Basic Statistics for Health Economics and Health Service Management

This is an independent study guide for postgraduate learners working with statistics in health economics and health-service management. It was prepared in the context of MHEC 51023, but it is not an official University of Kelaniya course document and does not represent institutional guidance or endorsement.

The 2026 migration retained the useful exercises and recalculated every numerical answer. It also corrected distribution choices, removed invented context and marked questions that cannot be answered from the information given. All datasets and rates in the exercises are hypothetical unless a source is explicitly identified.

These exercises are not clinical evidence or medical advice. Their health-related language provides practice in statistical reasoning; it cannot support diagnosis, treatment or policy decisions.

Calculation review date: 14 August 2026.

Module information

Field	Programme context
Programme	Master of Health Economics / Master of Health Service Management
Module	Basic Statistics
Module code	MHEC 51023
Referencing convention in the legacy outline	APA 7
Status of this page	Independent study material

Aim

Develop the statistical knowledge needed to summarise data, represent uncertainty, select an appropriate probability model and interpret quantitative evidence in health-related settings.

Learning outcomes

After using this guide, a learner should be able to:

classify variables by measurement type;
present data using suitable tables, graphs and summary measures;
translate a substantive question into a statistical quantity;
calculate probabilities for common discrete and continuous models;
distinguish individual observations from sampling distributions;
construct and interpret basic confidence intervals; and
identify when a prompt lacks the assumptions or inputs needed for a numerical answer.

Module content

The guide covers probability rules; conditional probability and Bayes’ theorem; discrete random variables; binomial, hypergeometric and Poisson models; the normal distribution; sampling distributions; and confidence intervals for a mean and proportion. The separate Probability Theory lesson provides a slower introduction to the probability axioms and diagnostic-test terminology.

Select the model from the data-generating process

Begin by defining the random variable in words. “Number of positive results among 16 people” identifies a count, while “waiting time in minutes” identifies a continuous measurement. The label attached to an exercise section does not decide the distribution. The design, support of the variable and assumptions do.

Situation	Candidate model	Questions to ask before using it
Success count across a fixed number of trials	Binomial	Is the number of trials fixed? Are there two defined outcomes? Is the success probability constant? Are trials independent?
Success count from a finite population sampled without replacement	Hypergeometric	Are the population size, number of successes and sample size fixed? Are sampled units not replaced?
Event count during a fixed exposure	Poisson	Is the rate reasonably stable over the exposure? Are events independent at the scale being modelled? Can multiple events occur?
Continuous measurement with an approximately symmetric bell-shaped distribution	Normal	Is a normal model defensible for individual observations, or only for a sample mean through a large-sample argument? Are outliers or bounds important?
Mean across repeated random samples	Sampling distribution of the mean	What are the population mean and variance? Is the sample independent or is a finite-population correction needed?

These are candidate models, not automatic classifications. Real hospital arrivals may vary by shift, weekday or season, violating a constant-rate Poisson assumption. Patient outcomes within the same facility may be correlated, violating an independent-binomial assumption. A model may remain useful after those limitations are recognised, but its results should not be reported with stronger certainty than its assumptions allow.

Keep parameters, estimates and observations separate

A parameter describes a population or probability model: examples include a population mean μ, probability p or Poisson rate λ. A statistic is calculated from a sample, such as a sample mean or sample proportion. An estimate uses a statistic to learn about an unknown parameter. An observed sample mean is therefore evidence about μ; it is not automatically equal to μ.

The distinction becomes especially important in sampling-distribution questions. The standard deviation of individual observations describes variation between people or units. The standard error describes variation between estimates from repeated samples. For independent observations with population standard deviation σ, the sample mean has standard error σ/√n. Substituting the individual standard deviation where a standard error is required can change a probability substantially.

