In the print window, choose “Save as PDF” to keep a copy on your device.
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 study, exposed people contributed 1600 person-years with 34 new cases, and unexposed people 1400 person-years with 10 new cases. Find the incidence rate ratio IRR (exposed versus unexposed). Round to two decimal places.
exposed: 34/1600 PYnot exposed: 10/1400 PY
IRR =
2.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A treatment reduces the risk of an event by 35% in relative terms (RRR = 35%). In a patient group the risk without treatment (CER) is 19% 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 = 35%CER = 19%
NNT =
3.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 38, 33, 58, 20, 51, 38, 39.
n = 7
s =
4.Descriptive statisticsHard
A quantity is normally distributed with mean 70 and standard deviation 8. What proportion P of values is above 54?
μ = 70σ = 8x = 54
P =
5.Confidence intervalsMedium
The blood pressure of 9 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −7.5 and standard deviation 0.5. Find the upper limit of the 99% confidence interval for the mean difference.
n = 9d̄ = −7.5sd = 0.599%t(8) = 3.355
upper limit = mmHg
6.Confidence intervalsHard
The population standard deviation is 31. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 1?
σ = 31E = 190%
n =
7.Hypothesis tests & confidence intervalsMedium
Two teaching methods are compared. Group 1: 13 students, mean score 73, standard deviation 8.5. Group 2: 9 students, mean score 67.5, standard deviation 2. 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 = 13x̄₁ = 73s₁ = 8.5n₂ = 9x̄₂ = 67.5s₂ = 2
t =
8.Hypothesis tests & confidence intervalsHard
A one-sided z test of H₀: μ = 200 against H₁: μ < 200 is done at α = 0.01 with a sample of 64; the standard deviation is known to be 24. In reality the mean is 189.8. Find the probability of a Type II error β in percent.
μ₀ = 200μ = 189.8σ = 24n = 64α = 0.01
β = %
3Solutions
1.Evidence-based practice statistics
Answer: IRR = 2.98
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.
Incidence rates
exposed: 34/1600 = 21.25 per 1000 person-years
not exposed: 10/1400 = 7.143 per 1000 person-years
Rate ratio
IRR = 21.25/7.143 = 2.975 ≈ 2.98
Answer
IRR = 2.98
2.Evidence-based practice statistics
Answer: NNT = 16
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 = 0.19·0.35 = 0.0665
Number needed to treat
NNT = 1/0.0665 = 15.04 → 16
Answer
NNT = 16
The same relative effect gives a smaller NNT when the baseline risk is higher.
3.Descriptive statistics
Answer: s = 12.26
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 = −2
Standard normal table
Φ(−2) = 0.02275
P = 1 − 0.02275 = 0.9772
Answer
P = 0.9772
5.Confidence intervals
Answer: upper limit = −6.941 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 = 8, t = 3.355
Margin of error
E = 3.355·0.5/√9 = 0.5592
Interval
−7.5 ± 0.5592 → [−8.059, −6.941]
Answer
upper limit = −6.941 mmHg
6.Confidence intervals
Answer: n = 2601
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 = (1.645·31/1)² = 2600
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 2601
Answer
n = 2601
7.Hypothesis tests & confidence intervals
Answer: t = 1.892
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 = (12·8.5² + 8·2²)/20 = 44.95
sp = 6.704
Standard error
SE = 6.704·√(1/13 + 1/9) = 2.907
Test statistic
t = (73 − 67.5)/2.907 = 1.892
Answer
t = 1.892
With df = 20, the two-sided 5% critical value is 2.086; |t| is smaller, so H₀ is not rejected.
8.Hypothesis tests & confidence intervals
Answer: β = 14.14 %
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).