Did you know? For the same hypothesis test, the critical-region method and the p-value method should lead to exactly the same reject-or-do-not-reject decision when the same alternative hypothesis and significance level are used. If the two approaches disagree, the maths has not revealed a surprising new fact: the student has most likely chosen different tails, used an inconsistent threshold or made a calculator error.
This JC2 H2 Mathematics Tutor Bukit Timah guide answers a focused A-Level syllabus 9758 question: should a learner decide a hypothesis test using a critical value or a p-value, and how can they check that both methods agree? We begin with a complete normal-mean test, set a sensible significance level and then practise upper-tail, lower-tail and two-tailed cases. The aim is to explain the decision in a sentence that a reader can understand, not merely report the word “reject”.
For Bukit Timah JC2 H2 Mathematics preparation, the problem is not simply learning a z-test formula. A solution must state the population claim, define the hypotheses, identify the correct sampling model, compare evidence with the preselected significance threshold, and interpret the outcome. The 2027 H2 syllabus includes hypothesis tests for one population mean using a normal population with known variance or an appropriate large sample. It does not require testing a difference between two population means. Every numerical case here is fictional.
The quick answer: two views of one rejection rule
Suppose the test statistic Z follows the standard normal reference distribution under H₀. The critical-region method chooses the extreme Z values whose total probability under H₀ is the significance level α. The p-value method calculates how extreme the observed statistic is, in the direction(s) specified by H₁, and compares that probability with α. Both methods use the same null model and the same rejection area.
- Upper-tailed H₁: μ > μ₀. Reject for sufficiently large positive Z, or when the upper-tail p-value is at most α.
- Lower-tailed H₁: μ < μ₀. Reject for sufficiently negative Z, or when the lower-tail p-value is at most α.
- Two-tailed H₁: μ ≠ μ₀. Reject for a sufficiently large absolute value of Z, or when the two-tailed p-value is at most α.
- Fix H₁ before viewing the outcome. Do not choose a tail after discovering whether the sample mean was high or low.
- Do not reject H₀ does not mean “H₀ has been proved”. It means the evidence did not meet the specified rejection rule.
The normal-mean test statistic used in our examples
Suppose a random sample of size n is taken from a normal population with known standard deviation σ. To test H₀: μ=μ₀, the sample mean X̄ has null mean μ₀ and standard deviation σ/√n. The resulting statistic is z = (x̄ − μ₀)/(σ/√n). The positive or negative sign tells you which side of μ₀ the observed sample mean lies; the absolute size tells you how many standard errors separate the result from the null value.
For an upper-tail claim, a large negative z cannot be compelling support, regardless of its magnitude, because it points away from the stated alternative. For a two-tailed claim, either direction can count as evidence, and the p-value accounts for both extremes. That is why the question’s verbs “greater than”, “less than” and “different from” belong in the hypotheses before any probability button is pressed.
Fully worked example: z = 2 at the 5% level
A fictional setting states that numerical readings come from a normal population of known standard deviation σ=12. The null claim is H₀: μ=50; the investigation set in advance is H₁: μ > 50. A random sample of n=64 produces x̄=53. The standard error under H₀ is 12/√64=1.5. Therefore z=(53−50)/1.5=2.00. Use the supplied 5% significance level, α=0.05.
The critical-region approach uses the upper 5% of the standard normal curve. Its boundary is approximately z=1.64485. Since 2.00 is greater than 1.64485, the statistic falls in the rejection region. Therefore reject H₀ at the 5% level. This conclusion is about a single population mean under the stated normal sampling model, not about comparing two different groups.
The p-value approach calculates P(Z≥2.00)≈0.02275 under H₀. This is less than 0.05, so it also says reject H₀. In context: there is statistically significant evidence at the 5% level, under the stated model, that the population mean exceeds 50. The methods agree because they are equivalent ways of locating the same extreme region on the null distribution.
What changes at the 1% level?
Keep the observed statistic z=2.00 and upper-tail alternative H₁: μ>50, but now specify α=0.01. The upper 1% critical boundary is approximately z=2.32635. Since 2.00 does not reach that boundary, do not reject H₀ at 1%. The p-value is still 0.02275, but now 0.02275 is greater than 0.01, producing the same decision. The evidence did not change; the preselected evidential threshold did.
This illustrates why a statement that a result is simply “significant” is incomplete. State the level at which the decision is being made. A result can cross a 5% threshold and not cross a 1% threshold without any contradiction. Report the significance level alongside the decision and its substantive interpretation.
An alternative fixed before the data: two-sided versus one-sided
Suppose instead the investigation asked whether the mean differs from 50, setting H₁: μ≠50 before the sample was obtained. With z=2.00, the two-tailed normal p-value is approximately 2×0.02275=0.04550. At 5%, it still leads to rejecting H₀, because 0.04550<0.05; the corresponding two-tailed critical boundaries are approximately ±1.95996 and |2.00| exceeds 1.95996.
It would be wrong to change a two-sided alternative to a one-sided one merely because the observed x̄ happens to be greater than 50. Such a switch changes the problem being tested and can make a p-value look more convincing after the fact. A separate guide explains how to choose one-tailed or two-tailed tests. This page concentrates on checking the equivalence of the two decision methods once the test is specified.
A practical table of normal critical values
| Significance α | Upper-tail rejection boundary | Lower-tail rejection boundary | Two-tailed rejection boundaries |
|---|---|---|---|
| 10% | +1.28155 | −1.28155 | ±1.64485 |
| 5% | +1.64485 | −1.64485 | ±1.95996 |
| 1% | +2.32635 | −2.32635 | ±2.57583 |
A one-tailed boundary places the entire rejection area α on one side of the normal curve. The two-tailed boundary splits α between two sides, which is why its absolute z threshold is larger at the same α. Use the table as a sense check, but follow the graphing calculator and question instructions if greater precision is required. Boundaries shown here are approximate normal critical values and do not apply unchanged to an unrelated statistical reference distribution.
