← Common Core coursesBMS 1110 · Applied Mathematics and Medical Biostatistics
BMS 1110 · Common Core (Tronc commun)Section 6.4 · Sampling, estimation and inference
Section 30 of 30

Estimation, hypotheses and significance

Compute a point estimate of a population mean or proportion from sample data, state the sampling error, and use a null hypothesis, an alternative hypothesis and a p-value at α = 0.05 to decide what a study result does and does not show.

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Why this matters

Almost every number in a research paper or a hospital report is an estimate, and an estimate without its uncertainty is only half a fact. When a colleague says “20% of our patients have hypertension”, you need to know whether that 20% came from 40 patients or 4,000, and whether it describes a sample or the whole population. When a trial reports p = 0.03, you need to know that this is a statement about data, not proof that the treatment works — and that a statistically significant 0.2 mmHg change may still be clinically useless. These three habits — estimate, state the uncertainty, make the decision — are what let you read evidence instead of being persuaded by it.

What you need to know first

You should be comfortable with the mean and the standard deviation from Unit 5, and with the difference between a population and a sample and between a parameter and a statistic from section 6.1. No test formula is used here. This section is about what estimates and p-values mean and how to read them safely.

Words to know (6)
Parameter
A numerical characteristic of a whole population, such as the population mean μ or the population proportion π. It is usually unknown because measuring everyone is impractical. The true mean systolic blood pressure of every adult in the city is a parameter; no one can measure it exactly.
Point estimate
A single numerical value calculated from sample data and used as the estimate of a population parameter. A sample mean of 70 kg, used to estimate the population mean weight, is a point estimate.
Sampling error
The difference between an estimate and the true value of the parameter, expressed in the units of the estimate. It comes from which individuals the sample happens to contain. A sample mean of 71 kg when the true mean is 70 kg gives a sampling error of 71 − 70 = +1 kg.
Null hypothesis H0
The baseline assumption tested by a study: no difference, no association, no effect. It is written before the data are analysed. H0: the mean systolic pressure is the same on the new drug and the standard drug.
Significance level α
The threshold, chosen before the results are seen, that decides when evidence against H0 is strong enough. A common value is α = 0.05. With α = 0.05 fixed in advance, a result of p = 0.02 leads to rejecting H0 and p = 0.21 does not.
p-value
The probability of obtaining data at least as extreme as those observed if H0 were true under the statistical model. It is not the probability that H0 is true. p = 0.02 means data this extreme would appear about 2% of the time if H0 were true.
One idea at a timeIdea 1 / 7

In one sentence Estimation means using data from a sample to make an approximate statement about an unknown characteristic of the whole population.

Estimation uses a sample to describe a population you cannot measure

Estimation is the process of using information obtained from a sample to make an approximate statement about an unknown characteristic of a population. A population parameter — a mean, a proportion, a variance or a standard deviation — is usually unknown, because studying every individual is expensive, slow or simply impossible. Suppose a researcher wants the average systolic blood pressure of every adult in a city. Measuring all of them is not realistic, so the researcher selects a sample of 500 adults and calculates their average. That sample average is then used as the estimate of the population average. The logic is empirical: the evidence comes from observed data rather than from a theory. The estimate is honest only if the sample was selected so that it represents the population, which is why the sampling methods of the earlier sections matter here.

For example

A hospital wants the mean waiting time in its outpatient clinic. It times 120 consecutive patients, obtains a mean of 48 minutes, and reports that mean as its estimate of the waiting time for all outpatients.

Pause and check yourself

A researcher measures the haemoglobin of 200 students and reports a mean of 13.5 g/dL. What exactly is being estimated, and why is the true value unknown?

Show the answer

The sample mean of 13.5 g/dL is estimating the population mean haemoglobin of all students in the group being studied. The true value is unknown because not every student was measured; only a sample of 200 was, and each different sample would give a slightly different mean.

