One-Sample Z Tests for Means With Known Variance

Why Test a Mean With a Known Variance?

A one-sample z test assesses whether data provide evidence that a population mean differs from, exceeds, or falls below a specified value. Its defining feature is that the population variance is known. This is a distinct assumption: the sample variance is not substituted for the known population variance in the test statistic.

In actuarial and statistical work, a mean may represent an average claim amount, processing time, or measurement for a defined population. The hypothesis test gives a disciplined way to assess a stated value against sample evidence. The procedure in this tutorial is aligned with the SOA VEE Mathematical Statistics syllabus.

Definition: Suppose $X_1,\ldots,X_n$ are independent observations from a population with mean $\mu$ and known variance $\sigma^2$. For testing a null value $\mu_0$, the one-sample z statistic is $$ Z=\frac{\bar{X}-\mu_0}{\sigma/\sqrt{n}}, $$ where $\bar{X}$ is the sample mean and $\sigma/\sqrt{n}$ is the standard error of the sample mean under the null hypothesis.

Model, Hypotheses, and Test Statistic

For an exact one-sample z test, assume $X_1,\ldots,X_n$ are a random sample from a normal population $N(\mu,\sigma^2)$, with the population variance $\sigma^2$ known. Then $\bar{X}$ has a normal distribution with mean $\mu$ and variance $\sigma^2/n$. If the population is not normal, the same z procedure may be used as an approximation when the sample is large enough for the central limit theorem to be credible. There is no universal sample-size cutoff that guarantees a good approximation; strong skewness, heavy tails, or dependent observations can make it unreliable.

The null hypothesis specifies the reference value, usually $H_0:\mu=\mu_0$. The alternative hypothesis describes the direction of departure relevant to the question. A test is one-sided if the alternative specifies a direction and two-sided if departures in either direction matter.

Question of interest	Null hypothesis	Alternative hypothesis
Is the mean greater than $\mu_0$?	$H_0:\mu=\mu_0$	$H_A:\mu>\mu_0$
Is the mean less than $\mu_0$?	$H_0:\mu=\mu_0$	$H_A:\mu<\mu_0$
Does the mean differ from $\mu_0$?	$H_0:\mu=\mu_0$	$H_A:\mu\ne\mu_0$

The significance level $\alpha$ is the probability of rejecting the null hypothesis when it is true, for a test designed at that level. The test statistic measures the distance between the observed sample mean and the null mean in standard-error units. Under $H_0$, it has the standard normal distribution for the exact normal model; in a sufficiently large-sample approximation, it is treated as approximately standard normal.

Why the Standard Error Appears in the Denominator

The standardization follows from the sampling distribution of the sample mean. Under the model, each observation has variance $\sigma^2$, and independence makes the variance of the sum equal to the sum of the variances. Dividing that sum by $n^2$ gives the variance of the average:

\operatorname{Var}(\bar{X}) =\operatorname{Var}\left(\frac{1}{n}\sum_{i=1}^{n}X_i\right) =\frac{1}{n^2}\sum_{i=1}^{n}\operatorname{Var}(X_i) =\frac{n\sigma^2}{n^2} =\frac{\sigma^2}{n}.

Under $H_0:\mu=\mu_0$, the mean of $\bar{X}$ is $\mu_0$, so subtracting $\mu_0$ centers the sample mean at zero. Dividing by its standard deviation, $\sqrt{\sigma^2/n}=\sigma/\sqrt{n}$, standardizes it. Thus $Z=(\bar{X}-\mu_0)/(\sigma/\sqrt{n})$ has a standard normal distribution under the null for a normal population.

Critical Values and p-Values

A critical-value procedure compares the observed z statistic with a cutoff determined by $\alpha$ and the alternative hypothesis. A p-value instead measures the probability, assuming the null hypothesis is true, of a test statistic at least as extreme in the direction specified by the alternative. Both procedures give the same reject-or-do-not-reject decision when they use the same significance level.

Theorem: Let $\Phi(z)$ denote the standard normal cumulative distribution function. For an upper-tailed test, the p-value is $1-\Phi(z_{\mathrm{obs}})$; for a lower-tailed test, it is $\Phi(z_{\mathrm{obs}})$; and for a two-sided test, it is $2[1-\Phi(|z_{\mathrm{obs}}|)]$. Reject $H_0$ when the p-value is at most $\alpha$.