The interpretation problem: p is not the probability that H₀ is true
A p-value is calculated under the assumption that H₀ is true. It measures how unusual the observed statistic, or a more extreme one in the specified test direction, would be under that null model. It does not tell you the probability that H₀ itself is true. When the p-value is large, it means the result has not provided sufficiently strong evidence to reject H₀ at the chosen α, not that the null hypothesis has somehow been proven.
Similarly, an extremely small p-value does not automatically mean the practical effect is large, important or guaranteed to repeat in a future sample. Test outcome, estimated effect size and practical significance are distinct concepts. An A-Level explanation should remain faithful to the question: provide a decision about the population mean claim and the significance level rather than inventing a broader scientific or educational claim.
Common calculator mistakes that create contradictory decisions
- Wrong tail: calculating P(Z≤2) when H₁ says μ>μ₀ gives about 0.9772, which is not the upper-tail p-value.
- Forgotten doubling: a two-tailed symmetric normal test uses both extremes; when appropriate, use 2P(Z≥|z|).
- Lost a negative sign: a large negative z supports a lower-tail claim, not an upper-tail claim.
- Variance entered as σ: if the question states N(μ,144), use σ=12, not 144.
- Wrong standard error: sample-mean spread is σ/√n under the stated independent normal model, not σ/n.
- Premature rounding: compare unrounded probabilities or z values when a result is close to a critical boundary.
A diagnostic sequence that takes one minute
First ask a learner to state H₁ before any sample figures. Then supply z=1.80 with H₁: μ>μ₀ and α=0.05. The upper-tail p-value is approximately 0.03593, so reject H₀. The upper critical boundary is 1.64485, again rejecting H₀. Next change only α to 0.01: now p>α and z<2.32635, so do not reject. Finally change only H₁ to μ≠μ₀ at α=0.05: the two-tailed p-value is about 0.07186, so do not reject. Each change probes one specific decision.
A student who gets different conclusions from the two methods in the same scenario should not be told simply to practise more calculator commands. Ask which tail is shaded, which α is used, and whether the p-value comes from the same distribution as the critical boundary. Repairing the exact inconsistent assumption is much more efficient than repeating unrelated numerical tests.
Thirty-six original decision clinics with two independent checks
Each clinic specifies a hypothetical one-population test, the null reference distribution and a prespecified alternative. The numerical means are fabricated instructional values, not collected school results or real research findings. For transparent arithmetic, all use a normal population with known σ=6 and a sample of n=36, so the null standard error is 1 and z=x̄−50. Work by both routes and confirm that the decision matches; do not use the examples to infer the probability of any hypothesis being true.
1. Upper-tailed normal test at 5%
Hypothetical test setup. H₀: μ=50 and H₁: μ > 50. The population is assumed normal with known standard deviation 6. A random sample of n=36 gives x̄=52. State the significance level α=0.05 before calculating. The reference standard error is 6/√36=1, so the observed statistic is z=(52−50)/1=2.
Critical-region route. For this upper-tailed test, the rejection condition is z ≥ 1.644854. The observed z=2 lies in that rejection region, so reject H₀. The chosen region uses the supplied alternative, not whichever side would make the test look more impressive.
p-value route. Under H₀, the corresponding one-sided normal p-value is approximately 0.02275. This is at most α=0.05, so the decision is again reject H₀. Explain why the observed sign matters for a directional alternative. A result pointing the wrong way cannot support the original directional claim through a small one-tailed p-value. In words, there is sufficient statistical evidence for the stated alternative at this level under the given model. A failed rejection does not establish H₀ as fact.
2. Lower-tailed normal test at 5%
Hypothetical test setup. H₀: μ=50 and H₁: μ < 50. The population is assumed normal with known standard deviation 6. A random sample of n=36 gives x̄=51.8. State the significance level α=0.05 before calculating. The reference standard error is 6/√36=1, so the observed statistic is z=(51.8−50)/1=1.8.
Critical-region route. For this lower-tailed test, the rejection condition is z ≤ −1.644854. The observed z=1.8 does not reach that rejection region, so do not reject H₀. The chosen region uses the supplied alternative, not whichever side would make the test look more impressive.
p-value route. Under H₀, the corresponding one-sided normal p-value is approximately 0.96407. This is greater than α=0.05, so the decision is again do not reject H₀. Compare the p-value with the critical region without rounding the statistic to a whole number first. The agreement is a built-in consistency test. In words, the evidence is insufficient to support the stated alternative at this level under the given model. A failed rejection does not establish H₀ as fact.
3. Two-tailed normal test at 5%
Hypothetical test setup. H₀: μ=50 and H₁: μ ≠ 50. The population is assumed normal with known standard deviation 6. A random sample of n=36 gives x̄=51.2. State the significance level α=0.05 before calculating. The reference standard error is 6/√36=1, so the observed statistic is z=(51.2−50)/1=1.2.
Critical-region route. For this two-tailed test, the rejection condition is |z| ≥ 1.959964. The observed z=1.2 does not reach that rejection region, so do not reject H₀. The chosen region uses the supplied alternative, not whichever side would make the test look more impressive.
p-value route. Under H₀, the corresponding two-sided normal p-value is approximately 0.230139. This is greater than α=0.05, so the decision is again do not reject H₀. Distinguish the strength of evidence under the null model from the magnitude or practical importance of a population difference. In words, the evidence is insufficient to support the stated alternative at this level under the given model. A failed rejection does not establish H₀ as fact.
4. Upper-tailed normal test at 1%
Hypothetical test setup. H₀: μ=50 and H₁: μ > 50. The population is assumed normal with known standard deviation 6. A random sample of n=36 gives x̄=48. State the significance level α=0.01 before calculating. The reference standard error is 6/√36=1, so the observed statistic is z=(48−50)/1=-2.
