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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 randomised trial, 710 patients received a new treatment and 850 received the control treatment. The table shows how many in each group had the outcome (an unwanted event such as a stroke). Find the number needed to treat NNT to prevent one event. Round up to a whole number of patients.
treatment: 108/710control: 323/850
NNT =
2.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). 10 000 people are screened with a test of sensitivity 92% and specificity 80%. The prevalence of the disease is 10%. How many false-positive results FP are expected?
N = 10 000p = 10%Se = 92%Sp = 80%
FP =
3.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 56, 60, 43, 52, 49, 60, 25.
n = 7
s =
4.Descriptive statisticsHard
Exam scores are normally distributed with mean 170 and standard deviation 12. Find the z-score z of a score of 164.
μ = 170σ = 12x = 164
z =
5.Confidence intervalsMedium
The blood pressure of 15 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 7.1 and standard deviation 5. Find the lower limit of the 99% confidence interval for the mean difference.
n = 15d̄ = 7.1sd = 599%t(14) = 2.977
lower limit = mmHg
6.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 11 plots, mean yield 27.5 kg, standard deviation 4.5 kg. Plot group 2: 5 plots, mean 29.5 kg, standard deviation 2.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the margin of error E of the 90% confidence interval for μ₁ − μ₂.
Two production lines make the same part. Line 1: a sample of 80 parts has mean length 49.5 mm, and the line's known standard deviation is 7 mm. Line 2: a sample of 30 parts has mean 45.8 mm, known standard deviation 18 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).
A sample of 16 measurements has mean 55. The population standard deviation is 20. Find the lower limit L of the 99% confidence interval for the mean.
x̄ = 55σ = 20n = 1699%
L =
3Solutions
1.Evidence-based practice statistics
Answer: NNT = 5
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.
Event rates
EER = 108/710 = 0.1521
CER = 323/850 = 0.38
Number needed to treat
ARR = 0.38 − 0.1521 = 0.2279
NNT = 1/0.2279 = 4.388 → 5
Answer
NNT = 5
Treating 5 patients prevents about one event.
2.Evidence-based practice statistics
Answer: FP = 1800
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.
With and without the disease
10 000·0.1 = 1000
10 000 − 1000 = 9000
False positives
FP = 9000·0.2 = 1800
Answer
FP = 1800
Of the 2720 positives only 920 have the disease (PPV 33.8%).
3.Descriptive statistics
Answer: s = 12.32
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 = (164 − 170)/12 = −0.5
Answer
z = −0.5
5.Confidence intervals
Answer: lower limit = 3.257 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 = 14, t = 2.977
Margin of error
E = 2.977·5/√15 = 3.843
Interval
7.1 ± 3.843 → [3.257, 10.94]
Answer
lower limit = 3.257 mmHg
6.Confidence intervals
Answer: E = 3.829 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 = (10·4.5² + 4·2.5²)/14 = 16.25
sp = 4.031
Margin of error
E = 1.761·4.031·√(1/11 + 1/5) = 3.829
Answer
E = 3.829 kg
7.Hypothesis tests & confidence intervals
Answer: z = 1.095
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.
Standard error of the difference
SE = √(7²/80 + 18²/30) = √11.41 = 3.378
Test statistic
z = (49.5 − 45.8)/3.378 = 1.095
Answer
z = 1.095
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
8.Hypothesis tests & confidence intervals
Answer: L = 42.12
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).