Read detailed explanations
What a confidence interval adds to a point estimate

A point estimate is a single value, and a single value hides how much it could be wrong. A confidence interval supplies the missing information by giving a range of plausible values for the parameter. The recipe uses three quantities: the estimate itself, the standard error of the estimate, and a multiplier that depends on the chosen confidence level. The standard error of a sample mean is s ÷ √n, where s is the sample standard deviation and n is the sample size; it measures how much the sample mean would wobble from one sample to the next. For 95% confidence the multiplier is about 1.96, so the interval runs from estimate − 1.96 × standard error to estimate + 1.96 × standard error. Suppose a sample of 100 patients has a mean systolic pressure of 128 mmHg and a standard deviation of 15 mmHg. The standard error is 15 ÷ √100 = 15 ÷ 10 = 1.5 mmHg, and the margin of error is 1.96 × 1.5 ≈ 2.9 mmHg, so the 95% interval is roughly 125.1 to 130.9 mmHg. Note carefully what the interval does and does not say. It does not say that 95% of patients have pressures in this range; that is the empirical rule for individual values, not for means. It says that if the study were repeated many times, 95% of the intervals so built would contain the true population mean. The width responds to sample size through the square root: to halve the margin of error you must quadruple n, from 100 to 400. That square-root rule is why very large studies are expensive, and why a narrow interval is the real currency of estimation. In nursing research, an interval that crosses a difference of zero — or a value that matters clinically — tells the reader that the data are compatible with no effect, even when the point estimate looks impressive.

The two mistakes a test can make: false positives, false negatives and power

A decision to reject H0 is a judgement made under uncertainty, so it can be wrong in two directions, and naming those directions is what lets you read a test result honestly. The first mistake is a Type I error, also called a false positive: H0 is true, but the data look extreme enough to reject it. The significance level α is exactly the long-run rate of these errors you are willing to accept; choosing α = 0.05 means that when H0 is genuinely true, about 5 in 100 identical studies would still produce p < 0.05. The second mistake is a Type II error, a false negative: H0 is false, a real effect exists, but the data are too weak to detect it. Its rate is written β, and the ability to avoid it is the power of the test, power = 1 − β. Conventionally researchers aim for power of 0.80 or more, and power is raised by increasing the sample size, reducing measurement error, or choosing a design with less variability. Power is calculated before a study begins, which is the real reason the sample size matters: an underpowered study can miss a genuine and important effect and then be reported as “no difference”. Two further details change the arithmetic. A two-sided test splits α across both directions of the effect, while a one-sided test puts it all in one direction and is easier on the same data; a one-sided test is only legitimate when a change in the opposite direction would be uninteresting, which is rare in clinical work. Finally, because a researcher could compute a p-value and then choose a threshold to fit it, α must be declared in advance. Fixing the threshold before the data are opened, and reporting the test used and the effect size with its units, is what keeps the procedure honest rather than merely arithmetical.

Matching the statistical test to the research situation

No single test fits every question. The appropriate test depends on the research question, the study design, the type of variable measured and the assumptions the data must satisfy. The common matches are these. When one sample mean is compared with a fixed reference value — a ward’s mean waiting time against a national target — the one-sample t-test is used. When the means of two separate groups are compared — patients on a new drug against patients on a standard drug — the independent-samples t-test is used, and the two groups must be genuinely independent. When the same people are measured twice — blood pressure before and after a treatment, or a matched pair of twins — the paired t-test is used, because each person acts as their own control and the two measurements are linked. When two categorical variables are cross-tabulated — sex against whether a patient has hypertension — the chi-square test is used, and it works on the counts in the cells rather than on means. When the association between two quantitative variables is examined — height and weight, or dose and blood level — correlation is used. When one quantitative variable is used to predict another — a regression of blood pressure on age — regression is used. When more than two group means are compared in one analysis — three diets, or four wards — analysis of variance, shortened to ANOVA, is used, because running several t-tests instead would inflate the chance of a false positive. The choice must be justified by the design and not by familiarity. It also carries assumptions worth knowing: t-tests and ANOVA assume roughly normal data and independent observations, chi-square assumes adequate expected counts in every cell, and correlation assumes a straight-line relationship. Checking those assumptions is part of the analysis, and a wrong test produces a p-value that answers a question nobody asked.