Critical-region route. For this upper-tailed test, the rejection condition is z ≥ 2.326348. The observed z=-2 does not reach that rejection region, so do not reject H₀. The chosen region uses the supplied alternative, not whichever side would make the test look more impressive.
p-value route. Under H₀, the corresponding one-sided normal p-value is approximately 0.97725. This is greater than α=0.01, so the decision is again do not reject H₀. State the complete inference in the context of an unknown population mean, rather than writing a bare numerical inequality. In words, the evidence is insufficient to support the stated alternative at this level under the given model. A failed rejection does not establish H₀ as fact.
5. Lower-tailed normal test at 1%
Hypothetical test setup. H₀: μ=50 and H₁: μ < 50. The population is assumed normal with known standard deviation 6. A random sample of n=36 gives x̄=48.2. State the significance level α=0.01 before calculating. The reference standard error is 6/√36=1, so the observed statistic is z=(48.2−50)/1=-1.8.
Critical-region route. For this lower-tailed test, the rejection condition is z ≤ −2.326348. The observed z=-1.8 does not reach that rejection region, so do not reject H₀. The chosen region uses the supplied alternative, not whichever side would make the test look more impressive.
p-value route. Under H₀, the corresponding one-sided normal p-value is approximately 0.03593. This is greater than α=0.01, so the decision is again do not reject H₀. Imagine a different study choosing a stricter α in advance. The evidence would be identical, but the decision threshold might change. In words, the evidence is insufficient to support the stated alternative at this level under the given model. A failed rejection does not establish H₀ as fact.
6. Two-tailed normal test at 1%
Hypothetical test setup. H₀: μ=50 and H₁: μ ≠ 50. The population is assumed normal with known standard deviation 6. A random sample of n=36 gives x̄=48.8. State the significance level α=0.01 before calculating. The reference standard error is 6/√36=1, so the observed statistic is z=(48.8−50)/1=-1.2.
Critical-region route. For this two-tailed test, the rejection condition is |z| ≥ 2.575829. The observed z=-1.2 does not reach that rejection region, so do not reject H₀. The chosen region uses the supplied alternative, not whichever side would make the test look more impressive.
p-value route. Under H₀, the corresponding two-sided normal p-value is approximately 0.230139. This is greater than α=0.01, so the decision is again do not reject H₀. If the sample mean lies close to the null mean, the test statistic is close to zero and is unlikely to cross a standard extreme-tail boundary. In words, the evidence is insufficient to support the stated alternative at this level under the given model. A failed rejection does not establish H₀ as fact.
7. Upper-tailed normal test at 10%
Hypothetical test setup. H₀: μ=50 and H₁: μ > 50. The population is assumed normal with known standard deviation 6. A random sample of n=36 gives x̄=52.5. State the significance level α=0.1 before calculating. The reference standard error is 6/√36=1, so the observed statistic is z=(52.5−50)/1=2.5.
Critical-region route. For this upper-tailed test, the rejection condition is z ≥ 1.281552. The observed z=2.5 lies in that rejection region, so reject H₀. The chosen region uses the supplied alternative, not whichever side would make the test look more impressive.
p-value route. Under H₀, the corresponding one-sided normal p-value is approximately 0.00621. This is at most α=0.1, so the decision is again reject H₀. Never choose a new H₁ merely because the observed x̄ points in an unexpected direction. The research question should determine the direction. In words, there is sufficient statistical evidence for the stated alternative at this level under the given model. A failed rejection does not establish H₀ as fact.
8. Lower-tailed normal test at 10%
Hypothetical test setup. H₀: μ=50 and H₁: μ < 50. The population is assumed normal with known standard deviation 6. A random sample of n=36 gives x̄=47.5. State the significance level α=0.1 before calculating. The reference standard error is 6/√36=1, so the observed statistic is z=(47.5−50)/1=-2.5.
Critical-region route. For this lower-tailed test, the rejection condition is z ≤ −1.281552. The observed z=-2.5 lies in that rejection region, so reject H₀. The chosen region uses the supplied alternative, not whichever side would make the test look more impressive.
p-value route. Under H₀, the corresponding one-sided normal p-value is approximately 0.00621. This is at most α=0.1, so the decision is again reject H₀. A two-sided alternative has two rejection regions. Both together have total probability α under the null model. In words, there is sufficient statistical evidence for the stated alternative at this level under the given model. A failed rejection does not establish H₀ as fact.
9. Two-tailed normal test at 10%
Hypothetical test setup. H₀: μ=50 and H₁: μ ≠ 50. The population is assumed normal with known standard deviation 6. A random sample of n=36 gives x̄=50.8. State the significance level α=0.1 before calculating. The reference standard error is 6/√36=1, so the observed statistic is z=(50.8−50)/1=0.8.
Critical-region route. For this two-tailed test, the rejection condition is |z| ≥ 1.644854. The observed z=0.8 does not reach that rejection region, so do not reject H₀. The chosen region uses the supplied alternative, not whichever side would make the test look more impressive.
p-value route. Under H₀, the corresponding two-sided normal p-value is approximately 0.423711. This is greater than α=0.1, so the decision is again do not reject H₀. Explain why the observed sign matters for a directional alternative. A result pointing the wrong way cannot support the original directional claim through a small one-tailed p-value. In words, the evidence is insufficient to support the stated alternative at this level under the given model. A failed rejection does not establish H₀ as fact.
10. Upper-tailed normal test at 5%
Hypothetical test setup. H₀: μ=50 and H₁: μ > 50. The population is assumed normal with known standard deviation 6. A random sample of n=36 gives x̄=49.2. State the significance level α=0.05 before calculating. The reference standard error is 6/√36=1, so the observed statistic is z=(49.2−50)/1=-0.8.