A reproducible calculation sequence

Use the same six-step sequence for each exercise:

Define the variable and event. State, for example, that X counts damaged fruits and that the event of interest is X ≤ 5.
Record the inputs and their source. Separate figures supplied by the question from quantities you calculate. Do not turn an example threshold into a recovered fact.
Name the model and assumptions. State whether sampling is with or without replacement, whether outcomes are independent, and whether a rate is assumed constant.
Write the symbolic probability. Expressions such as P(X ≥ 8) expose off-by-one errors before software is used.
Substitute and calculate at full precision. Keep unrounded intermediate values. Round only the displayed result unless the assessment directs otherwise.
Interpret in the units and scope of the question. A probability generated by a hypothetical model is not an observed prevalence, causal effect or forecast.

Basic quality checks

Every probability must be between 0 and 1, and a complete discrete distribution must sum to 1. A complement calculation should agree with direct summation. Expected values should be plausible given the possible range. Increasing a threshold should not increase an upper-tail probability. For a symmetric standard normal distribution, P(Z < 0) = 0.5, and probabilities equidistant from zero should reflect that symmetry.

When software is used, preserve enough information for another learner to reproduce the result: distribution, parameters, tail direction and threshold. Reporting only a calculator display hides the statistical decision that produced it.

Finally, compare the numerical answer with the wording of the event. “At least”, “more than”, “no more than” and “less than” place the boundary value in different tails. Many avoidable errors come from translating these phrases incorrectly rather than from the arithmetic itself.

Reference-manager notes

The legacy page included a hand-written BibTeX block for seven textbooks. That block is not reproduced because catalogue metadata should be imported from the publisher, library catalogue or ISBN record and checked against the edition actually consulted. The reading list above supplies enough information to locate each book without creating duplicate or inaccurate Zotero records.

Practice format

The previously linked PDF and Google Drive answer template are deliberately excluded from this migration. The exercises are presented below in accessible HTML text and tables. Learners should place the assumptions, formula, substitution, result and interpretation on separate lines when preparing their own answers.

Exercises and audited answers

Attempt each problem before reading the answer. A correct numerical result does not compensate for using the wrong probability model or leaving an assumption unstated.

The answers use more digits than would normally appear in prose so that rounding can be checked. Percentages are obtained by multiplying a probability by 100; a probability of 0.01886 is 1.886%, not 0.01886%. When two reported answers differ slightly, first check whether one rounded a z-score or distribution parameter before evaluating the cumulative probability. Larger discrepancies usually indicate a different model, tail or threshold.

Probability theory

The exercise set covers simple probability, set operations, conditional probability and Bayes’ theorem. For formal definitions, first read the Probability Theory lesson.

Audited answers: probability theory

(a) Simple probability

For equally likely outcomes, P(A) = favourable outcomes / total outcomes. If one ticket is selected uniformly from 1,000,000 distinct tickets, the probability that a specified ticket is selected is 1/1,000,000 = 0.000001, or 0.0001%. The calculation is conditional on the selection being uniform; it is not evidence about any particular lottery.

(b) Card probabilities

A standard 52-card deck contains four jacks and four aces.

P(jack) = 4/52 = 1/13 ≈ 0.076923, or 7.6923%.
P(not an ace) = 48/52 = 12/13 ≈ 0.923077, or 92.3077%.

(c) Set operations with a die

For one fair die, Ω = {1, 2, 3, 4, 5, 6}. Let X = {2, 3, 5} be the prime outcomes and Y = {1, 3, 5} the odd outcomes.

Operation	Result
`X ∪ Y`	`{1, 2, 3, 5}`
`X ∩ Y`	`{3, 5}`
`X \ Y`	`{2}`
`Xᶜ`	`{1, 4, 6}`

(d) Conditional probability with two dice

Given that the sum is 9, the ordered outcomes are (3,6), (4,5), (5,4) and (6,3). The first die is larger in two of the four outcomes. Therefore:

P(first die > second die | sum = 9) = 2/4 = 0.5.

(e) Blood-group counts

From the stated group of 200 people, 70 have type O and 15 have type AB.