Explore the diagram

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Estimation, hypotheses and significance

A four-panel teaching map. Panel 1 turns five observed weights into a point estimate of the population mean and states its sampling error. Panel 2 turns 80 hypertensive patients out of 400 into an estimated proportion and percentage, then states the sampling error against a known truth. Panel 3 lays out the hypothesis-testing sequence: H0 and H1, the significance level α fixed in advance, the p-value, and the decision rule at 0.05. Panel 4 matches six common research situations to the statistical test each one calls for, and lists the six factors that set sample size.

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01 / 5Point estimate from sample data

Estimation, hypotheses and significanceA four-panel teaching map. Panel 1 turns five observed weights into a point estimate of the population mean and states its sampling error. Panel 2 turns 80 hypertensive patients out of 400 into an estimated proportion and percentage, then states the sampling error against a known truth. Panel 3 lays out the hypothesis-testing sequence: H0 and H1, the significance level α fixed in advance, the p-value, and the decision rule at 0.05. Panel 4 matches six common research situations to the statistical test each one calls for, and lists the six factors that set sample size.From sample data to a point estimateSample weights (kg): 60, 65, 70, 75, 80x̄ = (60+65+70+75+80) ÷ 5 = 350 ÷ 5= 70 kgPoint estimate of μ = 70 kgIf the true μ is 71 kg, sampling error =70 − 71 = −1 kgEstimating an unknown percentageSurvey of 400 patients: 80 have hypertensionp̂ = x ÷ n = 80 ÷ 400 = 0.20Percentage = 0.20 × 100 = 20%If the true proportion is 0.18,sampling error = 0.20 − 0.18 = 0.02,that is 2 percentage pointsHypotheses, α and the p-valueH0: baseline — no difference, no associationH1: the effect, difference or association studiedSet α before the results: α = 0.05p = chance of data this extreme if H0 were truep < 0.05 → reject H0p ≥ 0.05 → do not reject H0p is not the probability that H0 is trueMatching a test to the situationOne mean vs a reference → one-sample t-testTwo independent means → independent-samples t-testPaired measurements → paired t-testTwo categorical variables → chi-square testTwo quantitative variables → correlationMore than two group means → ANOVASample size: precision, confidence, variability,expected proportion, design and resources

If the drawing is wider than this panel, scroll it sideways. Labels stay readable; the numbered explanations below show the same ideas.

Point estimate from sample data

A point estimate is one number standing in for an unknown population parameter. Here five weights, 60, 65, 70, 75 and 80 kg, are added to 350 kg and divided by five, giving a sample mean of 70 kg that estimates the population mean μ. The estimate carries the unit of the data, and on its own it gives no sense of how far it might be from the truth.

Schematic teaching map drawn to a regular layout; the four panels are independent and are not to scale with one another. The numbers are the worked values used in the section.

Key points to remember

  • Estimation uses observed sample data to make an approximate statement about an unknown population parameter.
  • A point estimate is one number: the sample mean estimates μ, the sample proportion estimates π, and the estimated percentage is the proportion × 100.
  • Sampling error is estimate − true value and shrinks as the sample grows; bias is a systematic pull in one direction and does not.
  • Sample size is fixed by desired precision, confidence level, expected variability, expected proportion, study design and resources.
  • H0 is the baseline of no difference or no association; H1 is the effect investigated, and failing to reject H0 is not proof that H0 is true.
  • α is fixed before the results; the p-value is the probability of data at least as extreme as those observed if H0 were true, and with α = 0.05, p < 0.05 leads to rejecting H0.
  • Statistical significance is not clinical significance: a tiny effect can be statistically significant in a large study and still be irrelevant to practice.
Go deeper

Worked example

Estimating a mean and stating the sampling error. A student nurse weighs five patients on a ward: 60, 65, 70, 75 and 80 kg. The sample mean will be used as the estimate of the mean weight of all patients on the ward, whose true value is believed to be 71 kg. Find the point estimate and the sampling error.