Critical-region route. For this upper-tailed test, the rejection condition is z ≥ 1.644854. The observed z=-0.8 does not reach that rejection region, so do not reject H₀. The chosen region uses the supplied alternative, not whichever side would make the test look more impressive.
p-value route. Under H₀, the corresponding one-sided normal p-value is approximately 0.788145. This is greater than α=0.05, so the decision is again do not reject H₀. Compare the p-value with the critical region without rounding the statistic to a whole number first. The agreement is a built-in consistency test. In words, the evidence is insufficient to support the stated alternative at this level under the given model. A failed rejection does not establish H₀ as fact.
11. Lower-tailed normal test at 5%
Hypothetical test setup. H₀: μ=50 and H₁: μ < 50. The population is assumed normal with known standard deviation 6. A random sample of n=36 gives x̄=51.5. State the significance level α=0.05 before calculating. The reference standard error is 6/√36=1, so the observed statistic is z=(51.5−50)/1=1.5.
Critical-region route. For this lower-tailed test, the rejection condition is z ≤ −1.644854. The observed z=1.5 does not reach that rejection region, so do not reject H₀. The chosen region uses the supplied alternative, not whichever side would make the test look more impressive.
p-value route. Under H₀, the corresponding one-sided normal p-value is approximately 0.933193. This is greater than α=0.05, so the decision is again do not reject H₀. Distinguish the strength of evidence under the null model from the magnitude or practical importance of a population difference. In words, the evidence is insufficient to support the stated alternative at this level under the given model. A failed rejection does not establish H₀ as fact.
12. Two-tailed normal test at 5%
Hypothetical test setup. H₀: μ=50 and H₁: μ ≠ 50. The population is assumed normal with known standard deviation 6. A random sample of n=36 gives x̄=48.5. State the significance level α=0.05 before calculating. The reference standard error is 6/√36=1, so the observed statistic is z=(48.5−50)/1=-1.5.
Critical-region route. For this two-tailed test, the rejection condition is |z| ≥ 1.959964. The observed z=-1.5 does not reach that rejection region, so do not reject H₀. The chosen region uses the supplied alternative, not whichever side would make the test look more impressive.
p-value route. Under H₀, the corresponding two-sided normal p-value is approximately 0.133614. This is greater than α=0.05, so the decision is again do not reject H₀. State the complete inference in the context of an unknown population mean, rather than writing a bare numerical inequality. In words, the evidence is insufficient to support the stated alternative at this level under the given model. A failed rejection does not establish H₀ as fact.
13. Upper-tailed normal test at 1%
Hypothetical test setup. H₀: μ=50 and H₁: μ > 50. The population is assumed normal with known standard deviation 6. A random sample of n=36 gives x̄=52.7. State the significance level α=0.01 before calculating. The reference standard error is 6/√36=1, so the observed statistic is z=(52.7−50)/1=2.7.
Critical-region route. For this upper-tailed test, the rejection condition is z ≥ 2.326348. The observed z=2.7 lies in that rejection region, so reject H₀. The chosen region uses the supplied alternative, not whichever side would make the test look more impressive.
p-value route. Under H₀, the corresponding one-sided normal p-value is approximately 0.003467. This is at most α=0.01, so the decision is again reject H₀. Imagine a different study choosing a stricter α in advance. The evidence would be identical, but the decision threshold might change. In words, there is sufficient statistical evidence for the stated alternative at this level under the given model. A failed rejection does not establish H₀ as fact.
14. Lower-tailed normal test at 1%
Hypothetical test setup. H₀: μ=50 and H₁: μ < 50. The population is assumed normal with known standard deviation 6. A random sample of n=36 gives x̄=47.3. State the significance level α=0.01 before calculating. The reference standard error is 6/√36=1, so the observed statistic is z=(47.3−50)/1=-2.7.
Critical-region route. For this lower-tailed test, the rejection condition is z ≤ −2.326348. The observed z=-2.7 lies in that rejection region, so reject H₀. The chosen region uses the supplied alternative, not whichever side would make the test look more impressive.
p-value route. Under H₀, the corresponding one-sided normal p-value is approximately 0.003467. This is at most α=0.01, so the decision is again reject H₀. If the sample mean lies close to the null mean, the test statistic is close to zero and is unlikely to cross a standard extreme-tail boundary. In words, there is sufficient statistical evidence for the stated alternative at this level under the given model. A failed rejection does not establish H₀ as fact.
15. Two-tailed normal test at 1%
Hypothetical test setup. H₀: μ=50 and H₁: μ ≠ 50. The population is assumed normal with known standard deviation 6. A random sample of n=36 gives x̄=52.1. State the significance level α=0.01 before calculating. The reference standard error is 6/√36=1, so the observed statistic is z=(52.1−50)/1=2.1.
Critical-region route. For this two-tailed test, the rejection condition is |z| ≥ 2.575829. The observed z=2.1 does not reach that rejection region, so do not reject H₀. The chosen region uses the supplied alternative, not whichever side would make the test look more impressive.
p-value route. Under H₀, the corresponding two-sided normal p-value is approximately 0.035729. This is greater than α=0.01, so the decision is again do not reject H₀. Never choose a new H₁ merely because the observed x̄ points in an unexpected direction. The research question should determine the direction. In words, the evidence is insufficient to support the stated alternative at this level under the given model. A failed rejection does not establish H₀ as fact.
16. Upper-tailed normal test at 10%
Hypothetical test setup. H₀: μ=50 and H₁: μ > 50. The population is assumed normal with known standard deviation 6. A random sample of n=36 gives x̄=47.9. State the significance level α=0.1 before calculating. The reference standard error is 6/√36=1, so the observed statistic is z=(47.9−50)/1=-2.1.
Critical-region route. For this upper-tailed test, the rejection condition is z ≥ 1.281552. The observed z=-2.1 does not reach that rejection region, so do not reject H₀. The chosen region uses the supplied alternative, not whichever side would make the test look more impressive.
p-value route. Under H₀, the corresponding one-sided normal p-value is approximately 0.982136. This is greater than α=0.1, so the decision is again do not reject H₀. A two-sided alternative has two rejection regions. Both together have total probability α under the null model. In words, the evidence is insufficient to support the stated alternative at this level under the given model. A failed rejection does not establish H₀ as fact.