P(O) = 70/200 = 0.35, or 35%.
P(AB) = 15/200 = 0.075, or 7.5%.

These are sample proportions for the hypothetical group, not population prevalence estimates.

(f) Addition rule

For events that can overlap:

P(A ∪ B) = P(A) + P(B) − P(A ∩ B).

If P(A) = 0.30, P(B) = 0.25 and P(A ∩ B) = 0.10, then P(A ∪ B) = 0.45. Subtracting the intersection prevents double counting.

(g) General multiplication rule

For P(A) > 0, P(A ∩ B) = P(A) × P(B | A).

For independent events, P(B | A) = P(B), giving P(A ∩ B) = P(A)P(B). Sampling without replacement usually creates dependence because the composition of the remaining population changes.

(h) Sampling without replacement

A bag contains five white and three black balls. For two draws without replacement:

P(both black) = (3/8) × (2/7) = 3/28 ≈ 0.107143, or 10.7143%.

(i) Bayes’ theorem: three plants

The plants produce 500, 1,000 and 2,000 units, with defect probabilities 0.005, 0.008 and 0.010. Expected defective units per day are 2.5, 8 and 20. Thus:

P(plant 1 | defective) = 2.5 / (2.5 + 8 + 20) = 0.081967.

The probability is about 8.20% under the stated production mix and stable defect rates.

(j) Bayes’ theorem: rare-condition test

Assume prevalence 0.1%, sensitivity 100% and false-positive rate 5%.

P(positive) = (0.001 × 1.00) + (0.999 × 0.05) = 0.05095.

P(condition | positive) = 0.001 / 0.05095 = 0.019627.

The positive predictive value is about 1.96%. The 100% sensitivity and 5% false-positive rate are hypothetical inputs, not a claim about a real test.

(k) Bayes’ theorem: second test example

Assume prevalence 8%, sensitivity 95% and false-positive rate 7%.

P(positive) = (0.08 × 0.95) + (0.92 × 0.07) = 0.1404.

P(condition | positive) = 0.076 / 0.1404 = 0.541311.

The positive predictive value is about 54.13%. Sensitivity, specificity and predictive value must remain distinct; the FDA guidance defines these measures and their denominators.

Random variables and probability distributions

Let Y be the number of tails in three fair coin tosses. Let X be the sum of two fair dice.

Audited answers: random variables

Question (a): three coins

There are 2³ = 8 equally likely ordered outcomes.

(i) Values of `Y`

`Y` tails	Number of outcomes	`P(Y)`
0	1	1/8
1	3	3/8
2	3	3/8
3	1	1/8

The probabilities sum to 1.

(ii) Expected value and variance of `Y`

E(Y) = Σ yP(Y=y) = 0 + 3/8 + 6/8 + 3/8 = 1.5.

E(Y²) = 0 + 3/8 + 12/8 + 9/8 = 3.

Var(Y) = E(Y²) − [E(Y)]² = 3 − 2.25 = 0.75.

The standard deviation is √0.75 ≈ 0.8660 tails.

Question (b): two fair dice

Two dice produce 6 × 6 = 36, not 2⁶ = 64, equally likely ordered outcomes.

(i) Values of `X`

Sum `X`	2	3	4	5	6	7	8	9	10	11	12
Number of ways	1	2	3	4	5	6	5	4	3	2	1
Probability	1/36	2/36	3/36	4/36	5/36	6/36	5/36	4/36	3/36	2/36	1/36

(ii) Expected value and standard deviation of `X`

The sum of two independent fair dice has E(X) = 3.5 + 3.5 = 7 and Var(X) = 35/12 + 35/12 = 35/6 ≈ 5.833333.

SD(X) = √(35/6) ≈ 2.415229.

Discrete probability distributions

Use a binomial model for a fixed number of independent Bernoulli trials with constant success probability. Use a hypergeometric model for sampling without replacement from a finite population with a fixed number of successes. A Poisson model represents event counts over a fixed exposure under its rate and independence assumptions.

