (X \leq x)$ ? ::: .cdf, i.e. $F (x)$ Which method for $P (X > x)$ ? ::: .sf, i.e. $1 - F (x)$ Given a percentile, find the value? ::: .ppf, i.e. $F^{- 1}$ What is a z-score? ::: $z = (x - μ) / σ$ — how many spreads $σ$ the value $x$ is from centre $μ$ norm.ppf(0) equals? ::: $- \infty$ For expon, scale equals? ::: $1/ λ$ , not $λ$ Chi-square statistic formula? ::: $\sum (O - E)^{2} / E$ Why divide by $n - 1$ for $s$ ? ::: one degree of freedom is spent computing $\overset{x}{ˉ}$ , leaving $df = n - 1$ ; it de-biases the estimate ttest_ind default assumption? ::: equal_var=True (pooled); use equal_var=False for Welch when spreads differ