For a two-sided test, the critical values are $-z_{1-\alpha/2}$ and $z_{1-\alpha/2}$, where $z_q$ is the standard normal quantile with cumulative probability $q$. For an upper-tailed test, reject when $Z>z_{1-\alpha}$; for a lower-tailed test, reject when $Z<-z_{1-\alpha}$. For example, at $\alpha=0.05$, the one-sided critical value is approximately $1.6449$, while the two-sided critical values are approximately $-1.9600$ and $1.9600$.

Choose hypotheses.
Use the research question to select a greater-than, less-than, or not-equal alternative.

Check the model and known variance.
Confirm independent observations and a known $\sigma$; use an exact normal model or justify a large-sample approximation.

Compute and standardize.
Calculate $\sigma/\sqrt{n}$, then $z_{\mathrm{obs}}=(\bar{x}-\mu_0)/(\sigma/\sqrt{n})$.

Make and interpret the decision.
Compare the p-value with $\alpha$, or compare $z_{\mathrm{obs}}$ with the appropriate critical value. State the conclusion in the context of the population mean.

Worked Applications

Worked Example: Testing Whether a Mean Processing Time Increased

A hypothetical claims unit has historically recorded a mean processing time of 100 minutes. For a current random sample of 36 independent files, the sample mean is 104 minutes. Assume processing times follow $N(\mu,144)$, so the known population variance is $\sigma^2=144$ minutes squared and $\sigma=12$ minutes. At significance level $\alpha=0.05$, test whether the mean processing time has increased.

Method and hypotheses. This is an upper-tailed one-sample z test. Set $H_0:\mu=100$ minutes and $H_A:\mu>100$ minutes.

Calculate the standard error. $$ \operatorname{SE}(\bar{X})=\frac{\sigma}{\sqrt{n}} =\frac{12}{\sqrt{36}} =\frac{12}{6} =2\text{ minutes}. $$

Calculate the test statistic. $$ z_{\mathrm{obs}}=\frac{104-100}{2} =\frac{4}{2} =2.0000. $$

Use a critical value and p-value. For an upper-tailed test at $\alpha=0.05$, the critical value is $1.6449$. Since $2.0000>1.6449$, reject $H_0$. The p-value is $1-\Phi(2.0000)=0.0228$, which is also less than $0.05$.

Conclusion and check. The sample provides evidence at the 5% level that the mean processing time is greater than 100 minutes. The calculation is consistent: the observed mean is two standard errors above the null mean, and the upper-tail probability is about 0.0228. A common mistake is to use the variance of the sample mean, $144/36=4$, as the denominator. That would give $4/4=1$, but the statistic requires the standard error, $\sqrt{4}=2$, so the correct statistic is $4/2=2$.

Worked Example: A Two-Sided Test of an Average Sensor Reading

A hypothetical monitoring system is calibrated to have a mean reading of 50 units. A random sample of 64 independent readings has mean 49.1 units. Assume readings follow $N(\mu,64)$, with known population variance $\sigma^2=64$ units squared and known standard deviation $\sigma=8$ units. At $\alpha=0.10$, test whether the population mean differs from 50 units.

Method and hypotheses. Because either an increase or a decrease is relevant, use a two-sided test: $H_0:\mu=50$ and $H_A:\mu\ne50$.

Calculate the standard error and z statistic. $$ \operatorname{SE}(\bar{X})=\frac{8}{\sqrt{64}}=\frac{8}{8}=1\text{ unit}, \qquad z_{\mathrm{obs}}=\frac{49.1-50}{1}=-0.9000. $$

Use the two-sided critical values. With $\alpha=0.10$, each tail has probability $0.05$, so the critical values are approximately $-1.6449$ and $1.6449$. The observed statistic is between them: $-1.6449<-0.9000<1.6449$. Do not reject $H_0$.

Check with the p-value and interpret. $$ p=2[1-\Phi(|-0.9000|)] =2[1-\Phi(0.9000)] =2(0.1841) =0.3681. $$ Since $0.3681>0.10$, the p-value gives the same decision. The sample does not provide sufficient evidence at the 10% level that the mean differs from 50 units. The sign of the statistic indicates that the sample mean is below 50, but the two-sided test also counts equally extreme results above 50. The common mistake of using only the lower-tail probability would not answer the two-sided question.

Worked Example: A Lower-Tailed Test Near a Strict Cutoff

A hypothetical laboratory is required to maintain a mean measurement of at least 100 units. A random sample of 25 independent measurements has mean 93.4 units. Assume measurements follow $N(\mu,225)$, with known population variance $\sigma^2=225$ units squared and known standard deviation $\sigma=15$ units. At $\alpha=0.01$, test whether the mean is below 100 units.