Audited answers: discrete distributions

(a) Finding `n`, `p` and `q`

For a binomial distribution, mean = np = 45 and variance = npq = 6² = 36. Therefore q = 36/45 = 0.8, p = 0.2 and n = 45/0.2 = 225.

(b) At least one admission

With n = 5 and p = 0.6, assuming independent outcomes with the same probability:

P(X ≥ 1) = 1 − P(X=0) = 1 − 0.4⁵ = 0.98976.

The answer is 98.976%. The independence and constant-probability assumptions would need justification in a real admissions study.

(c) Binomial upper tail

With X ~ Binomial(16, 0.235):

P(X ≥ 8) = Σ from x=8 to 16 of C(16,x)(0.235)ˣ(0.765)¹⁶⁻ˣ = 0.0188577.

The audited probability is 1.886%. The legacy answer of 5.53% overstated the tail because several component probabilities were calculated incorrectly. The rate is a hypothetical exercise input and should not be presented as an observed HIV prevalence.

(d) Medication side effects

The source prompt asks for the probability that “more than a certain number” of 100 patients experiences side effects, but it does not supply the threshold. A unique numerical answer is therefore impossible.

For a stated threshold k, with X ~ Binomial(100, 0.05), calculate P(X > k) = 1 − P(X ≤ k). If k = 8 is introduced purely as an illustration, P(X > 8) = 0.0630896, or 6.309%. That illustrative threshold is not recovered from the question.

(e) Pages sampled without replacement

There are 50 error pages among 500, and 25 distinct pages are sampled without replacement. The exact model is hypergeometric:

P(X ≥ 2) = 1 − [C(50,0)C(450,25) + C(50,1)C(450,24)] / C(500,25)

= 0.736406, or 73.641%.

The legacy binomial calculation, 72.879%, treated draws as independent. It is a close approximation because the sampling fraction is 5%, but it is not the exact model stated by the question.

(f) At most five damaged fruits

If each of 100 fruits is independently damaged with probability 0.02, the exact binomial result is:

P(X ≤ 5) = Σ from x=0 to 5 of C(100,x)(0.02)ˣ(0.98)¹⁰⁰⁻ˣ = 0.984516.

The answer is 98.452%. A Poisson approximation with λ = 2 gives 98.344%; it is useful for comparison but should be labelled as an approximation.

(g) Hospital deliveries

Assuming 3,000 deliveries are distributed through a 365-day year according to a stationary Poisson process, the daily rate is λ = 3000/365 = 8.21918.

P(X ≥ 5) = 1 − P(X ≤ 4) = 0.912247.

The expected number of days with at least five deliveries is 365 × 0.912247 = 332.97, or about 333 days.

The phrase “greatest number expected in any night” is ambiguous and cannot yield a single maximum: a Poisson distribution has no finite upper bound. If the purpose is one-night capacity planning, the model’s 95th and 99th percentiles are 13 and 16 deliveries, respectively. Those are quantiles under the assumed model, not guaranteed annual maxima.

Continuous probability distributions

The normal distribution is continuous, whereas the first piston exercise below remains binomial. This distinction matters because a section heading does not determine the correct model; the data-generating process does.

Audited answers: normal and sampling distributions

Question (a): distribution names and characteristics

(i) `X ~ Binomial(12, 0.5)`

X counts successes in 12 independent Bernoulli trials with constant success probability 0.5. Its mean is np = 6, variance is np(1−p) = 3, and standard deviation is √3 ≈ 1.7321. Because p = 0.5, the probability mass function is symmetric about 6.

(ii) `X ~ Normal(36, 16)`

Under the common notation Normal(mean, variance), the mean is 36, variance 16 and standard deviation 4. The distribution is continuous and symmetric. The 68–95–99.7 rule gives approximate intervals of 32–40, 28–44 and 24–48.

Question (b): standard normal distribution

The standard normal variable Z has mean 0 and variance 1. A normal variable X with mean μ and standard deviation σ is standardised as Z = (X−μ)/σ. A z-score states how many standard deviations an observation lies above or below the mean.