  1. Add the five observations to obtain the total weight: 60 + 65 + 70 + 75 + 80 = 350 kg.
  2. Divide the total by the number of observations, n = 5, because a mean shares the total equally: 350 ÷ 5 = 70 kg.
  3. State the point estimate with its unit. The sample mean is x̄ = 70 kg, and it estimates the unknown population mean μ.
  4. Subtract the true value from the estimate to obtain the sampling error: 70 − 71 = −1 kg. The negative sign means this sample under-estimates the true mean by 1 kg.
  5. Check the interpretation. Another sample of five patients would give a different mean and a different sampling error; the variation is random, not a fault in the method, and it would shrink if the sample were larger.

The point estimate is 70 kg and the sampling error against a true mean of 71 kg is −1 kg. A point estimate is a value with uncertainty attached: reporting 70 kg alone hides the possibility that the truth is one kilogram away, and a larger sample would narrow that gap.

Tutorial 1 — estimating a proportion, its percentage and the sampling error

A maternal clinic wants to estimate how common anaemia is among pregnant women in its district. It surveys 250 pregnant women and finds that 45 of them are anaemic. The district health office believes the true proportion is 0.15. Work out the point estimate, the estimated percentage, and the sampling error.

Rule used: Estimated proportion = number with the characteristic ÷ sample size. Estimated percentage = proportion × 100. Sampling error = the estimate − the true value, in the units of the estimate; here the unit is a proportion, and it can also be read in percentage points.

  1. Write down x, the number with the characteristic, and n, the number in the sample.

    Reveal this step

    x = 45 women with anaemia and n = 250 women surveyed. The rule is p̂ = x ÷ n, so the two numbers must be identified before any division is done.

  2. Why must the denominator be the 250 women surveyed and not the 45 who are anaemic?

    Reveal the answer

    Because a proportion compares a part with the whole. Dividing 45 by 45 would give 1, and dividing 250 by 45 would give more than 5; only 45 ÷ 250 expresses the share of the sample that is anaemic, which is what estimates the population proportion.

  3. Divide x by n to obtain the point estimate of the proportion.

    Reveal this step

    45 ÷ 250 = 0.18. The point estimate is p̂ = 0.18, a proportion between 0 and 1, and it estimates the unknown population proportion π.

  4. Convert the proportion to an estimated percentage and say what it means in one sentence.

    Reveal the answer

    0.18 × 100 = 18%. The estimate is that 18% of pregnant women in the district are anaemic. This is subject to sampling uncertainty, because it comes from 250 women and not from all pregnant women in the district.

  5. Compare the estimate with the believed true proportion of 0.15 and state the sampling error.

    Reveal this step

    Sampling error = estimate − true value = 0.18 − 0.15 = 0.03. In percentage terms, this sample over-estimates the true proportion by 3 percentage points.

  6. A second sample of 250 women gives 34 anaemic cases instead of 45. Is this change sampling error or bias, and why?

    Reveal the answer

    Sampling error: a different set of women was selected, so the estimate moved. Bias would be a systematic pull in one direction caused by the method, such as refusing to include women who attend the clinic only at night; random change between samples is not bias.

Answer: p̂ = 45 ÷ 250 = 0.18, so the estimated percentage is 18%, and the sampling error against a true proportion of 0.15 is 0.03, that is 3 percentage points.

Keep this habit: A proportion always divides the part by the whole sample, and the percentage is the proportion multiplied by 100. State the sampling error in the same units as the estimate, and keep it mentally separate from bias: sampling error moves with each sample, bias persists across all of them.

Tutorial 2 — from hypotheses to a decision, then to clinical judgement

A randomised trial compares a new antihypertensive drug with the standard drug on mean systolic blood pressure. The new drug lowers the mean by 1.5 mmHg more than the standard drug, and the test reports p = 0.02. The protocol fixed α = 0.05, and clinicians agreed in advance that a difference of at least 5 mmHg would be clinically important.

Rule used: State H0 as no difference and H1 as the difference being investigated. Fix α before seeing the results. Decision rule: if p < α, reject H0; if p ≥ α, do not reject H0. Then compare the effect size, with its units, against the clinically important difference.

