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1Formulas and methods
Evidence-based practice statistics
Sensitivity and specificity
Se = a/(a + c), Sp = d/(b + d)
Predictive values
PPV = a/(a + b), NPV = d/(c + d)
Accuracy
(a + d)/n
Likelihood ratios
LR+ = Se/(1 − Sp), LR− = (1 − Se)/Sp
PPV at prevalence p
Se·p / (Se·p + (1 − Sp)(1 − p))
Odds
odds = p/(1 − p); p = odds/(1 + odds); post-test odds = pre-test odds × LR
Risk measures
RR = EER/CER, ARR = CER − EER, RRR = ARR/CER
Number needed to treat
NNT = 1/ARR, rounded up; NNH = 1/(EER − CER)
Odds ratio
OR = ad/(bc)
Incidence rate
new cases / person-time
Method
Draw the 2 × 2 table and label a, b, c and d before you calculate.
Read sensitivity and specificity down the columns, predictive values along the rows.
For another prevalence, imagine 10 000 people and fill the table with Se, Sp and the prevalence.
For a likelihood ratio, convert the probability to odds, multiply, and convert back.
For a trial, find EER and CER first; then RR = EER/CER, ARR = CER − EER and NNT = 1/ARR rounded up.
Descriptive statistics
Mean
x̄ = Σx/n
Sample standard deviation
s = √(Σ(x − x̄)²/(n − 1))
Median
middle value of the sorted data
z-score
z = (x − μ)/σ
Normal distribution
68% within ±1σ, 95% within ±2σ
Confidence intervals
σ known
x̄ ± z·σ/√n
σ unknown
x̄ ± t·s/√n, df = n − 1
Critical z
90%: 1.645, 95%: 1.96, 99%: 2.576
Sample size
n = (z·σ/E)², rounded up
Method
Decide between z (population σ known) and t (only the sample s, with n − 1 degrees of freedom).
Find the critical value for the confidence level.
Margin of error: E = critical value × standard deviation/√n.
Interval: x̄ ± E.
Hypothesis tests & confidence intervals
z-test for a mean
z = (x̄ − μ0)/(σ/√n)
t-test for a mean
t = (x̄ − μ0)/(s/√n), df = n − 1
Two-sided at 5%
reject H0 if |z| > 1.96
Confidence interval
x̄ ± z·σ/√n
Method
State H₀ and H₁ and decide whether the test is one- or two-sided.
Calculate the test statistic: (x̄ − μ₀)/(σ/√n), or with s and t.
Compare with the critical value, or find the p-value.
Reject H₀ if the statistic is beyond the critical value (p below α), and state the conclusion in words.
2Practice problems
1.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). In a cohort, groups of people were followed for different lengths of time (see the table: number of people and years of follow-up for each). During follow-up, 176 new cases occurred. Find the incidence rate IR per 1000 person-years. Round to one decimal place.
Practice exercise (made-up values, not clinical guidance). A patient has a pre-test probability of 40% for a disease. A test with sensitivity 91% and specificity 95% comes back positive. Find the post-test probability p′ of the disease as a percentage. Round to one decimal place.
pre-test 40%Se = 91%Sp = 95%positive
p′ = %
3.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 44, 34, 35, 25, 28, 52.
n = 6
s =
4.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 8. What proportion P of values is above 53.2?
μ = 50σ = 8x = 53.2
P =
5.Confidence intervalsMedium
A sample of 12 items has mean 16.5 and standard deviation s = 14. The population is roughly normal. Find the upper limit of the 95% confidence interval for the mean.
n = 12x̄ = 16.5s = 1495%t(11) = 2.201
upper limit =
6.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 10 plots, mean yield 44.5 kg, standard deviation 7.5 kg. Plot group 2: 12 plots, mean 35.5 kg, standard deviation 6 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 90% confidence interval for μ₁ − μ₂.