Question (c): rejected pistons

Let X ~ Binomial(10, 0.12) be the number rejected in a batch of ten, assuming independent and stable rejection outcomes.

(i) No more than two rejects

P(X ≤ 2) = P(0) + P(1) + P(2) = 0.891318, or 89.132%.

(ii) At least two rejects

P(X ≥ 2) = 1 − [P(0) + P(1)] = 0.341725, or 34.172%. “At least two” includes two; it is not the complement of “no more than two”.

Question (d): standard normal probabilities

The values below use the unrounded z-scores and the standard normal cumulative distribution function Φ.

(i) `P(Z < 1.5)`

Φ(1.5) = 0.933193.

(ii) `P(−1.5 < Z < 1.5)`

Φ(1.5) − Φ(−1.5) = 0.866386.

(iii) `P(Z > 1.046)`

1 − Φ(1.046) = 0.147781.

(iv) `P(0 < Z < 1.05)`

Φ(1.05) − 0.5 = 0.353141.

(v) `P(|Z| ≤ 1.86)`

Φ(1.86) − Φ(−1.86) = 0.937114.

Question (e): battery life

Assume individual battery life is Normal(50, 6²) months.

(i) One battery lasts less than 48 months

z = (48−50)/6 = −0.333333; therefore P(X < 48) = 0.369441, or 36.944%. Rounding the z-score to −0.33 before using a table produces 37.07%; the audited value retains the original z-score.

(ii) Mean of 36 batteries is less than 48 months

The standard error is 6/√36 = 1 month. Thus z = (48−50)/1 = −2 and P(mean < 48) = 0.022750, or 2.275%.

Question (f): student marks

For marks distributed as Normal(60, 100), the standard deviation is 10. If a mark below 40 is a fail, z = (40−60)/10 = −2, giving 2.275% below 40. The result depends on the assumed normal model; it is not an observed failure rate.

Question (g): sample mean income

Assume population mean 600, population standard deviation 100 and a random sample of 25. The standard error is 100/√25 = 20. For a sample mean below 550, z = (550−600)/20 = −2.5, so the probability is 0.006210, or 0.621% under the model.

Question (h): student heights

The prompt gives a sample of 400, an observed sample mean of 67.39 inches and a variance of 1.69, but it does not state a population model or clarify whether 1.69 is a population or sample variance. It then asks both about a sampling distribution and the number of observed students above 67.5 inches.

(i) Distribution of the sample mean

A sampling distribution describes the means from repeated samples. If the population mean were μ, population variance were 1.69 and the observations met the conditions for the central limit theorem, the approximate sampling distribution would have mean μ and standard error √1.69/√400 = 0.065. The observed sample mean 67.39 does not establish that the population mean equals 67.39.

(ii) Number above 67.5 inches

The exact number must be counted from the 400 observed heights, which are not supplied. A model-based expected count could be calculated only after specifying a population distribution and its parameters. The legacy answer of about 186 silently treated the observed sample mean and variance as known normal-population parameters, so it is not retained as an answer to the stated question.

Estimation

An interval estimate combines a point estimate with a margin of error. Its validity depends on the sampling design, estimator and model assumptions, not only on the formula.

Audited answers: confidence intervals

(a) Interval estimates

A confidence interval is a procedure that generates a range of plausible parameter values. Under repeated sampling, a 95% confidence-interval procedure captures the fixed population parameter in 95% of repetitions when its assumptions hold. It is not generally correct to say that the fixed parameter has a 95% probability of lying inside a particular interval after that interval has been computed.

(b) Interval width and sample size

For a normal-theory mean interval with fixed 95% confidence and known standard deviation 15,000, the full width is 2 × 1.959964 × 15,000/√n.

