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). A case-control study compares 290 patients with a disease (cases) and 760 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 = 112b = 152c = 178d = 608
OR =
2.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A treatment reduces the risk of an event by 25% in relative terms (RRR = 25%). In a patient group the risk without treatment (CER) is 16% 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 = 25%CER = 16%
NNT =
3.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 23, 55, 50, 39, 41, 16.
n = 6
s =
4.Descriptive statisticsHard
A quantity is normally distributed with mean 170 and standard deviation 20. What proportion P of values is below 210?
μ = 170σ = 20x = 210
P =
5.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 60 phones, mean 106.5 h, known standard deviation 8 h. Model 2: 60 phones, mean 92 h, known standard deviation 7 h. Find the upper limit of the 95% confidence interval for μ₁ − μ₂.
Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 90% confidence interval for the proportion has a margin of error of at most 0.01?
p = 0.5E = 0.0190%
n =
7.Hypothesis tests & confidence intervalsMedium
A sample of 64 students has a mean score of 201.4 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 200.
μ₀ = 200x̄ = 201.4s = 4n = 64
t =
8.Hypothesis tests & confidence intervalsHard
In a trial, 140 of 200 patients on treatment A recover, and 65 of 80 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 140n₁ = 200x₂ = 65n₂ = 80
z =
3Solutions
1.Evidence-based practice statistics
Answer: OR = 2.52
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: 112/178 = 0.6292
controls: 152/608 = 0.25
Odds ratio
OR = (112·608)/(152·178) = 2.517 ≈ 2.52
Answer
OR = 2.52
OR above 1: exposure is more common among the cases.
2.Evidence-based practice statistics
Answer: NNT = 25
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 = 0.16·0.25 = 0.04
Number needed to treat
NNT = 1/0.04 = 25 → 25
Answer
NNT = 25
The same relative effect gives a smaller NNT when the baseline risk is higher.
3.Descriptive statistics
Answer: s = 15.16
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) = 0.9772
P = 0.9772
Answer
P = 0.9772
5.Confidence intervals
Answer: upper limit = 17.19 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 = 106.5 − 92 = 14.5
Standard error
SE = √(8²/60 + 7²/60) = 1.372
Margin of error
E = 1.96·1.372 = 2.69
Interval
14.5 ± 2.69 → [11.81, 17.19]
Answer
upper limit = 17.19 h
The interval does not contain 0, so the means differ significantly.
6.Confidence intervals
Answer: n = 6766
Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.
Solve for n
n = 1.645²·0.5·0.5/0.01² = 6765
Round up
n = 6766
Answer
n = 6766
7.Hypothesis tests & confidence intervals
Answer: t = 2.8
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 = 4/√64 = 4/8 = 0.5
Test statistic
t = (201.4 − 200)/0.5 = 2.8
Answer
t = 2.8
t0.025; df = 63 = 1.998
|t| > 1.998, so H₀ is rejected at the 5% level (two-sided).
8.Hypothesis tests & confidence intervals
Answer: z = −1.92
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 = 140/200 = 0.7
p̂₂ = 65/80 = 0.8125
p̂ = 205/280 = 0.7321
Standard error under H₀
SE = √(0.7321·0.2679·(1/200 + 1/80)) = 0.05858
Test statistic
z = (0.7 − 0.8125)/0.05858 = −1.92
Answer
z = −1.92
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).