One-Sample Z-Tests

The z-test is the simplest hypothesis test. It uses the standard normal distribution to decide whether sample data provides evidence against a null hypothesis. If you have already learned the logic of hypothesis testing — null and alternative hypotheses, p-values, significance levels, and Type I/II errors — then you are ready to put that framework into action. The z-test is the natural starting point because its reference distribution (the standard normal curve) is the one you already know from z-scores. In this lesson, you will learn two forms: the z-test for a population proportion and the z-test for a population mean when the population standard deviation $\sigma$ is known.

Z-Test for a Population Proportion

Use this test when you are testing a claim about a population proportion $p$ — for example, whether a defect rate is 3%, whether a majority of voters favor a policy, or whether a medication’s success rate matches its label.

Conditions (all must be met):

• The sample is a simple random sample from the population of interest
• Individual observations are independent (sampling less than 10% of the population, or observations are naturally independent)
• The sample size is large enough that $np_0 \geq 10$ and $n(1 - p_0) \geq 10$, where $p_0$ is the null value. This ensures the sampling distribution of $\hat{p}$ is approximately normal.

Test statistic:

$z = \frac{\hat{p} - p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}$

Here $\hat{p}$ is the sample proportion, $p_0$ is the hypothesized proportion from $H_0$, and $n$ is the sample size. The denominator is the standard error calculated under the null hypothesis — this is important because we are asking “If $p_0$ were the true proportion, how surprising is $\hat{p}$?”

Example 1: Defective Rate (One-Sided Right)

A manufacturer claims its defect rate is 3%. A quality inspector randomly tests 500 items and finds 22 defective, giving $\hat{p} = 22/500 = 0.044$. Is there evidence the defect rate exceeds 3%? Test at $\alpha = 0.05$.

Step 1: State the hypotheses.

$H_0: p = 0.03 \quad \text{(defect rate is 3\% as claimed)}$

$H_a: p > 0.03 \quad \text{(defect rate is higher — one-sided right)}$

Step 2: Choose significance level: $\alpha = 0.05$.

Step 3: Check conditions.

• Random sample ✓
• Independence: 500 items is far less than 10% of total production ✓
• Large-sample check: $500 \times 0.03 = 15 \geq 10$ ✓ and $500 \times 0.97 = 485 \geq 10$ ✓

Step 4: Calculate the test statistic.

$SE = \sqrt{\frac{0.03 \times 0.97}{500}} = \sqrt{\frac{0.0291}{500}} = \sqrt{0.0000582} = 0.00763$

$z = \frac{0.044 - 0.03}{0.00763} = \frac{0.014}{0.00763} = 1.835$

Step 5: Find the p-value. Since this is a one-sided right test:

$\text{p-value} = P(Z > 1.835) \approx 0.0333$

Step 6: Make a decision. Since $0.0333 < 0.05$, we reject $H_0$.

Step 7: Conclusion in context. There is statistically significant evidence at the 5% level that the manufacturer’s defect rate exceeds 3%. The observed rate of 4.4% is significantly higher than the claimed rate. The inspector should flag this batch for further review.

Example 2: Patient Satisfaction (Two-Sided)

A hospital claims an 80% patient satisfaction rate. A survey of 300 randomly selected patients finds 225 who are satisfied, giving $\hat{p} = 225/300 = 0.75$. Test whether the satisfaction rate differs from 80% at $\alpha = 0.05$.

Step 1: State the hypotheses.

$H_0: p = 0.80 \quad \text{(satisfaction rate is 80\%)}$

$H_a: p \neq 0.80 \quad \text{(satisfaction rate is different — two-sided)}$

Step 2: Choose significance level: $\alpha = 0.05$.

Step 3: Check conditions.

• Random sample ✓
• Independence: 300 is less than 10% of all patients ✓
• Large-sample check: $300 \times 0.80 = 240 \geq 10$ ✓ and $300 \times 0.20 = 60 \geq 10$ ✓

Step 4: Calculate the test statistic.

$SE = \sqrt{\frac{0.80 \times 0.20}{300}} = \sqrt{\frac{0.16}{300}} = \sqrt{0.000533} = 0.02309$

$z = \frac{0.75 - 0.80}{0.02309} = \frac{-0.05}{0.02309} = -2.165$

Step 5: Find the p-value. Since this is a two-sided test:

$\text{p-value} = 2 \times P(Z < -2.165) \approx 2 \times 0.0152 = 0.0304$

Step 6: Make a decision. Since $0.0304 < 0.05$, we reject $H_0$.

Step 7: Conclusion in context. There is statistically significant evidence that the patient satisfaction rate differs from the claimed 80%. At 75%, satisfaction is significantly lower. Hospital leadership should investigate root causes and develop improvement strategies.

Z-Test for a Population Mean ($\sigma$ Known)

Use this test when you are testing a claim about a population mean $\mu$ and the population standard deviation $\sigma$ is known. This scenario is uncommon in practice — it arises mainly when a manufacturing process has a well-established historical standard deviation, or in textbook problems designed to teach the z-test framework. In most real-world situations, $\sigma$ is unknown and you would use a t-test instead.