Two teaching methods are compared. Group 1: 12 students, mean score 33.5, standard deviation 3. Group 2: 14 students, mean score 24.5, standard deviation 9. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).
n₁ = 12x̄₁ = 33.5s₁ = 3n₂ = 14x̄₂ = 24.5s₂ = 9
t =
8.Hypothesis tests & confidence intervalsHard
A sample of 64 measurements has mean 203.8. The population standard deviation is 15. Find the lower limit L of the 95% confidence interval for the mean.
x̄ = 203.8σ = 15n = 6495%
L =
3Solutions
1.Evidence-based practice statistics
Answer: IR = 47 per 1000 person-years
Likelihood ratios and risk measures: LR+ = sensitivity/(1 − specificity), LR− = (1 − sensitivity)/specificity. In a trial, EER and CER are the event rates with and without treatment: RR = EER/CER, ARR = CER − EER, RRR = ARR/CER and NNT = 1/ARR, rounded up. In a case-control study only the odds ratio OR = ad/(bc) can be used. An incidence rate = new cases/person-time.
Person-years
PY = 450·1.5 + 590·3 + 260·5 = 3745
Incidence rate
IR = 176/3745·1000 = 47 per 1000 person-years
Answer
IR = 47 per 1000 person-years
2.Evidence-based practice statistics
Answer: p′ = 92.4 %
Predictive values depend on prevalence: PPV = Se·p/(Se·p + (1 − Sp)(1 − p)) and NPV = Sp(1 − p)/(Sp(1 − p) + (1 − Se)p). With likelihood ratios: pre-test odds = p/(1 − p), post-test odds = pre-test odds × LR, probability = odds/(1 + odds). For a population of N, true positives = N·p·Se and false positives = N·(1 − p)·(1 − Sp). NNT = 1/(CER × RRR), rounded up.
Likelihood ratio
LR+ = 0.91/(1 − 0.95) = 18.2
Pre-test odds
0.4/(1 − 0.4) = 0.6667
Post-test odds
0.6667·18.2 = 12.13
Post-test probability
p′ = 12.13/(1 + 12.13) = 0.9239 ≈ 92.4%
Answer
p′ = 92.4%
3.Descriptive statistics
Answer: s = 10.09
Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.
Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.
Standardise
z = 0.4
Standard normal table
Φ(0.4) = 0.6554
P = 1 − 0.6554 = 0.3446
Answer
P = 0.3446
5.Confidence intervals
Answer: upper limit = 25.4
Confidence interval for a mean: x̄ ± margin of error. With a known population σ the margin is z·σ/√n; with only the sample standard deviation s, use t with n − 1 degrees of freedom: t·s/√n. Critical values: z = 1.645 (90%), 1.96 (95%), 2.576 (99%). To get a margin E or smaller, you need n ≥ (z·σ/E)², rounded up.
Critical value
df = n − 1 = 11, t = 2.201
Margin of error
E = 2.201·14/√12 = 8.895
Interval
16.5 ± 8.895 → [7.605, 25.4]
Answer
upper limit = 25.4
6.Confidence intervals
Answer: lower limit = 4.039 kg
Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.
Pooled standard deviation
sp² = (9·7.5² + 11·6²)/20 = 45.11
sp = 6.717
Margin of error
E = 1.725·6.717·√(1/10 + 1/12) = 4.961
Interval
9 ± 4.961 → [4.039, 13.96]
Answer
lower limit = 4.039 kg
7.Hypothesis tests & confidence intervals
Answer: t = 3.302
Hypothesis test: the test statistic = (estimate − value under H₀)/(standard error under H₀). The p-value is the probability, assuming H₀ is true, of a statistic at least as extreme as the one observed: the area in one tail for a one-sided H₁, twice the tail area for a two-sided H₁. Reject H₀ when p ≤ α.
Pooled variance
sp² = (11·3² + 13·9²)/24 = 48
sp = 6.928
Standard error
SE = 6.928·√(1/12 + 1/14) = 2.726
Test statistic
t = (33.5 − 24.5)/2.726 = 3.302
Answer
t = 3.302
With df = 24, the two-sided 5% critical value is 2.064; |t| is larger, so H₀ is rejected.
8.Hypothesis tests & confidence intervals
Answer: L = 200.1
Testing a mean: the test statistic measures how many standard errors the sample mean lies from the value claimed in H₀: z = (x̄ − μ₀)/(σ/√n) when σ is known, t = (x̄ − μ₀)/(s/√n) when it is estimated from the sample. A confidence interval is x̄ ± (critical value)·(standard error).