Method and hypotheses. Use a lower-tailed test, with $H_0:\mu=100$ and $H_A:\mu<100$.

Calculate the standard error. $$ \operatorname{SE}(\bar{X})=\frac{15}{\sqrt{25}} =\frac{15}{5} =3\text{ units}. $$

Calculate the test statistic. $$ z_{\mathrm{obs}}=\frac{93.4-100}{3} =\frac{-6.6}{3} =-2.2000. $$

Compare with the correct lower-tail cutoff. At $\alpha=0.01$, the lower critical value is approximately $-2.3263$. Since $-2.2000>-2.3263$, do not reject $H_0$. The lower-tail p-value is $\Phi(-2.2000)=0.0139$, which exceeds $0.01$.

Conclusion and check. Although the sample mean is below 100, it is not far enough below the null value, in standard-error units, to meet the 1% rejection rule. The evidence is not sufficient at that significance level to conclude that the population mean is below 100 units. A common mistake is to compare $|z_{\mathrm{obs}}|=2.2000$ with the positive one-sided cutoff $2.3263$ and then treat the absolute value as the decision rule. For this lower-tailed test the comparison is $z_{\mathrm{obs}}<-2.3263$; it is not satisfied.

Interpreting Decisions and Avoiding Traps

A rejection means the observed data would be relatively unusual under the null model, according to the chosen significance level and alternative. It does not prove the alternative hypothesis, establish the practical importance of the difference, or show that the null value is impossible. A failure to reject means the evidence did not meet the rejection rule; it does not prove that the population mean equals $\mu_0$.

Do not confuse variance and standard error. The variance of $\bar{X}$ is $\sigma^2/n$; the standard error used in the z statistic is its square root, $\sigma/\sqrt{n}$.
Do not choose the tail after seeing the sample mean. The alternative should reflect the question being tested, not the direction that makes the observed result look more significant.
Do not split $\alpha$ in a one-sided test. The entire rejection probability is in the specified tail. It is split between two tails only for a two-sided test.
Do not substitute the sample standard deviation. This procedure relies on a known population standard deviation. If it is not known, the assumptions and reference distribution for a different procedure are needed.
Keep the conclusion tied to the setting. State what the data indicate about the population mean, and name the significance level; avoid saying that the null hypothesis has been proved.

Key takeaway: For a one-sample z test, the known population standard deviation determines the standard error. The direction of the alternative determines the tail, and that tail determines the p-value or critical region. A sound conclusion follows from all three choices: model, hypotheses, and significance level.

Check Your Understanding

For each problem, identify the test direction, calculate the z statistic, and make a decision at the stated significance level. Answers use probabilities rounded to four decimals.

A random sample of 49 independent observations from a normal population has mean 52. The known population standard deviation is 14. Test $H_0:\mu=50$ against $H_A:\mu>50$ at $\alpha=0.05$. (Answer: $z=1.0000$, p-value $=0.1587$; do not reject $H_0$.)
A random sample of 100 independent observations from a normal population has mean 103. The known population standard deviation is 20. Test $H_0:\mu=100$ against $H_A:\mu\ne100$ at $\alpha=0.05$. (Answer: $z=1.5000$, p-value $=0.1336$; do not reject $H_0$.)
A random sample of 36 independent observations from a normal population has mean 18.4. The known population standard deviation is 6. Test $H_0:\mu=20$ against $H_A:\mu<20$ at $\alpha=0.05$. (Answer: $z=-1.6000$, p-value $=0.0548$; do not reject $H_0$.)
A random sample of 25 independent observations from a normal population has mean 104. The known population standard deviation is 10. Test $H_0:\mu=100$ against $H_A:\mu\ne100$ at $\alpha=0.10$. (Answer: $z=2.0000$, p-value $=0.0455$; reject $H_0$.)
A random sample of 64 independent observations from a normal population has mean 98.5. The known population standard deviation is 8. Test $H_0:\mu=100$ against $H_A:\mu<100$ at $\alpha=0.01$. (Answer: $z=-1.5000$, p-value $=0.0668$; do not reject $H_0$.)

Question of interest	Null hypothesis	Alternative hypothesis
Is the mean greater than \(\mu_0\)?	\(H_0:\mu=\mu_0\)	\(H_A:\mu>\mu_0\)
Is the mean less than \(\mu_0\)?	\(H_0:\mu=\mu_0\)	\(H_A:\mu<\mu_0\)
Does the mean differ from \(\mu_0\)?	\(H_0:\mu=\mu_0\)	\(H_A:\mu\ne\mu_0\)

What You'll Learn