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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, 950 patients received a new treatment and 810 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 reduction RRR as a percentage. Round to one decimal place.
treatment: 93/950control: 113/810
RRR = %
2.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A patient has a pre-test probability of 10% for a disease. A test with sensitivity 88% and specificity 93% comes back negative. Find the post-test probability p′ of the disease as a percentage. Round to one decimal place.
pre-test 10%Se = 88%Sp = 93%negative
p′ = %
3.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 58, 58, 20, 18, 13, 26.
n = 6
s =
4.Descriptive statisticsHard
A quantity is normally distributed with mean 100 and standard deviation 10. What proportion P of values is below 92?
μ = 100σ = 10x = 92
P =
5.Confidence intervalsMedium
A sample of 20 items has mean 45.5 and standard deviation s = 16.5. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.
n = 20x̄ = 45.5s = 16.590%t(19) = 1.729
lower limit =
6.Confidence intervalsHard
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 0.025?
p = 0.5E = 0.02590%
n =
7.Hypothesis tests & confidence intervalsMedium
Two teaching methods are compared. Group 1: 12 students, mean score 41.5, standard deviation 10.5. Group 2: 5 students, mean score 36.6, standard deviation 4. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).
n₁ = 12x̄₁ = 41.5s₁ = 10.5n₂ = 5x̄₂ = 36.6s₂ = 4
t =
8.Hypothesis tests & confidence intervalsHard
In a trial, 113 of 250 patients on treatment A recover, and 83 of 250 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 113n₁ = 250x₂ = 83n₂ = 250
z =
3Solutions
1.Evidence-based practice statistics
Answer: RRR = 29.8 %
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 = 93/950 = 0.09789
CER = 113/810 = 0.1395
Relative risk reduction
ARR = 0.1395 − 0.09789 = 0.04161
RRR = 0.04161/0.1395 = 0.2983 ≈ 29.8%
Answer
RRR = 29.8%
2.Evidence-based practice statistics
Answer: p′ = 1.4 %
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.
Likelihood ratio
LR− = (1 − 0.88)/0.93 = 0.129
Pre-test odds
0.1/(1 − 0.1) = 0.1111
Post-test odds
0.1111·0.129 = 0.01434
Post-test probability
p′ = 0.01434/(1 + 0.01434) = 0.01413 ≈ 1.4%
Answer
p′ = 1.4%
3.Descriptive statistics
Answer: s = 20.44
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 = −0.8
Standard normal table
Φ(−0.8) = 0.2119
P = 0.2119
Answer
P = 0.2119
5.Confidence intervals
Answer: lower limit = 39.12
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 = 19, t = 1.729
Margin of error
E = 1.729·16.5/√20 = 6.379
Interval
45.5 ± 6.379 → [39.12, 51.88]
Answer
lower limit = 39.12
6.Confidence intervals
Answer: n = 1083
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 = 1.645²·0.5·0.5/0.025² = 1082
Round up
n = 1083
Answer
n = 1083
7.Hypothesis tests & confidence intervals
Answer: t = 0.9978
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.
Pooled variance
sp² = (11·10.5² + 4·4²)/15 = 85.12
sp = 9.226
Standard error
SE = 9.226·√(1/12 + 1/5) = 4.911
Test statistic
t = (41.5 − 36.6)/4.911 = 0.9978
Answer
t = 0.9978
With df = 15, the two-sided 5% critical value is 2.131; |t| is smaller, so H₀ is not rejected.
8.Hypothesis tests & confidence intervals
Answer: z = 2.748
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 = 113/250 = 0.452
p̂₂ = 83/250 = 0.332
p̂ = 196/500 = 0.392
Standard error under H₀
SE = √(0.392·0.608·(1/250 + 1/250)) = 0.04367
Test statistic
z = (0.452 − 0.332)/0.04367 = 2.748
Answer
z = 2.748
|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).