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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, 590 patients received a new treatment and 100 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: 25/590control: 7/100
NNT =
2.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A treatment reduces the risk of an event by 33% in relative terms (RRR = 33%). In a patient group the risk without treatment (CER) is 14% 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 = 33%CER = 14%
NNT =
3.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 55, 26, 41, 53, 26, 15.
n = 6
s =
4.Descriptive statisticsHard
Exam scores are normally distributed with mean 500 and standard deviation 60. Find the z-score z of a score of 590.
μ = 500σ = 60x = 590
z =
5.Confidence intervalsMedium
The blood pressure of 9 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 6.4 and standard deviation 5.2. Find the margin of error E of the 95% confidence interval for the mean difference.
n = 9d̄ = 6.4sd = 5.295%t(8) = 2.306
E = mmHg
6.Confidence intervalsHard
The population standard deviation is 20. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 2.5?
σ = 20E = 2.595%
n =
7.Hypothesis tests & confidence intervalsMedium
A sample of 49 students has a mean score of 195.6 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 200.
μ₀ = 200x̄ = 195.6s = 10n = 49
t =
8.Hypothesis tests & confidence intervalsHard
In a trial, 50 of 250 patients on treatment A recover, and 13 of 100 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 50n₁ = 250x₂ = 13n₂ = 100
z =
3Solutions
1.Evidence-based practice statistics
Answer: NNT = 37
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 = 25/590 = 0.04237
CER = 7/100 = 0.07
Number needed to treat
ARR = 0.07 − 0.04237 = 0.02763
NNT = 1/0.02763 = 36.2 → 37
Answer
NNT = 37
Treating 37 patients prevents about one event.
2.Evidence-based practice statistics
Answer: NNT = 22
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 = 0.14·0.33 = 0.0462
Number needed to treat
NNT = 1/0.0462 = 21.65 → 22
Answer
NNT = 22
The same relative effect gives a smaller NNT when the baseline risk is higher.
3.Descriptive statistics
Answer: s = 16.22
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 = (590 − 500)/60 = 1.5
Answer
z = 1.5
5.Confidence intervals
Answer: E = 3.997 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 = 2.306
Margin of error
E = 2.306·5.2/√9 = 3.997
Answer
E = 3.997 mmHg
6.Confidence intervals
Answer: n = 246
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 = (1.96·20/2.5)² = 245.9
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 246
Answer
n = 246
7.Hypothesis tests & confidence intervals
Answer: t = −3.08
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).
Standard error
SE = 10/√49 = 10/7 = 1.429
Test statistic
t = (195.6 − 200)/1.429 = −3.08
Answer
t = −3.08
t0.025; df = 48 = 2.011
|t| > 2.011, so H₀ is rejected at the 5% level (two-sided).
8.Hypothesis tests & confidence intervals
Answer: z = 1.54
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 = 50/250 = 0.2
p̂₂ = 13/100 = 0.13
p̂ = 63/350 = 0.18
Standard error under H₀
SE = √(0.18·0.82·(1/250 + 1/100)) = 0.04546
Test statistic
z = (0.2 − 0.13)/0.04546 = 1.54
Answer
z = 1.54
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).