The one-proportion z-test

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z = (p̂ − p₀)/√(p₀(1 − p₀)/n): the test’s SE is built from the null value p₀, not from p̂ — the line the exam marks — then the critical value or the p-value from the cumulative table.

The SE comes from the null

A test of $H_{0} : p = p_{0}$ asks how far $\overset{p}{^}$ sits from $p_{0}$ if $H_{0}$ is true. If it is true, the SE of $\overset{p}{^}$ is known exactly — it is $p_{0} (1 - p_{0}) / n$ , with no estimate needed. So the statistic is

$z = \frac{p ^ - p _{0}}{\frac{p _{0} ( 1 - p _{0} )}{n}}$

This is the one place proportions differ from means in shape. The interval (L12.1) has no null value and must use $\overset{p}{^}$ in its SE; the test has $p_{0}$ and uses it. Writing $\overset{p}{^}$ in a test’s denominator is the error the marker looks for. Conditions, under the null: $n p_{0} \geq 5$ and $n (1 - p_{0}) \geq 5$ .

Worked example — does the error rate exceed the tolerable 10 %?

Maple Ledger’s tolerable invoice error rate is 10 %. The audit found 60 errors in 500 invoices. Is the true rate above 10 %? $α = 0.05$ .

Step 1. $H_{0} : p = 0.10$ , $H_{a} : p > 0.10$ , where $p$ is the error rate for all invoices. Upper-tailed, from exceed.

Step 2. $\overset{p}{^} = 0.12$ .

$SE_{0} = \frac{0.10 \times 0.90}{500} = 0.00018 = 0.01342, z = \frac{0.12 - 0.10}{0.01342} = 1.49$

Step 3 — critical value. Upper tail at 5 %: $z_{0.05} = 1.645$ . Step 3 — p-value. $p = P (Z > 1.49) = 1 - Φ (1.49) = 1 - 0.9319 = 0.0681$ .

Step 4. $1.49 < 1.645$ , equivalently $0.0681 > 0.05$ : do not reject $H_{0}$ at the 5 % level. There is insufficient evidence, at the 5 % level, that the error rate exceeds 10 %.

Say what that does and does not mean: the sample rate of 12 % is above the benchmark, and the interval $(0.0915, 0.1485)$ from L12.1 reaches almost to 15 %. The rate may well exceed 10 %; the test simply did not establish it with 500 invoices. At $α = 0.10$ the same $z$ would reject ( $1.49 > 1.282$ ; $0.0681 < 0.10$ ) — the p-value lets the reader see that.

Two-sided and lower-tailed versions

Does the rate differ from 10 %? — $H_{a} : p \neq = 0.10$ , compare $∣ z ∣ = 1.49$ with 1.96 and double the tail: $p = 2 \times 0.0681 = 0.1362$ . Is the rate below 10 %? — $H_{a} : p < 0.10$ ; with $\overset{p}{^} = 0.12$ the estimate leans the wrong way, $p = Φ (1.49) = 0.9319$ , and $H_{0}$ stands without further arithmetic. The tail comes from the question, as always.

The interval and the test disagree slightly, by design

The interval’s SE was 0.01453 (from $\overset{p}{^}$ ); the test’s is 0.01342 (from $p_{0}$ ). They differ because they answer different questions — how precise is my estimate? versus how surprising is my estimate if $p_{0}$ is true? — so Module 10’s exact duality holds only approximately for proportions. Here both agree that 0.10 is plausible at 5 % two-sided; near the boundary they can differ, and then the test is the answer to a test question and the interval to an estimation question.