Chapter 76 · Hypothesis Testing & Inference
z-Tests 🔔
When the sample is large or the outcome is yes/no, the normal distribution is the reference and the z-test is the workhorse. We build the one-sample z, the one-proportion z, and the two-proportion z from the formula, then use statsmodels' proportions_ztest on a real approval poll.
Statistics, Data Science and AI: A Visual Handbook · John Fisher · 2026
⚙️ Setup¶
In [1]:
import numpy as np, pandas as pd, matplotlib.pyplot as plt
from scipy import stats
# statsmodels = the R/SAS-style stats library (pre-installed on Colab): it computes the
# standard errors, test statistics, intervals, and post-hoc comparisons, so we write less by hand.
import statsmodels.api as sm
from statsmodels.formula.api import ols
from statsmodels.stats.proportion import proportions_ztest, confint_proportions_2indep, proportion_confint
from statsmodels.stats.multicomp import pairwise_tukeyhsd
from statsmodels.stats.weightstats import DescrStatsW, CompareMeans
from scipy.stats.contingency import association # Cramer's V in one call
IND="#4f46e5"; DEEP="#4338ca"; LIGHT="#818cf8"; INK="#1a2138"; GRID="#e6e9f2"; GREEN="#059669"; RED="#ef4444"; AMBER="#d97706"
plt.rcParams.update({"figure.facecolor":"white","axes.facecolor":"white","figure.dpi":110,"font.size":11,
"axes.edgecolor":GRID,"axes.grid":True,"grid.color":GRID,"axes.axisbelow":True,"axes.spines.top":False,
"axes.spines.right":False,"axes.titlesize":12,"axes.titleweight":"bold","legend.frameon":False})
BASE="https://raw.githubusercontent.com/johnfisher-ai/Statistics-Data-Science-AI-Visual-Book/main/data/"
rng = np.random.default_rng(76)
DEMO 1 · WHEN z, NOT t
Large samples and proportions
The z-test uses the normal distribution as the reference. It is the right tool when the population sigma is known, or when the sample is large enough that the sample sd is essentially sigma, and for proportions, where the variance is fixed by p itself.
In [2]:
mu0=100; n=400
sample=rng.normal(101.2, 15, n)
xbar=sample.mean(); s=sample.std(ddof=1); se=s/np.sqrt(n)
z=(xbar-mu0)/se; p=2*(1-stats.norm.cdf(abs(z)))
print(f"large-sample z-test of H0: mu=100 (n={n})")
print(f"xbar={xbar:.2f}, SE={se:.3f}, z={z:.2f}, two-sided p={p:.4f}")
print(f"compare t-test p (nearly identical at this n): {stats.ttest_1samp(sample,mu0).pvalue:.4f}")
large-sample z-test of H0: mu=100 (n=400) xbar=101.67, SE=0.728, z=2.29, two-sided p=0.0221 compare t-test p (nearly identical at this n): 0.0226
At large n the t and z tests agree to the decimal, the t distribution has essentially converged to the normal. So z is the natural default for big samples and, especially, for proportions, where the standard error is determined entirely by the proportion under H0.
DEMO 2 · THE ONE-PROPORTION z-TEST
Does a rate differ from a claimed value?
For a yes/no outcome we test H0: p = p0. Under H0 the standard error is sqrt(p0(1-p0)/n), a KNOWN number, so z = (phat - p0)/SE is exactly standard normal for large n. We build it once from the formula, then confirm with statsmodels' proportions_ztest, the one-line library version.
In [3]:
successes, n, p0 = 545, 1000, 0.5
phat=successes/n; se0=np.sqrt(p0*(1-p0)/n)
z=(phat-p0)/se0; p=2*(1-stats.norm.cdf(abs(z)))
print(f"by hand: phat={phat:.3f}, SE0={se0:.4f}, z={z:.2f}, two-sided p={p:.4f}")
# the library way: prop_var=p0 tells it to use the NULL value for the standard error
z2,p2 = proportions_ztest(successes, n, value=p0, prop_var=p0)
print(f"statsmodels: z={z2:.2f}, p={p2:.4f} (identical)")
print(f"success-failure check: n*p0={n*p0:.0f}, n*(1-p0)={n*(1-p0):.0f} (both >= 10? OK)")
by hand: phat=0.545, SE0=0.0158, z=2.85, two-sided p=0.0044 statsmodels: z=2.85, p=0.0044 (identical) success-failure check: n*p0=500, n*(1-p0)=500 (both >= 10? OK)
Note the standard error uses p0 (the null value), not p̂, because we compute everything in the world where H0 is true. A 54.5% result in 1,000 trials is about 2.8 SEs from 50%, easily significant. Once you understand the formula, proportions_ztest is the practical tool.
DEMO 3 · THE TWO-PROPORTION z-TEST
Comparing two groups' rates
To compare two proportions we test H0: pA = pB. Under that null the best estimate of the common rate is the POOLED proportion, and the SE uses it. Again we show the formula once, then the one-line statsmodels call.
