Evidence-Based Practice: cram pack

Formula sheets, the key method steps for every topic and 8 practice problems, with full solutions at the back.

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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

  1. Draw the 2 × 2 table and label a, b, c and d before you calculate.
  2. Read sensitivity and specificity down the columns, predictive values along the rows.
  3. For another prevalence, imagine 10 000 people and fill the table with Se, Sp and the prevalence.
  4. For a likelihood ratio, convert the probability to odds, multiply, and convert back.
  5. 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

  1. Decide between z (population σ known) and t (only the sample s, with n − 1 degrees of freedom).
  2. Find the critical value for the confidence level.
  3. Margin of error: E = critical value × standard deviation/√n.
  4. 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

  1. State H₀ and H₁ and decide whether the test is one- or two-sided.
  2. Calculate the test statistic: (x̄ − μ₀)/(σ/√n), or with s and t.
  3. Compare with the critical value, or find the p-value.
  4. 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
Se85%Sp71%LR+ = ?

Practice exercise (made-up values, not clinical guidance). A test has a sensitivity of 85% and a specificity of 71%. Find the positive likelihood ratio LR+. Round to two decimal places.

Se = 85%Sp = 71%
LR+ =
2.Evidence-based practice statisticsHard
CER20%EER?RRR 50%

Practice exercise (made-up values, not clinical guidance). A treatment reduces the risk of an event by 50% in relative terms (RRR = 50%). In a patient group the risk without treatment (CER) is 20% 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 = 50%CER = 20%
NNT =
3.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 15, 58, 58, 28, 51, 32.

n = 6
s =
4.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 170 and standard deviation 15. Find the z-score z of a score of 162.5.

μ = 170σ = 15x = 162.5
z =
5.Confidence intervalsMedium
n = 13, s = 14.5L = ?Ux̄ = 65.5E95%

A sample of 13 items has mean 65.5 and standard deviation s = 14.5. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.

n = 13x̄ = 65.5s = 14.595%t(12) = 2.179
lower limit =
6.Confidence intervalsHard
σ = 37, n = ?x̄ − Ex̄ + Ex̄E = 299%

The population standard deviation is 37. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 2?

σ = 37E = 299%
n =
7.Hypothesis tests & confidence intervalsMedium
−2.0862.086t = ?df = 20

Two teaching methods are compared. Group 1: 15 students, mean score 43.5, standard deviation 6.5. Group 2: 7 students, mean score 34.4, standard deviation 11.5. 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 = 15x̄₁ = 43.5s₁ = 6.5n₂ = 7x̄₂ = 34.4s₂ = 11.5
t =
8.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 2.5α = 0.05βσ = 13, power 95%, n = ?

A one-sided z test at α = 0.05 must detect a shift of the mean of 2.5 units with a power of 95%. The standard deviation is 13. Find the minimum sample size n.

α = 0.05power = 95%δ = 2.5σ = 13zα = 1.645zβ = 1.645
n =

3Solutions

1.Evidence-based practice statistics

Answer: LR+ = 2.93

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.

Positive likelihood ratio

LR+ = 0.85/(1 − 0.71) = 2.931 ≈ 2.93

Answer

LR+ = 2.93
2.Evidence-based practice statistics

Answer: NNT = 10

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 = 0.2·0.5 = 0.1

Number needed to treat

NNT = 1/0.1 = 10 → 10

Answer

NNT = 10

The same relative effect gives a smaller NNT when the baseline risk is higher.

3.Descriptive statistics

Answer: s = 17.9

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.

Mean

x̄ = 242/6 = 40.33

Squared deviations

Σ(x − x̄)² = (−25.33)² + 17.67² + 17.67² + (−12.33)² + 10.67² + (−8.333)² = 1601

Standard deviation

s = √(1601/5) = 17.9

Answer

s = 17.9
4.Descriptive statistics

Answer: z = −0.5

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 = (162.5 − 170)/15 = −0.5

Answer

z = −0.5
5.Confidence intervals

Answer: lower limit = 56.74

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 = 12, t = 2.179

Margin of error

E = 2.179·14.5/√13 = 8.763

Interval

65.5 ± 8.763 → [56.74, 74.26]

Answer

lower limit = 56.74
6.Confidence intervals

Answer: n = 2272

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 = (2.576·37/2)² = 2271

Round up

A sample size is a whole number, and rounding down would make the margin too large.

n = 2272

Answer

n = 2272
7.Hypothesis tests & confidence intervals

Answer: t = 2.389

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 = (14·6.5² + 6·11.5²)/20 = 69.25
sp = 8.322

Standard error

SE = 8.322·√(1/15 + 1/7) = 3.809

Test statistic

t = (43.5 − 34.4)/3.809 = 2.389

Answer

t = 2.389

With df = 20, the two-sided 5% critical value is 2.086; |t| is larger, so H₀ is rejected.

8.Hypothesis tests & confidence intervals

Answer: n = 293

Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.

Solve for n

n = ((1.645 + 1.645)·13/2.5)² = 292.7

Round up

n = 293

Answer

n = 293

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