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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). A test has a sensitivity of 61% and a specificity of 66%. Find the positive likelihood ratio LR+. Round to two decimal places.
Se = 61%Sp = 66%
LR+ =
2.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A treatment reduces the risk of an event by 21% in relative terms (RRR = 21%). In a patient group the risk without treatment (CER) is 29% over five years. Find the number needed to treat NNT for five years to prevent one event. Round up to a whole number of patients.
RRR = 21%CER = 29%
NNT =
3.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 14, 55, 45, 16, 59, 21.
n = 6
s =
4.Descriptive statisticsHard
Exam scores are normally distributed with mean 500 and standard deviation 75. Find the z-score z of a score of 650.
μ = 500σ = 75x = 650
z =
5.Confidence intervalsMedium
The blood pressure of 7 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −9.9 and standard deviation 7.4. Find the upper limit of the 95% confidence interval for the mean difference.
n = 7d̄ = −9.9sd = 7.495%t(6) = 2.447
upper limit = mmHg
6.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 7 plots, mean yield 99.5 kg, standard deviation 7.5 kg. Plot group 2: 5 plots, mean 92 kg, standard deviation 7.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the margin of error E of the 90% confidence interval for μ₁ − μ₂.
The mean of a population is claimed to be 500; the population standard deviation is known to be 30. A random sample of 100 observations has a mean of 503.78. Test H₀: μ = 500 against H₁: μ > 500. Find the p-value p.
μ₀ = 500σ = 30n = 100x̄ = 503.78H₁: μ > μ₀
p =
8.Hypothesis tests & confidence intervalsHard
A one-sided z test of H₀: μ = 100 against H₁: μ < 100 is done at α = 0.01 with a sample of 100; the standard deviation is known to be 15. In reality the mean is 95.95. Find the power 1 − β of the test in percent.
μ₀ = 100μ = 95.95σ = 15n = 100α = 0.01
1 − β = %
3Solutions
1.Evidence-based practice statistics
Answer: LR+ = 1.79
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.
Positive likelihood ratio
LR+ = 0.61/(1 − 0.66) = 1.794 ≈ 1.79
Answer
LR+ = 1.79
2.Evidence-based practice statistics
Answer: NNT = 17
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.
Absolute risk reduction
ARR = 0.29·0.21 = 0.0609
Number needed to treat
NNT = 1/0.0609 = 16.42 → 17
Answer
NNT = 17
The same relative effect gives a smaller NNT when the baseline risk is higher.
3.Descriptive statistics
Answer: s = 20.37
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 = (650 − 500)/75 = 2
Answer
z = 2
5.Confidence intervals
Answer: upper limit = −3.056 mmHg
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.
Critical value
df = 6, t = 2.447
Margin of error
E = 2.447·7.4/√7 = 6.844
Interval
−9.9 ± 6.844 → [−16.74, −3.056]
Answer
upper limit = −3.056 mmHg
6.Confidence intervals
Answer: E = 7.957 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 = (6·7.5² + 4·7.5²)/10 = 56.25
sp = 7.5
Margin of error
E = 1.812·7.5·√(1/7 + 1/5) = 7.957
Answer
E = 7.957 kg
7.Hypothesis tests & confidence intervals
Answer: p = 0.1038
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.
Test statistic
SE = σ/√n = 30/√100 = 3
z = (503.78 − 500)/3 = 1.26
Tail area
Φ(1.26) = 0.8962
1 − Φ(1.26) = 0.1038
p-value
p = 0.1038
Answer
p = 0.1038
p > 0.05, so H₀ is not rejected at the 5% level.
8.Hypothesis tests & confidence intervals
Answer: 1 − β = 64.58 %
Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).