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Hypothesis tests for a mean · Use a critical region

Choose a rejection region before the observation

Use alongside the distribution plots. These cutoffs assume a justified standard-normal null model; an approximate model gives an approximate test.

  1. Allocate the fixed significance level

    Follow the pre-specified alternative. Put all α in one tail for a directional test, or α/2 in each tail for a two-sided test.

    Let Φ\Phi be the standard normal CDF. Write c=Φ−1(1−α)c=\Phi^{-1}(1-\alpha) and d=Φ−1(1−α/2)d=\Phi^{-1}(1-\alpha/2).

    Decrease · H₁: μ < μ₀Use the lower critical tail

    Reject when z≤−cz\le-c. The lower-tail cutoff is Φ−1(α)=−c\Phi^{-1}(\alpha)=-c by normal symmetry.

    Increase · H₁: μ > μ₀Use the upper critical tail

    Reject when z≥cz\ge c. The upper-tail area beyond cc is α\alpha.

    Either direction · H₁: μ ≠ μ₀Split α between both tails

    Reject when ∣z∣≥d|z|\ge d, equivalently z≤−dz\le-d or z≥dz\ge d. Each tail has area α/2\alpha/2.

    Use unrounded cutoffs and the declared boundary rule. For a continuous reference model, a single boundary point has zero probability; equality is included here to match the p ≤ α rule.

  2. Compare the observation with the fixed region

    Calculate the statistic from the observed sample, then check the region chosen before observing it. Do not move the region to match the sign of z.

    For this same model, alternative and level, rejection-region membership is equivalent to p≤αp\le\alpha.

    In the rejection regionReject the null hypothesis

    There is sufficient evidence, at the stated level and under the model assumptions, in favour of the pre-specified alternative. State the conclusion about the population mean in context.

    Outside the rejection regionDo not reject the null hypothesis

    There is insufficient evidence against the null at this level. This does not prove or accept the null hypothesis, nor assign it a probability of being true.

    Both methods must use the same null model, alternative, significance level and precision. Non-rejection does not establish the null hypothesis.

Try a worked example

Read a case, make a prediction, then reveal the reasoning. The examples keep their stated values.

Does the sign of z change the test?

Illustration: an upper-sided standard-normal test was fixed at 5%. Compare observations z=2z=2 and z=−2z=-2. Then compare a separately pre-specified two-sided test.

Show reasoning

  1. 1. Fix the upper cutoff

    All 5% belongs in the upper tail. Use the full quantile in the decision; the value shown here is rounded.

    c=Φ−1(0.95)≈1.64485c=\Phi^{-1}(0.95)\approx1.64485

  2. 2. Keep the upper-sided alternative

    The observation 2 is inside the upper rejection region, but −2 is outside. Reject for the first and do not reject for the second. A negative observation does not justify switching to the lower tail.

    2>c,−2<c2>c,\qquad -2<c

  3. 3. Compare a different, pre-specified two-sided rule

    Split the level into 2.5% in each tail. Both observations have magnitude 2 and lie in this two-sided rejection region. That is a different test, not a test chosen after seeing the result.

    d=Φ−1(0.975)≈1.95996,∣2∣=∣−2∣>dd=\Phi^{-1}(0.975)\approx1.95996,\qquad |2|=|-2|>d

  4. 4. State the limited conclusion

    For the upper-sided test with z = −2, there is insufficient evidence that the population mean exceeds the null value. This neither proves the null nor establishes that the mean is smaller; that would require its own pre-specified inference.