In [4]:
sA,nA = 120, 800; sB,nB = 160, 820
pA,pB = sA/nA, sB/nB; pool=(sA+sB)/(nA+nB)
se=np.sqrt(pool*(1-pool)*(1/nA+1/nB))
z=(pB-pA)/se; p=2*(1-stats.norm.cdf(abs(z)))
print(f"by hand: pA={pA:.3f}, pB={pB:.3f}, pooled={pool:.3f}, z={z:.2f}, p={p:.4f}")
# the library way: pass the two counts and the two sample sizes
z2,p2 = proportions_ztest([sB, sA], [nB, nA])
print(f"statsmodels: z={z2:.2f}, p={p2:.4f} (identical, pooled SE by default)")
by hand: pA=0.150, pB=0.195, pooled=0.173, z=2.40, p=0.0163 statsmodels: z=2.40, p=0.0163 (identical, pooled SE by default)
The pooled standard error is the signature of the two-proportion test: under H0 both groups share one rate, so we estimate it from the combined data. This is the engine behind comparing conversion rates, approval by subgroup, and pass rates between sites, and proportions_ztest handles it in a line.
REAL-WORLD EXAMPLE · APPROVAL POLL
Does approval beat 60%, and does it differ by age?
A poll of 820 respondents (z-tests--approval_poll.xlsx) records whether each approves of a policy. The campaign claims approval is above 60%. We explore first, then use statsmodels for a one-proportion z-test of the claim and a two-proportion z-test comparing the youngest and oldest age bands.
In [5]:
try: poll = pd.read_excel("../../data/z-tests--approval_poll.xlsx", sheet_name="Poll")
except FileNotFoundError: poll = pd.read_excel(BASE+"z-tests--approval_poll.xlsx", sheet_name="Poll")
# EXPLORE FIRST: size, missing, overall rate, and approval by age band
print("shape:", poll.shape, "| missing:", poll.isna().sum().sum())
print(f"overall approval = {poll.approve.mean():.1%}")
print(poll.groupby("age_group").approve.agg(["size","mean"]).round(3))
shape: (820, 5) | missing: 0
overall approval = 64.3%
size mean
age_group
18-34 270 0.707
35-54 328 0.640
55+ 222 0.568
In [6]:
# one-proportion z-test of H0: p = 0.60 (library, null value for the SE)
a=poll.approve
z,p = proportions_ztest(a.sum(), len(a), value=0.60, prop_var=0.60)
print(f"approval {a.mean():.1%} vs claim 60%: z={z:.2f}, two-sided p={p:.4f}",
"-> reject H0 (higher)" if p<0.05 else "-> fail to reject")
phi = lambda pp: 2*np.arcsin(np.sqrt(pp))
print(f"Cohen h vs 60% = {phi(a.mean())-phi(0.60):+.2f} (proportion effect size; |0.2| small, |0.5| medium, |0.8| large)")
# two-proportion z-test: youngest vs oldest band
y=poll[poll.age_group=="18-34"].approve; o=poll[poll.age_group=="55+"].approve
z2,p2 = proportions_ztest([y.sum(), o.sum()], [len(y), len(o)])
print(f"young {y.mean():.1%} (n={len(y)}) vs old {o.mean():.1%} (n={len(o)}): z={z2:.2f}, p={p2:.4f}",
"-> reject H0 (differ by age)" if p2<0.05 else "-> fail to reject")
fig,ax=plt.subplots(figsize=(6.4,3.1))
ax.bar(["18-34","55+"],[y.mean(),o.mean()],color=[IND,LIGHT],width=0.55)
ax.axhline(0.60,color=INK,ls="--",lw=1.4,label="claim 60%")
ax.set_ylim(0,0.85); ax.set_ylabel("approval rate"); ax.set_title("Approval by age vs the 60% claim"); ax.legend()
for i,v in enumerate([y.mean(),o.mean()]): ax.text(i,v+0.02,f"{v:.0%}",ha="center",fontweight="bold")
plt.tight_layout(); plt.show()
approval 64.3% vs claim 60%: z=2.49, two-sided p=0.0126 -> reject H0 (higher) Cohen h vs 60% = +0.09 (proportion effect size; |0.2| small, |0.5| medium, |0.8| large) young 70.7% (n=270) vs old 56.8% (n=222): z=3.22, p=0.0013 -> reject H0 (differ by age)
Overall approval is about 64.3%, and the one-proportion z-test (z ≈ 2.49, p ≈ 0.013) rejects the 60% claim: approval is genuinely higher. The age gap is sharper still, the youngest band (70.7%) versus the oldest (56.8%) gives z ≈ 3.22, p ≈ 0.001, a clear, real difference. Large n plus the success-failure condition make the normal approximation trustworthy here.
✅ z-tests
The z-test uses the normal as its reference: right for large-sample means and for proportions, where the null fixes the variance. The one-proportion test uses SE from p0; the two-proportion test uses a POOLED rate, both available in one line from statsmodels' proportions_ztest. On real poll data, approval beats 60% (z=2.49, p=0.013) and differs by age (z=3.22, p=0.001). When sigma is unknown and n is small, the t-test, next chapter, is the honest choice.
Statistics, Data Science and AI: A Visual Handbook · © 2026 John Fisher