17. Lower-tailed normal test at 10%
Hypothetical test setup. H₀: μ=50 and H₁: μ < 50. The population is assumed normal with known standard deviation 6. A random sample of n=36 gives x̄=50.1. State the significance level α=0.1 before calculating. The reference standard error is 6/√36=1, so the observed statistic is z=(50.1−50)/1=0.1.
Critical-region route. For this lower-tailed test, the rejection condition is z ≤ −1.281552. The observed z=0.1 does not reach that rejection region, so do not reject H₀. The chosen region uses the supplied alternative, not whichever side would make the test look more impressive.
p-value route. Under H₀, the corresponding one-sided normal p-value is approximately 0.539828. This is greater than α=0.1, so the decision is again do not reject H₀. Explain why the observed sign matters for a directional alternative. A result pointing the wrong way cannot support the original directional claim through a small one-tailed p-value. In words, the evidence is insufficient to support the stated alternative at this level under the given model. A failed rejection does not establish H₀ as fact.
18. Two-tailed normal test at 10%
Hypothetical test setup. H₀: μ=50 and H₁: μ ≠ 50. The population is assumed normal with known standard deviation 6. A random sample of n=36 gives x̄=49.9. State the significance level α=0.1 before calculating. The reference standard error is 6/√36=1, so the observed statistic is z=(49.9−50)/1=-0.1.
Critical-region route. For this two-tailed test, the rejection condition is |z| ≥ 1.644854. The observed z=-0.1 does not reach that rejection region, so do not reject H₀. The chosen region uses the supplied alternative, not whichever side would make the test look more impressive.
p-value route. Under H₀, the corresponding two-sided normal p-value is approximately 0.920344. This is greater than α=0.1, so the decision is again do not reject H₀. Compare the p-value with the critical region without rounding the statistic to a whole number first. The agreement is a built-in consistency test. In words, the evidence is insufficient to support the stated alternative at this level under the given model. A failed rejection does not establish H₀ as fact.
19. Upper-tailed normal test at 5%
Hypothetical test setup. H₀: μ=50 and H₁: μ > 50. The population is assumed normal with known standard deviation 6. A random sample of n=36 gives x̄=53. State the significance level α=0.05 before calculating. The reference standard error is 6/√36=1, so the observed statistic is z=(53−50)/1=3.
Critical-region route. For this upper-tailed test, the rejection condition is z ≥ 1.644854. The observed z=3 lies in that rejection region, so reject H₀. The chosen region uses the supplied alternative, not whichever side would make the test look more impressive.
p-value route. Under H₀, the corresponding one-sided normal p-value is approximately 0.00135. This is at most α=0.05, so the decision is again reject H₀. Distinguish the strength of evidence under the null model from the magnitude or practical importance of a population difference. In words, there is sufficient statistical evidence for the stated alternative at this level under the given model. A failed rejection does not establish H₀ as fact.
20. Lower-tailed normal test at 5%
Hypothetical test setup. H₀: μ=50 and H₁: μ < 50. The population is assumed normal with known standard deviation 6. A random sample of n=36 gives x̄=47. State the significance level α=0.05 before calculating. The reference standard error is 6/√36=1, so the observed statistic is z=(47−50)/1=-3.
Critical-region route. For this lower-tailed test, the rejection condition is z ≤ −1.644854. The observed z=-3 lies in that rejection region, so reject H₀. The chosen region uses the supplied alternative, not whichever side would make the test look more impressive.
p-value route. Under H₀, the corresponding one-sided normal p-value is approximately 0.00135. This is at most α=0.05, so the decision is again reject H₀. State the complete inference in the context of an unknown population mean, rather than writing a bare numerical inequality. In words, there is sufficient statistical evidence for the stated alternative at this level under the given model. A failed rejection does not establish H₀ as fact.
21. Two-tailed normal test at 5%
Hypothetical test setup. H₀: μ=50 and H₁: μ ≠ 50. The population is assumed normal with known standard deviation 6. A random sample of n=36 gives x̄=51.65. State the significance level α=0.05 before calculating. The reference standard error is 6/√36=1, so the observed statistic is z=(51.65−50)/1=1.65.
Critical-region route. For this two-tailed test, the rejection condition is |z| ≥ 1.959964. The observed z=1.65 does not reach that rejection region, so do not reject H₀. The chosen region uses the supplied alternative, not whichever side would make the test look more impressive.
p-value route. Under H₀, the corresponding two-sided normal p-value is approximately 0.098943. This is greater than α=0.05, so the decision is again do not reject H₀. Imagine a different study choosing a stricter α in advance. The evidence would be identical, but the decision threshold might change. In words, the evidence is insufficient to support the stated alternative at this level under the given model. A failed rejection does not establish H₀ as fact.
22. Upper-tailed normal test at 1%
Hypothetical test setup. H₀: μ=50 and H₁: μ > 50. The population is assumed normal with known standard deviation 6. A random sample of n=36 gives x̄=48.35. State the significance level α=0.01 before calculating. The reference standard error is 6/√36=1, so the observed statistic is z=(48.35−50)/1=-1.65.
Critical-region route. For this upper-tailed test, the rejection condition is z ≥ 2.326348. The observed z=-1.65 does not reach that rejection region, so do not reject H₀. The chosen region uses the supplied alternative, not whichever side would make the test look more impressive.
p-value route. Under H₀, the corresponding one-sided normal p-value is approximately 0.950529. This is greater than α=0.01, so the decision is again do not reject H₀. If the sample mean lies close to the null mean, the test statistic is close to zero and is unlikely to cross a standard extreme-tail boundary. In words, the evidence is insufficient to support the stated alternative at this level under the given model. A failed rejection does not establish H₀ as fact.
