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
casescontrolsexposed8291abnot exposed78259cd

Practice exercise (made-up values, not clinical guidance). A case-control study compares 160 patients with a disease (cases) and 350 people without it (controls). The table shows how many in each group were exposed to a risk factor. Find the odds ratio OR for the exposure. Round to two decimal places.

a = 82b = 91c = 78d = 259
OR =
2.Evidence-based practice statisticsHard
CER9%EER?RRR 30%

Practice exercise (made-up values, not clinical guidance). A treatment reduces the risk of an event by 30% in relative terms (RRR = 30%). In a patient group the risk without treatment (CER) is 9% 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 = 30%CER = 9%
NNT =
3.Descriptive statisticsMedium
2025303540455055

Find the sample standard deviation s of the data: 21, 22, 24, 16, 57, 31, 45.

n = 7
s =
4.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 60 and standard deviation 8. Find the z-score z of a score of 53.6.

μ = 60σ = 8x = 53.6
z =
5.Confidence intervalsMedium
μ₁ − μ₂0L = ?Ux̄₁ − x̄₂E90%

The battery life of two phone models is compared. Model 1: 50 phones, mean 116.5 h, known standard deviation 4 h. Model 2: 60 phones, mean 115 h, known standard deviation 11 h. Find the lower limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 50x̄₁ = 116.5 hσ₁ = 4 hn₂ = 60x̄₂ = 115 hσ₂ = 11 h90%
lower limit = h
6.Confidence intervalsHard
μ₁ − μ₂, t(14) = 2.9770LUx̄₁ − x̄₂E = ?99%

Two fertilisers are tested. Plot group 1: 8 plots, mean yield 59.5 kg, standard deviation 7.5 kg. Plot group 2: 8 plots, mean 52 kg, standard deviation 2 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the margin of error E of the 99% confidence interval for μ₁ − μ₂.

n₁ = 8x̄₁ = 59.5s₁ = 7.5n₂ = 8x̄₂ = 52s₂ = 299%t(14) = 2.977
E = kg
7.Hypothesis tests & confidence intervalsMedium
t −1.9901.99

A sample of 81 students has a mean score of 200.4 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 200.4s = 5n = 81
t =
8.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 4α = 0.01βσ = 11, power 90%, n = ?

A one-sided z test at α = 0.01 must detect a shift of the mean of 4 units with a power of 90%. The standard deviation is 11. Find the minimum sample size n.

α = 0.01power = 90%δ = 4σ = 11zα = 2.326zβ = 1.282
n =

3Solutions

1.Evidence-based practice statistics

Answer: OR = 2.99

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.

Odds of exposure

cases: 82/78 = 1.051
controls: 91/259 = 0.3514

Odds ratio

OR = (82·259)/(91·78) = 2.992 ≈ 2.99

Answer

OR = 2.99

OR above 1: exposure is more common among the cases.

2.Evidence-based practice statistics

Answer: NNT = 38

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 = 0.09·0.3 = 0.027

Number needed to treat

NNT = 1/0.027 = 37.04 → 38

Answer

NNT = 38

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

3.Descriptive statistics

Answer: s = 14.87

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 = 216/7 = 30.86

Squared deviations

Σ(x − x̄)² = (−9.857)² + (−8.857)² + (−6.857)² + (−14.86)² + 26.14² + 0.1429² + 14.14² = 1327

Standard deviation

s = √(1327/6) = 14.87

Answer

s = 14.87
4.Descriptive statistics

Answer: z = −0.8

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 = (53.6 − 60)/8 = −0.8

Answer

z = −0.8
5.Confidence intervals

Answer: lower limit = −1.015 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 = 116.5 − 115 = 1.5

Standard error

SE = √(4²/50 + 11²/60) = 1.529

Margin of error

E = 1.645·1.529 = 2.515

Interval

1.5 ± 2.515 → [−1.015, 4.015]

Answer

lower limit = −1.015 h

The interval contains 0, so the data do not show a significant difference.

6.Confidence intervals

Answer: E = 8.17 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 = (7·7.5² + 7·2²)/14 = 30.13
sp = 5.489

Margin of error

E = 2.977·5.489·√(1/8 + 1/8) = 8.17

Answer

E = 8.17 kg
7.Hypothesis tests & confidence intervals

Answer: t = 0.72

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 = 5/√81 = 5/9 = 0.5556

Test statistic

t = (200.4 − 200)/0.5556 = 0.72

Answer

t = 0.72
t0.025; df = 80 = 1.99

|t| < 1.99, so H₀ is not rejected at the 5% level (two-sided).

8.Hypothesis tests & confidence intervals

Answer: n = 99

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 = ((2.326 + 1.282)·11/4)² = 98.45

Round up

n = 99

Answer

n = 99

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