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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 case-control study compares 190 patients with a disease (cases) and 580 people without it (controls). The table shows how many in each group were exposed to a risk factor. Find the odds ratio OR for the exposure. Round to two decimal places.
a = 141b = 284c = 49d = 296
OR =
2.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A treatment reduces the risk of an event by 15% in relative terms (RRR = 15%). In a patient group the risk without treatment (CER) is 4% 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 = 15%CER = 4%
NNT =
3.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 40, 60, 40, 33, 51, 48.
n = 6
s =
4.Descriptive statisticsHard
Exam scores are normally distributed with mean 50 and standard deviation 5. Find the z-score z of a score of 47.5.
μ = 50σ = 5x = 47.5
z =
5.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 60 phones, mean 40.5 h, known standard deviation 7 h. Model 2: 100 phones, mean 46 h, known standard deviation 15 h. Find the upper limit of the 90% confidence interval for μ₁ − μ₂.
The population standard deviation is 31. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 2.5?
σ = 31E = 2.599%
n =
7.Hypothesis tests & confidence intervalsMedium
Two teaching methods are compared. Group 1: 7 students, mean score 107, standard deviation 4.5. Group 2: 14 students, mean score 101.9, standard deviation 2.5. 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 = 7x̄₁ = 107s₁ = 4.5n₂ = 14x̄₂ = 101.9s₂ = 2.5
t =
8.Hypothesis tests & confidence intervalsHard
In a trial, 36 of 120 patients on treatment A recover, and 34 of 80 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 36n₁ = 120x₂ = 34n₂ = 80
z =
3Solutions
1.Evidence-based practice statistics
Answer: OR = 3
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.
Odds of exposure
cases: 141/49 = 2.878
controls: 284/296 = 0.9595
Odds ratio
OR = (141·296)/(284·49) = 2.999 ≈ 3
Answer
OR = 3
OR above 1: exposure is more common among the cases.
2.Evidence-based practice statistics
Answer: NNT = 167
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 = 0.04·0.15 = 0.006
Number needed to treat
NNT = 1/0.006 = 166.7 → 167
Answer
NNT = 167
The same relative effect gives a smaller NNT when the baseline risk is higher.
3.Descriptive statistics
Answer: s = 9.626
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 = (47.5 − 50)/5 = −0.5
Answer
z = −0.5
5.Confidence intervals
Answer: upper limit = −2.619 h
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.
Point estimate
x̄₁ − x̄₂ = 40.5 − 46 = −5.5
Standard error
SE = √(7²/60 + 15²/100) = 1.751
Margin of error
E = 1.645·1.751 = 2.881
Interval
−5.5 ± 2.881 → [−8.381, −2.619]
Answer
upper limit = −2.619 h
The interval does not contain 0, so the means differ significantly.
6.Confidence intervals
Answer: n = 1021
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.
Solve for n
n = (2.576·31/2.5)² = 1020
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 1021
Answer
n = 1021
7.Hypothesis tests & confidence intervals
Answer: t = 3.373
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 = (6·4.5² + 13·2.5²)/19 = 10.67
sp = 3.267
Standard error
SE = 3.267·√(1/7 + 1/14) = 1.512
Test statistic
t = (107 − 101.9)/1.512 = 3.373
Answer
t = 3.373
With df = 19, the two-sided 5% critical value is 2.093; |t| is larger, so H₀ is rejected.
8.Hypothesis tests & confidence intervals
Answer: z = −1.816
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.
Sample proportions
p̂₁ = 36/120 = 0.3
p̂₂ = 34/80 = 0.425
p̂ = 70/200 = 0.35
Standard error under H₀
SE = √(0.35·0.65·(1/120 + 1/80)) = 0.06884
Test statistic
z = (0.3 − 0.425)/0.06884 = −1.816
Answer
z = −1.816
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).