23. Lower-tailed normal test at 1%
Hypothetical test setup. H₀: μ=50 and H₁: μ < 50. The population is assumed normal with known standard deviation 6. A random sample of n=36 gives x̄=52.33. State the significance level α=0.01 before calculating. The reference standard error is 6/√36=1, so the observed statistic is z=(52.33−50)/1=2.33.
Critical-region route. For this lower-tailed test, the rejection condition is z ≤ −2.326348. The observed z=2.33 does not reach that rejection region, so do not reject H₀. The chosen region uses the supplied alternative, not whichever side would make the test look more impressive.
p-value route. Under H₀, the corresponding one-sided normal p-value is approximately 0.990097. This is greater than α=0.01, so the decision is again do not reject H₀. Never choose a new H₁ merely because the observed x̄ points in an unexpected direction. The research question should determine the direction. In words, the evidence is insufficient to support the stated alternative at this level under the given model. A failed rejection does not establish H₀ as fact.
24. Two-tailed normal test at 1%
Hypothetical test setup. H₀: μ=50 and H₁: μ ≠ 50. The population is assumed normal with known standard deviation 6. A random sample of n=36 gives x̄=47.67. State the significance level α=0.01 before calculating. The reference standard error is 6/√36=1, so the observed statistic is z=(47.67−50)/1=-2.33.
Critical-region route. For this two-tailed test, the rejection condition is |z| ≥ 2.575829. The observed z=-2.33 does not reach that rejection region, so do not reject H₀. The chosen region uses the supplied alternative, not whichever side would make the test look more impressive.
p-value route. Under H₀, the corresponding two-sided normal p-value is approximately 0.019806. This is greater than α=0.01, so the decision is again do not reject H₀. A two-sided alternative has two rejection regions. Both together have total probability α under the null model. In words, the evidence is insufficient to support the stated alternative at this level under the given model. A failed rejection does not establish H₀ as fact.
25. Upper-tailed normal test at 10%
Hypothetical test setup. H₀: μ=50 and H₁: μ > 50. The population is assumed normal with known standard deviation 6. A random sample of n=36 gives x̄=50. State the significance level α=0.1 before calculating. The reference standard error is 6/√36=1, so the observed statistic is z=(50−50)/1=0.
Critical-region route. For this upper-tailed test, the rejection condition is z ≥ 1.281552. The observed z=0 does not reach that rejection region, so do not reject H₀. The chosen region uses the supplied alternative, not whichever side would make the test look more impressive.
p-value route. Under H₀, the corresponding one-sided normal p-value is approximately 0.5. This is greater than α=0.1, so the decision is again do not reject H₀. Explain why the observed sign matters for a directional alternative. A result pointing the wrong way cannot support the original directional claim through a small one-tailed p-value. In words, the evidence is insufficient to support the stated alternative at this level under the given model. A failed rejection does not establish H₀ as fact.
26. Lower-tailed normal test at 10%
Hypothetical test setup. H₀: μ=50 and H₁: μ < 50. The population is assumed normal with known standard deviation 6. A random sample of n=36 gives x̄=51.4. State the significance level α=0.1 before calculating. The reference standard error is 6/√36=1, so the observed statistic is z=(51.4−50)/1=1.4.
Critical-region route. For this lower-tailed test, the rejection condition is z ≤ −1.281552. The observed z=1.4 does not reach that rejection region, so do not reject H₀. The chosen region uses the supplied alternative, not whichever side would make the test look more impressive.
p-value route. Under H₀, the corresponding one-sided normal p-value is approximately 0.919243. This is greater than α=0.1, so the decision is again do not reject H₀. Compare the p-value with the critical region without rounding the statistic to a whole number first. The agreement is a built-in consistency test. In words, the evidence is insufficient to support the stated alternative at this level under the given model. A failed rejection does not establish H₀ as fact.
27. Two-tailed normal test at 10%
Hypothetical test setup. H₀: μ=50 and H₁: μ ≠ 50. The population is assumed normal with known standard deviation 6. A random sample of n=36 gives x̄=48.6. State the significance level α=0.1 before calculating. The reference standard error is 6/√36=1, so the observed statistic is z=(48.6−50)/1=-1.4.
Critical-region route. For this two-tailed test, the rejection condition is |z| ≥ 1.644854. The observed z=-1.4 does not reach that rejection region, so do not reject H₀. The chosen region uses the supplied alternative, not whichever side would make the test look more impressive.
p-value route. Under H₀, the corresponding two-sided normal p-value is approximately 0.161513. This is greater than α=0.1, so the decision is again do not reject H₀. Distinguish the strength of evidence under the null model from the magnitude or practical importance of a population difference. In words, the evidence is insufficient to support the stated alternative at this level under the given model. A failed rejection does not establish H₀ as fact.
28. Upper-tailed normal test at 5%
Hypothetical test setup. H₀: μ=50 and H₁: μ > 50. The population is assumed normal with known standard deviation 6. A random sample of n=36 gives x̄=51.97. State the significance level α=0.05 before calculating. The reference standard error is 6/√36=1, so the observed statistic is z=(51.97−50)/1=1.97.
Critical-region route. For this upper-tailed test, the rejection condition is z ≥ 1.644854. The observed z=1.97 lies in that rejection region, so reject H₀. The chosen region uses the supplied alternative, not whichever side would make the test look more impressive.
p-value route. Under H₀, the corresponding one-sided normal p-value is approximately 0.024419. This is at most α=0.05, so the decision is again reject H₀. State the complete inference in the context of an unknown population mean, rather than writing a bare numerical inequality. In words, there is sufficient statistical evidence for the stated alternative at this level under the given model. A failed rejection does not establish H₀ as fact.
29. Lower-tailed normal test at 5%
Hypothetical test setup. H₀: μ=50 and H₁: μ < 50. The population is assumed normal with known standard deviation 6. A random sample of n=36 gives x̄=48.03. State the significance level α=0.05 before calculating. The reference standard error is 6/√36=1, so the observed statistic is z=(48.03−50)/1=-1.97.
