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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, 120 patients received a new treatment and 1000 received the control treatment. The table shows how many in each group had the outcome (an unwanted event such as a stroke). Find the relative risk RR of the event with treatment. Round to two decimal places.
treatment: 15/120control: 310/1000
RR =
2.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). Without a drug, 25% of patients have a certain side effect during a year. The drug raises this risk with a relative risk of RR = 1.65. Find the number needed to harm NNH: how many patients must take the drug for a year for one extra patient to have the side effect? Round up to a whole number of patients.
control 25%RR = 1.65
NNH =
3.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 56, 51, 11, 15, 28, 38, 53.
n = 7
s =
4.Descriptive statisticsHard
Exam scores are normally distributed with mean 500 and standard deviation 50. Find the z-score z of a score of 520.
μ = 500σ = 50x = 520
z =
5.Confidence intervalsMedium
A sample of 8 items has mean 41.5 and standard deviation s = 2. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.
n = 8x̄ = 41.5s = 299%t(7) = 3.499
upper limit =
6.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 16 plots, mean yield 19 kg, standard deviation 1 kg. Plot group 2: 6 plots, mean 12 kg, standard deviation 9.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the upper limit of the 99% confidence interval for μ₁ − μ₂.
A sample of 16 students has a mean score of 494.8 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 500.
μ₀ = 500x̄ = 494.8s = 15n = 16
t =
8.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 3? The standard deviation is 10. Find the smallest sample size n.
σ = 10E = 390%
n =
3Solutions
1.Evidence-based practice statistics
Answer: RR = 0.4
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 = 15/120 = 0.125
CER = 310/1000 = 0.31
Relative risk
RR = 0.125/0.31 = 0.4032 ≈ 0.4
Answer
RR = 0.4
2.Evidence-based practice statistics
Answer: NNH = 7
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.
Risk with the drug
0.25·1.65 = 0.4125
Absolute risk increase
ARI = 0.4125 − 0.25 = 0.1625
Number needed to harm
NNH = 1/0.1625 = 6.154 → 7
Answer
NNH = 7
3.Descriptive statistics
Answer: s = 18.48
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 = (520 − 500)/50 = 0.4
Answer
z = 0.4
5.Confidence intervals
Answer: upper limit = 43.97
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.
Critical value
df = n − 1 = 7, t = 3.499
Margin of error
E = 3.499·2/√8 = 2.474
Interval
41.5 ± 2.474 → [39.03, 43.97]
Answer
upper limit = 43.97
6.Confidence intervals
Answer: upper limit = 13.58 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 = (15·1² + 5·9.5²)/20 = 23.31
sp = 4.828
Margin of error
E = 2.845·4.828·√(1/16 + 1/6) = 6.576
Interval
7 ± 6.576 → [0.4241, 13.58]
Answer
upper limit = 13.58 kg
7.Hypothesis tests & confidence intervals
Answer: t = −1.387
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 = 15/√16 = 15/4 = 3.75
Test statistic
t = (494.8 − 500)/3.75 = −1.387
Answer
t = −1.387
t0.025; df = 15 = 2.131
|t| < 2.131, so H₀ is not rejected at the 5% level (two-sided).
8.Hypothesis tests & confidence intervals
Answer: n = 31
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).