  1. State H0 and H1 for this trial in words.

    Reveal the answer

    H0: the mean systolic blood pressure is the same on the new drug and on the standard drug — no difference. H1: the mean systolic blood pressures differ between the two drugs. Both statements are written before the data are examined.

  2. Write the decision rule using the protocol value α = 0.05.

    Reveal this step

    Reject H0 if p < 0.05; do not reject H0 if p ≥ 0.05. The threshold is fixed by the protocol before the analysis, not chosen after seeing the result.

  3. The reported value is p = 0.02. What is the decision, and what does it mean in words?

    Reveal the answer

    p = 0.02 is less than 0.05, so we reject H0. The data are unlikely if there were truly no difference in mean systolic pressure between the two drugs. This is statistical significance: the result is far enough from the baseline of no effect.

  4. A colleague says the result proves the new drug works and that there is a 2% chance the two drugs are equal. Correct both statements.

    Reveal the answer

    Neither is right. Rejecting H0 is not proof that H1 is true, and the p-value is not the probability that H0 is true. The value p = 0.02 is the probability of data at least as extreme as these if H0 were true: it is a statement about the data under H0.

  5. Compare the observed difference of 1.5 mmHg with the clinically important difference of 5 mmHg.

    Reveal this step

    1.5 mmHg is well below the 5 mmHg that clinicians agreed would matter, so the effect is not clinically important even though the test is statistically significant. The trial was large enough to detect a small difference that no one would act on.

  6. How would you report this result to a ward colleague in two sentences?

    Reveal the answer

    The trial found a statistically significant difference in mean systolic pressure of 1.5 mmHg with p = 0.02, but this is smaller than the 5 mmHg we agreed would be clinically important. So the evidence does not justify changing practice for this drug yet.

Answer: p = 0.02 < 0.05, so H0 is rejected and the difference is statistically significant; the effect of 1.5 mmHg is nevertheless below the 5 mmHg clinical threshold, so it is not clinically important.

Keep this habit: Fix α before the analysis and apply the rule mechanically; then put the p-value down and look at the size of the effect in its units. Statistically significant and clinically important are two separate questions, and a study can answer yes to the first and no to the second.

Now you try

Attempt each one on paper first. Write the formula, substitute the numbers, then check the unit of your answer.

  1. A sample of five patients has weights 62, 68, 70, 74 and 76 kg. Compute the point estimate of the population mean weight, and state the sampling error if the true population mean is 69 kg.

    Hint

    Add the five weights first, then divide by five. Sampling error is the estimate minus the true value, and it carries kilograms.

    Show the worked answer

    62 + 68 + 70 + 74 + 76 = 350 kg. The sample mean is 350 ÷ 5 = 70 kg, so 70 kg is the point estimate of the population mean weight. If the true mean is 69 kg, the sampling error is 70 − 69 = +1 kg, meaning this sample over-estimates the true mean by 1 kg.

  2. A survey of 600 patients finds that 90 have been vaccinated against influenza. Compute the estimated proportion and the estimated percentage, and state the sampling error if the true proportion is 0.14.

    Hint

    Divide the 90 vaccinated patients by the 600 surveyed, multiply by 100 for the percentage, then subtract 0.14 from the proportion.

    Show the worked answer

    p̂ = 90 ÷ 600 = 0.15. The estimated percentage is 0.15 × 100 = 15%. If the true proportion is 0.14, the sampling error is 0.15 − 0.14 = 0.01, that is 1 percentage point; the sample slightly over-estimates vaccination coverage.

  3. A study of a new dressing reports p = 0.21 for the difference in healing time, with α fixed at 0.05. A second study of the same dressing reports p = 0.03. State the decision in each study and explain what the p-value does not say.

    Hint

    Compare each p with 0.05 before interpreting. The p-value is computed assuming H0 is true, so it cannot be the probability that H0 is true.