Critical-region route. For this lower-tailed test, the rejection condition is z ≤ −1.644854. The observed z=-1.97 lies in that rejection region, so reject H₀. The chosen region uses the supplied alternative, not whichever side would make the test look more impressive.
p-value route. Under H₀, the corresponding one-sided normal p-value is approximately 0.024419. This is at most α=0.05, so the decision is again reject H₀. Imagine a different study choosing a stricter α in advance. The evidence would be identical, but the decision threshold might change. In words, there is sufficient statistical evidence for the stated alternative at this level under the given model. A failed rejection does not establish H₀ as fact.
30. Two-tailed normal test at 5%
Hypothetical test setup. H₀: μ=50 and H₁: μ ≠ 50. The population is assumed normal with known standard deviation 6. A random sample of n=36 gives x̄=52.6. State the significance level α=0.05 before calculating. The reference standard error is 6/√36=1, so the observed statistic is z=(52.6−50)/1=2.6.
Critical-region route. For this two-tailed test, the rejection condition is |z| ≥ 1.959964. The observed z=2.6 lies in that rejection region, so reject H₀. The chosen region uses the supplied alternative, not whichever side would make the test look more impressive.
p-value route. Under H₀, the corresponding two-sided normal p-value is approximately 0.009322. This is at most α=0.05, so the decision is again reject H₀. If the sample mean lies close to the null mean, the test statistic is close to zero and is unlikely to cross a standard extreme-tail boundary. In words, there is sufficient statistical evidence for the stated alternative at this level under the given model. A failed rejection does not establish H₀ as fact.
31. Upper-tailed normal test at 1%
Hypothetical test setup. H₀: μ=50 and H₁: μ > 50. The population is assumed normal with known standard deviation 6. A random sample of n=36 gives x̄=47.4. State the significance level α=0.01 before calculating. The reference standard error is 6/√36=1, so the observed statistic is z=(47.4−50)/1=-2.6.
Critical-region route. For this upper-tailed test, the rejection condition is z ≥ 2.326348. The observed z=-2.6 does not reach that rejection region, so do not reject H₀. The chosen region uses the supplied alternative, not whichever side would make the test look more impressive.
p-value route. Under H₀, the corresponding one-sided normal p-value is approximately 0.995339. This is greater than α=0.01, so the decision is again do not reject H₀. Never choose a new H₁ merely because the observed x̄ points in an unexpected direction. The research question should determine the direction. In words, the evidence is insufficient to support the stated alternative at this level under the given model. A failed rejection does not establish H₀ as fact.
32. Lower-tailed normal test at 1%
Hypothetical test setup. H₀: μ=50 and H₁: μ < 50. The population is assumed normal with known standard deviation 6. A random sample of n=36 gives x̄=52.2. State the significance level α=0.01 before calculating. The reference standard error is 6/√36=1, so the observed statistic is z=(52.2−50)/1=2.2.
Critical-region route. For this lower-tailed test, the rejection condition is z ≤ −2.326348. The observed z=2.2 does not reach that rejection region, so do not reject H₀. The chosen region uses the supplied alternative, not whichever side would make the test look more impressive.
p-value route. Under H₀, the corresponding one-sided normal p-value is approximately 0.986097. This is greater than α=0.01, so the decision is again do not reject H₀. A two-sided alternative has two rejection regions. Both together have total probability α under the null model. In words, the evidence is insufficient to support the stated alternative at this level under the given model. A failed rejection does not establish H₀ as fact.
33. Two-tailed normal test at 1%
Hypothetical test setup. H₀: μ=50 and H₁: μ ≠ 50. The population is assumed normal with known standard deviation 6. A random sample of n=36 gives x̄=47.8. State the significance level α=0.01 before calculating. The reference standard error is 6/√36=1, so the observed statistic is z=(47.8−50)/1=-2.2.
Critical-region route. For this two-tailed test, the rejection condition is |z| ≥ 2.575829. The observed z=-2.2 does not reach that rejection region, so do not reject H₀. The chosen region uses the supplied alternative, not whichever side would make the test look more impressive.
p-value route. Under H₀, the corresponding two-sided normal p-value is approximately 0.027807. This is greater than α=0.01, so the decision is again do not reject H₀. Explain why the observed sign matters for a directional alternative. A result pointing the wrong way cannot support the original directional claim through a small one-tailed p-value. In words, the evidence is insufficient to support the stated alternative at this level under the given model. A failed rejection does not establish H₀ as fact.
34. Upper-tailed normal test at 10%
Hypothetical test setup. H₀: μ=50 and H₁: μ > 50. The population is assumed normal with known standard deviation 6. A random sample of n=36 gives x̄=50.5. State the significance level α=0.1 before calculating. The reference standard error is 6/√36=1, so the observed statistic is z=(50.5−50)/1=0.5.
Critical-region route. For this upper-tailed test, the rejection condition is z ≥ 1.281552. The observed z=0.5 does not reach that rejection region, so do not reject H₀. The chosen region uses the supplied alternative, not whichever side would make the test look more impressive.
p-value route. Under H₀, the corresponding one-sided normal p-value is approximately 0.308538. This is greater than α=0.1, so the decision is again do not reject H₀. Compare the p-value with the critical region without rounding the statistic to a whole number first. The agreement is a built-in consistency test. In words, the evidence is insufficient to support the stated alternative at this level under the given model. A failed rejection does not establish H₀ as fact.
35. Lower-tailed normal test at 10%
Hypothetical test setup. H₀: μ=50 and H₁: μ < 50. The population is assumed normal with known standard deviation 6. A random sample of n=36 gives x̄=49.5. State the significance level α=0.1 before calculating. The reference standard error is 6/√36=1, so the observed statistic is z=(49.5−50)/1=-0.5.