Conditions:

• The sample is a simple random sample
• Observations are independent
• The population distribution is normal, OR the sample size is large ($n \geq 30$) so that the Central Limit Theorem applies

Test statistic:

$z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}}$

Here $\bar{x}$ is the sample mean, $\mu_0$ is the hypothesized population mean from $H_0$, $\sigma$ is the known population standard deviation, and $n$ is the sample size.

Example 3: Bottling Machine Fill Volume (One-Sided Left)

A bottling machine is designed to fill each bottle to $\mu = 500$ mL. From historical calibration data, the fill standard deviation is known to be $\sigma = 5$ mL. A quality engineer draws a random sample of 40 bottles and measures $\bar{x} = 498.2$ mL. Is there evidence the machine is underfilling? Test at $\alpha = 0.01$.

Step 1: State the hypotheses.

$H_0: \mu = 500 \quad \text{(machine fills correctly)}$

$H_a: \mu < 500 \quad \text{(machine is underfilling — one-sided left)}$

Step 2: Choose significance level: $\alpha = 0.01$.

Step 3: Check conditions. Random sample ✓. Independence ✓. Sample size $n = 40 \geq 30$, so the CLT applies ✓. And $\sigma = 5$ is known ✓.

Step 4: Calculate the test statistic.

$SE = \frac{\sigma}{\sqrt{n}} = \frac{5}{\sqrt{40}} = \frac{5}{6.3246} = 0.7906$

$z = \frac{498.2 - 500}{0.7906} = \frac{-1.8}{0.7906} = -2.277$

Step 5: Find the p-value. Since this is a one-sided left test:

$\text{p-value} = P(Z < -2.277) \approx 0.0114$

Step 6: Make a decision. Since $0.0114 > 0.01$, we fail to reject $H_0$ at the 1% level.

Step 7: Conclusion in context. At the strict $\alpha = 0.01$ significance level, there is not sufficient evidence to conclude the machine is underfilling. However, the p-value of 0.0114 is quite close to the threshold. At the more common $\alpha = 0.05$ level, this result would be significant. The engineer may want to continue monitoring or collect a larger sample.

Important note: This example illustrates why the choice of $\alpha$ matters. The same data leads to different decisions depending on the threshold. In high-stakes manufacturing scenarios, a stricter $\alpha$ reduces the risk of a false alarm (Type I error) but increases the risk of missing a real problem (Type II error).

Making a Decision: The Complete Framework

Every z-test follows the same seven-step process. Here is a summary:

Decision rules at a glance:

• One-sided right ($H_a: >$): p-value $= P(Z > z)$
• One-sided left ($H_a: <$): p-value $= P(Z < z)$
• Two-sided ($H_a: \neq$): p-value $= 2 \times P(Z > |z|)$

Real-World Application: Nursing — Testing a Medication Compliance Rate

A nurse educator develops a new discharge instruction protocol aimed at improving medication compliance. The hospital’s historical compliance rate is 70%. After implementing the new protocol, a random sample of 150 discharged patients is followed up, and 115 are found to be compliant, giving $\hat{p} = 115/150 = 0.7667$.

Is there evidence that the new protocol improves compliance above 70%?

$H_0: p = 0.70 \quad H_a: p > 0.70$

Check conditions: $150 \times 0.70 = 105 \geq 10$ ✓ and $150 \times 0.30 = 45 \geq 10$ ✓.

$SE = \sqrt{\frac{0.70 \times 0.30}{150}} = \sqrt{\frac{0.21}{150}} = \sqrt{0.0014} = 0.03742$

$z = \frac{0.7667 - 0.70}{0.03742} = \frac{0.0667}{0.03742} = 1.782$

$\text{p-value} = P(Z > 1.782) \approx 0.0374$

At $\alpha = 0.05$, we reject $H_0$. There is evidence that the new protocol increases medication compliance above the baseline 70%. The observed rate of 76.7% represents a clinically meaningful improvement — roughly 10 additional compliant patients per 150 discharged — which could reduce hospital readmissions and improve patient outcomes.

Clinical consideration: In healthcare settings, both statistical and practical significance matter. A compliance increase from 70% to 77% has real clinical impact: fewer missed doses, fewer adverse events, and lower readmission rates. The nurse educator should recommend adopting the new protocol while continuing to monitor compliance rates.

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Key Takeaways

• The z-test for a proportion tests claims about a population proportion $p$ using the test statistic $z = \frac{\hat{p} - p_0}{\sqrt{p_0(1-p_0)/n}}$
• The z-test for a mean (when $\sigma$ is known) uses the test statistic $z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}}$
• Both tests require random sampling, independence, and sufficient sample size for the normal approximation to hold
• The z-test for a mean with known $\sigma$ is rare in practice — the one-sample t-test is far more common
• The p-value measures how surprising the sample result would be if $H_0$ were true — small p-values provide evidence against $H_0$
• Always check conditions before performing any test. Violating assumptions can lead to misleading results.
• State conclusions in context — a statistical decision (“reject $H_0$”) must be translated into a meaningful real-world statement
• The choice of $\alpha$ can change the decision. Always set $\alpha$ before looking at the data.

Return to Statistics for more topics in this section.