Normal population, known variance
For a random sample of size from a normal population with known variance , use .
H1 MATHEMATICS · HYPOTHESIS TESTING (SYLLABUS 8865)
A dedicated guide for Singapore A-Level H1 Mathematics. Formulate hypotheses for a population mean , apply the Central Limit Theorem when needed, calculate the p-value using your graphing calculator, and write a rigorous contextual conclusion.
THE TESTING LOGIC
The null hypothesis is the reference model. Under , calculate how unusual the observed sample mean would be.
If the p-value is no greater than the significance level, reject . Otherwise, do not reject ; a test does not prove that the null hypothesis is true.
SYLLABUS MODELS
For a random sample of size from a normal population with known variance , use .
When the parent distribution is not known to be normal and , invoke the Central Limit Theorem and use the unbiased estimate of the population variance.
If the given sample variance uses divisor , then .
Do not use tests for a population proportion, binomial hypothesis tests, two-sample tests or t-tests for an 8865 H1 hypothesis-testing question.
THE FOUR-PART CHECK
Define fully in context, including the quantity, units and population.
Use the population mean—not the observed sample mean—in and .
State the normal model and justify the CLT when the population is not known to be normal.
Compare the labelled p-value with , state the decision, then conclude about the population mean in context.
WORKED EXAMPLE
An energy drink manufacturer claims that each can contains a mean caffeine content of 120 mg. A consumer group suspects that the true mean is greater than 120 mg. A random sample of 40 cans has mean 121.8 mg and standard deviation 5.2 mg calculated using divisor 40. Test at the 5% level of significance.
The divisor matters. Here 5.2 mg is , so convert it to the unbiased estimate before using it in the large-sample test.
Let be the population mean caffeine content, in mg, per can of this energy drink.
Since the population distribution is unknown but , by the Central Limit Theorem,
On a graphing calculator, use a one-sample Z-Test with alternative .
Since p-value = 0.0153 < 0.05, reject at the 5% level of significance. Equivalently, , the right-tail critical value.
There is sufficient evidence at the 5% level of significance to conclude that the population mean caffeine content per can is greater than 120 mg.
COMMON MISTAKES
Hypotheses concern the population mean , never the observed sample mean.
If the population is not stated to be normal, justify the approximate normal model using the large sample.
The stated distribution and test statistic are based on the assumption that the null hypothesis is true.
When the result is not significant, write “do not reject” or “fail to reject” .
Name the population mean, its context and the significance level; do not write only “the claim is supported”.
PRACTISE WITH FEEDBACK
Ask Integrand to check your parameter, hypotheses, model, calculator result and contextual conclusion.
Try IntegrandDefine and state and .
Choose the correct normal model and obtain the p-value or critical value.
Compare, decide and conclude about the population mean in context.
QUICK QUESTIONS
One-sample tests for a population mean : either a sample from a normal population of known variance or a large sample from any population. One- and two-tailed tests are included. Tests for proportions, binomial hypothesis tests, two-sample tests and t-tests are not included.
Use the CLT when the parent population is not known to be normal and the sample is sufficiently large, such as . If the population is already stated to be normal, a CLT justification is unnecessary.
Both are within the syllabus. The p-value method using a graphing calculator Z-Test is often faster, while a question may explicitly ask for a critical value or critical region.
It is the probability, assuming is true, of obtaining a result at least as extreme as the observation in the direction of .
Syllabus reference: SEAB 2026 H1 Mathematics 8865, sections 3.4–3.5.
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