Critical-region route. For this lower-tailed test, the rejection condition is z ≤ −1.281552. The observed z=-0.5 does not reach that rejection region, so do not reject H₀. The chosen region uses the supplied alternative, not whichever side would make the test look more impressive.
p-value route. Under H₀, the corresponding one-sided normal p-value is approximately 0.308538. This is greater than α=0.1, so the decision is again do not reject H₀. Distinguish the strength of evidence under the null model from the magnitude or practical importance of a population difference. In words, the evidence is insufficient to support the stated alternative at this level under the given model. A failed rejection does not establish H₀ as fact.
36. Two-tailed normal test at 10%
Hypothetical test setup. H₀: μ=50 and H₁: μ ≠ 50. The population is assumed normal with known standard deviation 6. A random sample of n=36 gives x̄=51.9. State the significance level α=0.1 before calculating. The reference standard error is 6/√36=1, so the observed statistic is z=(51.9−50)/1=1.9.
Critical-region route. For this two-tailed test, the rejection condition is |z| ≥ 1.644854. The observed z=1.9 lies in that rejection region, so reject H₀. The chosen region uses the supplied alternative, not whichever side would make the test look more impressive.
p-value route. Under H₀, the corresponding two-sided normal p-value is approximately 0.057433. This is at most α=0.1, so the decision is again reject H₀. State the complete inference in the context of an unknown population mean, rather than writing a bare numerical inequality. In words, there is sufficient statistical evidence for the stated alternative at this level under the given model. A failed rejection does not establish H₀ as fact.
Six reader questions that distinguish genuine understanding
- 1. For an upper 5% test, z=2.0: p≈0.02275, reject. The critical value 1.64485 agrees.
- 2. For an upper 1% test, the same z=2.0: p≈0.02275, do not reject. The boundary is about 2.32635.
- 3. For a lower 5% test, z=−2.0: p≈0.02275, reject. The boundary is about −1.64485.
- 4. For a two-tailed 5% test, z=1.8: p≈0.07186, do not reject. The two-sided boundary is about ±1.95996.
- 5. If p=0.04 and α=0.05 for the same correctly specified test, reject H₀.
- 6. If p=0.04 and α=0.01, do not reject H₀. The p-value did not change; the threshold did.
A four-step exam-ready conclusion paragraph
A compact written solution can follow this order: (1) state H₀ and H₁ with the population parameter and correct inequality; (2) compute and label the appropriate test statistic under H₀; (3) compare the statistic with the critical region or the p-value with α, with the other route as a check; and (4) write a sentence about whether there is sufficient evidence for the population claim in context. If a calculator gives a very small p-value, write it precisely enough for the decision, rather than rounding it to zero and implying absolute certainty.
For an upper-tail claim such as “the population mean exceeds 50”, a satisfactory final sentence is: “At the 5% significance level, there is sufficient evidence, under the stated normal-population model, that the population mean exceeds 50.” If there is insufficient evidence, use that phrase instead of saying the population mean equals 50. The statistical decision is a disciplined interpretation of data under assumptions, not proof of a physical or educational law.
A five-day mixed practice path
Day one: translate twenty short claims into H₀ and H₁, distinguishing upper, lower and two-tailed questions without any numbers. Day two: calculate z for a normal population with known variance and confirm the sign relative to μ₀. Day three: use only critical regions and sketch the rejection areas. Day four: use only p-values for the same questions and compare decisions. Day five: mix all three alternatives and significance levels, then explain why the two approaches must agree when used consistently.
For a small-group mathematics tutorial of three learners, one can write the hypotheses, one can calculate the statistic and critical boundary, and one can compute and interpret the p-value. Swap roles on the next question. This creates a natural audit: if the two decisions disagree, the group must identify which assumption or tail selection changed. The goal is independent reasoning under exam pressure rather than memorising a calculator button order.
Frequently asked questions
Which method should I use in a timed examination?
Use the method that the question requires or that you can carry out accurately with the permitted tools. The p-value method can be concise, while a critical region offers a visible threshold. They must agree for the same specified test and α.
Does p=0.03 mean H₀ has a 3% chance of being true?
No. The p-value is calculated from the probability of data at least as extreme under the assumed null model. It is not a posterior probability that H₀ is true.
Does not rejecting H₀ prove it?
No. It means the observed evidence did not meet the chosen rejection threshold for the specified alternative.
Why is the two-tailed critical value larger?
The total significance area α is split between two tails, so each tail is smaller. The corresponding absolute z boundaries are farther from zero.
Can I switch from two-tailed to upper-tailed after seeing x̄?
That would change the test post hoc. The alternative should be selected from the investigation’s question before observing the result.
Does a small p-value measure practical importance?
No. Statistical evidence and practical importance are distinct; a numerically small departure can be statistically clear in a suitable large sample.
What if variance is unknown?
Use the assumptions and method the question supplies. The 2027 H2 syllabus includes a normal population with known variance or an appropriately large-sample population-mean test, not arbitrary tests that require unintroduced distributions.
Is testing two separate population means part of 9758 for 2027?
No. The official 2027 H2 9758 hypothesis-testing section excludes testing the difference between two population means.
Related learning routes and syllabus source
- Complex Conjugation or Multiplying by i?
- Regression Transforms Back to Original Units
- Critical Region or p-Value? — this guide
- Recovering n and p from Binomial Mean and Variance
For general hypothesis-testing background, read the JC H2 hypothesis testing and p-values guide; for the choice of test direction, use one-tailed or two-tailed hypothesis tests. The Bukit Timah JC2 tuition pathway places this lesson within a broader study programme; the eduKateSG small-group mathematics tutorial reference illustrates the method of instruction at a different year level.
Curriculum authority: SEAB Singapore–Cambridge H2 Mathematics (9758) 2027 examination syllabus, Hypothesis Testing. All numbers and illustrative claims are fictional, and the calculations are not official past-year exam questions or records of pupil performance. The numerical p-values use standard-normal approximations with sufficient precision for the stated examples; decisions near a boundary should be made without premature rounding.