Population-mean hypothesis tests

State a claim about the population parameter

A process is claimed to have mean 50 units. A sample mean can differ from 50 through sampling variation even if that population claim is correct. A hypothesis test asks whether the observed result is sufficiently unusual under a specified null model to provide evidence for a chosen alternative.

The null hypothesis, H₀, states the population value used to calculate the test’s sampling distribution. The alternative hypothesis, H₁, states the change being investigated:

Hypotheses concern the population mean μ, not the observed sample mean x̄. Once the sample is collected, x̄ is already a known number. Also, a population mean of 50 does not claim that every individual measurement equals 50.

Choose the alternative and the significance level α before using the sample result. For an exact test, α specifies the null-model probability of the critical region that leads to rejection. It sets the evidence threshold; it is not the probability that H₀ is true. Switching from a two-tailed test to a favourable one-tailed test after seeing the result changes the question and its planned evidence standard.

Standardise with the null mean and sampling spread

For an independent random sample of size n from a normal population with known variance σ², the null hypothesis H₀:μ = μ₀ gives

The test statistic compares the observed mean with the null value in standard-error units:

Under H₀, its corresponding random statistic is standard normal. Use the claimed μ₀ in the numerator and the sampling spread σ/√n in the denominator. Replacing σ/√n by σ would compare the mean with individual-observation spread.

Suppose the population is normal with known standard deviation 8, and an independent random sample of 64 has mean 52.2. The planned test is H₀: μ = 50 against H₁: μ > 50, at α = 0.05. The standard error is 8/√64 = 1, so

For the upper-tail alternative, the p-value is

This is the probability, assuming the null model, of a result at least as far in the specified upper direction. Since p < 0.05, reject H₀. There is sufficient evidence at the 5% level that the population mean exceeds 50 units, under the stated sampling and population assumptions.

It is not a 1.39% probability that μ = 50. Nor does it establish that the actual population mean is exactly the observed 52.2.

A critical region is the same decision expressed as a boundary

For the planned 5% upper-tail test, the standard normal boundary with lower-tail area 0.95 is z_(0.95) ≈ 1.64485363. The critical value is that boundary; the critical region is the whole set of statistic values leading to rejection:

In this example the equivalent sample-mean boundary is 50 + z_(0.95)(1) ≈ 51.64485363. An observed mean at or above that boundary leads to rejection. The observed 52.2 is in this region, agreeing with the p-value method.

In general, compare p with α: reject when p ≤ α, otherwise fail to reject. Use unrounded probabilities or critical values for a boundary decision. A displayed 0.050 might conceal a value just above or below 0.05. For a continuous exact model, including an exact critical boundary does not alter the region’s probability.

Let the alternative determine the tail

For a normal test statistic, the planned alternative determines what “at least as extreme” means:

The two-tailed formula includes both equally distant tails by symmetry. At the 5% level, each tail gets 2.5%, giving critical region |z obs| ≥ z_(0.975) ≈ 1.95996398. Using the original observed z = 2.2 with this different, prespecified alternative gives p ≈ 0.0278069. It still leads to rejection at 5%, but answers a different population question.

For a lower-tail test at level α, reject for z obs ≤ zα. At 5%, this lower critical value is approximately −1.64485363. The sign and tail must match the alternative, rather than whichever gives the smaller p-value.

Say what the decision establishes

A small p-value describes incompatibility with the specified null model in the planned direction or directions. It does not measure the probability that H₀ is true, and it does not by itself measure how important a change is in practice.

If p > α, fail to reject H₀: there is insufficient evidence for the stated alternative. Do not turn that into proof that μ equals μ₀, that the process has not changed, or that two individual measurements must agree. State the population, the direction of the claim and the chosen level in the conclusion.

Use an estimated spread only with its approximation

For a sufficiently large independent random sample from a finite-variance population, the Central Limit Theorem can justify an approximately normal sample mean even when the individual population is not normal. If σ² is unknown, estimate it with

and use estimated standard error s/square root of n. The statistic

then supplies a large-sample approximate normal test. The computed ratio is the observed statistic; the approximation is treating its null distribution as standard normal. It is not made exact by the calculator’s decimal places.

If the estimated standard deviation s is zero, this standardisation cannot be used: its denominator is zero. Identical recorded values do not establish that the population has no variation. Check the measurement resolution and model assumptions rather than reporting an infinite statistic or treating an undefined ratio as evidence.

A small sample from a normal population with unknown variance does not meet the known-variance exact normal-test conditions. Simply replacing σ by s does not make its standardised statistic exactly standard normal. Likewise, a large dependent or systematically biased sample does not acquire a valid test just by using s or citing n ≥ 30.