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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). A new test was compared with the reference standard in a study (see the 2 × 2 table). Find the positive likelihood ratio LR+. Round to two decimal places.
a = 289b = 76c = 81d = 1014
LR+ =
2.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A screening test has a sensitivity of 75% and a specificity of 90%. It is used in a population where the prevalence is 20%. Find the positive predictive value PPV as a percentage. Round to one decimal place.
Se = 75%Sp = 90%p = 20%
PPV = %
3.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 18, 53, 32, 19, 45, 59.
n = 6
s =
4.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 8. What proportion P of values is below 62?
μ = 50σ = 8x = 62
P =
5.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 40 phones, mean 72 h, known standard deviation 5 h. Model 2: 60 phones, mean 73.5 h, known standard deviation 19 h. Find the upper limit of the 99% confidence interval for μ₁ − μ₂.
The population standard deviation is 24. 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 = 24E = 1.590%
n =
7.Hypothesis tests & confidence intervalsMedium
Two teaching methods are compared. Group 1: 7 students, mean score 86.5, standard deviation 11.5. Group 2: 7 students, mean score 71.8, 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 = 7x̄₁ = 86.5s₁ = 11.5n₂ = 7x̄₂ = 71.8s₂ = 4
t =
8.Hypothesis tests & confidence intervalsHard
In a trial, 150 of 250 patients on treatment A recover, and 175 of 250 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 150n₁ = 250x₂ = 175n₂ = 250
z =
3Solutions
1.Evidence-based practice statistics
Answer: LR+ = 11.2
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.
Sensitivity and specificity
Se = 289/370 = 0.7811
Sp = 1014/1090 = 0.9303
Positive likelihood ratio
LR+ = 0.7811/(1 − 0.9303) = 11.2
Answer
LR+ = 11.2
2.Evidence-based practice statistics
Answer: PPV = 65.2 %
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.
True and false positives (per person)
Se·p = 0.75·0.2 = 0.15
(1 − Sp)·(1 − p) = 0.1·0.8 = 0.08
Positive predictive value
PPV = 0.15/(0.15 + 0.08) = 0.6522 ≈ 65.2%
Answer
PPV = 65.2%
The lower the prevalence, the lower the PPV of the same test.
3.Descriptive statistics
Answer: s = 17.39
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 = 1.5
Standard normal table
Φ(1.5) = 0.9332
P = 0.9332
Answer
P = 0.9332
5.Confidence intervals
Answer: upper limit = 5.139 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 = 72 − 73.5 = −1.5
Standard error
SE = √(5²/40 + 19²/60) = 2.577
Margin of error
E = 2.576·2.577 = 6.639
Interval
−1.5 ± 6.639 → [−8.139, 5.139]
Answer
upper limit = 5.139 h
The interval contains 0, so the data do not show a significant difference.
6.Confidence intervals
Answer: n = 693
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 = (1.645·24/1.5)² = 692.7
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 693
Answer
n = 693
7.Hypothesis tests & confidence intervals
Answer: t = 3.194
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 = (6·11.5² + 6·4²)/12 = 74.13
sp = 8.61
Standard error
SE = 8.61·√(1/7 + 1/7) = 4.602
Test statistic
t = (86.5 − 71.8)/4.602 = 3.194
Answer
t = 3.194
With df = 12, the two-sided 5% critical value is 2.179; |t| is larger, so H₀ is rejected.
8.Hypothesis tests & confidence intervals
Answer: z = −2.344
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 = 150/250 = 0.6
p̂₂ = 175/250 = 0.7
p̂ = 325/500 = 0.65
Standard error under H₀
SE = √(0.65·0.35·(1/250 + 1/250)) = 0.04266
Test statistic
z = (0.6 − 0.7)/0.04266 = −2.344
Answer
z = −2.344
|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).