Worked comparison

Sample size `n`	Margin of error	Full interval width
50	4,157.71	8,315.42
200	2,078.86	4,157.71
500	1,314.78	2,629.57

The width is proportional to 1/√n, not 1/n. Quadrupling the sample size halves the margin of error when the other inputs remain fixed.

Why confidence intervals matter

Confidence intervals report an effect or estimate together with sampling precision. They help a reader judge whether estimates compatible with the data include values that would lead to different practical decisions. A narrow interval does not remove bias, confounding, measurement error or model misspecification. Precision and validity are separate questions.

(c) Mean weight example

The hypothetical sample contains 1,000 adult men, with sample mean 80 kg and sample standard deviation 30 kg, drawn from a population stated to contain 1,000,000 men. Because the population standard deviation is unknown, a one-sample t interval is the direct calculation:

SE = 30/√1000 = 0.948683.

With 999 degrees of freedom, t₀.₉₇₅ = 1.962341. The margin of error is 1.962341 × 0.948683 = 1.861641 kg.

The 95% interval is 78.138 kg to 81.862 kg. Applying the finite-population correction changes it only to about 78.139 kg to 81.861 kg because the sample is 0.1% of the stated population.

This interval describes sampling precision under random, representative sampling. It cannot correct undercoverage of age, geography or socioeconomic groups, and the hypothetical figures should not be reported as measured Colombo population values.

(d) Proportion of overdue accounts

In the hypothetical sample, 300/500 = 0.60. A 90% Wilson score interval is 0.56352 to 0.63541, or 56.35% to 63.54%.

The simpler Wald interval is 56.40% to 63.60%, close here because the sample is large and the estimate is away from 0 and 1. The interval estimates the population proportion under representative sampling; it does not establish why accounts are overdue or support an industry comparison.

(e) Relative risks and confidence intervals

Hypothetical pill	Reported relative risk	Reported 95% interval	Statistical reading
A	0.80	0.62–0.91	Compatible with a relative risk reduction; the interval excludes 1
B	0.60	0.50–1.10	Imprecise; compatible with benefit, no effect or a small increase
C	1.02	1.04–1.33	Internally impossible because the point estimate lies outside its interval

Pill C must be checked against the source data before interpretation. Pill A has the clearest statistical signal of the three reported rows, but none can be recommended from these figures alone. Decision-making also requires absolute baseline risk, adverse effects, study validity, certainty of evidence, costs and patient-relevant outcomes.

Sources, limitations and disclosure

Sources

Illowsky, B. and Dean, S. (2023) Introductory Statistics 2e. OpenStax: probability formulas, binomial distribution, Poisson distribution, Student t intervals and population proportions.
Pennsylvania State University, Eberly College of Science (n.d.) STAT 414: Introduction to Probability Theory.
US Food and Drug Administration (2007) Statistical Guidance on Reporting Results from Studies Evaluating Diagnostic Tests.
Kirkwood, B.R. and Sterne, J.A.C. (2003) Essential Medical Statistics. 2nd edn. Blackwell Science.
Rosner, B. (2016) Fundamentals of Biostatistics. 8th edn. Cengage Learning.

Limitations

The guide recalculates inherited exercises; it does not validate the origin of their hypothetical inputs.
Distributional assumptions are stated locally. Real data require design checks, diagnostics and subject-matter justification.
Several prompts are incomplete or internally inconsistent. Those defects are reported instead of being repaired with undisclosed assumptions.
Rounded displayed values may differ slightly from software output that retains more digits, but every calculation here used full precision before final rounding.
This introductory guide does not cover causal inference, missing data, complex survey designs, regression, economic evaluation or decision modelling.

Academic integrity and AI disclosure

Use these answers to check an independent attempt. Do not submit the wording, interpretations or calculations as your own assessed work without following the applicable academic-integrity and citation rules.

AI assistance was used during migration to extract the legacy exercises, identify calculation discrepancies and restructure the material. Numerical answers were independently recomputed with standard probability distributions and checked against the cited statistical references. The author remains responsible for the assumptions, corrections and final publication.