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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, 130 patients received a new treatment and 960 received the control treatment. The table shows how many in each group had a side effect. Find the number needed to harm NNH: how many patients must be treated for one extra patient to have the side effect? Round up to a whole number of patients.
treatment: 42/130control: 240/960
NNH =
2.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). 10 000 people are screened with a test of sensitivity 71% and specificity 77%. The prevalence of the disease is 14%. How many positive results n₊ are expected in total?
N = 10 000p = 14%Se = 71%Sp = 77%
n₊ =
3.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 18, 38, 39, 22, 34, 26.
n = 6
s =
4.Descriptive statisticsHard
A quantity is normally distributed with mean 100 and standard deviation 12. What proportion P of values is above 112?
μ = 100σ = 12x = 112
P =
5.Confidence intervalsMedium
A 99% confidence interval for a mean, based on 36 observations and a known σ, has a margin of error of 4.1. What is the margin of error E₂ if the same study uses 9 observations instead?
99%n₁ = 36E₁ = 4.1n₂ = 9
E₂ =
6.Confidence intervalsHard
The population standard deviation is 15. 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 1.5?
σ = 15E = 1.590%
n =
7.Hypothesis tests & confidence intervalsMedium
The mean of a population is claimed to be 50; the population standard deviation is known to be 4. A random sample of 64 observations has a mean of 50.48. Test H₀: μ = 50 against H₁: μ > 50. Find the p-value p.
μ₀ = 50σ = 4n = 64x̄ = 50.48H₁: μ > μ₀
p =
8.Hypothesis tests & confidence intervalsHard
A sample of 16 measurements has mean 48.3. The population standard deviation is 4. Find the lower limit L of the 90% confidence interval for the mean.
x̄ = 48.3σ = 4n = 1690%
L =
3Solutions
1.Evidence-based practice statistics
Answer: NNH = 14
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 = 42/130 = 0.3231
CER = 240/960 = 0.25
Number needed to harm
ARI = 0.3231 − 0.25 = 0.07308
NNH = 1/0.07308 = 13.68 → 14
Answer
NNH = 14
About one in every 14 treated patients has the side effect because of the treatment.
2.Evidence-based practice statistics
Answer: n₊ = 2972
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
10 000·0.14 = 1400
10 000 − 1400 = 8600
All positive results
TP = 1400·0.71 = 994
FP = 8600·0.23 = 1978
n₊ = 994 + 1978 = 2972
Answer
n₊ = 2972
Of the 2972 positives only 994 have the disease (PPV 33.4%).
3.Descriptive statistics
Answer: s = 8.758
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 = 1
Standard normal table
Φ(1) = 0.8413
P = 1 − 0.8413 = 0.1587
Answer
P = 0.1587
5.Confidence intervals
Answer: E₂ = 8.2
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.
Ratio of margins
E₂/E₁ = (2.576/2.576)·√(36/9) = 2
New margin
E₂ = 4.1·2 = 8.2
Answer
E₂ = 8.2
6.Confidence intervals
Answer: n = 271
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 = (1.645·15/1.5)² = 270.6
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 271
Answer
n = 271
7.Hypothesis tests & confidence intervals
Answer: p = 0.1685
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.
Test statistic
SE = σ/√n = 4/√64 = 0.5
z = (50.48 − 50)/0.5 = 0.96
Tail area
Φ(0.96) = 0.8315
1 − Φ(0.96) = 0.1685
p-value
p = 0.1685
Answer
p = 0.1685
p > 0.05, so H₀ is not rejected at the 5% level.
8.Hypothesis tests & confidence intervals
Answer: L = 46.65
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).