FCPS Part 1 Biostatistics and epidemiology

Confidence intervals

Probability, Standard Error and Confidence Intervals in Biostatistics

Addition and multiplication rules of probability, genetic risk calculations, standard error, and how to calculate and interpret a 95% confidence interval. Includes worked examples on blood groups, carrier frequency, CI width and significance for FCPS Part 1.

Probability basics

  • Probability runs from 0 (impossible) to 1 (certain); it may be written as a fraction, decimal or percentage.
  • The probabilities of all possible outcomes add up to 1, so P(not A) = 1 − P(A).
  • Odds = p ÷ (1 − p); probability = odds ÷ (1 + odds). A probability of 0.2 equals odds of 0.25 (1 to 4).

The addition rule: "or"

  • For mutually exclusive events (cannot happen together): P(A or B) = P(A) + P(B).
  • Blood groups are mutually exclusive: if 40% of people are group O and 45% group A, the chance that a random person is O or A = 0.40 + 0.45 = 0.85.
  • For events that can happen together, subtract the overlap: P(A or B) = P(A) + P(B) − P(A and B). If 30% smoke and 20% are diabetic, and 6% are both, P(smoker or diabetic) = 0.30 + 0.20 − 0.06 = 0.44.

Exam trap Before adding, check whether the events can occur together; if they can, subtract the overlap.

The multiplication rule: "and"

  • For independent events (one does not affect the other): P(A and B) = P(A) × P(B).
  • Two children both boys = 0.5 × 0.5 = 0.25; three girls in a row = 0.5³ = 0.125.
  • Conditional probability: P(A given B) = P(A and B) ÷ P(B). Bayes' theorem turns pre-test probability into post-test probability and underlies predictive values.

Remember "Or" means add (if exclusive); "and" means multiply (if independent).

Probability in genetics

SituationRisk for each child
Autosomal dominant, one heterozygous parent (Marfan, Aa × aa)1 in 2 (50%) affected
Autosomal recessive, both parents carriers (cystic fibrosis, Aa × Aa)1 in 4 affected, 1 in 2 carriers, 1 in 4 unaffected non-carriers
Unaffected sibling of an AR-affected child2 in 3 chance of being a carrier
X-linked recessive, carrier mother and normal fatherHalf of sons affected; half of daughters carriers
  • Each pregnancy is independent: having one affected child does not change the risk for the next. Two carrier parents have a 1/4 × 1/4 = 1/16 chance that two children are both affected.
  • Carrier frequency from disease frequency (Hardy–Weinberg): if an autosomal recessive disease affects 1 in 10,000, q² = 1/10,000, so q = 1/100, and carrier frequency 2pq is about 2q = 1 in 50. For CF at 1 in 2,500, q = 1/50 and carriers are about 1 in 25.

Estimation

  • A point estimate is the single best value from the sample (mean, proportion, RR). An interval estimate (the confidence interval) gives a range around it showing how precise it is.
  • Sample statistics estimate population parameters; they vary from sample to sample by chance.

Standard error

  • Standard error of the mean (SE) = SD ÷ √n. It is the SD of the means you would get if you repeated the sampling many times, so it measures the precision of the sample mean.
  • SD describes the scatter of individuals; SE describes the uncertainty in the estimate. A study of 10,000 people has the same SD as one of 100 but a much smaller SE.
  • SE of a proportion = √[p(1 − p) ÷ n].
  • SE falls with the square root of n: quadrupling the sample halves the SE.
  • The main use of the SE is to calculate confidence intervals (and test statistics).
  • Central limit theorem: means of large samples (over about 30) are approximately normally distributed even when the data are not, which is why ±1.96 SE works.

Confidence intervals

  • 95% CI = estimate ± 1.96 × SE; 99% CI = estimate ± 2.58 × SE; 90% CI = estimate ± 1.645 × SE.
  • Meaning: if the study were repeated many times, 95% of the intervals calculated would contain the true population value. Informally, we are 95% confident the true value lies inside.
  • A CI is narrower with a larger sample, less variability and a lower confidence level (a 90% CI is narrower than a 99% CI).
  • The CI shows both statistical significance and clinical importance: its position shows the size of the effect and its width shows precision, which a p value alone cannot.
Estimate typeNo-effect valueSignificant at 5% if 95% CI
Difference (means, proportions, ARR)0Excludes 0
Ratio (RR, OR, HR)1Excludes 1
  • Non-overlapping CIs for two groups imply a significant difference; overlapping CIs do not by themselves prove there is none, and a formal test is needed.

Worked examples

  • Mean BP: sample mean 130 mmHg, SD 20, n = 100. SE = 20 ÷ 10 = 2; 95% CI = 130 ± 1.96 × 2 = 130 ± 3.9, so about 126 to 134 mmHg. With n = 400, SE = 1 and the CI narrows to about 128 to 132.
  • Prevalence: 20% in a sample of 400. SE = √(0.2 × 0.8 ÷ 400) = 0.02; 95% CI = 20% ± 3.9%, about 16% to 24%.
  • Ratios: RR 1.5 (95% CI 0.9 to 2.4) is not significant because it includes 1; RR 1.5 (1.1 to 2.0) is significant.
  • Differences: mean fall in BP 5 mmHg (95% CI −1 to 11) is not significant because it includes 0.

Exam trap A confidence interval is about the population value, not about where 95% of individual patients lie; that is mean ± 1.96 SD.

Degrees of freedom

  • The number of values free to vary once a statistic is fixed: n − 1 for a single sample SD or one-sample t-test; (rows − 1) × (columns − 1) for chi-squared, so a 2 × 2 table has 1 degree of freedom.

How FCPS asks about this

  • Probability of group O or group A when 40% are O and 45% are A: 0.85 (mutually exclusive, so add).
  • Two independent events together: multiply the probabilities.
  • Both parents CF carriers: each child has a 1 in 4 chance of being affected.
  • Autosomal recessive disease at 1 in 10,000: carrier frequency about 1 in 50.
  • Primary use of the standard error of the mean: calculating confidence intervals; SE = SD ÷ √n.
  • 95% CI from mean 130, SD 20, n 100: about 126 to 134.
  • RR with 95% CI including 1: not statistically significant.

Test yourself on this topic

5 of the 8 questions in our bank on this topic. Choose an answer to see why every option is right or wrong.

Question 1Medium

A paper reports birthweights as 'mean 3.1 kg (SD 0.5 kg)' and the precision of the mean as 'SE 0.05 kg'. What does the standard deviation describe, as opposed to the standard error?

Question 2Medium

A trial reports a relative risk of 0.8 with a 95% confidence interval of 0.6 to 1.1. What can be concluded?

Question 3Medium

The probability that a baby is a boy is 0.5, and the probability of blood group O is 0.4, independently. What is the probability that a baby is a boy with blood group O?

Question 4Medium

A study of 400 term babies in Multan reports a mean birth weight of 2.9 kg with a 95% confidence interval of 2.85 to 2.95 kg. What does this confidence interval mean?

Question 5Easy

A couple have two sons and hope their next two babies will both be girls. Assuming each birth is independent and the chance of a girl is 0.5, what is the probability that both of their next two babies are girls?

Practise all 8 questions on this topic

Where this sits in the CPSP syllabus

  • Paper 1: Biostatistics, Probability, estimation and confidence intervals

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