    Show the worked answer

    In the first study p = 0.21 ≥ 0.05, so we do not reject H0; the data are compatible with no difference. In the second p = 0.03 < 0.05, so we reject H0 and call the result statistically significant. Neither p-value states the probability that H0 is true, nor the probability that the dressing works; each is the probability of data at least as extreme as those observed if H0 were true.

  4. Match each situation to the test named in this section: (a) the mean HbA1c of the same 40 patients before and after a diet programme; (b) three wards compared on mean waiting time; (c) sex and hypertension status in one sample; (d) one ward’s mean waiting time compared with a national target of 30 minutes.

    Hint

    Ask first whether the comparison is between independent groups, the same people twice, categories counted in cells, or one mean against a fixed value.

    Show the worked answer

    (a) Paired t-test, because the same patients are measured twice and the two measurements are linked. (b) ANOVA, because more than two group means are compared and several t-tests would inflate the risk of a false positive. (c) Chi-square test, because both variables are categorical and the data are counts in a table. (d) One-sample t-test, because one sample mean is compared with a fixed reference value.

Common misconception

The most common misreading of a p-value is to treat it as the probability that the null hypothesis is true. A trial reporting p = 0.02 does not mean “there is a 2% chance the treatments are equal”; it means that if the treatments really were equal, data at least as extreme as these would appear about 2% of the time. The probability runs from the hypothesis to the data, not from the data back to the hypothesis. The same confusion produces two further errors: believing that p < 0.05 proves the alternative hypothesis, and believing that p ≥ 0.05 proves there is no effect. Neither follows. A p-value says how unusual the observed data are under a stated model, and rejecting H0 is a decision made against a threshold fixed in advance, not a demonstration of truth.

Check your reasoning

A colleague reports p = 0.04 for a new wound-care protocol and says: “This proves the protocol works, and there is only a 4% chance it does not.” Write the two questions you would ask to test that claim, and say what extra number you would want to see.

Reveal an answer

I would ask what the null hypothesis was and what size the effect had, in its units. The p-value is the probability of data at least this extreme if H0 were true, so it does not measure the probability that the protocol is ineffective; the 4% figure is not that. I would also want to see the effect size with a confidence interval, and the clinically important difference agreed in advance, because a small effect can reach p = 0.04 in a large study and still be too small to change practice.

Where this comes from

  • BMS 1110 Applied Mathematics and Medical Biostatistics lecture notes (supplied) · Estimation and hypothesis-testing block, lines 1106–1158: empirical estimation (60 of 500 → p = 0.12 → 12%), parameters and statistics, the point estimate of a mean (60, 65, 70, 75, 80 → 350 ÷ 5 = 70 kg), estimating an unknown percentage (80 ÷ 400 = 0.20 → 20%), sampling error versus bias (71 − 70 = 1 kg), and the six factors that set sample size. Lines 1212–1257: the null and alternative hypotheses, the significance level α = 0.05 chosen before the results, the p-value and the decision rule at 0.05, the caveat that a p-value is not the probability that H0 is true, the table of statistical tests, and statistical versus clinical significance (the 0.2 mmHg example).
  • Note on the course code · The supplied cover page reads “Course Number: BMS 110”. The curriculum catalogue in this project records “BMS 1110”; the catalogue is treated as authoritative and the notes’ cover value is retained here for provenance.
  • Beyond the supplied notes: confidence intervals and test error rates · The standard-error formula s ÷ √n, the 1.96 multiplier and the 95% confidence interval, the square-root relationship between sample size and margin of error, the distinction between the interval for a mean and the empirical rule for individuals, Type I and Type II errors, β, power = 1 − β, one-sided versus two-sided tests, and the assumptions of the named tests are standard biostatistics that the supplied notes do not develop.

Lesson progress is saved on this device for the current learner. Practice answers stay separate and are not stored.

About this course and its sources

Bilingual lessons organised from the supplied BMS 1110 Applied Mathematics and Medical Biostatistics lecture notes (Mr. Ebangan Prosper Achuanoh). Diagrams are original educational schematics; sections that go beyond the supplied notes say so in their sources. Mathematics, clinical and French-language review are pending.